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diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch1.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch1.ipynb new file mode 100644 index 00000000..1956140e --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch1.ipynb @@ -0,0 +1,369 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 1:Semiconductor Marerials and Crystal Properties" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.1 Page No.23" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Miller indices of the given plane are 3.0 2.0 3.0\n" + ] + } + ], + "source": [ + "#Example 1.1\n", + "#Find the miller indices for a plane.\n", + "\n", + "#Given\n", + "#Length of intercept\n", + "l1=2.0\n", + "l2=3.0\n", + "l3=2.0\n", + "\n", + "#Calcuation\n", + "#reciprocal of intercept\n", + "r1=1/l1\n", + "r2=1/l2\n", + "r3=1/l3\n", + "m1=6*r1\n", + "m2=6*r2\n", + "m3=6*r3\n", + "\n", + "#Result\n", + "print\"Miller indices of the given plane are\",m1,m2,m3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.2 Page No.24" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Miller indices of the given plane are 2.0 1.0 0\n" + ] + } + ], + "source": [ + "#Example 1.2\n", + "#Find the miller indices for a plane.\n", + "\n", + "#Given\n", + "#Length of intercept\n", + "l1=1.0\n", + "l2=2.0\n", + "l3=0\n", + "\n", + "#Calcuation\n", + "#reciprocal of intercept\n", + "r1=1/l1\n", + "r2=1/l2\n", + "r3=0\n", + "m1=2*r1\n", + "m2=2*r2\n", + "m3=2*r3\n", + "\n", + "#Result\n", + "print\"Miller indices of the given plane are\",m1,m2,m3\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.3 Page No.24" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lattice constant is 3.22 A\n", + "radius of simple lattice is 1.61 A\n" + ] + } + ], + "source": [ + "#Example 1.3\n", + "#Obtain lattice constant and radius of the atom.\n", + "\n", + "#Given\n", + "V=3*(10**22) #kg/m**3, density of SCC lattice\n", + "p=(1/3.0)*10**-22\n", + "\n", + "#Calculation\n", + "n=1 #no. of lattice point \n", + "a=(n*p)**(1/3.0) #lattice constant\n", + "r=(a*10**8/2)\n", + "\n", + "#Result\n", + "print\"Lattice constant is\",round(a*10**8,2),\"A\"\n", + "print\"radius of simple lattice is\",round(r,2),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.4 Page no.25" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Density of crystal is 8928.8 Kg/m**3\n" + ] + } + ], + "source": [ + "#Exampe 1.4\n", + "#Determine the density of crystal\n", + "\n", + "#given data\n", + "import math\n", + "r=1.278 #in Angstrum\n", + "AtomicWeight=63.5 #constant\n", + "AvogadroNo=6.023*10**23 #constant\n", + "\n", + "#Calculation\n", + "#For FCC structure a=4*r/math.sqrt(2)\n", + "a=4*r*10**-10/math.sqrt(2) #in meter\n", + "V=a**3 #in meter**3\n", + "#mass of one atom = m\n", + "m=AtomicWeight/AvogadroNo #in gm\n", + "m=m/1000 #in Kg\n", + "n=4 # no. of atoms per unit cell for FCC structure\n", + "rho=m*n/V #in Kg/m**3\n", + "\n", + "#Result\n", + "print \"Density of crystal is\",round(rho,2),\"Kg/m**3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.5 Page no.26" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Density of silicon crystal is 1249.0 Kg/m**3\n" + ] + } + ], + "source": [ + "#Example 1.5\n", + "#What is Density of silicon crystal .\n", + "\n", + "#given data\n", + "n=4 # no. of atoms per unit cell of silicon\n", + "AtomicWeight=28 #constant\n", + "AvogadroNo=6.021*10**23 #constant\n", + "\n", + "#calculation\n", + "m=AtomicWeight/AvogadroNo #in gm\n", + "m=m/1000 #in Kg\n", + "a=5.3 #lattice constant in Angstrum\n", + "a=a*10**-10 #in meter\n", + "V=a**3 #in meter**3\n", + "rho=m*n/V #in Kg/m**3\n", + "\n", + "#result\n", + "print\"Density of silicon crystal is\",round(rho,0),\"Kg/m**3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.6 Page no.26" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Surface density in FCC on (111)Plane is %.e 1.02382271468e+13 atoms/mm**2\n" + ] + } + ], + "source": [ + "#Example 1.5\n", + "#What is Surface density in FCC .\n", + "\n", + "#given data\n", + "a=4.75 #lattice constant in Angstrum\n", + "a=a*10**-10 #in meter\n", + "\n", + "#Calculation\n", + "dp=2.31/a**2 #in atom/m**2\n", + "dp=dp/10**6 #in atom/mm**2\n", + "\n", + "#Result\n", + "print \"Surface density in FCC on (111)Plane is %.e\",dp,\"atoms/mm**2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.7 Page no. 28" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Interpolar distance in Angstrum 2.01 A\n" + ] + } + ], + "source": [ + "#Example 1.7\n", + "#find the Interpolar distance\n", + "\n", + "#given data\n", + "import math\n", + "l=1.539 #in Angstrum\n", + "theta=22.5 #in degree\n", + "n=1 #order unitless\n", + "\n", + "#Calculation\n", + "d=n*l/(2*math.sin(theta*math.pi/180)) #in Angstrum\n", + "\n", + "#result\n", + "print \"Interpolar distance in Angstrum \",round(d,2),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 1.8 Page no. 28" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "wavelength of X-rays in Angstrum 0.584 A\n" + ] + } + ], + "source": [ + "#Example 1.8\n", + "#Find the wavelength of X-rays \n", + "\n", + "#given data\n", + "import math\n", + "\n", + "theta=16.8/2.0 #in degree\n", + "n=2.0 #order unitless\n", + "d=0.4 #in nm\n", + "\n", + "#Calculation\n", + "l=(2*d*10**-9*sin(theta*math.pi/180.0))/n #in Angstrum\n", + "\n", + "#result\n", + "print \"wavelength of X-rays in Angstrum \",round(l*10**10,3),\"A\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch2.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch2.ipynb new file mode 100644 index 00000000..74bcb44f --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch2.ipynb @@ -0,0 +1,1510 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter2 : Energy Bands and Charge Carriers in semiconductor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.1 Page No.58" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Probability when the energy of the state is above 0.1 eV 0.02\n", + "Probability when the energy of the state is below 0.1 eV 0.98\n" + ] + } + ], + "source": [ + "#Example 2.1\n", + "#Find probability of an electronic state\n", + "\n", + "#Given\n", + "dE1=0.1 #eV\n", + "dE2=-0.1 #eV\n", + "k=8.61*10**-5 #Boltzman constant\n", + "T=300 #K\n", + "\n", + "#Calcualtion\n", + "import math\n", + "FE1=1/(1+math.exp(dE1/(k*T)))\n", + "FE2=1/(1+math.exp(dE2/(k*T)))\n", + "\n", + "#Result\n", + "print\"Probability when the energy of the state is above 0.1 eV\",round(FE1,2)\n", + "print\"Probability when the energy of the state is below 0.1 eV\",round(FE2,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.2 Page No. 58" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Temprature is 758.3 K\n" + ] + } + ], + "source": [ + "#Calculate the temprature at which there is 1 percent probability\n", + "#that a state of 0.30 eV below the fermi energy level will not contain electrons.\n", + "import math\n", + "#Exa 2.2\n", + "Ef=6.25 #EV fermi energy level\n", + "dE=-0.30 #eV\n", + "k=8.61*10**-5 #Boltzman constant\n", + "fE=0.99\n", + "\n", + "#calculation\n", + "#From the probability formula fE=1/(1+math.exp(dE/(k*T)))\n", + "T=(dE)/(k*math.log(1/fE-1))\n", + "\n", + "#result\n", + "print\"The Temprature is\",round(T,1),\"K\" " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.3 Page No. 64" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the fraction of total no. of electron is 8.85e-07\n" + ] + } + ], + "source": [ + "#Example 2.3\n", + "#Determine the fraction of total no. of electron\n", + "\n", + "#Given\n", + "Eg=0.72 #eV\n", + "Ef=0.5*Eg\n", + "dE=Eg-Ef #eV\n", + "k=8.61*10**-5 #Boltzman constant\n", + "T=300 #K\n", + "\n", + "#Calcualtion\n", + "import math\n", + "N=1/(1+math.exp(dE/(k*T)))\n", + "\n", + "\n", + "#Result\n", + "print\"the fraction of total no. of electron is \",round(N,9)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.4 Page No. 64" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The wavwlength is 1.416 A\n" + ] + } + ], + "source": [ + "#Example 2.4\n", + "#Calculate the wave length\n", + "import math\n", + "#Given\n", + "E=300*1.602*10**-19 #eV Energy\n", + "m=9.108*10**-31 #kg, mass of electron\n", + "h=6.626*10**-34 #Planck constant\n", + "\n", + "#Calculation\n", + "v=math.sqrt(2*E/m)\n", + "lam=h*v/E\n", + "\n", + "#Result\n", + "print\"The wavwlength is\",round(lam*10**10,3),\"A\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.5 Page No. 70" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Ratio of electron to hole concentration : 1e+12\n" + ] + } + ], + "source": [ + "#Exa 2.5\n", + "#Find the ratio of electron to hole concentration ratio\n", + "\n", + "#given data\n", + "ni=1.4*10**18\t\t\t#in atoms/m**3\n", + "Nd=1.4*10**24\t\t\t#in atoms/m**3\n", + "n=Nd\t\t\t\t#in atoms/m**3\n", + "\n", + "#Calculation\n", + "p=ni**2/n\t\t\t#in atoms/m**3\n", + "ratio=n/p\t\t\t#unitless\n", + "\n", + "#Result\n", + "print\"Ratio of electron to hole concentration : \",round(ratio,2)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.7 Page no 74" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of current is 0.24 A\n" + ] + } + ], + "source": [ + "#Example 2.7\n", + "#Calculate the magnitude of current\n", + "\n", + "#Given\n", + "n=10**24 #Electron density\n", + "e=1.6*10**-19 #Electron charge\n", + "v=0.015 #m/s drift velocity\n", + "A=10**-4 #m**2 area\n", + "\n", + "#Calculation\n", + "I=n*e*v/A\n", + "\n", + "#Result\n", + "print\"The magnitude of current is\",round(I/10**8,2),\"A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.8 Page No. 74" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Relaxation time in sec : 4.004e-14 s\n", + "Resistivity of conductor in ohm-m : 1.531e-08 ohm m\n", + "velocity of electron with fermi energy is 1390707.0 m/s\n" + ] + } + ], + "source": [ + "#Exa 2.8\n", + "#calculate (i) Relaxation time (ii)Resistivity of conductor (iii) velocity of electron \n", + "\n", + "#given data\n", + "Ef=5.5\t\t\t#in eV\n", + "MUe=7.04*10**-3\t\t#in m**2/V-s\n", + "n=5.8*10**28\t\t#in m**-3\n", + "e=1.6*10**-19\t\t#constant\n", + "m=9.1*10**-31\t\t#in Kg\n", + "\n", + "#calculation\n", + "#part (i)\n", + "import math\n", + "tau=MUe*m/e\t\t#in sec\n", + "rho=1/(n*e*MUe)\t\t#in ohm-m\n", + "vF=math.sqrt(2*Ef*1.6*10**-19/m)\n", + "\n", + "#Result\n", + "print\"Relaxation time in sec : \",tau,\"s\"\n", + "print\"Resistivity of conductor in ohm-m : \",round(rho,11),\"ohm m\"\n", + "print\"velocity of electron with fermi energy is \",round(vF,0),\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.9 Page No. 75" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "NO. of free electrons are 8.49e+28\n", + "mobility of electrons is 0.004 m**2/Vs\n" + ] + } + ], + "source": [ + "#Example 2.9\n", + "#Find (i)the valence electrons per unit volume (ii) mobility\n", + "\n", + "#Given\n", + "rho=1.73*10**-8 #resistivity\n", + "Tav=2.42*10**-14 #Average Time\n", + "e=1.6*10**-19\t\t#constant\n", + "m=9.1*10**-31\t\t#in Kg\n", + "\n", + "#Calculation\n", + "n=m/(e**2*Tav*rho)\n", + "mu=(e*Tav)/m\n", + "\n", + "#Result\n", + "print\"NO. of free electrons are\",round(n,-26)\n", + "print\"mobility of electrons is\",round(mu,3),\"m**2/Vs\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.10 page No. 75" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Relaxation time in sec : 3.95e-14 s\n", + "velocity of electron with fermi energy is 0.7 m/s\n" + ] + } + ], + "source": [ + "#Example 2.10\n", + "#calculate Relaxation time and drift velocity\n", + "\n", + "Ef=100\t\t\t#in V/m Applied electric field\n", + "n=6*10**28\t\t#in m**-3\n", + "e=1.6*10**-19\t\t#constant electronic charge\n", + "m=9.1*10**-31\t\t#in Kg mass of electron\n", + "rho=1.5*10**-8 #Density\n", + "\n", + "#calculation\n", + "import math\n", + "tau=m/(n*e**2*rho)\t\t#in sec\n", + "vF=e*Ef*tau/m\n", + "\n", + "#Result\n", + "print\"Relaxation time in sec : \",round(tau,16),\"s\"\n", + "print\"velocity of electron with fermi energy is \",round(vF,1),\"m/s\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.11 Page No.75" + ] + }, + { + "cell_type": "code", + "execution_count": 69, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Charge density is 1.133e+29 m**-3\n", + "current density is 1160000.0 A/m**2\n", + "curret flowing is 3.644 A\n", + "electron drift velocityis 6.4e-05 m/s\n" + ] + } + ], + "source": [ + "#Exampl 2.11\n", + "#Determine charge density, current density ,Current flowing in the wire, Electron drift velocity\n", + "\n", + "#Given\n", + "d=0.002 #m, diameter of pipe\n", + "s=5.8*10**7 #Conductivity S/m\n", + "mu=0.0032 #m**2/Vs, Electron mobility\n", + "e=1.6*10**-19\t\t#constant electronic charge\n", + "m=9.1*10**-31\t\t#in Kg mass of electron\n", + "E=0.02 #V/m Electric field\n", + "\n", + "#Calculation\n", + "import math\n", + "#From eq 2.62\n", + "n=s/(e*mu)\n", + "J=s*E\n", + "I=J*(math.pi*d**2/4.0)\n", + "v=mu*E\n", + "\n", + "#Result\n", + "print\"Charge density is\",round(n,-26),\"m**-3\"\n", + "print\"current density is\",round(J,6),\"A/m**2\"\n", + "print\"curret flowing is\",round(I,3),\"A\"\n", + "print\"electron drift velocityis\",round(v,6),\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.12 page no 76" + ] + }, + { + "cell_type": "code", + "execution_count": 156, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drift velocity is 20.0 m/s\n", + "Time taken by the electron is 5e-07 s\n" + ] + } + ], + "source": [ + "#example 2.12\n", + "#calculate the drift velocity and time\n", + "\n", + "#Given\n", + "rho=0.5 #ohm-m Resistivity\n", + "J=100 #A/m**2 Current density\n", + "mue=0.4 #m**2/Vs Electron mobility\n", + "d=10*10**-6 #m distance\n", + "\n", + "#calculation\n", + "Ve=mue*J*rho\n", + "t=d/Ve\n", + "\n", + "#Result\n", + "print\"The drift velocity is \",Ve,\"m/s\"\n", + "print\"Time taken by the electron is\",round(t,8),\"s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.13 Page No.76" + ] + }, + { + "cell_type": "code", + "execution_count": 76, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Concentration of electron is 4.45e+16 /cm**3\n", + "Concentration of holes is 14040000000.0 /cm**3\n" + ] + } + ], + "source": [ + "#Example 2.13\n", + "#Calculate drift velocity and time\n", + "\n", + "#Given\n", + "e=1.6*10**-19\t\t#constant electronic charge\n", + "m=9.1*10**-31\t\t#in Kg mass of electron\n", + "rho=0.039 #ohm-cm resistivity\n", + "mu=3600 #cm**2/Vs Carrier mobility\n", + "ni=2.5*10**13\n", + "\n", + "#Calculation \n", + "Nd=(1/(rho*e*mu))\n", + "n=Nd\n", + "p=(ni**2/n)\n", + "\n", + "#Result\n", + "print\"Concentration of electron is\",round(n,-14),\"/cm**3\"\n", + "print\"Concentration of holes is\",round(p,0),\"/cm**3\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.14 page No 76" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Concentration of electrons is 4.41e+14 atoms/cm**3\n", + "Concentration of holes is 1.42e+12 atoms/cm**3\n", + "Conductivity of N-type germanium 26.8 /ohm-m\n" + ] + } + ], + "source": [ + "#Example 2.14\n", + "#Determine concentration of holes and electrons\n", + "\n", + "#Given\n", + "rho=5.32 #kg/m**3, density\n", + "Aw=72.6 #kg/K kmol atomic weight\n", + "ni=2.5*10**13\n", + "di=10**8 #Donor impurity\n", + "e=1.6*10**-19 #Electronic charge\n", + "mue=0.38 #m**/Vs\n", + "muh=0.18 #m**/Vs\n", + "\n", + "#CAlculation\n", + "N=6.023*10**23*rho/Aw #No 0f germanium atoms per cm**3\n", + "Nd=N/di\n", + "n=Nd\n", + "p=(ni**2/n)\n", + "s=n*e*mue*10**4\n", + "\n", + "#Result\n", + "print\"Concentration of electrons is\",round(n,-12),\"atoms/cm**3\"\n", + "print\"Concentration of holes is\",round(p,-10),\"atoms/cm**3\"\n", + "\n", + "if n > p:\n", + " \n", + " print\"Conductivity of N-type germanium\",round(s*100,1),\"/ohm-m\" \n", + "else:\n", + " print \"Calculate p-type germanium conductivity\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.15 Page no.79" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Density of electrons is 2.29e+19 /m**3\n", + "Drift velocity for electrons 3900.0 m/s\n", + "Drift velocity for holes 1900.0 m/s\n" + ] + } + ], + "source": [ + "#Example 2.15\n", + "#Calculate the density and drift velocity\n", + "\n", + "#Given\n", + "e=1.6*10**-19 #Electronic charge\n", + "mue=0.39 #m**/Vs\n", + "muh=0.19 #m**/Vs\n", + "rhoi=0.47 #ohm-m, intrinsic resistivity\n", + "E=10**4 #Electric field\n", + "\n", + "#Calculation\n", + "sigmai=1/rhoi\n", + "ni=sigmai/(e*(mue+muh))\n", + "Vn=mue*E\n", + "Vh=muh*E\n", + "\n", + "#Result\n", + "print\"Density of electrons is\",round(ni,-17),\"/m**3\"\n", + "print\"Drift velocity for electrons\",round(Vn,0),\"m/s\"\n", + "print\"Drift velocity for holes\",round(Vh,0),\"m/s\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.16 page No.80" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Intrinsic conductivity of Ge is 0.0224 ohm-cm**-1\n", + "Conductivity of N type Ge semiconductor is 2.68 ohm-cm**-1\n" + ] + } + ], + "source": [ + "#Example 2.16\n", + "#Calculate conductivity\n", + "\n", + "#Given\n", + "i=10**7 #IMpurity in Ge atom\n", + "ni=2.5*10**13 #/cm**3\n", + "N=4.4*10**22 #No. of atoms of Ge\n", + "mue=3800 #cm**2/Vs\n", + "muh=1800 #cm**2/Vs\n", + "e=1.6*10**-19 #Electronic charge\n", + "E=400 #Electric field\n", + "\n", + "#Calculation\n", + "sigmai=ni*e*(mue+muh)\n", + "Nd=N/i\n", + "n=Nd\n", + "p=ni**2/(Nd)\n", + "sigman=e*Nd*mue\n", + "\n", + "print\"Intrinsic conductivity of Ge is \",sigmai,\"ohm-cm**-1\"\n", + "print\"Conductivity of N type Ge semiconductor is\",round(sigman,2),\"ohm-cm**-1\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.17 Page No. 80" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Electron drift velocity is 152.0 m/s\n", + " hole drift velocity is 72.0 m/s\n", + "(ii)Intrinsic Conductivity of Ge is 2.24 ohm-m**-1\n", + "(iii)The total current is 5.376 mA\n" + ] + } + ], + "source": [ + "#Example 2.17\n", + "#(i)Electron drift velocity & hole drift velocity .\n", + "#(ii)Intrinsic Conductivity of Ge,(iii)The total current .\n", + "\n", + "#Given\n", + "V=10 #Volt\n", + "l=0.025 #m, length\n", + "w=0.004 #m width\n", + "t=0.0015 #m thickness\n", + "\n", + "ni=2.5*10**19 #/cm**3\n", + "mue=0.38 #m**2/Vs\n", + "muh=0.18 #m**2/Vs\n", + "e=1.6*10**-19 #Electronic charge\n", + "E=400 #Electric field\n", + "\n", + "#Calculation\n", + "E=V/l\n", + "Ve=mue*E\n", + "Vh=muh*E\n", + "sigmai=ni*e*(mue+muh)\n", + "I=sigmai*E*w*t\n", + "\n", + "#Result\n", + "print\"(i)Electron drift velocity is \",Ve,\"m/s\"\n", + "print\" hole drift velocity is \",Vh,\"m/s\"\n", + "print\"(ii)Intrinsic Conductivity of Ge is\",sigmai,\"ohm-m**-1\"\n", + "print\"(iii)The total current is \",I*1000,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.18 page no.80" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ratio of electrons to holes drift velocity is 1.0\n" + ] + } + ], + "source": [ + "#Example 2.18\n", + "#What is ratio of electrons to holes\n", + "\n", + "#Given\n", + "Ie=3/4.0 #Current due to electron\n", + "Ih=1-Ie #Current due to holes\n", + "Vh=1 #Hole velocity\n", + "Ve=3 #Electron velocity 3 times the hole velocity\n", + "\n", + "#ccalculation\n", + "R=(Ie*Vh/(Ih*Ve))\n", + "\n", + "#Result\n", + "print\"The ratio of electrons to holes drift velocity is \",R" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.19 Page No.81" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Diffusion constant of electron is 43.99 (in cm**2/s)\n", + "Diffusion constant of hole is 6.47 (in cm**2/s)\n" + ] + } + ], + "source": [ + "#Exa 2.19\n", + "#Find the diffusion coefficients of electrons and holes\n", + "\n", + "#given data\n", + "e=1.6*10**-19\t\t\t#in coulamb\n", + "T=300\t\t\t\t#in Kelvin\n", + "MUh=0.025\t\t\t#in m**2/V-s\n", + "MUe=0.17\t\t\t#in m**2/V-s\n", + "k=1.38*10**-23\t\t\t#in J/K\n", + "De=MUe*k*T/e\t\t\t#in cm**2/s\n", + "Dh=MUh*k*T/e\t\t\t#in cm**2/s\n", + "\n", + "#Result\n", + "print\"Diffusion constant of electron is \",round(De*10000,2),\"(in cm**2/s)\"\n", + "print\"Diffusion constant of hole is \",round(Dh*10000,2),\"(in cm**2/s)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.20 Page no. 81 " + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The intrinsic carries concentration is 1.76e+16 /m**3\n", + "The conductivity of Si is 0.00054 S/m\n" + ] + } + ], + "source": [ + "#Example 2.20\n", + "#Find intrinsic carries cncentration and conductivity\n", + "import math\n", + "#Given\n", + "N=3*10**25 #No of atoms\n", + "e=1.6*10**-19\n", + "Eg=1.1*e #eV\n", + "k=1.38*10**-23 #j/k boltzman's constant\n", + "T=300 #K\n", + "mue=0.14\n", + "muh=0.05\n", + "\n", + "#Calculation\n", + "ni=N*math.exp(-Eg/(2*k*T))\n", + "sigma=ni*e*(mue+muh)\n", + "\n", + "#Result\n", + "print\"The intrinsic carries concentration is \",round(ni,-14),\"/m**3\"\n", + "print\"The conductivity of Si is \",round(sigma,5),\"S/m\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.21 Page No.84" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The effective density is 4.6e+25 /m**3\n" + ] + } + ], + "source": [ + "#Example 2.21\n", + "#Find the effective density\n", + "\n", + "#Given\n", + "a=1.5 #a=me/mo\n", + "T=300 #K\n", + "\n", + "#calculation\n", + "#from eq. 2.29\n", + "Nc=4.82*10**21*(a)**(1.5)*T**(1.5)\n", + "\n", + "#Result\n", + "print\"The effective density is\",round(Nc,-23),\"/m**3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.22 page no. 84" + ] + }, + { + "cell_type": "code", + "execution_count": 55, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The intrinsic concentration of charge carrier is 2.27e+18 /m**3\n" + ] + } + ], + "source": [ + "#Example 2.22\n", + "#Calculate the intrinsic concentration\n", + "\n", + "#Given\n", + "a=0.07 #a=me/mo\n", + "b=0.4 #b=mh/mo\n", + "T=300 #K\n", + "Eg=0.7 #eV\n", + "k=8.62*10**-5 # Boltzman constant\n", + "\n", + "#calculation\n", + "import math\n", + "#From eq 2.101\n", + "ni=math.sqrt(2.33*10**43*(a*b)**(1.5)*T**3*math.exp(-Eg/(k*T)))\n", + "\n", + "#Result\n", + "print\"The intrinsic concentration of charge carrier is\",round(ni,-16),\"/m**3\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.23 Page no. 85" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The absolute temprature is 0.14 K\n" + ] + } + ], + "source": [ + "#Example 2.23\n", + "#Find the value of absolute temprature\n", + "\n", + "#Given\n", + "C=5*10**28 #atom/m**3, concentration of Si atoms\n", + "DL=2*10**8 #Doping level \n", + "m=1\n", + "me=m\n", + "#calculation\n", + "Nd=C/DL\n", + "nc=Nd\n", + "T=((nc/(4.82*10**21))*(m/me)**(1.5))**(2/3.0)\n", + "\n", + "#Result\n", + "print\"The absolute temprature is\",round(T,2),\"K\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.24 Page No. 85" + ] + }, + { + "cell_type": "code", + "execution_count": 110, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The effective density at temprature 300 K is 3.2e+20 /m**3\n", + "The effective density at temprature 400 K is 8.98e+21 /m**3\n" + ] + } + ], + "source": [ + "#Example 2.24\n", + "#Determine the effective density\n", + "\n", + "#Given\n", + "T1=300.0 #K temprature\n", + "T2=400.0\n", + "k=1.38*10**-23 #J/k\n", + "m=1.25*9.107*10**-31\n", + "h=6.625*10**-34\n", + "dE=0.3 #eV\n", + "k_=8.62*10**-5\n", + "\n", + "#calculation\n", + "import math\n", + "nc1=2*(2*math.pi*m*k*T1/(h**2))**(1.5)\n", + "n1=nc1*math.exp(-(0.3/(k_*T1)))\n", + "\n", + "nc2=2*(2*math.pi*m*k*T2/(h**2))**(1.5)\n", + "n2=nc2*math.exp(-(0.3/(k_*T2)))\n", + "\n", + "#result\n", + "print\"The effective density at temprature 300 K is\",round(n1,-19),\"/m**3\"\n", + "print\"The effective density at temprature 400 K is\",round(n2,-19),\"/m**3\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.25 Page no.86" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of fermi level is -0.0079 eV\n" + ] + } + ], + "source": [ + "#example 2.25\n", + "#determine the position of intrinsic fermi level\n", + "import math\n", + "#Given\n", + "T=300.0\n", + "k=8.62*10**-5 #J/k\n", + "m=9.107*10**-31\n", + "me=0.6*m\n", + "mh=0.4*m\n", + "\n", + "\n", + "#calculation\n", + "dE=-3*k*T*math.log((me/mh)**(1))/4.0 #dE=Ef-Emidgap\n", + "\n", + "#Result\n", + "print\"The position of fermi level is\",round(dE,4),\"eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.26 Page no 86" + ] + }, + { + "cell_type": "code", + "execution_count": 131, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of fermi level is 0.3912 eV\n" + ] + } + ], + "source": [ + "#example 2.26\n", + "#determine the position of intrinsic fermi level\n", + "\n", + "#Given\n", + "T=300.0\n", + "Eg=0.72 #eV Energy gap\n", + "k=8.62*10**-5 #J/k\n", + "me=1\n", + "mh=5.0\n", + "\n", + "#calculation\n", + "#from Ef=Ec-kTlog(nc/Nd)\n", + "import math\n", + "dE=(Eg/2.0)-3*k*T*math.log(me/mh)/4.0 #dE=Ef-Emidgap\n", + "\n", + "#Result\n", + "print\"The position of fermi level is\",round(dE,4),\"eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.27 Page no 87" + ] + }, + { + "cell_type": "code", + "execution_count": 134, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of fermi level is 0.28 eV\n" + ] + } + ], + "source": [ + "#example 2.27\n", + "#determine the position of intrinsic fermi level\n", + "\n", + "#Given\n", + "T1=300.0\n", + "T2=350\n", + "Eg=0.24 #eV Energy gap\n", + "\n", + "#calculation\n", + "#from Ef=Ev+kTlog(nc/Nd)\n", + "import math\n", + "dE=(T2/T1)*Eg\n", + "\n", + "#Result\n", + "print\"The position of fermi level is\",round(dE,4),\"eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.28 Page no.88" + ] + }, + { + "cell_type": "code", + "execution_count": 133, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of fermi level is 0.36 eV\n" + ] + } + ], + "source": [ + "#Example 2.28\n", + "#determine the position of intrinsic fermi level\n", + "\n", + "#Given\n", + "T1=300.0\n", + "T2=400\n", + "Eg=0.27 #eV Energy gap\n", + "\n", + "#calculation\n", + "import math\n", + "dE=(T2/T1)*Eg\n", + "\n", + "#Result\n", + "print\"The position of fermi level is\",round(dE,4),\"eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.29 page no.88" + ] + }, + { + "cell_type": "code", + "execution_count": 137, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of fermi level is 0.258 eV\n" + ] + } + ], + "source": [ + "##Example 2.29\n", + "#determine the position of intrinsic fermi level\n", + "\n", + "#Given\n", + "dE1=0.3 #eV Energy gap\n", + "kT=0.026 #eV\n", + "\n", + "#calculation\n", + "import math\n", + "x=math.exp(-dE1/kT) #x=Nd/nc\n", + "y=5 #y=Nd2/Nd1\n", + "dE2=-math.log(y)*kT+dE1\n", + "\n", + "#Result\n", + "print\"The position of fermi level is\",round(dE2,3),\"eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.30 Page no.89" + ] + }, + { + "cell_type": "code", + "execution_count": 143, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The position of fermi level is 0.36 eV\n" + ] + } + ], + "source": [ + "##Example 2.30\n", + "#determine the position of intrinsic fermi level\n", + "\n", + "#Given\n", + "dE1=0.39 #eV Energy gap\n", + "kT=0.026 #eV\n", + "\n", + "#calculation\n", + "import math\n", + "x=math.exp(-dE1/kT) #x=NA1/nV\n", + "y=3 #y=NA2/NA1\n", + "dE2=((dE1/kT)-math.log(y))*kT\n", + "\n", + "\n", + "#Result\n", + "print\"The position of fermi level is\",round(dE2,2),\"eV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.31 Page no.91" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The electron density is 6.25e+22 /m**3\n", + "The mobility is 1e-04 /m**3\n" + ] + } + ], + "source": [ + "#example 2.31\n", + "#Determine electron density and mobility\n", + "\n", + "#Given\n", + "rho=1 #ohm-m Resistivity\n", + "Rh=100.0 #cm**3/coulomb\n", + "e=1.6*10**-19\n", + "\n", + "#calculation\n", + "con=1/rho #Conductivity\n", + "R=1/Rh #Charge density\n", + "ED=R*10**6/e\n", + "mu=con/(R*10**6)\n", + "\n", + "#Result\n", + "print\"The electron density is\",ED,\"/m**3\"\n", + "print\"The mobility is %.e\"%mu,\"/m**3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.32 Page no. 92" + ] + }, + { + "cell_type": "code", + "execution_count": 146, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hall Voltage is 22.8 micro V\n" + ] + } + ], + "source": [ + "#Example 2.32\n", + "#Calculate Hall Voltage\n", + "\n", + "#Given\n", + "w=0.1 #m width\n", + "t=0.01 #m thickness\n", + "F=0.6 #T, field\n", + "Rh=3.8*10**-4 #Hall Coefficient\n", + "I=10 #mA\n", + "\n", + "#calculation\n", + "Vh=(Rh*F*I/w)\n", + "\n", + "#Result\n", + "print\"Hall Voltage is\",Vh*1000,\"micro V\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.33 Page No. 92" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnitude of hall voltage is 76.0 mV\n" + ] + } + ], + "source": [ + "#Exa 2.33\n", + "#What is magnitude of Hall Voltage\n", + "\n", + "#given data\n", + "e=1.6*10**-19\t\t\t#in coulamb\n", + "ND=10**17\t\t\t#in cm**-3\n", + "Bz=0.1\t\t\t\t#in Wb/m**2\n", + "w=4\t\t\t\t#in mm\n", + "d=4\t\t\t\t#in mm\n", + "Ex=5\t\t\t\t#in V/cm\n", + "MUe=3800\t\t\t#in cm**2/V-s\n", + "\n", + "#calculation\n", + "v=MUe*Ex\t\t\t#in cm/s\n", + "v=v*10**-2\t\t\t#in m/s\n", + "VH=Bz*v*d\t\t\t#in mV\n", + "\n", + "#Result\n", + "print\"Magnitude of hall voltage is\",VH,\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.34 Page No.92" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Magnitude of hall voltage is 3.0 mV\n" + ] + } + ], + "source": [ + "#Exa 2.34\n", + "#What is magnitude of hall voltage\n", + "\n", + "#given data\n", + "e=1.6*10**-19\t\t\t#in coulamb\n", + "ND=10**21\t\t\t#in m**-3\n", + "Bz=0.2\t\t\t\t#in T\n", + "d=4\t\t\t\t#in mm\n", + "d=d*10**-3\t\t\t#in meter\n", + "J=600\t\t\t\t#in A/m**2\n", + "n=ND\t\t\t\t#in m**-3\n", + "\n", + "#calculation\n", + "#formula : VH*w/(B*I)=1/(n*e)\n", + "VH=Bz*J*d/(n*e)\t\t\t#in V\n", + "\n", + "#Result\n", + "print\"Magnitude of hall voltage is \",VH*10**3,\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 2.35 Page No." + ] + }, + { + "cell_type": "code", + "execution_count": 169, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hall angle is 1.0709 degree\n" + ] + } + ], + "source": [ + "#Exa 2.35\n", + "#Calculate hall angle\n", + "\n", + "#given data\n", + "e=1.6*10**-19\t\t\t#in coulamb\n", + "rho=0.00912\t\t\t#in ohm-m\n", + "B=0.48\t\t\t\t#in Wb/m**2\n", + "RH=3.55*10**-4\t\t\t#in m**3-coulamb**-1\n", + "SIGMA=1/rho\t\t\t#in (ohm=m)**-1\n", + "\n", + "#calculation\n", + "import math\n", + "THETAh=math.atan(SIGMA*B*RH)\t#in Degree\n", + "\n", + "#result\n", + "print\"Hall angle is\",round(THETAh*180/3.14,4),\"degree\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch3.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch3.ipynb new file mode 100644 index 00000000..6c538996 --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch3.ipynb @@ -0,0 +1,332 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter3 : Excess Carriers in Semiconductor" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.2 Page No 111" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum required energy is 2.06 eV \n" + ] + } + ], + "source": [ + "#Example 3.2\n", + "#What is Minimum required energy \n", + "\n", + "#given data\n", + "l=6000 #in Angstrum\n", + "h=6.6*10**(-34) #Planks constant\n", + "c=3*10**8 #speed of light in m/s\n", + "e=1.602*10**(-19) #Constant\n", + "\n", + "#calculation\n", + "phi=c*h/(e*l*10**(-10))\n", + "\n", + "#result\n", + "print\"Minimum required energy is\",round(phi,2),\"eV \"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.3 Page No 112" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work function of the cathode material is 2.39 eV\n" + ] + } + ], + "source": [ + "#Exa 3.3\n", + "#calculate Work function of the cathode material\n", + "\n", + "#given data\n", + "Emax=2.5 #maximum energy of emitted electrons in eV \n", + "l=2537.0 #in Angstrum\n", + "\n", + "#Calculation\n", + "EeV=12400.0/l #in eV\n", + "phi=EeV-Emax #in eV\n", + "\n", + "#result\n", + "print \"Work function of the cathode material is \",round(phi,2),\"eV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.4 Page No 115" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Thus power absorbed is 0.009 J/s\n", + "(ii)Energy converted into heat is 0.0026 J/s\n", + "(iii)Number of photons per second given off from recombination events 2.81e+16\n" + ] + } + ], + "source": [ + "#Example 3.4\n", + "#Find (i)The fraction of each photon energy unit which is converted into heat\",f\n", + "#(ii)Energy converted into heat in ,((2-1.43)/2)*0.009,\"J/s\"\n", + "#(iii)Number of photons per second given off from recombination events \",0.009/(e*2)\n", + "\n", + "#given data\n", + "t=0.46*10**-4 #in centi meters\n", + "hf1=2 #in ev\n", + "hf2=1.43\n", + "Pin=10 #in mW\n", + "alpha=50000 # in per cm\n", + "e=1.6*10**-19 #constant\n", + "Io=0.01 #in mW\n", + "\n", + "import math\n", + "\n", + "#Calculation\n", + "It=Io*math.exp(-alpha*t) #in mW\n", + "Iabs=Io-It\n", + "f=(hf1-hf2)/hf1\n", + "E=f*Iabs\n", + "N=Iabs/(e*hf1)\n", + "\n", + "#result\n", + "print\"(i)Thus power absorbed is \",round(Iabs,3),\"J/s\"\n", + "print\"(ii)Energy converted into heat is\",round(E,4),\"J/s\"\n", + "print\"(iii)Number of photons per second given off from recombination events \",round(N,-14)\n", + "#In book there is calculation mistake in Number of photons." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.5 Page No 123" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Electron transit time in sec is 6.4e-09 s\n", + "Photoconductor gain is 216.0\n" + ] + } + ], + "source": [ + "#Example 3.5\n", + "#What is Photoconductor gain \n", + "#Electron transit time.\n", + "\n", + "#given data\n", + "L=100 #in uM\n", + "A=10&-7 #in cm**2\n", + "th=10**-6 #in sec\n", + "V=12 #in Volts\n", + "ue=0.13 #in m**2/V-s\n", + "uh=0.05 #in m**2/V-s\n", + "\n", + "#Calculation\n", + "E=V/(L*10**-6) #in V/m\n", + "tn=(L*10**-6)/(ue*E)\n", + "Gain=(1+uh/ue)*(th/tn)\n", + "\n", + "#result\n", + "print\"Electron transit time in sec is \",round(tn,10),\"s\"\n", + "print\"Photoconductor gain is \",Gain" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.6 Page No128" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current flowing through diode is 15.0 micra A\n" + ] + } + ], + "source": [ + "#Example3.6\n", + "#Calculate Current flowing through diode .\n", + "\n", + "#given datex\n", + "import math\n", + "Io=0.15 #in uA\n", + "V=0.12 #in mVolt\n", + "Vt=26 #in mVolt\n", + "\n", + "#calculation\n", + "I=Io*10**-6*(math.exp(V/(Vt*10**-3))-1) #in A\n", + "\n", + "#result\n", + "print\"Current flowing through diode is \",round(I*10**6,2),\"micra A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.7 Page No 128" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Forward voltage is 0.43 V\n" + ] + } + ], + "source": [ + "#Exa 3.7\n", + "#Determine the Forward voltage \n", + "\n", + "#given data\n", + "import math\n", + "Io=2.5 #in uA\n", + "I=10 #in mA\n", + "Vt=26 #in mVolt\n", + "n=2 #for silicon\n", + "\n", + "#Calculation\n", + "V=n*Vt*10**-3*math.log((I*10**-3)/(Io*10**-6))\n", + "\n", + "#Result\n", + "print \"Forward voltage is \",round(V,2),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 3.8 Page No 128" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Reverse saturation current density is 0.16 uA \n" + ] + } + ], + "source": [ + "#Example 3.8\n", + "#What is Reverse saturation current density \n", + "\n", + "#given data\n", + "ND=10**21 #in m**-3\n", + "NA=10**22 #in m**-3\n", + "De=3.4*10**-3 #in m**2-s**-1\n", + "Dh=1.2*10**-3 #in m**2-s**-1\n", + "Le=7.1*10**-4 #in meters\n", + "Lh=3.5*10**-4 #in meters\n", + "ni=1.6*10**16 #in m**-3\n", + "e=1.602*10**-19 #constant\n", + "\n", + "#calculation\n", + "IoA=e*ni**2*(Dh/(Lh*ND)+De/(Le*NA))\n", + "\n", + "#Result\n", + "print\"Reverse saturation current density is \",round(IoA*10**6,2),\"uA \"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch4.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch4.ipynb new file mode 100644 index 00000000..07dd71a0 --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch4.ipynb @@ -0,0 +1,1059 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 4: Junction Properties" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.1 page No. 146" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Majority carrier electron concentration is 5.97e+13 cm**-3\n", + "Minority carrier hole concentration is 9.7e+12 cm**-3\n" + ] + } + ], + "source": [ + "#Exa4.1\n", + "#find the Majority and Minority carrier hole concentration\n", + "\n", + "#given data\n", + "import math\n", + "T=300\t\t\t #in Kelvin\n", + "ND=5*10**13\t\t #in cm**-3\n", + "NA=0\t\t\t #in cm**-3\n", + "ni=2.4*10**13\t\t#in cm**-3\n", + "\n", + "#Calculation\n", + "no=ND/2.0+math.sqrt((ND/2.0)**2+ni**2)\t#in cm**-3\n", + "po=ni**2/no\t\t#in cm**-3\n", + "\n", + "#Result\n", + "print\"Majority carrier electron concentration is \",round(no,-11),\"cm**-3\"\n", + "print\"Minority carrier hole concentration is \",round(po,-11),\" cm**-3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.2 Page No.146" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Majority carrier electron concentration is 1e+16 cm**-3\n", + "Minority carrier hole concentration is 22500.0 cm**-3\n" + ] + } + ], + "source": [ + "#Exa4.2\n", + "#find the Majority and Minority carrier hole concentration\n", + "\n", + "#given data\n", + "import math\n", + "T=300\t\t\t#in Kelvin\n", + "ND=10**16\t\t#in cm**-3\n", + "NA=0\t\t\t #in cm**-3\n", + "ni=1.5*10**10\t\t#in cm**-3\n", + "\n", + "#Calculation\n", + "no=ND/2.0+math.sqrt((ND/2.0)**2+ni**2)\t#in cm**-3\n", + "po=ni**2/no\t\t#in cm**-3\n", + "\n", + "#result\n", + "print\"Majority carrier electron concentration is \",no,\"cm**-3\"\n", + "print\"Minority carrier hole concentration is \",round(po,0),\" cm**-3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.3 Page No. 147" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Majority carrier hole concentration is 7e+15 cm**-3\n", + "Minority carrier electron concentration is 36571.0 cm**-3\n" + ] + } + ], + "source": [ + "#Exa4.3\n", + "#find the Majority and Minority carrier hole concentration\n", + "\n", + "#given data\n", + "import math\n", + "T=300\t\t\t#in Kelvin\n", + "ND=3*10**15\t\t#in cm**-3\n", + "NA=10**16\t\t#in cm**-3\n", + "ni=1.6*10**10\t\t#in cm**-3\n", + "\n", + "#Calculation\n", + "po=(NA-ND)/2+math.sqrt(((NA-ND)/2.0)**2+ni**2.0)\t#in cm**-3\n", + "no=ni**2/po\t\t#in cm**-3\n", + "\n", + "#Result\n", + "print\"Majority carrier hole concentration is\",round(po,-8),\" cm**-3\"\n", + "print\"Minority carrier electron concentration is \",round(no,0),\" cm**-3\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.4 Page No. 147" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum Temprature is 642.0 K\n" + ] + } + ], + "source": [ + "#Example 4.4\n", + "#What is maximum Temprature\n", + "\n", + "#Given \n", + "import math\n", + "ND=3*10**15\t\t#in cm**-3\n", + "Eg=1.12 #eV\n", + "k=8.62*10**-5 #eV/k\n", + "Nc=2.8*10**19\n", + "Nv=1.04*10**19\n", + "\n", + "#Calculation\n", + "import math\n", + "# from the equation po=(NA-ND)/2+math.sqrt(((NA-ND)/2.0)**2+ni**2.0)\t#in cm**-3\n", + "No=1.05*ND\n", + "ni=math.sqrt((No-ND/2.0)**2-0.25*ND**2)\n", + "#From ni**2=Nc*Nv*exp(-Eg/(k*t))\n", + "T=Eg/(-math.log(ni**2/(Nc*Nv))*k)\n", + "\n", + "#Result\n", + "print \"The maximum Temprature is \",round(T,1),\"K\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.5 Page No. 151" + ] + }, + { + "cell_type": "code", + "execution_count": 47, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Built in potential barrier is 0.7532 V\n" + ] + } + ], + "source": [ + "#Exa4.5\n", + "#determine the built in potential\n", + "\n", + "#given data\n", + "import math\n", + "T=300\t\t#in Kelvin\n", + "ND=10**15\t#in cm**-3\n", + "NA=10**18\t#in cm**-3\n", + "ni=1.5*10**10\t#in cm**-3\n", + "VT=T/11600.0\t#in Volts\n", + "\n", + "#Calculation\n", + "Vbi=VT*math.log(NA*ND/ni**2)\t#in Volts\n", + "\n", + "#result\n", + "print\"Built in potential barrier is\",round(Vbi,4),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.6 Page No.151" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Contact potential is 0.5745 V\n" + ] + } + ], + "source": [ + "#Exa4.6\n", + "#What is Contact Potential.\n", + "import math\n", + "#given data\n", + "T=300\t\t #in Kelvin\n", + "ND=10**21\t #in m**-3\n", + "NA=10**21\t #in m**-3\n", + "ni=1.5*10**16 #in m**-3\n", + "VT=T/11600.0\t#in Volts\n", + "\n", + "#Calculation\n", + "import math\n", + "Vo=VT*math.log(NA*ND/ni**2)\t#in Volts\n", + "\n", + "#result\n", + "print\"Contact potential is\",round(Vo,4),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.7 Page No. 154" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Space charge width is 0.95 micro meter\n", + "At metallurgical junction, i.e for x=0 the electric field is -13345.0 V\n" + ] + } + ], + "source": [ + "#Exa4.7\n", + "#Determine the space charge.\n", + "\n", + "#given data\n", + "import math\n", + "T=300\t\t\t#in Kelvin\n", + "ND=10**15\t\t#in cm**-3\n", + "NA=10**16\t\t#in cm**-3\n", + "ni=1.5*10**10\t\t#in cm**-3\n", + "VT=T/11600.0\t\t#in Volts\n", + "e=1.6*10**-19\t #in Coulamb\n", + "\n", + "#calculation\n", + "epsilon=11.7*8.854*10**-14\t #constant\n", + "Vbi=VT*math.log(NA*ND/ni**2)\t\t#in Volts\n", + "SCW=math.sqrt((2*epsilon*Vbi/e)*(NA+ND)/(NA*ND))#in cm\n", + "SCW=SCW*10**4 #in uMeter\n", + "xn=0.864\t\t#in uM\n", + "xp=0.086\t\t#in uM\n", + "Emax=-e*ND*xn/epsilon\t#in V/cm\n", + "\n", + "#result\n", + "print\"Space charge width is\",round(SCW,2),\"micro meter\"\n", + "print\"At metallurgical junction, i.e for x=0 the electric field is \",round(Emax/10000,0),\"V\"#Note : Ans in the book is wrong" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.8 Page No.160" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "New position of fermi level is 0.328 V\n" + ] + } + ], + "source": [ + "#Exa4.8\n", + "#Find the new position of fermi level\n", + "\n", + "#given data\n", + "import math\n", + "Ecf=0.3 #in Volts\n", + "T=27.0+273.0 #in Kelvin\n", + "delT=55 #in degree centigrade\n", + "\n", + "#calculation\n", + "#formula : Ecf=Ec-Ef=K*T*math.log(nc/ND)\n", + "#let K*math.log(nc/ND)=y\n", + "#Ecf=Ec-Ef=T*y\n", + "y=Ecf/T #assumed\n", + "Tnew=273+55 #in Kelvin\n", + "EcfNEW=y*Tnew #in Volts\n", + "\n", + "#result\n", + "print\"New position of fermi level is \",round(EcfNEW,4),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.9 Page No. 161" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Contact potential is 0.19 V\n" + ] + } + ], + "source": [ + "#Exa4.9\n", + "#Determine the Contact Potential\n", + "\n", + "#given data\n", + "import math\n", + "T=300\t\t\t#in Kelvin\n", + "ND=8*10**14\t\t#in cm**-3\n", + "NA=8*10**14\t\t#in cm**-3\n", + "ni=2*10**13\t\t#in cm**-3\n", + "k=8.61*10**-5\t\t#in eV/K\n", + "\n", + "#calculation\n", + "Vo=k*T*math.log(NA*ND/ni**2)\t#in Volts\n", + "\n", + "#Result\n", + "print\"Contact potential is \",round(Vo,2),\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.10 page No.161" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Hole concentration in cm**-3 : 1.3e+04 /cm**3\n", + "electron concentration in cm**-3 :5.3e+04 /cm**3\n", + "\n", + "NOTE:\n", + "Slight Variation in answer due to wrong value of ni in book as 1.6*10**16 instead of 1.63166259315e+16\n", + "\n", + "The given Si is of N-type\n" + ] + } + ], + "source": [ + "#Example 4.10\n", + "#(i)Find the hole and electron concentration \n", + "#Is this Silicon P or N type\n", + "from math import e\n", + "#given data\n", + "ND=2*10**16 #in cm**-3\n", + "NA=5*10**15 #in cm**-3\n", + "Ao=4.83*10**21 \t#constant\n", + "T=300.0\t\t\t #in Kelvin\n", + "EG=1.1\t \t \t #in eV\n", + "kT=0.026 \t\t#in eV\n", + "\n", + "#Calculation\n", + "ni=Ao*T**(1.5)*math.exp(-EG/(2*kT))\t\t#in m**-3\n", + "p=(ni/10**6)**2/ND\t\t\t#in cm**-3\n", + "n=((ni/10**6)**2)/NA\t\t\t#in cm**-3\n", + "\n", + "#Result\n", + "\n", + "print\"Hole concentration in cm**-3 : %.1e\"%round(p,0),\"/cm**3\"\n", + "print\"electron concentration in cm**-3 :%.1e\"%round(n,0),\"/cm**3\"\n", + "print\"\\nNOTE:\\nSlight Variation in answer due to wrong value of ni in book as 1.6*10**16 instead of\",ni\n", + "if n < e:\n", + " \n", + " print\"\\n\\nthe given Si is of P-type\" \n", + "else:\n", + " print \"\\nThe given Si is of N-type\"\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.11 Page No. 168" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current flowing through the circuit is 43.0 mA\n" + ] + } + ], + "source": [ + "#Exa4.11\n", + "#Determine current\n", + "\n", + "#In given circuit \n", + "V=5\t\t #in volts\n", + "Vo=0.7\t #in Volts\n", + "R=100\t\t#in Kohm\n", + "\n", + "#Calculation\n", + "I=(V-Vo)/R\t#in Ampere\n", + "\n", + "#result\n", + "print\"Current flowing through the circuit is\",round(I*1000,0),\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.12 Page No. 168" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltagee VA is 13.6 V\n" + ] + } + ], + "source": [ + "#Exa4.12\n", + "#Find the Voltage VA\n", + "\n", + "#In given circuit \n", + "V=15\t\t\t #in volts\n", + "Vo=0.7\t\t\t#in Volts\n", + "R=7\t \t \t#in Kohm\n", + "\n", + "#Calculation\n", + "I=(V-2*Vo)/R\n", + "I=(V-2*Vo)/R\t\t#in mAmpere\n", + "VA=I*R\t \t\t#in Volts\n", + "\n", + "#result\n", + "print\"Voltagee VA is \",VA,\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.13 Page No.169" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The Voltage VA is 14.7 V\n" + ] + } + ], + "source": [ + "#Example 4.13\n", + "#Determine the Voltage VA\n", + "\n", + "#Given\n", + "V=15 #V, voltage\n", + "Vb=0.3 #V, Barrier Potential #When supply is switched on\n", + "\n", + "#Calculation\n", + "VA=V-Vb\n", + "\n", + "#Result\n", + "print\"The Voltage VA is \",VA,\"V\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.14 Page No.172" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Temperature coefficient f zener diode is -0.053 percent\n" + ] + } + ], + "source": [ + "#Exa4.14\n", + "#find Temperature coefficient f zener diode\n", + "\n", + "#given data\n", + "Vz=5\t\t\t#in volts\n", + "to=25\t\t\t#in degree centigrade\n", + "t=100\t\t\t#in degree centigrade\n", + "Vdrop=4.8\t\t#in Volts\n", + "\n", + "#calculation\n", + "delVz=Vdrop-Vz\t\t#in Volts\n", + "delt=t-to\t\t#in degree centigrade\n", + "TempCoeff=delVz*100/(Vz*delt)\n", + "\n", + "#result\n", + "print\"Temperature coefficient f zener diode is \",round(TempCoeff,3),\"percent\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.15 Page No. 174" + ] + }, + { + "cell_type": "code", + "execution_count": 47, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(a)Output voltage will be equal to Vout= 8.0 Volts\n", + "(b)Voltage across Rs is Rs= 4.0 V\n", + "(c)Current through zener diode is Iz= 0.0 mA\n" + ] + } + ], + "source": [ + "#Exa4.15\n", + "#Find (a)output Voltage (b) Voltage across Rs (c) Current\n", + "\n", + "#given data\n", + "Vz=8.0\t\t\t#in volts\n", + "VS=12.0\t\t\t#in volts\n", + "RL=10.0\t\t\t#in Kohm\n", + "Rs=5.0\t\t\t#in Kohm\n", + "\n", + "#part (a)\n", + "Vout=Vz\t\t\t#in volts\n", + "\n", + "#part (b)\n", + "Vrs=VS-Vout\t\t#in volts\n", + "IL=Vout/RL \t\t#in mAmpere\n", + "Is=(VS-Vout)/Rs\t#in mAmpere\n", + "\n", + "#part c\n", + "Iz=Is-IL\t \t#in mAmpere\n", + "\n", + "#result\n", + "print\"(a)Output voltage will be equal to Vout=\",Vout,\" Volts\"\n", + "print\"(b)Voltage across Rs is Rs=\",Vrs,\"V\"\n", + "print\"(c)Current through zener diode is Iz=\",round(Iz,1),\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.16 Page No. 175" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum zener diode current is 9.0 mA\n", + "Minimum zener diode current is 1.0 mA\n" + ] + } + ], + "source": [ + "#Exa4.16\n", + "#Find the min and max value of zener diode current\n", + "\n", + "#given data\n", + "Vz=50.\t\t\t#in volts\n", + "VSmax=120.0\t\t#in volts\n", + "VSmin=80.0\t\t#in volts\n", + "RL=10.0\t\t\t#in Kohm\n", + "Rs=5.0\t\t\t#in Kohm\n", + "\n", + "#Calculation\n", + "Vout=Vz\t\t\t#in Volts\n", + "IL=Vout/RL\t\t#in mAmpere\n", + "\n", + "ISmax=(VSmax-Vout)/Rs\t#in mAmpere\n", + "Izmax=ISmax-IL\t\t#in mA\n", + "Ismin=(VSmin-Vout)/Rs#in mAmpere\n", + "Izmin=Ismin-IL#in mA\n", + "\n", + "#Result\n", + "print\"Maximum zener diode current is \",Izmax,\"mA\"\n", + "print\"Minimum zener diode current is \",Izmin,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.17 Page No. 175" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "sereis Resistance is 192.3 ohm\n", + "The zener current will be minimum i.e. Izk = 6mA when load current is maximum i.e. ILmax = 20mA\n", + "when the load current will decrease and become 10 mA, the zener current will increase and become 6+10 i.e. 16 mA. \n", + "Thus the current through series resistance Rs will remain unchanged at 6+20 i.e. 26 mA. \n", + "Thus voltage drop in series resistance Rs will remain constant. Consequently, the output voltage will also remain constant. \n" + ] + } + ], + "source": [ + "#Exa4.17\n", + "#Design a regulator\n", + "\n", + "#given data\n", + "Vz=15\t\t#in volts\n", + "Izk=6.0\t\t#in mA\n", + "Vout=15\t\t#in Volts\n", + "Vs=20\t\t#in Volts\n", + "ILmin=10.0\t#in mA\n", + "ILmax=20.0\t#in mA\n", + "RS=(Vs-Vz)*1000/(ILmax+Izk)\t#in ohm\n", + "\n", + "#result\n", + "print\"sereis Resistance is \",round(RS,1),\"ohm\"\n", + "print\"The zener current will be minimum i.e. Izk = 6mA when load current is maximum i.e. ILmax = 20mA\"\n", + "print\"when the load current will decrease and become 10 mA, the zener current will increase and become 6+10 i.e. 16 mA. \\nThus the current through series resistance Rs will remain unchanged at 6+20 i.e. 26 mA. \\nThus voltage drop in series resistance Rs will remain constant. Consequently, the output voltage will also remain constant. \"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.18 Page No. 175" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When zener open circuited Voltage across load is 8.73 V\n", + "Zener current is 0 mA\n", + "Power is 0.0 watt\n" + ] + } + ], + "source": [ + "#Exa4.18\n", + "#Determine Vl,Iz,Pz\n", + "\n", + "#given data\n", + "Vs=16.0\t\t #in volts\n", + "RL=1.2\t\t\t#in Kohm\n", + "Rs=1.0\t\t\t#in Kohm\n", + "\n", + "#calculation\n", + "#If zener open circuited\n", + "VL=Vs*RL/(Rs+RL)\t#in Volts\n", + "Iz=0\t\t\t#in mA\n", + "Pz=VL*Iz\t\t#in watts\n", + "\n", + "#result\n", + "print\"When zener open circuited Voltage across load is \",round(VL,2),\"V\"\n", + "print\"Zener current is \",Iz,\"mA\"\n", + "print\"Power is\",Pz,\"watt\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.19 Page No. 126" + ] + }, + { + "cell_type": "code", + "execution_count": 64, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Zener diode will not conduct and VL= 9.5 V\n", + "When RL=200 ohm\n", + "IL is 47.62 mA\n", + "IR is 47.62 mA\n", + "Iz in mA: 0.0 mA\n", + "Zener diode will not conduct and VL= 3.7 V\n", + "When RL=50 ohm\n", + "IL is 74.07 mA\n", + "IR is 74.07 mA\n", + "Iz in mA: 0 mA\n" + ] + } + ], + "source": [ + "#Exa4.19\n", + "#determine VL,IL,IZ,IR\n", + "\n", + "#given data\n", + "Vin=20\t\t\t#in volts\n", + "Rs=220.0\t\t\t#in Kohm\n", + "Vz=10\t\t \t#in volts\n", + "RL2=50.0\t\t\t#in Kohm\n", + "RL1=200\t\t\t#in Kohm\n", + "\n", + "#calculation\n", + "# part (i) RL=50\t#in Kohm\n", + "VL1=Vin*RL1/(RL+Rs)\n", + "IR=Vin/(Rs+RL)\t#in mA\n", + "IL=IR\t\t \t#in mA\n", + "IZ=0\t\t\t #in mA\n", + "\n", + "if VL1< Vz:\n", + " \n", + " print\"Zener diode will not conduct and VL=\",round(VL1,1),\"V\" \n", + "else:\n", + " print \"Zener diode will conduct\"\n", + "\n", + " \n", + "#Result\n", + "print\"When RL=200 ohm\"\n", + "print\"IL is\",round(IL*1000,2),\"mA\"\n", + "print\"IR is\",round(IR*10**3,2),\"mA\"\n", + "print\"Iz in mA: \",round(IZ,0),\"mA\"\n", + "\n", + "# part (ii) RL=200#in Kohm\n", + "RL=200\t\t\t#in Kohm\n", + "VL2=Vin*RL2/(RL2+Rs)\n", + "IR=Vin/(Rs+RL2)\t\t#in mA\n", + "IL=IR\t\t\t#in mA\n", + "IZ=0\t\t\t#in mA\n", + "\n", + "#result\n", + "if VL2< Vz:\n", + " \n", + " print\"Zener diode will not conduct and VL=\",round(VL2,1),\"V\" \n", + "else:\n", + " print \"Zener diode will conduct\"\n", + "\n", + "print\"When RL=50 ohm\"\n", + "print\"IL is\",round(IL*1000,2),\"mA\"\n", + "print\"IR is\",round(IR*10**3,2),\"mA\"\n", + "print\"Iz in mA: \",IZ,\"mA\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.20 Page No. 176" + ] + }, + { + "cell_type": "code", + "execution_count": 67, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "zener diode is ON state\n", + "Hence the voltage dropp across the 5 Kohm resistor in Volts is 50 V\n" + ] + } + ], + "source": [ + "#Exa4.20\n", + "#Find the voltage drop across the resistance\n", + "\n", + "#given data\n", + "RL=10.0\t\t\t #in Kohm\n", + "Rs=5.0 #in Kohm\n", + "Vin=100\t\t\t #in Volts\n", + "\n", + "#Calculation\n", + "V=Vin*RL/(RL+Rs)\t#in Volt\n", + "VZ=50\t\t\t#in Volts\n", + "VL=VZ\t\t\t#in volts\n", + "#Apply KVL\n", + "VR=100-50\t\t#in Volts\n", + "VR=50\t\t\t#in Volts\n", + "\n", + "if V< VZ:\n", + " \n", + " print\"Zener diode is OFF state\" \n", + "else:\n", + " print \"zener diode is ON state\"\n", + "\n", + "print\"Hence the voltage dropp across the 5 Kohm resistor in Volts is \",VR,\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.21 Page No. 176" + ] + }, + { + "cell_type": "code", + "execution_count": 72, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance Ri is 25.0 ohm\n" + ] + } + ], + "source": [ + "#Exa 4.21\n", + "#Find the input resistance\n", + "\n", + "#given data\n", + "RL=120.0\t\t\t#in ohm, load resistance\n", + "Izmin=20\t\t#in mA min. diode current\n", + "Izmax=200\t\t#in mA max. diode current\n", + "VL=12\t\t\t#in Volts\n", + "VDCmin=15\t\t#in Volts\n", + "VDCmax=19.5\t\t#in Volts\n", + "Vz=12\t\t\t#in Volts\n", + "IL=VL/RL\t\t#in Ampere\n", + "IL=IL*1000\t\t#in mAmpere\n", + "\n", + "#calculation\n", + "#For VDCmin = 15 volts\n", + "VSmin=VDCmin-Vz\t\t#in Volts\n", + "#For VDCmax = 19.5 volts\n", + "VSmax=VDCmax-Vz\t\t#in Volts\n", + "ISmin=Izmin+IL\t\t#in mA\n", + "Ri=VSmin/ISmin\t\t#in Kohm\n", + "Ri=Ri*10**3\t\t#in ohm\n", + "\n", + "#result\n", + "print\"The resistance Ri is \",Ri,\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 4.22 Page No. 177" + ] + }, + { + "cell_type": "code", + "execution_count": 71, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Range of RL in Kohm : From 250.0 ohm to 1.25 kohm\n", + "Range of IL in mA : From 8.0 mA to 40.0 mA\n" + ] + } + ], + "source": [ + "#Exa4.22\n", + "#Determine the range of Rl and Il\n", + "\n", + "#given data\n", + "VRL=10\t\t\t#in Volts Diode resistance\n", + "Vi=50\t\t\t#in Volts\n", + "R=1.0\t\t\t#in Kohm Resistance\n", + "Vz=10\t\t\t#in Volts\n", + "VL=Vz\t\t\t#in Volts\n", + "Izm=32\t\t\t#in mA\n", + "IR=(Vi-VL)/R\t\t#in mA\n", + "\n", + "Izmin=0\t\t\t #in mA\n", + "ILmax=IR-Izmin\t\t#in mA\n", + "RLmin=VL/ILmax\t\t#in Ohm\n", + "Izmax=32\t\t #in mA\n", + "ILmin=IR-Izmax\t\t#in mA\n", + "VL=Vz\t\t\t #in Volts\n", + "RLmax=VL/ILmin\t\t#in Ohm\n", + "\n", + "#Result\n", + "print\"Range of RL in Kohm : From \",RLmin*1000,\"ohm to \",RLmax,\"kohm\"\n", + "print\"Range of IL in mA : From \",ILmin,\"mA to \",ILmax,\"mA\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch5.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch5.ipynb new file mode 100644 index 00000000..3795cda6 --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch5.ipynb @@ -0,0 +1,168 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 5: Junction Properties (Continued)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.1 Page No 191" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When no external voltage is applied, Junction width is 3.9e-07 m\n", + "When external voltage of -10 Volt is applied, Junction width is 1.5e-06 m\n" + ] + } + ], + "source": [ + "#Exa 5.1\n", + "#Estimate the junction width in two cases.\n", + "\n", + "#given data\n", + "import math\n", + "ND=10**17 #in atoms/cm**3\n", + "NA=0.5*10**16 #in atoms/cm**3\n", + "Vo=0.7 #in Volts\n", + "V=-10.0 #in Volts\n", + "ND=ND*10**6 #in atoms/m**3\n", + "NA=NA*10**6 #in atoms/m**3\n", + "epsilon=8.85*10**-11 #in F/m\n", + "e=1.6*10**-19 #coulamb\n", + "\n", + "#Calculation\n", + "#part (i)\n", + "#print \"When no external voltage is applied i.e. V=0\"\n", + "#print\"VB = 0.7 volts\"\n", + "VB=0.7 #in Volts\n", + "W1=math.sqrt(2*epsilon*VB*(1/NA+1/ND)/e) #in m\n", + "\n", + "#part (ii)\n", + "#print\"When external voltage of -10 volt is applied\"\n", + "#print\"VB = Vo-V volts\"\n", + "VB=Vo-V #in Volts\n", + "W2=math.sqrt(2*epsilon*VB*(1/NA+1/ND)/e) #in m\n", + "\n", + "#result\n", + "print \"When no external voltage is applied, Junction width is \",round(W1,8),\"m\"\n", + "print\"When external voltage of -10 Volt is applied, Junction width is \",round(W2,7),\"m\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.3 Page No 195" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Junction capacitance is 21.59 pF\n" + ] + } + ], + "source": [ + "#Exa 5.3\n", + "#Determine the junction capacitance\n", + "\n", + "#given data\n", + "CTzero=50 #in pF\n", + "VR=8 #in Volt\n", + "VK=0.7 #in Volt\n", + "n=1/3.0 #for Si\n", + "\n", + "#calculation\n", + "CT=CTzero/((1+VR/VK)**n) #in pF\n", + "\n", + "#result\n", + "print\"Junction capacitance is\",round(CT,2),\"pF\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 5.4 Page No.196" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The tuning range of circuit lies between 318.31 khz and 1007.0 Mhz\n" + ] + } + ], + "source": [ + "#Example 5.4\n", + "#Determine the tuning range of the circuit\n", + "import math\n", + "#Given\n", + "L=12.5*10**-3 #mH inductance\n", + "C1=4.0 #pF Capacitance\n", + "C2=40.0 #pF Capacitance\n", + "\n", + "#Calculation\n", + "Ctmin=(C1*C1)/(C1+C1) #Min value of total Capacitance\n", + "Ctmax=(C2*C2)/(C2+C2) #Max value of total Capacitance\n", + "Fmax=1/(2*math.pi*math.sqrt(L*Ctmin*10**-12))\n", + "Fmin=1/(2*math.pi*math.sqrt(L*Ctmax*10**-12))\n", + "\n", + "#result\n", + "print\"The tuning range of circuit lies between\",round(Fmin/1000,2),\"khz and\",round(Fmax/1000,0),\"Mhz\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch6.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch6.ipynb new file mode 100644 index 00000000..5a300573 --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch6.ipynb @@ -0,0 +1,757 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 6: Bipolar junction Transistors (BJTs)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.1 page No.215" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Emitter current is 0.05 mA\n" + ] + } + ], + "source": [ + "#Exa 6.1\n", + "#find the Base current\n", + "\n", + "#given data\n", + "Ic=9.95\t\t\t#in mA\n", + "Ie=10 \t\t#in mA\n", + "\n", + "#Calculation\n", + "Ib=Ie-Ic\t\t#in mA\n", + "\n", + "#result\n", + "print\"Emitter current is \",Ib,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.2 page No. 216" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Emitter current is 1.0 mA\n", + "Current amplification factor is 0.98\n", + "Current gain factor is 49.0\n" + ] + } + ], + "source": [ + "#Exa 6.2\n", + "#Find (i)Emitter current (ii)Current amplification factor (iii)Current gain factor \n", + "\n", + "#given data\n", + "IC=0.98\t\t\t#in mA\n", + "IB=20.0\t\t\t#in uA\n", + "IB=IB*10**-3\t\t#in mA\n", + "\n", + "#Calculation\n", + "#part (i)\n", + "IE=IB+IC\t\t#in mA\n", + "\n", + "#part (ii)\n", + "alpha=IC/IE\t\t#unitless\n", + "#part (iii)\n", + "Beta=IC/IB\t\t#unitless\n", + "\n", + "#Result\n", + "print\"Emitter current is\",IE,\"mA\"\n", + "print\"Current amplification factor is \",alpha\n", + "print\"Current gain factor is \",Beta" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.3 page No.216" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Emitter current is 2.7 mA\n", + "Collector current is 2.65 mA\n" + ] + } + ], + "source": [ + "#Exa 6.3\n", + "#Emitter current and Collector current\n", + "\n", + "#given data\n", + "alfaDC=0.98\t\t\t#unitless\n", + "ICBO=4\t\t\t\t#in uA\n", + "ICBO=ICBO*10**-3\t\t#in mA\n", + "IB=50\t\t\t\t#in uA\n", + "IB=IB*10**-3\t\t\t#in mA\n", + "\n", + "#calculation\n", + "#Formula : IC=alfaDC*(IB+IC)+ICBO\n", + "IC=alfaDC*IB/(1-alfaDC)+ICBO/(1-alfaDC)\t#in mA\n", + "IE=IC+IB\t\t\t#in mA\n", + "\n", + "#Result\n", + "print\"Emitter current is \",IE,\"mA\"\n", + "print\"Collector current is \",IC,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.4 page No. 216" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector current in mA : 1.09 mA\n" + ] + } + ], + "source": [ + "#Exa 6.4\n", + "#Find the collector current\n", + "\n", + "#given data\n", + "IB=10\t\t\t#in uA\n", + "IB=IB*10**-3\t\t#in mA\n", + "Beta=99\t\t\t#Unitless\n", + "ICO=1\t\t\t#in uA\n", + "ICO=ICO*10**-3\t\t#in mA\n", + "\n", + "#calculation\n", + "#Formula : IC=alfa*(IB+IC)+ICO\n", + "IC=Beta*IB+(1+Beta)*ICO\t#in mA\n", + "\n", + "#Result\n", + "print\"Collector current in mA : \",IC,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.5 Page No.216" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Current gain factor is 98.0\n", + " Current amplification factor is 0.99\n", + " Emitter Current is 5.05 mA\n", + "(ii)New level of Ib is 101.0 micro A\n" + ] + } + ], + "source": [ + "#Example 6.5\n", + "#Find (i) alpha , beta and Ie \n", + "#(ii)New level of Ib\n", + "\n", + "#Given\n", + "Ic=5*10**-3 #mA collector current\n", + "Ic_=10*10**-3 #mA collector current\n", + "Ib=50*10**-6 #mA, Base current\n", + "Icbo=1*10**-6 #micro A, Current to base open current\n", + "\n", + "#Calculation\n", + "beta=(Ic-Icbo)/(Ib+Icbo)\n", + "alpha=(beta/(1+beta))\n", + "Ie=Ib+Ic\n", + "\n", + "Ib=(Ic_-(beta+1)*Icbo)/(beta)\n", + "\n", + "#Result\n", + "print\"(i) Current gain factor is\",round(beta,0)\n", + "print\" Current amplification factor is\",round(alpha,2)\n", + "print\" Emitter Current is\",Ie*1000,\"mA\"\n", + "print\"(ii)New level of Ib is\",round(Ib*10**6,0),\"micro A\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.6 page No. 222" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Dynamic input resistance is 40 mohm\n" + ] + } + ], + "source": [ + "#Exa 6.6\n", + "#Find the dynamic input resistance\n", + "\n", + "#given data\n", + "delVEB=200\t\t\t#in Volts\n", + "delIE=5\t\t\t\t#in mA\n", + "\n", + "#calculation\n", + "rin=delVEB/delIE\t\t#in ohm\n", + "\n", + "#Result\n", + "print\"Dynamic input resistance is \",rin,\"mohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.7 page No. 222" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current gain : 0.979\n", + "Base current is 0.03 mA\n" + ] + } + ], + "source": [ + "#Exa 6.7\n", + "#Determine Current gain and base current\n", + "\n", + "\n", + "#given data\n", + "ICBO=12.5 \t\t\t#in uA\n", + "ICBO=ICBO*10**-3 \t\t#in mA\n", + "IE=2 \t\t\t\t#in mA\n", + "IC=1.97 \t\t\t#in mA\n", + "\n", + "#calculation\n", + "alfa=(IC-ICBO)/IE \t\t#unitless\n", + "IB=IE-IC \t\t\t#in mA\n", + "\n", + "#result\n", + "print\"Current gain : \",round(alfa,3)\n", + "print\"Base current is \",IB,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.8 page No. 222" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Base current ia 0.03 mA\n" + ] + } + ], + "source": [ + "#Exa 6.8\n", + "#given data\n", + "RL=4.0 \t\t\t#in Kohm\n", + "VL=3.0\t\t\t#in volt\n", + "alfa=0.96 \t\t#unitless\n", + "IC=VL/RL \t\t#in mA\n", + "\n", + "#calculation\n", + "IE=IC/alfa \t\t#in mA\n", + "IB=IE-IC \t\t#in mA\n", + "\n", + "#result\n", + "print\"Base current ia\",round(IB,2),\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.9 page No.227" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector-emitter Voltage is 9.2 V\n", + "Base current in uA : 41.67 microA\n" + ] + } + ], + "source": [ + "#Exa 6.9\n", + "#Determine Collector emitter voltage and base current\n", + "\n", + "#given data\n", + "VCC=10\t\t\t #in volt\n", + "RL=800\t\t\t #in ohm\n", + "VL=0.8\t\t\t #in volt\n", + "alfa=0.96\t\t #unitless\n", + "\n", + "#calculation\n", + "#VR=IC*RL\n", + "VCE=VCC-VL \t\t#in Volt\n", + "IC=VL*1000/RL \t\t#in mA\n", + "Beta=alfa/(1-alfa) \t#unitless\n", + "IB=IC/Beta \t\t#in mA\n", + "\n", + "#Result\n", + "print\"Collector-emitter Voltage is \",VCE,\"V\"\n", + "print\"Base current in uA : \",round(IB*1000,2),\"microA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.10 page No. 227" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector current is 11.28 mA\n" + ] + } + ], + "source": [ + "#Exa 6.10\n", + "#Determine Collector Current\n", + "\n", + "#given data\n", + "alfao=0.98 \t\t#unitless\n", + "ICO=10 \t\t\t#in uA\n", + "ICO=ICO*10**-3 \t\t#in mA\n", + "IB=0.22 \t\t#in mA\n", + "\n", + "#calculation\n", + "IC=(alfao*IB+ICO)/(1-alfao) \t#in mA\n", + "\n", + "#result\n", + "print\"Collector current is\",IC,\"mA\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.11 page No. 228" + ] + }, + { + "cell_type": "code", + "execution_count": 47, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Dynamic input resistance is 250 ohm\n" + ] + } + ], + "source": [ + "#Exa 6.11\n", + "#determine Dynamic input resistance \n", + "\n", + "#given data\n", + "delVEB=250 \t\t#in mVolts\n", + "delIE=1 \t\t#in mA\n", + "\n", + "#calculation\n", + "rin=delVEB/delIE \t#in ohm\n", + "\n", + "#result\n", + "print\"Dynamic input resistance is\",rin,\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.12 page No. 228" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Dynamic output resistance is 6.25 kohm\n" + ] + } + ], + "source": [ + "#Exa 6.12\n", + "#Determine Dynamic output resistance\n", + "\n", + "#given data\n", + "delVCE=10-5 \t\t#in Volts\n", + "delIC=5.8-5\t \t#in mA\n", + "\n", + "#calculation\n", + "rin=delVCE/delIC \t#in Kohm\n", + "\n", + "#result\n", + "print\"Dynamic output resistance is \",rin,\"kohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.13 page No.232" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point Q is ( 5.2 V, 0.6 mA)\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff0e0cf5890>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Exa 6.13\n", + "#Determine operating point\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "#given data\n", + "VCC=10 \t\t\t#in volt\n", + "RC=8 \t\t\t#in Kohm\n", + "Beta=40 \t\t#unitless\n", + "IB=15 \t\t\t#in uA\n", + "IB=IB*10**-3 \t\t#in mA\n", + "\n", + "#calculation\n", + "# For VCE = 0 Volts\n", + "IC=VCC/RC \t\t#in mA\n", + "#For IC=0 VCE=VCC=10V :\n", + "IC=Beta*IB \t\t#in mA\n", + "VCE=VCC-IC*RC \t\t#in Volts\n", + "\n", + "#result\n", + "print\"Operating point Q is (\",VCE,\"V,\",IC,\"mA)\"\n", + "\n", + "#Plot\n", + "import matplotlib.pyplot as plt\n", + "fig = plt.figure()\n", + "ax = fig.add_subplot(111)\n", + "\n", + "Vce=[0,10]\n", + "Ic=[1.25,0]\n", + "plt.xlabel('Vce,V')\n", + "plt.ylabel('Ic,mA')\n", + "ax.plot([5.2], [0.6], 'o')\n", + "ax.annotate('(5.2V,0.6 mA)', xy=(5.4,0.7))\n", + "\n", + "a=plt.plot(Vce,Ic)\n", + "plt.show(a)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.14 page No. 232" + ] + }, + { + "cell_type": "code", + "execution_count": 65, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point at load resistance 5 kohm is ( 6.0 V, 1.2 mA)\n", + "Operating point at load resistance 7.5 kohm is ( 3.0 V, 1.2 mA)\n" + ] + } + ], + "source": [ + "#Exa 6.14\n", + "#How will the Q point change when load resistance will be change\n", + "\n", + "#given data \n", + "Vcc=12 \t\t#in Volt collector supply voltage\n", + "Ic=1.2 #A, collector current\n", + "Rl=5 #kohm load resistance\n", + "\n", + "#calculation\n", + "Vce=Vcc-Ic*Rl #Collector emitter voltage\n", + "Rl1=7.5\n", + "Vce1=Vcc-Ic*Rl1\n", + "\n", + "#result\n", + "print\"Operating point at load resistance 5 kohm is (\",Vce,\"V,\",Ic,\"mA)\"\n", + "print\"Operating point at load resistance 7.5 kohm is (\",Vce1,\"V,\",Ic,\"mA)\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.15 Page No.233" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector to emitter voltage is (Vce) 20 V\n", + "Collector current at saturation point is (Ic) 6.0 mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff0b88bb790>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Example 6.15\n", + "#Given\n", + "Vcc=20 # V, collector voltage\n", + "Rc=3.3*10**3\n", + "\n", + "#calculation\n", + "#Appling kirchoff's Voltage Law\n", + "Ic=0 #for cut off point\n", + "Vce=Vcc\n", + "Ic=Vcc/Rc\n", + "print \"Collector to emitter voltage is (Vce)\",Vce,\"V\"\n", + "print \"Collector current at saturation point is (Ic)\",round(Ic*1000,0),\"mA\"\n", + "\n", + "#Plot\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "fig = plt.figure()\n", + "ax = fig.add_subplot(111)\n", + "\n", + "Vce=[0,20]\n", + "Ic=[6,0]\n", + "plt.xlabel(\"Vce (V)\") \n", + "plt.ylabel(\"Ic (mA)\") \n", + "plt.xlim((0,25))\n", + "plt.ylim((0,8))\n", + "ax.plot([0], [6], 'o')\n", + "ax.annotate('(0,6mA)', xy=(0,6))\n", + "\n", + "ax.plot([20], [0], 'o')\n", + "ax.annotate('(20V,0)', xy=(20,0))\n", + "a=plt.plot(Vce,Ic)\n", + "plt.show(a)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 6.16 page No. 233" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "collector voltage is -4.5 V\n", + "Base voltage is -8.3 V\n" + ] + } + ], + "source": [ + "#Exa 6.16\n", + "#find collector voltage and base voltage\n", + "\n", + "#given data \n", + "Beta=45 \t\t\t#Unitless\n", + "VBE=0.7 \t\t\t#in Volt\n", + "VCC=0 \t\t\t\t#in Volt\n", + "RB=10**5 \t\t\t#in ohm\n", + "RC=1.2*10**3 \t\t\t#in ohm\n", + "VEE=-9 \t\t\t\t#in Volt\n", + "\n", + "#calculation\n", + "#Applying Kirchoffs Voltage Law in input loop we have\n", + "#IB*RB+VBE+VEE=0\n", + "IB=-(VBE+VEE)/RB \t\t#in mA\n", + "IC=Beta*IB \t\t\t#in mA\n", + "VC=VCC-IC*RC \t\t\t#in Volts\n", + "VB=VBE+VEE \t\t\t#in Volts\n", + "\n", + "#Result\n", + "print\"collector voltage is \",round(VC,1),\"V\"\n", + "print\"Base voltage is \",VB,\"V\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch7.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch7.ipynb new file mode 100644 index 00000000..955aa6e3 --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch7.ipynb @@ -0,0 +1,468 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 7: Field effect Transistors" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.1 page no. 262" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance between gate and source is 10000.0 ohm\n" + ] + } + ], + "source": [ + "#Exa 7.1\n", + "#What is Resistance between gate and source\n", + "\n", + "#given data \n", + "VGS=10\t\t\t#in Volt\n", + "IG=0.001\t\t#in uA\n", + "IG=IG*10**-6\t\t#in A\n", + "\n", + "#calculation\n", + "RGS=VGS/IG\t\t#in ohm\n", + "\n", + "#result\n", + "print\"Resistance between gate and source is \",RGS/10**6,\"ohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.2 page no.262" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "AC drain resistance of JFET in Kohm 12.5 kohm\n" + ] + } + ], + "source": [ + "#Exa 7.2\n", + "#What is AC drain resistance of JFET\n", + "\n", + "#given data \n", + "delVDS=1.5\t\t\t#in Volt\n", + "delID=120\t\t\t#in uA\n", + "delID=120*10**-6\t\t#in A\n", + "\n", + "#Calculation\n", + "rd=delVDS/delID\t\t\t#in Ohm\n", + "\n", + "#Result\n", + "print\"AC drain resistance of JFET in Kohm \",rd*10**-3,\"kohm\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.3 page no. 262" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Transconductance is 2.22 mA/v\n" + ] + } + ], + "source": [ + "#Exa 7.3\n", + "#Determine Transconductance\n", + "import math\n", + "#given data \n", + "VP=-4.5\t\t\t#in Volt\n", + "IDSS=10.0\t\t\t#in mA\n", + "IDS=2.5\t\t\t#in mA\n", + "\n", + "#Calculation\n", + "VGS=VP*(1-math.sqrt(IDS/IDSS))\t\t#in Volt\n", + "gm=(-2*IDSS/VP)*(1-VGS/VP)\t\t#in mA/Volt\n", + "\n", + "#Result\n", + "print\"Transconductance is\",round(gm,2),\"mA/v\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.4 page no. 262" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VGS(OFF) is = -2.0 mV\n" + ] + } + ], + "source": [ + "#Exa 7.4\n", + "#calculate Vgs off\n", + "\n", + "#given data \n", + "gm=10\t\t\t#in mS\n", + "IDSS=10\t\t\t#in uA\n", + "IDSS=IDSS-10**-6\t#in Ampere\n", + "\n", + "#Calculation\n", + "VGS_OFF=-2*IDSS/gm\n", + "\n", + "#Result\n", + "print\"VGS(OFF) is =\",round(VGS_OFF),\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.5 page no. 262" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current= 2.5 mA\n", + "VDS(min) is : -4.0 V\n" + ] + } + ], + "source": [ + "#Exa 7.5\n", + "#Determine The minimum value of VDS for pinch-OFF region is equal to VP.\n", + "\n", + "#given data \n", + "VP=-4.0\t\t\t #in Volt\n", + "IDSS=10.0\t\t\t #in mA\n", + "IDSS=IDSS*10**-3\t#in Ampere\n", + "VGS=-2.0 #in Volt\n", + "\n", + "#Calculation\n", + "ID=IDSS*(1.0-VGS/VP)**2\t#in mA\n", + "\n", + "#result\n", + "print \"Drain current=\",ID*1000,\"mA\"\n", + "print\"VDS(min) is : \",VP,\"V\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.6 page no. 263" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ID is 3.9 mA\n", + "gmo is 5.8 mS\n", + "gm is 3.9 mS\n" + ] + } + ], + "source": [ + "#Exa 7.6\n", + "#Find the value of Id , gmo, gm\n", + "\n", + "#given data \n", + "VP=-3.0\t\t\t#in Volt\n", + "IDSS=8.7\t\t#in mA\n", + "IDSS=IDSS*10**-3\t#in mA\n", + "VGS=-1\t\t\t#in Volt\n", + "\n", + "#calculation\n", + "ID=IDSS*(1-VGS/VP)**2\t#in Ampere\n", + "gmo=-2*IDSS/VP\t\t#in mS\n", + "gm=gmo*(1-VGS/VP)\t#in mS\n", + "\n", + "#result\n", + "print\"ID is \",round(ID*1000,1),\"mA\"\n", + "print\"gmo is\",round(gmo*1000,1),\"mS\"\n", + "print\"gm is \",round(gm*1000,1),\"mS\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.7 page no.263" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current= 2.1 mA\n", + "Transconductance is 2.8 mS\n" + ] + } + ], + "source": [ + "#Exa 7.7\n", + "#Find gm\n", + "\n", + "#given data \n", + "VP=-3.0 \t\t#in Volt\n", + "IDSS=8.4 \t#in mA\n", + "VGS=-1.5 \t#in Volt\n", + "\n", + "#calculation\n", + "ID=IDSS*(1-VGS/VP)**2 \t\t#in mA\n", + "gmo=-2*IDSS/VP \t\t\t#in mS\n", + "gm=gmo*(1-VGS/VP) \t\t#in mS\n", + "\n", + "#result\n", + "print\"Drain current=\",ID,\"mA\"\n", + "print\"Transconductance is \",gm,\"mS\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.8 page no.263" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IDS = 3 mA when gm is 2.31 mS\n" + ] + } + ], + "source": [ + "#Exa 7.8\n", + "#What is gm \n", + "\n", + "#given data \n", + "VP=-4.5 \t\t #in Volt\n", + "IDSS=9 \t\t\t#in mA\n", + "IDSS=IDSS*10**-3 #in Ampere\n", + "IDS=3 \t\t\t #in mA\n", + "IDS=IDS*10**-3 \t\t#in Ampere\n", + "\n", + "#calculation\n", + "import math\n", + "VGS=VP*(1-math.sqrt(IDS/IDSS)) \t#in Volt\n", + "gm=(-2*IDSS/VP)*(1-VGS/VP) \t\t#in mS\n", + "\n", + "#result\n", + "print\"IDS = 3 mA when gm is \",round(gm*1000,2),\"mS\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.9 page no.271" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Transfer Characteristics are in mA 10.0 15.625 5.625 2.5 0.0\n", + "Transfer Characteristics for N channel MOSFET Type\n" + ] + }, + { + "data": { + "image/png": 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GAc9jdfOcBP4DPAZMLePniYS0sxfP0vs/vYmMiGR27GwqVVAJTMquuKTfDSvZ/xS4C0jz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\nRERERJzr/wFFhVeboXYOSQAAAABJRU5ErkJggg==\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7fb34db76e50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Exa 7.9\n", + "#given data :\n", + "Vp=-4.0 \t\t\t #in Volt\n", + "IDSS=10.0 \t\t #in mA\n", + "#From eq 7.1\n", + "Vgs1=0\n", + "Id1=IDSS # mA, at Vgs=0\n", + "Vgs2=1\n", + "Id2=Id1*(1-Vgs2/Vp)**2 #mA, at Vgs=1\n", + "Vgs3=-1\n", + "Id3=Id1*(1-Vgs3/Vp)**2 #mA, at Vgs=1\n", + "Vgs4=-2\n", + "Id4=Id1*(1-Vgs4/Vp)**2 #mA, at Vgs=-2\n", + "Vgs5=-4\n", + "Id5=Id1*(1-Vgs5/Vp)**2 #mA, at Vgs=-4\n", + "\n", + "print \"Transfer Characteristics are in mA \",Id1,Id2,Id3,Id4,Id5\n", + "\n", + "#Plot\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "fig = plt.figure()\n", + "ax = fig.add_subplot(111)\n", + "\n", + "Vgs=[-4,-2,-1,0,1]\n", + "Id=[0,2.5,5.625,10,15.625]\n", + "plt.xlabel(\"Vgs (V)\") \n", + "plt.ylabel(\"Id (mA)\") \n", + "plt.xlim((-4,2))\n", + "plt.ylim((0,18))\n", + "ax.plot([0], [10], 'o')\n", + "ax.annotate('(Idss)', xy=(0,10))\n", + "\n", + "a=plt.plot(Vgs,Id)\n", + "\n", + "print \"Transfer Characteristics for N channel MOSFET Type\"\n", + "plt.show(a)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 7.10 page no.275" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "When VGS=6V the drain current is 1.25 mA\n" + ] + } + ], + "source": [ + "#Exa 7.10\n", + "#Determine the drain current\n", + "\n", + "#given data \n", + "ID_on=5 \t\t#in mA\n", + "VGS=6 \t\t\t#in Volt\n", + "VGS_on=8.0 \t\t#in Volt\n", + "VGST=4 \t\t\t#in Volt\n", + "\n", + "#calculation\n", + "K=ID_on/(VGS_on-VGST)**2 \t\t#in mA/V**2\n", + "ID=K*(VGS-VGST)**2 \t\t\t#in mA\n", + "\n", + "#result\n", + "print\"When VGS=6V the drain current is \",ID,\"mA\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch8.ipynb b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch8.ipynb new file mode 100644 index 00000000..add2181f --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/Ch8.ipynb @@ -0,0 +1,226 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8: Photonic Devices" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.1 Page no 293" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steady state photocurrent density is 0.726 A/cm**2\n" + ] + } + ], + "source": [ + "#Exa 8.1\n", + "#Find Steady state photocurrent density\n", + "\n", + "#given data \n", + "NA=10**22 #in atoms/m**3\n", + "ND=10**22 #in atoms/m**3\n", + "De=25*10**-4 \t#in m**2/s\n", + "Dh=10**-3\t\t#in m**2/s\n", + "TAUeo=500\t\t#in ns\n", + "TAUho=100\t\t#in ns\n", + "ni=1.5*10**16\t\t#in atoms/m**3\n", + "VR=-10\t\t\t#in Volt\n", + "epsilon=11.6*8.854*10**-12\t#in F/m\n", + "e=1.6*10**-19\t\t\t#in Coulamb\n", + "VT=26\t\t\t\t#in mV\n", + "GL=10**27\t\t\t#in m**-3 s**-1\n", + "\n", + "\n", + "#calculation\n", + "import math\n", + "Le=math.sqrt(De*TAUeo*10**-9)\t#in um\n", + "Le=Le*10**6\t\t\t#in um\n", + "Lh=math.sqrt(Dh*TAUho*10**-9)\t#in um\n", + "Lh=Lh*10**6\t\t\t#in um\n", + "Vbi=VT*10**-3*math.log(NA*ND/ni**2)\t#in Volt\n", + "Vo=Vbi\t\t\t\t#in Volt\n", + "VB=Vo-VR\t\t\t#in Volt\n", + "W=math.sqrt((2*epsilon*VB/e)*(1/NA+1/ND))\t#in um\n", + "W=W*10**6\t\t\t#in um\n", + "JL=e*(W+Le+Lh)*10**-6*GL\t#in A/cm**2\n", + "\n", + "#Result\n", + "print \"Steady state photocurrent density is \",round(JL/10**4,3),\"A/cm**2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.2 Page no 295" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Steady state photocurrent density is 14.69 mA/cm**2\n" + ] + } + ], + "source": [ + "#Exa 8.2\n", + "#Find Steady state photocurrent density\n", + "\n", + "#given data \n", + "import math\n", + "W=25\t\t\t#in um\n", + "PhotonFlux=10**21\t#in m**2s**-1\n", + "alfa=10**5\t\t#in m**-1\n", + "e=1.6*10**-19\t\t#in Coulambs\n", + "\n", + "#calculation\n", + "GL1=alfa*PhotonFlux\t#in m**-3s**-1\n", + "GL2=alfa*PhotonFlux*math.exp(-alfa*W*10**-6)\t#in m**-3s**-1\n", + "JL=e*PhotonFlux*(1-math.exp(-alfa*W*10**-6))\t#in mA/cm**2\n", + "\n", + "#Result\n", + "print\"Steady state photocurrent density is \",round(JL/10,2),\"mA/cm**2\"" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.3 Page no 304" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Open circuit voltage is 0.522 V\n" + ] + } + ], + "source": [ + "#Exa 8.3\n", + "#DEtermine Open circuit voltage .\n", + "\n", + "#given data \n", + "NA=7.5*10**24\t\t#in atoms/m**3\n", + "ND=1.5*10**22\t\t#in atoms/m**3\n", + "De=25.0*10**-4\t\t#in m**2/s\n", + "Dh=10.0**-3\t\t#in m**2/s\n", + "TAUeo=500.0\t\t#in ns\n", + "TAUho=100.0\t\t#in ns\n", + "ni=1.5*10**16\t\t#in atoms/m**3\n", + "VR=-10.0\t\t\t#in Volt\n", + "epsilon=11.6*8.854*10**-12\t#in F/m\n", + "e=1.6*10**-19\t\t#in Coulamb\n", + "VT=26.0\t\t\t#in mV\n", + "GL=10.0**27\t\t#in m**-3 s**-1\n", + "\n", + "#Calculation\n", + "import math\n", + "Le=math.sqrt(De*TAUeo*10**-9)\t#in m\n", + "Le=Le*10**6\t\t\t#in um\n", + "Lh=math.sqrt(Dh*TAUho*10**-9)\t#in m\n", + "Lh=Lh*10**6\t\t\t#in um\n", + "JS=e*(ni**2)*(De/(Le*10**-6*NA)+Dh/(Lh*10**-6*ND))\t#in A/cm**2\n", + "JL=12.5\t\t\t\t#in mA/cm**2\n", + "VOC=VT*math.log(1.0+((JL*10**-3)/(JS*10**-4)))\t\t#in Volt\n", + "\n", + "#Result\n", + "print\"Open circuit voltage is\",round(VOC/1000,3),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Example 8.4 Page no 304" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total no. of cells required : 1244.0\n" + ] + } + ], + "source": [ + "#Exa 8.4\n", + "#Find The total no. of cells required\n", + "#given data \n", + "Vout=28\t\t\t#in Volts\n", + "Vcell=0.45\t\t#in Volt\n", + "n=Vout/Vcell\t\t#Unitless\n", + "Iout=1\t\t\t#in A\n", + "Icell=50\t\t#in mA\n", + "\n", + "#Calculation\n", + "m=Iout/(Icell*10**-3)\t#unitless\n", + "\n", + "#Result\n", + "print\"The total no. of cells required : \",round(m*n)\n", + "#Note : Answer in the book is wrong." + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/README.txt b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/README.txt new file mode 100644 index 00000000..f612609a --- /dev/null +++ b/Fundamentals_Of_Electronic_Devices_by_J._B._Gupta/README.txt @@ -0,0 +1,10 @@ +Contributed By: Rahul Garg +Course: btech +College/Institute/Organization: Gurgaon College of Engineering +Department/Designation: Electronics and Communication En +Book Title: Fundamentals Of Electronic Devices +Author: J. B. Gupta +Publisher: Katariya & Sons, New Delhi +Year of publication: 2009 +Isbn: 978-93-80027-74-6 +Edition: 1
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a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter1.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter1.ipynb new file mode 100644 index 00000000..e4a0b038 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter1.ipynb @@ -0,0 +1,220 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 1: Introduction to Operational Amplifiers" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 1.1_a" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector current Ic1 is 0.39 mA\n", + "Voltage Vc1 is 3.38 V\n", + "Voltage Ve4 is 2.68 V\n", + "Current Ie4 is 0.297 mA\n", + "Current Ic5 is 0.297 mA\n", + "Voltage Vc5 is 4.87 V\n", + "Voltage Ve6 is 4.17 V\n", + "Current Ie6 is 0.678 mA\n", + "Voltage Ve7 is 4.87 V\n", + "Current I1 is 2.82 mA\n", + "Current Ie8 is 2.82 mA\n", + "Voltage Ve8 at the output terminal is -0.35 V\n" + ] + } + ], + "source": [ + "\n", + "\n", + "#Example 1.1_a\n", + "#The equivalent circuit of the Motorola op-amp MC 1435 is shown in Figure.No-1.2\n", + "#Determine the collector current in each transistor and the dc voltage at the\n", + "#output terminal\n", + "\n", + "#Variable declaration\n", + "Vcc=6 #Voltage in volts\n", + "Vbe5=0.7 #Voltage in volts\n", + "Vee=6 #Voltage in volts\n", + "Vbe3=6.7 #Voltage in volts\n", + "Vbe6=0.7 #Voltage in volts\n", + "Vbe7=0.7 #Voltage in volts\n", + "Rc1=6.7*10**3 #Resistance in ohms\n", + "Ic1=0 #initialization\n", + "\n", + "#Calculation\n", + "Vc1=Vcc-Rc1*Ic1\n", + "Ve4=Vc1-Vbe5\n", + "I4=(Ve4+Vee)/(9.1*10**3+5.5*10**3)\n", + "Vb3=5.5*10**3*I4-Vee\n", + "Ve3=Vb3-Vbe3\n", + "Ie3=(Ve3+Vbe3)/3.3*10**3\n", + "Ic1=1.08*10**-3/2.765 #Since Ie3=2*Ic1\n", + "Vc1=Vcc-Rc1*Ic1\n", + "Ve4=Vc1-Vbe5\n", + "Ie4=(Ve4+Vee)/(29.2*10**3)\n", + "Ic5=Ie4\n", + "Vc5=Vcc-3.8*10**3*Ic5\n", + "Ve6=Vc5-Vbe6\n", + "Ie6=(Ve6+Vee)/(15*10**3)\n", + "Ve7=Ve6+Vbe7\n", + "I1=(Vcc-Ve7)/400\n", + "Ie8=I1\n", + "Ve8=-Vee+2*10**3*Ie8\n", + "\n", + "#Result\n", + "print \"Collector current Ic1 is\",round(Ic1*10**3,2),\"mA\"\n", + "print \"Voltage Vc1 is\",round(Vc1,2),\"V\" \n", + "print \"Voltage Ve4 is\",round(Ve4,2),\"V\"\n", + "print \"Current Ie4 is\",round(Ie4*10**3,3),\"mA\"\n", + "print \"Current Ic5 is\",round(Ic5*10**3,3),\"mA\"\n", + "print \"Voltage Vc5 is\",round(Vc5,2),\"V\"\n", + "print \"Voltage Ve6 is\",round(Ve6,2),\"V\"\n", + "print \"Current Ie6 is\",round(Ie6*10**3,3),\"mA\"\n", + "print \"Voltage Ve7 is\",round(Ve7,2),\"V\"\n", + "print \"Current I1 is\",round(I1*10**3,2),\"mA\"\n", + "print \"Current Ie8 is\",round(Ie8*10**3,2),\"mA\"\n", + "print \"Voltage Ve8 at the output terminal is\",round(Ve8,2),\"V\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 1.1_b" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain of the dual-input,balanced output-differential amplifier is 82.55\n", + "Voltage gain of the dual-input,unbalanced output-differential amplifier is 22.6\n", + "Overall gain of the op-amp is 1866.34\n" + ] + } + ], + "source": [ + "\n", + "\n", + "#Example 1.1_b, Figure.No-1.2\n", + "#Calculate the Voltage gain of the opamp\n", + "\n", + "#Variable decclaration\n", + "Ie1=0.39*10**-3 #Current in amps\n", + "Ie4=0.298*10**-3 #Current in amps\n", + "Ie6=0.678*10**-3 #Current in amps\n", + "Rc1=6.7*10**3 #Resistance in ohms\n", + "Rc5=3.8*10**3 #Resistance in ohms\n", + "beta_ac=150\n", + "\n", + "#Calculation\n", + "re1=(25*10**-3)/Ie1\n", + "re2=re1\n", + "re4=(25*10**-3)/Ie4\n", + "re5=re4\n", + "re6=(25*10**-3)/Ie6\n", + "k=(Rc1*2*beta_ac*re4)/(Rc1+2*beta_ac*re4)\n", + "Ad1=k/re1\n", + "k1=(Rc5*beta_ac*(re6+15*10**3))/(Rc5+beta_ac*(re6+15*10**3))\n", + "Ad2=k1/(2*re5)\n", + "Ad=Ad1*Ad2\n", + "\n", + "#Result\n", + "print \"Voltage gain of the dual-input,balanced output-differential amplifier is\",round(Ad1,2)\n", + "print \"Voltage gain of the dual-input,unbalanced output-differential amplifier is\",round(Ad2,1)\n", + "print \"Overall gain of the op-amp is\",round(Ad,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 1.1_c" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Input resistance Ri is 19.23 kilo ohms\n" + ] + } + ], + "source": [ + "\n", + "#Example 1.1_c, Figure.No-1.2\n", + "#Determine the Input resistance of the opamp\n", + "\n", + "#Variable declaration\n", + "beta_ac=150\n", + "re1=64.1 #Resistance in ohms\n", + "\n", + "#calculation\n", + "Ri=2*beta_ac*re1\n", + "\n", + "#result\n", + "print \"Input resistance Ri is\",round(Ri/10**3,2),\"kilo ohms\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter2.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter2.ipynb new file mode 100644 index 00000000..129cd7d8 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter2.ipynb @@ -0,0 +1,196 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 2: Interpretation of Data Sheets and Characteristics of an Op-Amp" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 2.1_a" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is vo 2.4 Volts\n" + ] + } + ], + "source": [ + "\n", + "\n", + "#Example 2.1_a\n", + "#Determine the Output voltage for open-loop differential amplifier for figure 2_9\n", + "\n", + "#Variable declaration\n", + "vin1=5*10**-6 #input voltage in volts\n", + "vin2=-7*10**-6 #input voltage in volts\n", + "A=200000 #Voltage gain\n", + "\n", + "#Calculation\n", + "vo=A*(vin1-vin2) #Output voltage in volts\n", + "\n", + "#Result\n", + "print \"Output voltage is vo\",vo,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 2.1_b" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is vo -2000.0 Volts\n" + ] + } + ], + "source": [ + "\n", + "\n", + "#Example 2.1_b\n", + "#Determine the Output voltage for open-loop differential amplifier for figure 2_9\n", + "\n", + "#Variable declaration\n", + "vin1=10*10**-3 #input voltage in volts\n", + "vin2=20*10**-3 #input voltage in volts\n", + "A=200000 #Voltage gain\n", + "\n", + "#Calculation\n", + "vo=A*(vin1-vin2) #Output voltage in volts\n", + "\n", + "#Result\n", + "print \"Output voltage is vo\",vo,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.2_a" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is vo -4000.0 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 2.2_a\n", + "#Determine the Output voltage for an inverting amplifier for figure 2_10\n", + "\n", + "#Variable declaration\n", + "vin1=20*10**-3 #input voltage in volts\n", + "A=200000 #Voltage gain\n", + "\n", + "#Calculation\n", + "vo=-A*(vin1) #Output voltage in volts\n", + "\n", + "#Result\n", + "print \"Output voltage is vo\",vo,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 2.2_b" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is vo 10.0 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 2.2_b\n", + "#Determine the Output voltage for an inverting amplifier for figure 2_10\n", + "\n", + "#Variable declaration\n", + "vin1=-50*10**-6 #input voltage in volts\n", + "A=200000 #Voltage gain\n", + "\n", + "#Calculation\n", + "vo=-A*(vin1) #Output voltage in volts\n", + "\n", + "#Result\n", + "print \"Output voltage is vo\",vo,\"Volts\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter3.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter3.ipynb new file mode 100644 index 00000000..1d73a725 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter3.ipynb @@ -0,0 +1,415 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 3: An Op-Amp with Negative Feedback" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain is 11.0\n", + "Input resistance with feedback is 36.37 Giga Ohm\n", + "Output resistance with feedback is 4.12 mOhm\n", + "Bandwidth with feedback is 90.91 KHz\n", + "Total output offset voltage with feedback is 0.715 mV\n" + ] + } + ], + "source": [ + "#Example 3.1\n", + "#Compute the following parameters of voltage-series feedback amplifier\n", + "#Af,Ri,Ro,fF,VooT\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=1000 #Resistance in ohms\n", + "Rf=10000 #Feedback Resistance in Ohms \n", + "A=200000 #Open-loop voltage gain\n", + "Ri=2*10**6 #Input resistance without feedback\n", + "Ro=75 #Output resistance without feedback\n", + "fo=5 #Break frequency of an Op-amp\n", + "Vsat=13 #Saturation voltage\n", + "\n", + "#calculation\n", + "B=R1/(R1+Rf) #Gain of the feedback circuit\n", + "Af=A/(1+A*B) #Closed-loop voltage gain\n", + "RiF=Ri*(1+A*B) #Input resistance with feedback\n", + "RoF=Ro/(1+A*B) #Output resistance with feedback\n", + "fF=fo*(1+A*B) #Bandwidth with feedback\n", + "VooT=Vsat/(1+A*B) #Total output offset voltage with feedback\n", + "\n", + "#Result\n", + "print \"Closed-loop voltage gain is\",round(Af,2)\n", + "print \"Input resistance with feedback is\",round(RiF/10**9,2),\"Giga Ohm\"\n", + "print \"Output resistance with feedback is\",round(RoF*10**3,2),\"mOhm\"\n", + "print \"Bandwidth with feedback is\",round(fF/10**3,2),\"KHz\"\n", + "print \"Total output offset voltage with feedback is \",round(VooT*10**3,3),\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain is 1.0\n", + "Input resistance with feedback is 400.0 Giga Ohm\n", + "Output resistance with feedback is 0.375 mOhm\n", + "Bandwidth with feedback is 1.0 MHz\n", + "Total output offset voltage with feedback is 65.0 uV\n" + ] + } + ], + "source": [ + "#Example 3.2\n", + "#Compute the following parameters of voltage follower circuit of figure 3-7\n", + "#Af,Ri,Ro,fF,VooT\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=1000 #Resistance in ohms\n", + "Rf=10000 #Feedback Resistance in Ohms \n", + "A=200000 #Open-loop voltage gain\n", + "Ri=2*10**6 #Input resistance without feedback\n", + "Ro=75 #Output resistance without feedback\n", + "fo=5 #Break frequency of an Op-amp\n", + "Vsat=13 #Saturation voltage\n", + "B=1 #Gain of the feedback circuit of voltage follower\n", + "\n", + "#calculation\n", + "\n", + "Af=A/(1+A*B) #Closed-loop voltage gain\n", + "RiF=Ri*(1+A*B) #Input resistance with feedback\n", + "RoF=Ro/(1+A*B) #Output resistance with feedback\n", + "fF=fo*(1+A*B) #Bandwidth with feedback\n", + "VooT=Vsat/(1+A*B) #Total output offset voltage with feedback\n", + "\n", + "#Result\n", + "print \"Closed-loop voltage gain is\",round(Af)\n", + "print \"Input resistance with feedback is\",round(RiF/10**9),\"Giga Ohm\"\n", + "print \"Output resistance with feedback is\",round(RoF*10**3,3),\"mOhm\"\n", + "print \"Bandwidth with feedback is\",round(fF/10**6,2),\"MHz\"\n", + "print \"Total output offset voltage with feedback is \",round(VooT*10**6,3),\"uV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.3" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain is -10.0\n", + "Input resistance with feedback is 470.0 Ohm\n", + "Output resistance with feedback is 4.12 mOhm\n", + "Bandwidth with feedback is 100.0 kHz\n", + "Total output offset voltage with feedback is 0.715 mV\n" + ] + } + ], + "source": [ + "#Example 3.3\n", + "#Compute the following parameters of inverting amplifierof figure 3-8\n", + "#Af,Ri,Ro,fF,VooT\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=470 #Resistance in ohms\n", + "Rf=4.7*10**3 #Feedback Resistance in Ohms \n", + "A=200000 #Open-loop voltage gain\n", + "Ri=2*10**6 #Input resistance without feedback\n", + "Ro=75 #Output resistance without feedback\n", + "fo=5 #Break frequency of an Op-amp\n", + "Vsat=13 #Saturation voltage\n", + "\n", + "\n", + "#calculation\n", + "\n", + "K=Rf/(R1+Rf) #Voltage attenuation factor\n", + "B=R1/(R1+Rf) #Gain of the feedback circuit\n", + "Af=-A*K/(1+A*B) #Closed-loop voltage gain\n", + "X=Rf/(1+A)\n", + "RiF=R1+(X*Ri)/(X+Ri) #Input resistance with feedback\n", + "RoF=Ro/(1+A*B) #Output resistance with feedback\n", + "fF=fo*(1+A*B)/K #Bandwidth with feedback\n", + "VooT=Vsat/(1+A*B) #Total output offset voltage with feedback\n", + "\n", + "#Result\n", + "print \"Closed-loop voltage gain is\",round(Af)\n", + "print \"Input resistance with feedback is\",round(RiF),\"Ohm\"\n", + "print \"Output resistance with feedback is\",round(RoF*10**3,2),\"mOhm\"\n", + "print \"Bandwidth with feedback is\",round(fF/10**3),\"kHz\"\n", + "print \"Total output offset voltage with feedback is \",round(VooT*10**3,3),\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.4" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f41141b6f90>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is -10.0 V peak to peak\n" + ] + } + ], + "source": [ + "%matplotlib inline\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show\n", + "import matplotlib.pyplot as plt\n", + "import math\n", + "import numpy as np\n", + "#Variable declaration\n", + "R1=470 #Resistance in ohms\n", + "Rf=4.7*10**3 #Feedback Resistance in Ohms \n", + "A=200000 #Open-loop voltage gain\n", + "vin=1 #input voltage in Volts\n", + "\n", + "\n", + "\n", + "#calculation\n", + "K=Rf/(R1+Rf) #Voltage attenuation factor\n", + "B=R1/(R1+Rf) #Gain of the feedback circuit\n", + "Af=-A*K/(1+A*B) #Closed-loop voltage gain\n", + "vo=Af*vin #output voltage\n", + "\n", + "x=np.arange(0,4*math.pi,0.1)\n", + "y=-5*np.sin(x)\n", + "plt.plot(x,y)\n", + "plt.ylabel('vo')\n", + "plt.xlabel('t')\n", + "plt.title(r'$output voltage$')\n", + "plt.show()\n", + "#Result\n", + "print \"Output voltage is\",round(vo),\"V peak to peak\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.5_a & 3.5_b" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain is -10.0\n", + "Input resistance of inverting amplifier is 1.0 kilo ohms\n", + "Input resistance of noninverting amplifier is 11.0 kilo ohms\n", + "Output voktage is 3.0 V peak to peak at 100 Hz\n" + ] + } + ], + "source": [ + "#Example 3.5_a & 3.5_b\n", + "#For the circuit of figure 3_14,R1=R2=1 kilo ohm and the opamp is 741 IC.\n", + "#a) What are the gain and input resistance of the amplifier?\n", + "#b) Calculate output voltage vo if vx=2.7 V pp and vy=3 V pp sine waves at 100 Hz\n", + "\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=1000 #Resistance in ohms\n", + "R2=1000 #Resistance in ohms\n", + "Rf=10*10**3 #Feedback Resistance in Ohms\n", + "R3=10*10**3\n", + "vx=2.7 #input voltage in Volts\n", + "vy=3 #input voltage in Volts\n", + "\n", + "\n", + "#calculation\n", + "#part a\n", + "AD=-Rf/R1 #voltage gain\n", + "RiFx=R1 #Input resistance of inverting amplifier\n", + "RiFy=R2+R3 #Input resistance of noninverting amplifier\n", + "#part b\n", + "vxy=vx-vy\n", + "vo=AD*vxy #output volatage\n", + "\n", + "#Result\n", + "print \"Voltage gain is\",AD\n", + "print \"Input resistance of inverting amplifier is\",RiFx/10**3,\"kilo ohms\"\n", + "print \"Input resistance of noninverting amplifier is\",round(RiFy/10**3),\"kilo ohms\"\n", + "print \"Output voktage is\",vo,\"V peak to peak at 100 Hz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 3.6_a & 3.6_b" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain is 11.0\n", + "Input resistance of first stage amplifier is 364.0 Giga ohms\n", + "Input resistance of second stage amplifier is 36.4 Giga ohms\n", + "Output voLtage is 5.5 V peak to peak at 1 KHz\n" + ] + } + ], + "source": [ + "#Example 3.6_a & 3.6_b\n", + "#For the differential amplifier of figure 3_16, R1=R3=680 ohm, Rf=R2=6.8 Kilo ohm\n", + "#vx=-1.5 V pp, vy=-2 V pp sine waves at 1 KHz and the opamp is 741 IC.\n", + "#a) What are the gain and input resistance of the amplifier?\n", + "#b) Calculate output voltage of the amplifier.(Assume vooT=0V)\n", + "\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=680 #Resistance in ohms\n", + "R2=6800 #Resistance in ohms\n", + "Rf=6800 #Feedback Resistance in Ohms\n", + "R3=680\n", + "Ri=2*10**6 #Open loop input resistance of the opamp\n", + "vx=-1.5 #input voltage in Volts\n", + "vy=-2 #input voltage in Volts\n", + "A=200000 #openloop gain\n", + "\n", + "\n", + "#calculation\n", + "#part a\n", + "AD=1+Rf/R1 #voltage gain\n", + "B=R2/(R2+R3)\n", + "RiFy=Ri*(1+A*B) #Input resistance of first stage amplifier\n", + "B=R1/(R1+Rf)\n", + "RiFx=Ri*(1+A*B) #Input resistance of second stage amplifier\n", + "#part b\n", + "vxy=vx-vy\n", + "vo=AD*vxy #output volatage\n", + "\n", + "#Result\n", + "print \"Voltage gain is\",AD\n", + "print \"Input resistance of first stage amplifier is\",round(RiFy/10**9),\"Giga ohms\"\n", + "print \"Input resistance of second stage amplifier is\",round(RiFx/10**9,1),\"Giga ohms\"\n", + "print \"Output voLtage is\",vo,\"V peak to peak at 1 KHz\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter4.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter4.ipynb new file mode 100644 index 00000000..d79d5385 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter4.ipynb @@ -0,0 +1,1036 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 4: The Practical Op-Amp" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.1" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance Rb is 10.0 kilo ohms\n", + "Resistance Ra is 4.0 kilo ohms\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.1\n", + "#Design a Compensating Network for the opamp LM307.\n", + "#The opamp uses +10 V and -10 V supply voltages.\n", + "\n", + "#Variable declaration\n", + "V=10 #Supply voltage\n", + "Vio=10*10**-3 #Input offset voltage\n", + "Rc=10 #Assumption\n", + "\n", + "#calculation\n", + "Rb=(V/Vio)*Rc\n", + "Ra=Rb/2.5 #Since Rb>Rmax,let us choose Rb=10*Rmax where Rmax=Ra/4\n", + "\n", + "#Result\n", + "print \"Resistance Rb is\",Rb/10**3,\"kilo ohms\"\n", + "print \"Resistance Ra is \",Ra/10**3,\"kilo ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.2" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max output offset voltage is 110.0 milli Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.2\n", + "#The opamp in the circuit of figure 4-13 is the LM307 with Vio=10 mV dc maximum.\n", + "#What is the maximum possible output offset voltage, Voo, caused by\n", + "#the input offset voltage Vio?\n", + "\n", + "#Variable declaration\n", + "R1=1*10**3\n", + "Rf=10*10**3\n", + "Vio=10*10**-3 #Input offset voltage\n", + "\n", + "#calculation\n", + "Aoo=1+Rf/R1 #To find max value of Voo,we reduce input voltage vin to zero.\n", + "Voo=Aoo*Vio #Max output offset voltage\n", + "\n", + "#Result\n", + "print \"Max output offset voltage is\",Voo*10**3,\"milli Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.3" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed loop gain of non-inverting amplifier is 11.0\n" + ] + } + ], + "source": [ + "#Example 4.3\n", + "#Design an input offset voltage-compensating network for the circuit in\n", + "#figure 4-13\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=1*10**3\n", + "Rf=10*10**3\n", + "Rc=10\n", + "\n", + "\n", + "#calculation\n", + "Af=1+Rf/(R1+Rc) #Closed loop gain of non-inverting amplifier\n", + "\n", + "#Result\n", + "print \"Closed loop gain of non-inverting amplifier is\",round(Af)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.4" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max output offset voltage due to Vio is 606.0 milli Volts\n", + "Max output offset voltage due to Ib is 23.5 milli Volts\n", + "Parallel combination of R1 and Rf,i.e ROM is 465.35 Ohms\n" + ] + } + ], + "source": [ + "#Example 4.4\n", + "#a) For the inverting amplifier of Figure 4-19, determine the maximum possible\n", + "#output offset voltage due to input offset voltage Vio and input bias current Ib\n", + "#The opamp is a type 741.\n", + "#b) What value of ROM is needed to reduce the effect of input bias current Ib.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=470\n", + "Rf=47*10**3\n", + "Vio=6*10**-3 #Input offset voltage\n", + "Ib=500*10**-9 #Input bias current\n", + "Vs=15 #Supply voltage\n", + "\n", + "#calculation\n", + "Voo=(1+Rf/R1)*Vio #Max output offset voltage due to input offset voltage Vio\n", + "VoIb=Rf*Ib #Max output offset voltage due to input offset voltage Ib\n", + "ROM=R1*Rf/(R1+Rf) #Parallel combination of R1 and Rf\n", + "\n", + "#Result\n", + "print \"Max output offset voltage due to Vio is\",round(Voo*10**3),\"milli Volts\"\n", + "print \"Max output offset voltage due to Ib is \",round(VoIb*10**3,1),\"milli Volts\"\n", + "print \"Parallel combination of R1 and Rf,i.e ROM is \",round(ROM,2),\"Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.5" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max output offset voltage due to Vio is 606.0 milli Volts\n", + "Max output offset voltage due to Ib is 50.0 milli Volts\n", + "Parallel combination of R1 and Rf,i.e ROM is 990.1 Ohms\n" + ] + } + ], + "source": [ + "#Example 4.5\n", + "#Repeat example 4.4 if R1 replaced by 1 kilo Ohm and Rf replaced by 100 kilo Ohm\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Vio=6*10**-3 #Input offset voltage\n", + "Ib=500*10**-9 #Input bias current\n", + "Vs=15 #Supply voltage\n", + "\n", + "#calculation\n", + "Voo=(1+Rf/R1)*Vio #Max output offset voltage due to input offset voltage Vio\n", + "VoIb=Rf*Ib #Max output offset voltage due to input offset voltage Ib\n", + "ROM=R1*Rf/(R1+Rf) #Parallel combination of R1 and Rf\n", + "\n", + "#Result\n", + "print \"Max output offset voltage due to Vio is\",round(Voo*10**3),\"milli Volts\"\n", + "print \"Max output offset voltage due to Ib is \",round(VoIb*10**3,1),\"milli Volts\"\n", + "print \"Parallel combination of R1 and Rf,i.e ROM is \",round(ROM,2),\"Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max output offset voltage due to Ib is 20.0 milli Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.6\n", + "#For the inverting amplifier in figure 4-21, determine the maximum output offset\n", + "#voltage VoIio caused by the input offset current Iio.\n", + "#The opamp is a type 741\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "Iio=200*10**-9 #Input offset current\n", + "Rf=100*10**3\n", + "\n", + "#calculation\n", + "VoIio=Rf*Iio #Max output offset voltage due to input offset voltage Ib\n", + "\n", + "#Result\n", + "print \"Max output offset voltage due to Ib is \",round(VoIio*10**3),\"milli Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Max output offset voltage due to Ib is 85.0 milli Volts\n", + "Max output offset voltage due to Iio is 83.0 milli Volts\n" + ] + } + ], + "source": [ + "#Example 4.7\n", + "#Compute the maximum possible output offset voltages in the amplifier circuits\n", + "#shown in the figure 4-22. The opamp is MC1536 with the following specifications.\n", + "#Vio=7.5 mV maximum, Iio=50 nA maximum,Ib=250 nA maximum at TA=25 degree celcius\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "R1=1*10**3\n", + "Rf=10*10**3\n", + "Vio=7.5*10**-3 #Max input offset voltage\n", + "Iio=50*10**-9 #Max input offset current\n", + "Ib=250*10**-9 #Max input bias current\n", + "\n", + "#calculation\n", + "# For figure 4.22(a)\n", + "VooT1=(1+Rf/R1)*Vio+(Rf*Ib) #Since the current generated output offset voltage is due to input bias current Ib\n", + "# For figure 4.22(b)\n", + "VooT2=(1+Rf/R1)*Vio+(Rf*Iio) #Since the current generated output offset voltage is due to input offset current Ib\n", + "\n", + "#Result\n", + "print \"Max output offset voltage due to Ib is \",VooT1*10**3,\"milli Volts\"\n", + "print \"Max output offset voltage due to Iio is \",VooT2*10**3,\"milli Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.8_a" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Error voltage is 30.6 mV\n", + "Output voltage 1 is -69.4 mV\n", + "Output voltage 2 is -130.6 mV\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.8_a\n", + "#Refer to the inverting amplifier in figure 4-24. The opamp is the LM307 with\n", + "#the following specifications.\n", + "#delta_Vio/delta_T =30 microVolt/degree celcius maximum\n", + "#delta_Iio/delta_T= 300 pA/degree celcius\n", + "#Vs=15 V, R1=1 Kilo Ohm, Rf=100 Kilo Ohm,Rl=10 Kilo Ohm\n", + "#Assume that the amplifier is nulled at 25 degree celcius. Calculate the value\n", + "#of the error voltage Ev and output voltage at 35 degree celcius if\n", + "#a) Vin= 1 mV dc\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=30*10**-6 #Change in input offset voltage\n", + "delta_T=1 #Unit change in temperature\n", + "delta_Iio=(300*10**-12) #Change in input offset current\n", + "Vs=15\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "Vin=1*10**-3 #Input voltage\n", + "k=25 #Amplifier is nulled at 25 deg\n", + "T=35-k #Change in temperature\n", + "\n", + "#calculation\n", + "Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T #Error voltage\n", + "Vo1=-(Rf/R1)*Vin+Ev #Output voltage\n", + "Vo2=-(Rf/R1)*Vin-Ev #Output voltage\n", + "\n", + "#Result\n", + "print \"Error voltage is \",round(Ev*10**3,1),\"mV\"\n", + "print \"Output voltage 1 is \",round(Vo1*10**3,1),\"mV\"\n", + "print \"Output voltage 2 is \",round(Vo2*10**3,1),\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.8_b" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Error voltage is 30.6 mV\n", + "Output voltage 1 is -969.4 mV\n", + "Output voltage 2 is -1030.6 mV\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.8_b\n", + "#Refer to the inverting amplifier in figure 4-24. The opamp is the LM307 with\n", + "#the following specifications.\n", + "#delta_Vio/delta_T =30 microVolt/degree celcius maximum\n", + "#delta_Iio/delta_T= 300 pA/degree celcius\n", + "#Vs=15 V, R1=1 Kilo Ohm, Rf=100 Kilo Ohm,Rl=10 Kilo Ohm\n", + "#Assume that the amplifier is nulled at 25 degree celcius. Calculate the value\n", + "#of the error voltage Ev and output voltage at 35 degree celcius if\n", + "#a) Vin= 10 mV dc\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=30*10**-6 #Change in input offset voltage\n", + "delta_T=1 #Unit change in temperature\n", + "delta_Iio=(300*10**-12) #Change in input offset current\n", + "Vs=15\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "Vin=10*10**-3 #Input voltage\n", + "k=25 #Amplifier is nulled at 25 deg\n", + "T=35-k #Change in temperature\n", + "\n", + "#calculation\n", + "Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T #Error voltage\n", + "Vo1=-(Rf/R1)*Vin+Ev #Output voltage\n", + "Vo2=-(Rf/R1)*Vin-Ev #Output voltage\n", + "\n", + "#Result\n", + "print \"Error voltage is \",round(Ev*10**3,1),\"mV\"\n", + "print \"Output voltage 1 is \",round(Vo1*10**3,1),\"mV\"\n", + "print \"Output voltage 2 is \",round(Vo2*10**3,1),\"mV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.8_a" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Error voltage is 0.0306 Volts\n", + "Output voltage 1 is -0.0694 Volts\n", + "Output voltage 2 is -0.1306 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.8_a\n", + "#Design of Compensating Network\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=30*10**-6 #Change in input offset voltage\n", + "delta_T=1 #Unit change in temperature\n", + "delta_Iio=(300*10**-12) #Change in input offset current\n", + "Vs=15\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "Vin=1*10**-3 #Input voltage\n", + "k=25 #Amplifier is nulled at 25 deg\n", + "T=35-k #Change in temperature\n", + "\n", + "#calculation\n", + "Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T #Error voltage\n", + "Vo1=-(Rf/R1)*Vin+Ev #Output voltage\n", + "Vo2=-(Rf/R1)*Vin-Ev #Output voltage\n", + "\n", + "#Result\n", + "print \"Error voltage is \",Ev,\"Volts\"\n", + "print \"Output voltage 1 is \",Vo1,\"Volts\"\n", + "print \"Output voltage 2 is \",Vo2,\"Volts\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.9_a" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Error voltage is 91.8 mV\n", + "Output voltage 1 is -908.2 mV\n", + "Output voltage 2 is -1091.8 mV\n" + ] + } + ], + "source": [ + "#Example 4.9_a\n", + "#Refer again to the amplifier circuit in figure 4-24.Use the same circuit\n", + "#specifications that are given in example 4-8. Assume that the amplifier is\n", + "#nulled at 25 degree celcius. If Vin is a 10 mV peak sine wave at 1 kilo Hz\n", + "#Calculate Ev and Vo at 55 degree celcius.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=30*10**-6 #Change in input offset voltage\n", + "delta_T=1 #Unit change in temperature\n", + "delta_Iio=(300*10**-12) #Change in input offset current\n", + "Vs=15\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "Vin=10*10**-3 #Input voltage\n", + "k=25 #Amplifier is nulled at 25 deg\n", + "T=55-k #Change in temperature\n", + "\n", + "#calculation\n", + "Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T #Error voltage\n", + "Vo1=-(Rf/R1)*Vin+Ev #Output voltage\n", + "Vo2=-(Rf/R1)*Vin-Ev #Output voltage\n", + "\n", + "#Result\n", + "print \"Error voltage is \",round(Ev*10**3,1),\"mV\"\n", + "print \"Output voltage 1 is \",round(Vo1*10**3,1),\"mV\"\n", + "print \"Output voltage 2 is \",round(Vo2*10**3,1),\"mV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.9_b" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f6982087550>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "\n", + "#Example 4.9_b\n", + "#Refer again to the amplifier circuit in figure 4-24.Use the same circuit\n", + "#specifications that are given in example 4-8. Assume that the amplifier is\n", + "#nulled at 25 degree celcius. If Vin is a 10 mV peak sine wave at 1 kilo Hz\n", + "#Draw the output voltage waveform at 55 degree celcius.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show\n", + "import matplotlib.pyplot as plt\n", + "import math\n", + "import numpy as np\n", + "\n", + "\n", + "\n", + "#calculation\n", + "x=np.arange(0,2*math.pi,0.1) #x coordinate\n", + "y=-1000*np.sin(x)+91.8 #y coordinate\n", + "\n", + "#result\n", + "plt.plot(x,y)\n", + "plt.ylabel('voltage')\n", + "plt.xlabel('time')\n", + "plt.title(r'$output waveform$')\n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.10_a" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Error voltage is 30.6 mV\n", + "Output voltage 1 is 131.6 mV\n", + "Output voltage 2 is 70.4 mV\n" + ] + } + ], + "source": [ + "#Example 4.10_a\n", + "#Repeat example 4-8 for the noniverting amplifier shown in figure 4-26.\n", + "#Assume that Rc<<R1.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=30*10**-6 #Change in input offset voltage\n", + "delta_T=1 #Unit change in temperature\n", + "delta_Iio=(300*10**-12) #Change in input offset current\n", + "Vs=15\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "Vin=1*10**-3 #Input voltage\n", + "k=25 #Amplifier is nulled at 25 deg\n", + "T=35-k #Change in temperature\n", + "\n", + "#calculation\n", + "Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T #Error voltage\n", + "Vo1=(1+Rf/R1)*Vin+Ev #Output voltage\n", + "Vo2=(1+Rf/R1)*Vin-Ev #Output voltage\n", + "\n", + "#Result\n", + "print \"Error voltage is \",round(Ev*10**3,1),\"mV\"\n", + "print \"Output voltage 1 is \",round(Vo1*10**3,1),\"mV\"\n", + "print \"Output voltage 2 is \",round(Vo2*10**3,1),\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.10_b" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Error voltage is 30.6 mV\n", + "Output voltage 1 is 1040.6 mV\n", + "Output voltage 2 is 979.4 mV\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.10_b\n", + "#Repeat example 4-8 for the noniverting amplifier shown in figure 4-26.\n", + "#Assume that Rc<<R1.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=30*10**-6 #Change in input offset voltage\n", + "delta_T=1 #Unit change in temperature\n", + "delta_Iio=(300*10**-12) #Change in input offset current\n", + "Vs=15\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "Vin=10*10**-3 #Input voltage\n", + "k=25 #Amplifier is nulled at 25 deg\n", + "T=35-k #Change in temperature\n", + "\n", + "#calculation\n", + "Ev=(1+Rf/R1)*(delta_Vio/delta_T)*T + Rf*(delta_Iio/delta_T)*T #Error voltage\n", + "Vo1=(1+Rf/R1)*Vin+Ev #Output voltage\n", + "Vo2=(1+Rf/R1)*Vin-Ev #Output voltage\n", + "\n", + "#Result\n", + "print \"Error voltage is \",round(Ev*10**3,1),\"mV\"\n", + "print \"Output voltage 1 is \",round(Vo1*10**3,1),\"mV\"\n", + "print \"Output voltage 2 is \",round(Vo2*10**3,1),\"mV\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.11_a" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in Output offset voltage is 3.2 mV\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.11_a\n", + "#The amplifier in figure 4-28(b) is nulled when the low dc supply voltage is\n", + "#20 mV.\n", + "#Because of poor regulation,low dc voltage varies with time from 18 V to 22 V.\n", + "#a) Determine the change in the output offset voltage caused by the change in\n", + "#supply voltages\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=15.85*10**-6 #Change in input offset voltage\n", + "delta_V=1 #Unit change in supply voltage\n", + "V=2 #Change in supply voltage\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "\n", + "#calculation\n", + "delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V #Change in output offset voltage\n", + " \n", + "\n", + "#Result\n", + "print \"Change in Output offset voltage is \",round(delta_Voo*10**3,2),\"mV\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.11_b" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total Output offset voltage 1 is -996.8 mV\n", + "Total Output offset voltage 2 is -1003.2 mV\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.11_b\n", + "#The amplifier in figure 4-28(b) is nulled when the low dc supply voltage is\n", + "#20 mV.\n", + "#Because of poor regulation,low dc voltage varies with time from 18 V to 22 V.\n", + "#b) Determine the output voltage Vo if Vin=10 mV dc. The opamp is the LM307\n", + "#with SVRR=96 dB.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=15.85*10**-6 #Change in input offset voltage\n", + "delta_V=1 #Unit change in supply voltage\n", + "V=2 #Change in supply voltage\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Vin=10*10**-3\n", + "\n", + "#calculation\n", + "delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V #Output offset voltage\n", + "Vo1=(-Rf/R1)*Vin+delta_Voo #Total output offset voltage 1\n", + "Vo2=(-Rf/R1)*Vin-delta_Voo #Total output offset voltage 2\n", + " \n", + "\n", + "#Result\n", + "print \"Total Output offset voltage 1 is \",round(Vo1*10**3,1),\"mV\"\n", + "print \"Total Output offset voltage 2 is \",round(Vo2*10**3,1),\"mV\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.11_b" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Total Output offset voltage 1 is -0.9968 Volts\n", + "Total Output offset voltage 2 is -1.0032 Volts\n" + ] + } + ], + "source": [ + "#Example 4.11_b\n", + "#Design of Compensating Network\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=15.85*10**-6 #Change in input offset voltage\n", + "delta_V=1 #Unit change in supply voltage\n", + "V=2 #Change in supply voltage\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Vin=10*10**-3\n", + "\n", + "#calculation\n", + "delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V #Output offset voltage\n", + "Vo1=(-Rf/R1)*Vin+delta_Voo #Total output offset voltage 1\n", + "Vo2=(-Rf/R1)*Vin-delta_Voo #Total output offset voltage 2\n", + " \n", + "\n", + "#Result\n", + "print \"Total Output offset voltage 1 is \",round(Vo1,4),\"Volts\"\n", + "print \"Total Output offset voltage 2 is \",round(Vo2,4),\"Volts\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.12" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in Output offset voltage 1 is 16.0 uV rms\n" + ] + } + ], + "source": [ + "#Example 4.12\n", + "#Referring to figure 4-28(b), suppose that the circuit is nulled when the voltage\n", + "#across terminals +Vcc and -Vee measures 20 V dc. Also suppose that because of\n", + "#poor filtering, 10 mV rms ac ripple is measured across terminals +Vcc and -Vee.\n", + "#While the input signal Vin=0V, how much ripple voltage can we expect at the\n", + "#output if the opamp is the LM307.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=15.85*10**-6 #Change in input offset voltage\n", + "delta_V=1 #Unit change in supply voltage\n", + "V=10*10**-3 #Change in supply voltage\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "\n", + "\n", + "#calculation\n", + "delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)*V #Output offset voltage\n", + "\n", + "#Result\n", + "print \"Change in Output offset voltage 1 is \",round(delta_Voo*10**6),\"uV rms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 4.13" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in Output offset voltage 1 is 2.82 mV\n" + ] + } + ], + "source": [ + "\n", + "#Example 4.13\n", + "#Suppose that the circuit in figure 4-24 is initially nulled. Assume also that\n", + "#room temperature and the voltage terminals +Vcc and -Vee remain constant.\n", + "#Determine the maximum possible chnage in output offset voltage after one month\n", + "#if the opamp is the LH0041C.\n", + "#Assume that R1=1kilo Ohm, Rf=100 kilo Ohm and Rl=10 kilo Ohm.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "\n", + "#Variable declaration\n", + "delta_Vio=5*10**-6 #Change in input offset voltage\n", + "delta_t=1 #Unit change in time\n", + "delta_Iio=2*10**-9 #Change in input offset current \n", + "t=4 #Time elapsed(weeks)\n", + "R1=1*10**3\n", + "Rf=100*10**3\n", + "Rl=10*10**3\n", + "\n", + "\n", + "#calculation\n", + "delta_Voot=(1+Rf/R1)*(delta_Vio/delta_t)*t+Rf*(delta_Iio/delta_t)*t #Output offset voltage\n", + "\n", + "#Result\n", + "print \"Change in Output offset voltage 1 is \",round(delta_Voot*10**3,2),\"mV\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter5.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter5.ipynb new file mode 100644 index 00000000..b5b33054 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter5.ipynb @@ -0,0 +1,267 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 5: Frequency response of an Op-Amp" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 5.1" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff51168c210>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum gain is 40 dB\n" + ] + }, + { + "data": { + "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff4f28db750>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Example 5.1\n", + "#The 741C is connected as a noniverting amplifier.What maximum gain can be used\n", + "#that will still keep the amplifier's response flat to 10kHz.\n", + "\n", + "%matplotlib inline\n", + "\n", + "from scipy import pi\n", + "import numpy as np\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, subplot, semilogx, savefig\n", + "import matplotlib.pyplot as plt\n", + "#Variable declaration\n", + "f=np.arange(0,1000000) #frequency range\n", + "s=2.0j*pi*f\n", + "A=200000 #Gain of opamp at 0 Hz\n", + "f0=5 #first break frequency in Hz\n", + "p=2.0*pi*f0\n", + "\n", + "#Calculation\n", + "\n", + "tf=A*p/(s+p) #open loop gain\n", + "\n", + "#Magnitude plot\n", + "clf() #clear the figure\n", + "subplot(211)\n", + "title('tf=p/(s+p)')\n", + "semilogx(f,20*log10(abs(tf)))\n", + "ylabel('Mag. Ratio (dB)')\n", + "\n", + "#Phase plot\n", + "subplot(212)\n", + "semilogx(f,arctan2(imag(tf),real(tf))*180.0/pi)\n", + "plt.ylabel('Phase (deg.)')\n", + "plt.xlabel('Freq (Hz)')\n", + "plt.show()\n", + "savefig('fig1.png') #savefig('fig1.eps')\n", + "\n", + "Amax=40 #from the graph\n", + "\n", + "#Result\n", + "print \"Maximum gain is\",Amax,\"dB\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 5.2" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gain equation is Aol(f)=A((1+(f/fo1)*j)*(1+(f/fo2)*j)\n", + "A,the gain of the opamp at 0 Hz is 140000\n", + "First break frequency fo1 is 6 Hz\n", + "Second break frequency fo2 is 1.24 MHz\n" + ] + } + ], + "source": [ + "#Example 5.2\n", + "#Using the frequency response and phase response curves obtained in figure 5-5\n", + "#Obtain the equation for the MC1556 opamp. Also determine the approximate values\n", + "#of the break frequencies.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "phase=-157.5 #Phase shift at about 3 MHz\n", + "f=3*10**6\n", + "fo1=6 #first break frequency,from the graph\n", + "A=140000 #Gain of the opamp at 0Hz\n", + "\n", + "\n", + "#calculation\n", + "k=-math.atan(f/fo1)*180/math.pi-phase\n", + "fo2=f/math.tan(k*math.pi/180) #second break frequency\n", + "\n", + "\n", + "#result\n", + "print \"Gain equation is Aol(f)=A((1+(f/fo1)*j)*(1+(f/fo2)*j)\"\n", + "print \"A,the gain of the opamp at 0 Hz is\",A\n", + "print \"First break frequency fo1 is\",fo1,\"Hz\" \n", + "print \"Second break frequency fo2 is\",round(fo2/10**6,2),\"MHz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 5.3" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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/Xn3X3wbMwJbSfD6DcohkTH+1iNQVwqZPeBybi6cD8J/AXGARUBVz\n7K3OsTOAp4BRCa53CrbK2c6YfclWu/sN7pPOR9jKaQDVwCXp/iIiIpJdO3G/pP8BdMKaqXo7n5+J\nrdgH9gdWDbai2fHAO8AewN7Yao+/THD90XH7HwM+oG4z2Ka4c1o7144++TTHAohIoNQkJaXuG+o2\nSYWANdgTBVjAOBP7YgdoCRyGBYnngW+dVzWJnxw6Am/GbEeAG6nbvLQ55n0Z8CTw+5h7bsOC1R7O\nvUQCoYAhsrstcdu/Bf4St+8G6gaIZM1MiT6r79gqYC3WJBZ/jhYEkkCpD0Okfq8CV2BPFmDLWrbB\nRi4NxW2SGkziL/Q1QNsU71UBnIYFo1jNsWaybekUXCTb9IQhpS7Rl3zsvtexhWZmOdubgcuw5qJn\nsY7wT4F5JH5yeBMbVlvfPaPbI4H2uM1hL2FPHD1j7i8iIgWuksSjpMqw4NIsg2vfAZyXwfkiWaEm\nKZHsSfa0Mh641OM1mwN9gBe9FkpEREREREREREREREREREREREREREREsub/AcvxV8ZEmg1JAAAA\nAElFTkSuQmCC\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff51a48b7d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "From the plot it is seen that phase angle is -90 degree\n", + "when the magnitude is o dB.Since the phase angle reaches >-180\n", + "when the magnitude is 0dB, voltage follower is stable at 0dB\n" + ] + }, + { + "data": { + "text/plain": [ + "<matplotlib.figure.Figure at 0x7ff4f2c08510>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "\n", + "#Example 5.3\n", + "#Determine the stability of the voltage follower shown in figure 3-7.\n", + "#Assume that the opamp is a 741 IC\n", + "\n", + "\n", + "%matplotlib inline\n", + "\n", + "from scipy import pi\n", + "import numpy as np\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, subplot, semilogx, savefig\n", + "import matplotlib.pyplot as plt\n", + "#Variable declaration\n", + "f=arange(10,1000000)\n", + "s=2.0j*pi*f\n", + "A=200000\n", + "f0=5\n", + "p=2.0*pi*f0\n", + "B=1 #For voltage follower B=1\n", + "\n", + "#Calculation\n", + "tf=A*p*B/(s+p) #open loop gain\n", + "\n", + "#Magnitude plot\n", + "clf() #clear the figure\n", + "subplot(211)\n", + "title('tf=p/(s+p)')\n", + "semilogx(f,20*log10(abs(tf)))\n", + "ylabel('Mag. Ratio (dB)')\n", + "\n", + "#Phase plot\n", + "subplot(212)\n", + "semilogx(f,arctan2(imag(tf),real(tf))*180.0/pi)\n", + "ylabel('Phase (deg.)')\n", + "xlabel('Freq (Hz)')\n", + "\n", + "show()\n", + "savefig('fig1.png') #savefig('fig1.eps')\n", + "\n", + "#Result\n", + "print \"From the plot it is seen that phase angle is -90 degree\"\n", + "print \"when the magnitude is o dB.Since the phase angle reaches >-180\"\n", + "print \"when the magnitude is 0dB, voltage follower is stable at 0dB\"" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter6.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter6.ipynb new file mode 100644 index 00000000..e940a81c --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter6.ipynb @@ -0,0 +1,876 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 6: General Linear Applications" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "\n", + " Low-freq cutoff is 10.61 kHz\n", + "\n", + " High-freq cutoff is 90.91 kHz\n", + "\n", + " Bandwidth is 80.3 kHz\n" + ] + } + ], + "source": [ + "#Example 6.1\n", + "#In the circuit of figure 6-3(a) Rin=50 Ohm,Ci=0.1 uF,R1=100 Ohm, Rf=1 KOhm\n", + "#Rl=10 kOhm and supply voltages +15, -15 V.\n", + "#Determine the bandwidth of the amplifier.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R1=100\n", + "Rf=1*10**3\n", + "Rin=50\n", + "Rl=10*10**3\n", + "Ci=0.1*10**-6 #Capacitance b/w 2 stages being coupled \n", + "RiF=R1 #ac input resistance of the second stage\n", + "Ro=Rin #ac output resistance of the 1st stage\n", + "UGB=10**6 #Unity gain bandwidth\n", + "\n", + "\n", + "#calculation\n", + "fl=1/(2*math.pi*Ci*(RiF+Ro)) #Low-freq cutoff\n", + "K=Rf/(R1+Rf)\n", + "Af=-Rf/R1 #closed loop voltage gain\n", + "fh=UGB*K/abs(Af) #High-freq cutoff\n", + "BW=fh-fl #Bandwidth\n", + "\n", + "#result\n", + "print \"\\n Low-freq cutoff is\",round(fl/10**3,2),\"kHz\"\n", + "print \"\\n High-freq cutoff is\",round(fh/10**3,2),\"kHz\"\n", + "print\"\\n Bandwidth is\",round(BW/10**3,2),\"kHz\" \n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Low-freq cutoff is 31.8 kHz\n", + "High-freq cutoff is 90.91 kHz\n", + "Bandwidth is 90.88 kHz\n", + "The ideal maximum output voltage swing is 15 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.2\n", + "#For the noninverting amplifier of figure 6-4(c), Rin=50 Ohm,Ci=C1=0.1 uF\n", + "#R1=R2=R3=100 kOhm,Rf=1 MOhm,Vcc=15 V.Determine\n", + "#a)the bandwidth of the amplifier\n", + "#b)the maximum output voltage swing\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R1=100*10**3\n", + "R2=100*10**3\n", + "R3=100*10**3\n", + "Rf=1*10**6\n", + "Rin=50\n", + "Ci=0.1*10**-6 #Capacitance b/w 2 stages being coupled\n", + "Ro=Rin #ac output resistance of the 1st stage\n", + "Vcc=15\n", + "UGB=10**6 #Unity gain bandwidth\n", + "Rif=R2*R3/(R2+R3) #since Ri*(1+A*B)>>R2 or R3\n", + "\n", + "#calculation\n", + "fl=1/(2*math.pi*Ci*(Rif+Ro)) #Low-freq cutoff\n", + "K=Rf/(R1+Rf)\n", + "Af=-Rf/R1 #closed loop voltage gain\n", + "fh=UGB*K/abs(Af) #High-freq cutoff\n", + "BW=fh-fl #Bandwidth\n", + "\n", + "#result\n", + "print \"Low-freq cutoff is\",round(fl,2),\"kHz\"\n", + "print \"High-freq cutoff is\",round(fh/10**3,2),\"kHz\"\n", + "print\"Bandwidth is\",round(BW/10**3,2),\"kHz\"\n", + "print \"The ideal maximum output voltage swing is\",Vcc,\"Volts\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Inductance is 9.9 mH\n", + "Figure of merit of the coil is 33.2\n", + "Parallel resistance of the tank circuit is 32.98 kHz\n", + "Feedback resistance is 1.03 kHz\n" + ] + } + ], + "source": [ + "#Example 6.3\n", + "#The circuit of figure 6-5(a) is to provide a gain of 10 at a peak frequency of\n", + "#16 kHz. Determine the value of all its components.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fp=16*10**3 #Peak frequency\n", + "Af=10 #Gain at peak frequency\n", + "C=0.01*10**-6 #Assume\n", + "R=30 #Assume the value of internal resistance of the inductor\n", + "R1=100 #Assume the value of internal resisrance of the coil\n", + "\n", + "#calculation\n", + "L=1/(((2*math.pi*fp)**2)*10**-8) #Simplifying fp=1/(2*pi*sqrt(L*C))\n", + "Xl=2*math.pi*fp*L #Inductive reactance\n", + "Qcoil=Xl/R #Figure of merit of the coil\n", + "Rp=((Qcoil)**2)*R #Parallel resistance of the tank circuit\n", + "Rf=-Rp/(1-(Rp/(Af*R1))) #Simplifying Af=(Rf||Rp)/R1\n", + "\n", + "\n", + "#result\n", + "print \"Inductance is\",round(L*10**3,1),\"mH\"\n", + "print \"Figure of merit of the coil is\",round(Qcoil,1)\n", + "print \"Parallel resistance of the tank circuit is\",round(Rp/10**3,2),\"kHz\"\n", + "print \"Feedback resistance is\",round(Rf/10**3,2),\"kHz\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.4" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is -2.0 Volts\n" + ] + } + ], + "source": [ + "#Example 6.4\n", + "#In the circuit of figure 6-6 Va=1V, Vb=2V, Vc=3V, Ra=Rb=Rc=3 kOhm,Rf=1 kOhm\n", + "#Supply voltages are 15V and -15V. Assuming that the opamp is initially nulled,\n", + "#determine the output voltage Vo\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Va=1 #Input voltage in Volts\n", + "Vb=2 #Input voltage in Volts\n", + "Vc=3 #Input voltage in Volts\n", + "Ra=3*10**3 #Resistance in ohms\n", + "Rb=3*10**3 #Resistance in ohms\n", + "Rc=3*10**3 #Resistance in ohms\n", + "Rf=1*10**3 #Resistance in ohms\n", + "\n", + "\n", + "#calculation\n", + "Vo=-((Rf/Ra)*Va+(Rf/Rb)*Vb+(Rf/Rc)*Vc) #Output voltage\n", + "\n", + "\n", + "#result\n", + "print \"Output voltage is\",Vo,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage at non-inverting terminal is 1.0 Volts\n", + "Output voltage is 3.0 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.5\n", + "#In the circuit of figure 6-7 Va=2V, Vb=-3V, Vc=4V, R=R1=1 kOhm,Rf=2 kOhm\n", + "#Supply voltages are 15V and -15V. Assuming that the opamp is initially nulled,\n", + "#determine the output voltage Vo and voltage V1 at the noninverting terminal.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Va=2 #Input voltage in volts\n", + "Vb=-3 #Input voltage in volts\n", + "Vc=4 #Input voltage in volts\n", + "R1=1*10**3 #Resistance in ohms \n", + "Rf=2*10**3 #Resistance in ohms\n", + "\n", + "\n", + "#calculation\n", + "V1=(Va+Vb+Vc)/3 #Voltage at non-inverting terminal\n", + "Vo=(1+Rf/R1)*V1 #Output voltage\n", + "\n", + "#result\n", + "print \"Voltage at non-inverting terminal is\",V1,\"Volts\"\n", + "print \"Output voltage is\",Vo,\"Volts\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is 4 Volts\n" + ] + } + ], + "source": [ + "#Example 6.6\n", + "#In the circuit of figure 6-9 Va=2V, Vb=3V,Vc=4V,Vc=4V,Vd=5V,R=1 kOhm\n", + "#Supply voltages are 15V and -15V. Assuming that the opamp is initially nulled,\n", + "#Determine the output voltage Vo\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Va=2 #Input voltage in volts\n", + "Vb=3 #Input voltage in volts\n", + "Vc=4 #Input voltage in volts\n", + "Vd=5 #Input voltage in volts\n", + "R=1*10**3 #Resistance in ohms\n", + "\n", + "#calculation\n", + "Vo=-Va-Vb+Vc+Vd #Output voltage\n", + "\n", + "#result\n", + "print \"Output voltage is\",Vo,\"Volts\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.7" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage at 0 degree is 1.47 Volts\n", + "Output voltage at 100 degree is -4.41 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.7\n", + "#In the circuit of figure 6-12 R1=1 kOhm, Rf=4.7 kOhm, Ra=Rb=Rc=100 kOhm.\n", + "#Vdc=5V and Supply voltages are 15V and -15V.\n", + "#The transducer is a thermistor with the following specifications.\n", + "#Rt=100 kOhm at a reference temperature of 25 degree celcius, temperature\n", + "#coefficient of resistance =-1 kOhm/ degree celcius or 1%/degree celcius.\n", + "#Determine the output voltage Vo at o degree C and 100 degree C.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R1=1*10**3 #Resistance in ohms\n", + "Rf=4.7*10**3 #Resistance in ohms\n", + "Ra=100*10**3 #Resistance in ohms\n", + "Rb=100*10**3 #Resistance in ohms\n", + "Rc=100*10**3 #Resistance in ohms\n", + "Vdc=5 #dc voltage in Volts\n", + "Rt=100*10**3 #Resistance of a thermistor\n", + "temp_coeff=1*10**3\n", + "R=Ra #Ra=Rb=Rc=R\n", + "\n", + "#calculation\n", + "delta_R=-temp_coeff*(0-25) #Change in resistance\n", + "Vo1=((Rf*delta_R)/(R1*4*R))*Vdc #Output voltage at degrees\n", + "delta_R=-temp_coeff*(100-25) #Change in resistance\n", + "Vo2=((Rf*delta_R)/(R1*4*R))*Vdc #Output voltage at 100 degrees\n", + "\n", + "#result\n", + "print \"Output voltage at 0 degree is\",round(Vo1,2),\"Volts\"\n", + "print \"Output voltage at 100 degree is\",round(Vo2,2),\"Volts\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.8" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Change in resistance is 0.1 ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.8\n", + "#The circuit of Figure 6-12 is used as an analog weight scale with the following\n", + "#specifications. The gain of the differential instrumentation amplifier = -100.\n", + "#Assume that Vdc= +10 V and the opamp supply voltages = +/- 10 V. The unstrained\n", + "#resistance of each of the four elements of the strain gage is 100 ohm Vo= 1 V.\n", + "#When a certain weight is placed on the scale platform,the output voltage Vo=1 V.\n", + "#Assuming that the output is initially 0,determine the change in the resistance\n", + "#of each strain-gage element.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "A=-100 #Gain of the differential instrumentation amplifier\n", + "Ra=100\n", + "Rb=100\n", + "Rc=100\n", + "Vdc=10\n", + "Vo=1\n", + "R=Ra #Ra=Rb=Rc=R\n", + "\n", + "#calculation\n", + "delta_R=(Vo*R)/(Vdc*abs(A)) #Change in resistance\n", + "\n", + "#result\n", + "print \"Change in resistance is\",delta_R,\"ohm\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Feedback resisrance is 1.8 kOhm\n", + "Gain of the differential amplifier is 38.0\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.9\n", + "#The differential input and output amplifier of figure 6-14(a) is used as a\n", + "#pre-amplifier and requires a differential output of atleast 3.7 V. Determine\n", + "#the gain of the circuit if the differential input Vin=100 mV.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Vo=3.7 #differential output voltage in Volts\n", + "Vin=100*10**-3 #differential input voltage in Volts\n", + "R1=100 #Assume\n", + "Rf=0.5*((Vo*R1)/Vin-1) #Feedback resisrance\n", + "\n", + "#calculation\n", + "A=(1+2*Rf/R1) #Gain of the differential amplifier\n", + "\n", + "#result\n", + "print \"Feedback resisrance is\",round(Rf/10**3,1),\"kOhm\"\n", + "print \"Gain of the differential amplifier is\",round(A,0)\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum input voltage is 1.1 Volts\n", + "Maximum input voltage is 7.48 Volts\n" + ] + } + ], + "source": [ + "#Example 6.10\n", + "#In the figure 6-17, for the indicated values of resistors, determine the full\n", + "#scale range for the input voltage.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R1min=1*10**3\n", + "R1max=6.8*10**3\n", + "io=1*10**-3 #Meter current for full-wave rectification\n", + "\n", + "\n", + "#calculation\n", + "vin_min=1.1*R1min*io #Minimum input voltage\n", + "vin_max=1.1*R1max*io #Maximum input voltage\n", + "\n", + "#result\n", + "print \"Minimum input voltage is\",vin_min,\"Volts\"\n", + "print \"Maximum input voltage is\",vin_max,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current through diode is 5.0 mA\n", + "Voltage drop across diode is 0.7 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.11\n", + "#The circuit of figure 6-18,when the switch is in position 1, Vin=0.5 V and\n", + "#Vo=1.2 V. Determine the current through the diode and the voltage drop across\n", + "#it.Assume that the opamp is initially nulled.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Vin=0.5 #Input voltage\n", + "Vo=1.2 #Output voltage\n", + "R1=100 \n", + "\n", + "\n", + "#calculation\n", + "Io=Vin/R1 #Current through diode\n", + "Vd=Vo-Vin #Voltage drop across diode\n", + "\n", + "#result\n", + "print \"Current through diode is\",Io*10**3,\"mA\"\n", + "print \"Voltage drop across diode is\",Vd,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.12" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Load Current is 0.5 mA\n", + "Output voltage is 2 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 6.12\n", + "#The circuit of figure 6-19,Vin=5 V, R=1 Kilo Ohm and V1=1 V. Find\n", + "#a) the load current.\n", + "#b) the output voltage Vo.\n", + "#Assume that the op-amp is initially nulled.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Vin=5 #Input voltage in Volts\n", + "V1=1 #Voltage in Volts\n", + "R1=10*10**3 #Resistance in ohms\n", + "\n", + "\n", + "#calculation\n", + "I1=Vin/R1 #Load current\n", + "Vo=2*V1 #Output voltage\n", + "\n", + "\n", + "#result\n", + "print \"Load Current is\",I1*10**3,\"mA\"\n", + "print \"Output voltage is\",Vo,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum output voltage is 0.0 Volts\n", + "Maximum output voltage is 5.38 Volts\n" + ] + } + ], + "source": [ + "#Example 6.13\n", + "#The circuit of figure 6-20, Vref=2V, R1=1 kilo Ohm. Rf=2.7 kilo Ohm. Assuming\n", + "#that the opamp is initially nulled, determine the range for the output voltage\n", + "#Vo.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "#Variable declaration\n", + "R1=1*10**3 #Resistance in ohms\n", + "Rf=2.7*10**3 #Resistance in ohms\n", + "Vref=2 #Voltage in Volts\n", + "Io=0 #Since all the binary inputs D0 to D7 are logic zero\n", + "\n", + "\n", + "#calculation\n", + "Vo_min=Io*Rf #Minimum output voltage\n", + "Io=(Vref/R1)*(1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256)\n", + "Vo_max=Io*Rf #Maximum output voltage\n", + "\n", + "#result\n", + "print \"Minimum output voltage is\",Vo_min,\"Volts\"\n", + "print \"Maximum output voltage is\",round(Vo_max,2),\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.14" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Minimum output voltage at darkness is -0.15 Volts\n", + "Maximum output voltage at illumination is -10.0 Volts\n" + ] + } + ], + "source": [ + "#Example 6.14\n", + "#The circuit of figure 6-21, Vdc=5 V and Rf=3 kilo Ohm. Determine the change in\n", + "#the output voltage if the photocell is exposed to light of 0.61 lux from a dark\n", + "#condition.Assume that the opamp is initially nulled.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Rf=3*10**3\n", + "Vdc=5\n", + "Rt1=100*10**3 #Resistance at darkness in ohms\n", + "Rt2=1.5*10**3 #Resistance at Illumination in ohms\n", + "\n", + "#calculation\n", + "Vomin=-(Vdc/Rt1)*Rf #Min output voltage at darkness\n", + "Vomax=-(Vdc/Rt2)*Rf #Max output voltage at Illumination\n", + "\n", + "#result\n", + "print \"Minimum output voltage at darkness is\",Vomin,\"Volts\"\n", + "print \"Maximum output voltage at illumination is\",round(Vomax,2),\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 6.15" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f75432e8a90>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "\n", + "\n", + "#Example 6.15\n", + "#In the figure 6-23, R1Cf=1 second, and the input is a step(dc) voltage, as\n", + "#shown in figure 6-26(a). Determine the output voltage and sketch it.\n", + "#Assume that the opamp is initially nulled.\n", + "\n", + "%matplotlib inline\n", + "import math\n", + "import scipy\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show\n", + "import scipy.integrate\n", + "\n", + "\n", + "#Variable declaration\n", + "Vin=2 #Input voltage in Volts\n", + "VoO=0 #Output offset voltage\n", + "\n", + "\n", + "#calculation\n", + "\n", + "def integrnd(x) :\n", + " return 2\n", + "val1, err = scipy.integrate.quad(integrnd, 0, 1)\n", + "val2, err = scipy.integrate.quad(integrnd, 1, 2)\n", + "val3, err = scipy.integrate.quad(integrnd, 2, 3)\n", + "val4, err = scipy.integrate.quad(integrnd, 3, 4)\n", + "\n", + "a=-val1\n", + "b=a+-val2\n", + "c=b+-val3\n", + "d=c+-val4\n", + "\n", + "import matplotlib.pyplot as plt\n", + "x=[0,1,2,3,4]\n", + "y=[VoO,a,b,c,d]\n", + "plt.plot(x,y)\n", + "title('Output voltage')\n", + "xlabel('time')\n", + "ylabel('Voutput')\n", + "plt.show()\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter7.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter7.ipynb new file mode 100644 index 00000000..924541b2 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter7.ipynb @@ -0,0 +1,1543 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 7: Active Filters and Oscillators" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R is 15.9 kOhms\n", + "Use 20 kohm POT as R\n", + "Resistance R1 is 10.0 kilo ohms\n", + "Resistance Rf is 10.0 kilo ohms\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.1\n", + "#Design a low pass filter at a cutoff frequency of 1kHz\n", + "#with a passband gain of 2\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fh=1*10**3 #Cut-off frequency\n", + "C=0.01*10**-6 #Assumption\n", + "R1=10*10**3 #Assumption\n", + "Rf=R1 #Since passband gain is 2,R1 and Rf must be equal\n", + "\n", + "\n", + "#calculation\n", + "R=1/(2*math.pi*fh*C)\n", + "\n", + "#result\n", + "print \"Resistance R is\",round(R/10**3,1),\"kOhms\"\n", + "print \"Use 20 kohm POT as R\"\n", + "print \"Resistance R1 is\",round(R1/10**3),\"kilo ohms\"\n", + "print \"Resistance Rf is\",round(Rf/10**3),\"kilo ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "New Resistance Rnew is 9937.5 Ohms\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.2\n", + "#Using the frequency scaling technique,convert the 1 kHz cutoff frequency of the\n", + "#lowpass filter of example 7-1 to a cutoff frequency of 1.6 kHz.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fc0=1*10**3 #Original cut-off frequency\n", + "fc1=1.6*10**3 #New cut-off frequency\n", + "R=15.9*10**3 #Original resistance value\n", + "\n", + "#calculation\n", + "k=fc0/fc1\n", + "Rnew=R*k\n", + "\n", + "#result\n", + "print \"New Resistance Rnew is\",Rnew,\"Ohms\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.3" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1f62117f50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gain magnitude av1 at f1 2.0\n", + "Gain magnitude av2 at f2 1.99\n", + "Gain magnitude av2 at f2 1.96\n", + "Gain magnitude av2 at f2 1.64\n", + "Gain magnitude av2 at f2 1.41\n", + "Gain magnitude av2 at f2 0.63\n", + "Gain magnitude av2 at f2 0.28\n", + "Gain magnitude av2 at f2 0.2\n", + "Gain magnitude av2 at f2 0.07\n", + "Gain magnitude av2 at f2 0.02\n" + ] + } + ], + "source": [ + "#Example 7.3\n", + "#Plot the frequency response of the lowpass filter of example 7.1\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "\n", + "from scipy import pi\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, semilogx\n", + "import math\n", + "import numpy as np\n", + "\n", + "#Variable declaration\n", + "Af=2 #Passband gain of the filter\n", + "fh=1000 #Cut-off frequency\n", + "\n", + "f1=10\n", + "f2=100\n", + "f3=200\n", + "f4=700\n", + "f5=1000\n", + "f6=3000\n", + "f7=7000\n", + "f8=10000\n", + "f9=30000\n", + "f10=100000\n", + "\n", + "\n", + "#calculation\n", + "av1=Af/math.sqrt(1+(f1/fh)**2)\n", + "av2=Af/math.sqrt(1+(f2/fh)**2)\n", + "av3=Af/math.sqrt(1+(f3/fh)**2)\n", + "av4=Af/math.sqrt(1+(f4/fh)**2)\n", + "av5=Af/math.sqrt(1+(f5/fh)**2)\n", + "av6=Af/math.sqrt(1+(f6/fh)**2)\n", + "av7=Af/math.sqrt(1+(f7/fh)**2)\n", + "av8=Af/math.sqrt(1+(f8/fh)**2)\n", + "av9=Af/math.sqrt(1+(f9/fh)**2)\n", + "av10=Af/math.sqrt(1+(f10/fh)**2)\n", + "\n", + "#Magnitude plot\n", + "f=np.arange(0,100000)\n", + "s=2.0j*pi*f\n", + "p=2.0*pi*fh\n", + "A=Af*p/(s+p)\n", + "\n", + "clf() #clear the figure\n", + "plot()\n", + "title('frequency response')\n", + "semilogx(f,20*np.log10(abs(A)))\n", + "ylabel('Voltage gain(dB)')\n", + "xlabel('frequency(Hz)')\n", + "show()\n", + "\n", + "\n", + "#result\n", + "print \"Gain magnitude av1 at f1\",round(av1,2)\n", + "print \"Gain magnitude av2 at f2\",round(av2,2)\n", + "print \"Gain magnitude av2 at f2\",round(av3,2)\n", + "print \"Gain magnitude av2 at f2\",round(av4,2)\n", + "print \"Gain magnitude av2 at f2\",round(av5,2)\n", + "print \"Gain magnitude av2 at f2\",round(av6,2)\n", + "print \"Gain magnitude av2 at f2\",round(av7,2)\n", + "print \"Gain magnitude av2 at f2\",round(av8,2)\n", + "print \"Gain magnitude av2 at f2\",round(av9,2)\n", + "print \"Gain magnitude av2 at f2\",round(av10,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.4.a" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 33.86 kOhm\n", + "Resistance R3 is 33.86 kOhm\n", + "Resistance Rf is 15.82 kOhm\n", + " Use 20 kohm POT as Rf\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.4.a\n", + "#Design a second order low-pass filter at a high cutoff frequency of 1 kHz.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fh=1*10**3 # Cut-off frequency\n", + "C2=0.0047*10**-6 # Assumption\n", + "C3=C2\n", + "R1=27*10**3 # Assumption\n", + "\n", + "#calculation\n", + "R2=1/(2*math.pi*fh*C2)\n", + "R3=R2\n", + "Rf=0.586*R1\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3,2),\"kOhm\"\n", + "print \"Resistance R3 is\",round(R3/10**3,2),\"kOhm\"\n", + "print \"Resistance Rf is\",round(Rf/10**3,2),\"kOhm\"\n", + "print \" Use 20 kohm POT as Rf\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.4.b" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1f625aac10>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gain magnitude av1 at f1 1.59\n", + "Gain magnitude av2 at f2 1.59\n", + "Gain magnitude av2 at f2 1.58\n", + "Gain magnitude av2 at f2 1.42\n", + "Gain magnitude av2 at f2 1.12\n", + "Gain magnitude av2 at f2 0.18\n", + "Gain magnitude av2 at f2 0.03\n", + "Gain magnitude av2 at f2 0.02\n", + "Gain magnitude av2 at f2 1.76 *10^-3\n", + "Gain magnitude av2 at f2 0.16 *10^-3\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.4.b\n", + "#Draw the frequency response of the network in example 7.4.a\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "\n", + "from scipy import pi\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, semilogx\n", + "import math\n", + "import numpy as np\n", + "#Variable declaration\n", + "Af=1.586 #Passband gain of the filter\n", + "fh=1000 #Cut-off frequency\n", + "\n", + "f1=10\n", + "f2=100\n", + "f3=200\n", + "f4=700\n", + "f5=1000\n", + "f6=3000\n", + "f7=7000\n", + "f8=10000\n", + "f9=30000\n", + "f10=100000\n", + "\n", + "\n", + "#calculation\n", + "av1=Af/math.sqrt(1+(f1/fh)**4)\n", + "av2=Af/math.sqrt(1+(f2/fh)**4)\n", + "av3=Af/math.sqrt(1+(f3/fh)**4)\n", + "av4=Af/math.sqrt(1+(f4/fh)**4)\n", + "av5=Af/math.sqrt(1+(f5/fh)**4)\n", + "av6=Af/math.sqrt(1+(f6/fh)**4)\n", + "av7=Af/math.sqrt(1+(f7/fh)**4)\n", + "av8=Af/math.sqrt(1+(f8/fh)**4)\n", + "av9=Af/math.sqrt(1+(f9/fh)**4)\n", + "av10=Af/math.sqrt(1+(f10/fh)**4)\n", + "\n", + "#Magnitude plot\n", + "f=np.arange(0,100000) #frequency range\n", + "s=2.0j*pi*f**2\n", + "p=2.0*pi*fh**2\n", + "A=Af*p/(s+p)\n", + "\n", + "clf() #clear the figure\n", + "plot()\n", + "title('frequency response')\n", + "semilogx(f,20*np.log10(abs(A)))\n", + "ylabel('Voltage gain(dB)')\n", + "xlabel('frequency(Hz)')\n", + "show()\n", + "\n", + "\n", + "#result\n", + "print \"Gain magnitude av1 at f1\",round(av1,2)\n", + "print \"Gain magnitude av2 at f2\",round(av2,2)\n", + "print \"Gain magnitude av2 at f2\",round(av3,2)\n", + "print \"Gain magnitude av2 at f2\",round(av4,2)\n", + "print \"Gain magnitude av2 at f2\",round(av5,2)\n", + "print \"Gain magnitude av2 at f2\",round(av6,2)\n", + "print \"Gain magnitude av2 at f2\",round(av7,2)\n", + "print \"Gain magnitude av2 at f2\",round(av8,2)\n", + "print \"Gain magnitude av2 at f2\",round(av9*10**3,2),\"*10^-3\"\n", + "print \"Gain magnitude av2 at f2\",round(av10*10**3,2),\"*10^-3\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.5.a" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R is 15.92 kOhm\n", + "Resistance R1 is 10.0 kOhm\n", + "Resistance Rf is 10.0 kOhm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.5.a\n", + "#Design a high pass filter at a cutoff frequency of 1 kHz with a passband gain\n", + "#of 2.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fh=1*10**3 #Cut-off frequency\n", + "C=0.01*10**-6 #Assumption\n", + "Af=2\n", + "\n", + "#calculation\n", + "R=1/(2*math.pi*fh*C)\n", + "R1=10*10**3\n", + "Rf=R1 #since passband gain is 2\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R is\",round(R/10**3,2),\"kOhm\"\n", + "print \"Resistance R1 is\",round(R1/10**3,2),\"kOhm\"\n", + "print \"Resistance Rf is\",round(Rf/10**3,2),\"kOhm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.5.b" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1f6224dcd0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gain magnitude av1 at f1 0.2\n", + "Gain magnitude av2 at f2 0.39\n", + "Gain magnitude av2 at f2 0.74\n", + "Gain magnitude av2 at f2 1.15\n", + "Gain magnitude av2 at f2 1.41\n", + "Gain magnitude av2 at f2 1.9\n", + "Gain magnitude av2 at f2 1.98\n", + "Gain magnitude av2 at f2 1.99\n", + "Gain magnitude av2 at f2 2.0\n", + "Gain magnitude av2 at f2 2.0\n" + ] + } + ], + "source": [ + "#Example 7.5.b\n", + "#The circuit of figure 6-17,for the indicated value of resistors\n", + "#determine the full scale range for the input voltage.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "import numpy as np\n", + "from scipy import pi\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, semilogx\n", + "import matplotlib.pyplot as plt\n", + "import math\n", + "\n", + "#Variable declaration\n", + "Af=2 #Passband gain of the filter\n", + "fl=1000 #Cut-off frequency\n", + "\n", + "f1=100\n", + "f2=200\n", + "f3=400\n", + "f4=700\n", + "f5=1000\n", + "f6=3000\n", + "f7=7000\n", + "f8=10000\n", + "f9=30000\n", + "f10=100000\n", + "\n", + "\n", + "#calculation\n", + "av1=(Af*(f1/fl))/math.sqrt(1+(f1/fl)**2)\n", + "av2=(Af*(f2/fl))/math.sqrt(1+(f2/fl)**2)\n", + "av3=(Af*(f3/fl))/math.sqrt(1+(f3/fl)**2)\n", + "av4=(Af*(f4/fl))/math.sqrt(1+(f4/fl)**2)\n", + "av5=(Af*(f5/fl))/math.sqrt(1+(f5/fl)**2)\n", + "av6=(Af*(f6/fl))/math.sqrt(1+(f6/fl)**2)\n", + "av7=(Af*(f7/fl))/math.sqrt(1+(f7/fl)**2)\n", + "av8=(Af*(f8/fl))/math.sqrt(1+(f8/fl)**2)\n", + "av9=(Af*(f9/fl))/math.sqrt(1+(f9/fl)**2)\n", + "av10=(Af*(f10/fl))/math.sqrt(1+(f10/fl)**2)\n", + "\n", + "#Magnitude plot\n", + "f=np.arange(100,100000)\n", + "s=2.0j*pi*f\n", + "p=2.0*pi*fl\n", + "A=Af*s/(s+p)\n", + "\n", + "clf() #clear the figure\n", + "plt.plot()\n", + "plt.title('frequency response')\n", + "semilogx(f,20*np.log10(abs(A)))\n", + "plt.ylabel('Voltage gain(dB)')\n", + "plt.xlabel('frequency(Hz)')\n", + "plt.show()\n", + "\n", + "\n", + "#result\n", + "print \"Gain magnitude av1 at f1\",round(av1,2)\n", + "print \"Gain magnitude av2 at f2\",round(av2,2)\n", + "print \"Gain magnitude av2 at f2\",round(av3,2)\n", + "print \"Gain magnitude av2 at f2\",round(av4,2)\n", + "print \"Gain magnitude av2 at f2\",round(av5,2)\n", + "print \"Gain magnitude av2 at f2\",round(av6,2)\n", + "print \"Gain magnitude av2 at f2\",round(av7,2)\n", + "print \"Gain magnitude av2 at f2\",round(av8,2)\n", + "print \"Gain magnitude av2 at f2\",round(av9,2)\n", + "print \"Gain magnitude av2 at f2\",round(av10,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.6.a" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Lower cutoff frequency fl is 1.0 kHz\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.6.a\n", + "#Determine the low cutoff frequency fl of the filter shown in figure 7-8(a)\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R2=33*10**3 #Resistance in ohms\n", + "R3=R2\n", + "C2=0.0047*10**-6 #Capacitance in Farads\n", + "C3=C2\n", + "\n", + "#calculation\n", + "fl=1/(2*math.pi*math.sqrt(R2*R3*C2*C3))\n", + "\n", + "\n", + "#result\n", + "print \"Lower cutoff frequency fl is\",round(fl/10**3,0),\"kHz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.6.b" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1f8840ec50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gain magnitude av1 at f1 0.0159\n", + "Gain magnitude av2 at f2 0.0634\n", + "Gain magnitude av2 at f2 0.2506\n", + "Gain magnitude av2 at f2 0.6979\n", + "Gain magnitude av2 at f2 1.1215\n", + "Gain magnitude av2 at f2 1.5763\n", + "Gain magnitude av2 at f2 1.5857\n", + "Gain magnitude av2 at f2 1.5859\n", + "Gain magnitude av2 at f2 1.586\n", + "Gain magnitude av2 at f2 1.586\n" + ] + } + ], + "source": [ + "#Example 7.6.b\n", + "#Draw the frequency response plot of the filter in example 7.6.a\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "\n", + "from scipy import pi\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, semilogx\n", + "import math\n", + "import numpy as np\n", + "#Variable declaration\n", + "Af=1.586 #Passband gain of the filter\n", + "fl=1000 #Cut-off frequency\n", + "\n", + "f1=100\n", + "f2=200\n", + "f3=400\n", + "f4=700\n", + "f5=1000\n", + "f6=3000\n", + "f7=7000\n", + "f8=10000\n", + "f9=30000\n", + "f10=100000\n", + "\n", + "\n", + "#calculation\n", + "av1=Af/math.sqrt(1+(fl/f1)**4)\n", + "av2=Af/math.sqrt(1+(fl/f2)**4)\n", + "av3=Af/math.sqrt(1+(fl/f3)**4)\n", + "av4=Af/math.sqrt(1+(fl/f4)**4)\n", + "av5=Af/math.sqrt(1+(fl/f5)**4)\n", + "av6=Af/math.sqrt(1+(fl/f6)**4)\n", + "av7=Af/math.sqrt(1+(fl/f7)**4)\n", + "av8=Af/math.sqrt(1+(fl/f8)**4)\n", + "av9=Af/math.sqrt(1+(fl/f9)**4)\n", + "av10=Af/math.sqrt(1+(fl/f10)**4)\n", + "\n", + "#Magnitude plot\n", + "f=np.arange(100,100000)\n", + "s=2.0j*pi*fl**2\n", + "p=2.0*pi*f**2\n", + "A=Af*p/(s+p)\n", + "\n", + "\n", + "clf() #clear the figure\n", + "plot()\n", + "title('frequency response')\n", + "semilogx(f,20*np.log10(abs(A)))\n", + "ylabel('Voltage gain(dB)')\n", + "xlabel('frequency(Hz)')\n", + "show()\n", + "\n", + "\n", + "#result\n", + "print \"Gain magnitude av1 at f1\",round(av1,4)\n", + "print \"Gain magnitude av2 at f2\",round(av2,4)\n", + "print \"Gain magnitude av2 at f2\",round(av3,4)\n", + "print \"Gain magnitude av2 at f2\",round(av4,4)\n", + "print \"Gain magnitude av2 at f2\",round(av5,4)\n", + "print \"Gain magnitude av2 at f2\",round(av6,4)\n", + "print \"Gain magnitude av2 at f2\",round(av7,4)\n", + "print \"Gain magnitude av2 at f2\",round(av8,4)\n", + "print \"Gain magnitude av2 at f2\",round(av9,4)\n", + "print \"Gain magnitude av2 at f2\",round(av10,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.7.a" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R1 is 10.0 kOhm\n", + "Resistance R is 15.92 kOhm\n", + "Bandpass Gain Af is 4\n", + "Resistance Rf is 10.0 kOhm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.7.a\n", + "#Design a wide band pass filter with fl=200 Hz, fh=1 kHz and passband gain=4.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fl=200 #Low cutoff freq in Hz\n", + "fh=1*10**3 #High cutoff freq in Hz\n", + "C=0.05*10**-6\n", + "\n", + "#calculation\n", + "R=1/(2*math.pi*fl*C)\n", + "R1=10*10**3\n", + "Rf=R1 #Since passband gain is 2,R1 and Rf must be equal\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R1 is\",round(R1/10**3,2),\"kOhm\"\n", + "print \"Resistance R is\",round(R/10**3,2),\"kOhm\"\n", + "print \"Bandpass Gain Af is 4\"\n", + "print \"Resistance Rf is\",round(Rf/10**3,2),\"kOhm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.7.b" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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gew6MyUg6oRBcdx3UqAGPFjakUETSXqdOsHKl/XDUuuD+iKbxoyo23qIM1oX2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\nrkJggg==\n", + "text/plain": [ + "<matplotlib.figure.Figure at 0x7f1f802ccc10>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Gain magnitude av1 at f1 0.1997\n", + "Gain magnitude av2 at f2 0.5931\n", + "Gain magnitude av2 at f2 1.78\n", + "Gain magnitude av2 at f2 2.7735\n", + "Gain magnitude av2 at f2 3.3333\n", + "Gain magnitude av2 at f2 3.1508\n", + "Gain magnitude av2 at f2 2.7735\n", + "Gain magnitude av2 at f2 1.78\n", + "Gain magnitude av2 at f2 0.5655\n", + "Gain magnitude av2 at f2 0.3979\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.7.b\n", + "#Draw the frequency response plot for the filter in example 7.7.a\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "\n", + "from scipy import pi\n", + "from matplotlib.pyplot import ylabel, xlabel, title, plot, show, clf, semilogx\n", + "import math\n", + "import numpy as np\n", + "#Variable declaration\n", + "Af=4 #Passband gain of the filter\n", + "fl=200 #Cut-off frequency\n", + "fh=1000 #Higher Cut-off frequency\n", + "\n", + "f1=10\n", + "f2=30\n", + "f3=100\n", + "f4=200\n", + "f5=447.2\n", + "f6=700\n", + "f7=1000\n", + "f8=2000\n", + "f9=7000\n", + "f10=10000\n", + "\n", + "\n", + "#calculation\n", + "av1=(Af*(f1/fl))/math.sqrt((1+(f1/fl)**2)*(1+(f1/fh)**2))\n", + "av2=(Af*(f2/fl))/math.sqrt((1+(f2/fl)**2)*(1+(f2/fh)**2))\n", + "av3=(Af*(f3/fl))/math.sqrt((1+(f3/fl)**2)*(1+(f3/fh)**2))\n", + "av4=(Af*(f4/fl))/math.sqrt((1+(f4/fl)**2)*(1+(f4/fh)**2))\n", + "av5=(Af*(f5/fl))/math.sqrt((1+(f5/fl)**2)*(1+(f5/fh)**2))\n", + "av6=(Af*(f6/fl))/math.sqrt((1+(f6/fl)**2)*(1+(f6/fh)**2))\n", + "av7=(Af*(f7/fl))/math.sqrt((1+(f7/fl)**2)*(1+(f7/fh)**2))\n", + "av8=(Af*(f8/fl))/math.sqrt((1+(f8/fl)**2)*(1+(f8/fh)**2))\n", + "av9=(Af*(f9/fl))/math.sqrt((1+(f9/fl)**2)*(1+(f9/fh)**2))\n", + "av10=(Af*(f10/fl))/math.sqrt((1+(f10/fl)**2)*(1+(f10/fh)**2))\n", + "\n", + "#Magnitude plot\n", + "f=np.arange(10,100000)\n", + "s=2.0j*pi*f\n", + "p1=2.0*pi*fl\n", + "p2=2.0*pi*fh\n", + "A=(Af*s)*p2/((s+p1)*(s+p2))\n", + "\n", + "clf() #clear the figure\n", + "plot()\n", + "title('frequency response')\n", + "semilogx(f,20*np.log10(abs(A)))\n", + "ylabel('Voltage gain(dB)')\n", + "xlabel('frequency(Hz)')\n", + "show()\n", + "\n", + "\n", + "#result\n", + "print \"Gain magnitude av1 at f1\",round(av1,4)\n", + "print \"Gain magnitude av2 at f2\",round(av2,4)\n", + "print \"Gain magnitude av2 at f2\",round(av3,4)\n", + "print \"Gain magnitude av2 at f2\",round(av4,4)\n", + "print \"Gain magnitude av2 at f2\",round(av5,4)\n", + "print \"Gain magnitude av2 at f2\",round(av6,4)\n", + "print \"Gain magnitude av2 at f2\",round(av7,4)\n", + "print \"Gain magnitude av2 at f2\",round(av8,4)\n", + "print \"Gain magnitude av2 at f2\",round(av9,4)\n", + "print \"Gain magnitude av2 at f2\",round(av10,4)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.7.c" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Center frequency fc is 447.21 Hz\n", + "Quality factor Q is 0.56\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.7.c\n", + "#Calculate the value of Q for the filter.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fh=1*10**3 #Higher cut-off frequency\n", + "fl=200 #Lower cut-off frequency\n", + "\n", + "\n", + "#calculation\n", + "fc=math.sqrt(fl*fh) #Center frequency\n", + "Q=fc/(fh-fl) #Quality factor\n", + "\n", + "\n", + "\n", + "#result\n", + "print \"Center frequency fc is\",round(fc,2),\"Hz\"\n", + "print \"Quality factor Q is\",round(Q,2)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.8.a" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R1 is 4.77 kilo ohm\n", + "Resistance R2 is 5.97 kilo ohm\n", + "Resistance R3 is 95.49 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.8.a\n", + "#Design the bandpass filter shown in figure 7-13(a) so that fc=1 kHz, Q=3 and\n", + "#Af=10.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fc=1*10**3 #Center frequency\n", + "Q=3 #Quality factor\n", + "Af=10 #Passband gain\n", + "C1=0.01*10**-6 #Assumption\n", + "\n", + "#calculation\n", + "C2=C1\n", + "R1=Q/(2*math.pi*fc*C1*Af)\n", + "R2=Q/(2*math.pi*fc*C1*(2*Q**2-Af))\n", + "R3=Q/(math.pi*fc*C1)\n", + "\n", + "#result\n", + "print \"Resistance R1 is\",round(R1/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R2 is\",round(R2/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R3 is\",round(R3/10**3,2),\"kilo ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.8.b" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R1 is 2.65 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.8.b\n", + "#Change the centre frequency of example 7.8.a to 1.5 kHz, keeping Af and\n", + "#bandwidth constant.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fc0=1*10**3 #Original center frequency\n", + "fc1=1.5*10**3 #New center frequency\n", + "R2=5.97*10**3 #Original resistance\n", + "\n", + "\n", + "\n", + "#calculation\n", + "R2new=R2*(fc0/fc1)**2\n", + "\n", + "#result\n", + "print \"Resistance R1 is\",round(R2new/10**3,2),\"kilo ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.9" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 of highpass section is 15.92 kilo ohm\n", + "Resistance R of lowpass section is 15.92 kilo ohm\n", + "Bandpass Gain Af is 4\n", + "Resistance R1 is 10.0 kilo ohm\n", + "Resistance Rf is 10.0 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.9\n", + "#Design a wide-band reject filter having fh=200 Hz and fl=1 KHz.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fh=200 # Low cutoff freq in Hz\n", + "fl=1*10**3 # High cutoff freq in Hz\n", + "C2=0.01*10**-6 # Assumption\n", + "R2=1/(2*math.pi*fl*C2)\n", + "C=0.05*10**-6\n", + "R1=10*10**3 # Assumption\n", + "Rf=R1 # Since passband gain is 2,R1 and Rf must be equal\n", + "Af=4 # Since gain of high pass and lowpass is set to 2\n", + "\n", + "\n", + "#calculation\n", + "R2=1/(2*math.pi*fl*C2)\n", + "R=1/(2*math.pi*fh*C)\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R2 of highpass section is\",round(R2/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R of lowpass section is\",round(R/10**3,2),\"kilo ohm\"\n", + "print \"Bandpass Gain Af is\",Af\n", + "print \"Resistance R1 is\",round(R1/10**3),\"kilo ohm\"\n", + "print \"Resistance Rf is\",round(Rf/10**3),\"kilo ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.10" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R is 39.01 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.10\n", + "#Design a 60 Hz active notch filter.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fn=60 #Notch-out frequency in Hz\n", + "C=0.068*10**-6 #Assumption\n", + "\n", + "\n", + "\n", + "#calculation\n", + "R=1/(2*math.pi*fn*C)\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R is\",round(R/10**3,2),\"kilo ohm\"\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.11" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Phase angle phi is -90.0 degree\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.11\n", + "#For the all-pass filter of figure 7-16(a),find the phase angle phi if the\n", + "#frequency of vin is 1 kHz.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "f=1*10**3 #Input frequency in Hz\n", + "C=0.01*10**-6 \n", + "R=15.9*10**3 #Resistance in ohms\n", + "\n", + "\n", + "\n", + "#calculation\n", + "phi=math.atan(2*math.pi*f*C*R) #Phase angle\n", + "phi1=-2*phi*180/math.pi\n", + "\n", + "#result\n", + "print \"Phase angle phi is\",round(phi1),\"degree\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.12" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R is 3.3 kilo ohm\n", + "Use Resistance R as 3.3 kohm\n", + "Resistance Rf is 957.0 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.12\n", + "#Design the phase shift oscillator of figure 7-18 so that fo=200 Hz.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fo=200 # Frequency of oscillation\n", + "C=0.1*10**-6 # Assumption\n", + "R=3.3*10**3\n", + "\n", + "#calculation\n", + "R=0.065/(fo*C)\n", + "R=3.3*10**3 #Using rounded value\n", + "R1=10*R # To prevent loading of amplifier\n", + "Rf=29*R1\n", + "\n", + "#result\n", + "print \"Resistance R is\",round(R/10**3,1),\"kilo ohm\"\n", + "print \"Use Resistance R as 3.3 kohm\"\n", + "print \"Resistance Rf is\",round(Rf/10**3),\"kilo ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.13" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R is 3.3 kilo ohm\n", + "Resistance Rf is 24.0 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.13\n", + "#Design the wein bridge oscillator of figure 7-19 so that fo=965 Hz.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fo=965 # Frequency of oscillation\n", + "C=0.05*10**-6 # Assumption\n", + "R1=12*10**3 # Assumption\n", + "\n", + "#calculation\n", + "R=0.159/(fo*C)\n", + "Rf=2*R1\n", + "\n", + "#result\n", + "print \"Resistance R is\",round(R/10**3,1),\"kilo ohm\"\n", + "print \"Resistance Rf is\",round(Rf/10**3),\"kilo ohm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.14" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance values R1,R2,R3 is 100.0 kilo ohm\n", + "Capacitance values C1,C2,C3 is 0.01 uF\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.14\n", + "#Design the quadrature oscillator of figure 7-20 so that fo=159 Hz.\n", + "#The opamp is the 1458/772.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fo=159 # Frequency of oscillation\n", + "C=0.01*10**-6 # Assumption\n", + "\n", + "#calculation\n", + "R=0.159/(fo*C)\n", + "\n", + "#result\n", + "print \"Resistance values R1,R2,R3 is\",round(R/10**3,1),\"kilo ohm\"\n", + "print \"Capacitance values C1,C2,C3 is\",round(C*10**6,2),\"uF\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.15" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 11.6 kilo ohm\n", + "Resistance R is 10.0 ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.15\n", + "#Design the square wave oscillator of figure 7-21(a) so that fo=1 kHz.\n", + "#The opamp is 741 with dc supply voltages = 15, -15 V.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fo=1*10**3 # Frequency of oscillation\n", + "C=0.05*10**-6 # Assumption\n", + "R1=10*10**3 # Assumption\n", + "\n", + "#calculation\n", + "R=1/(2*fo*C)\n", + "R2=1.16*R1\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3,1),\"kilo ohm\"\n", + "print \"Resistance R is\",round(R/10**3),\"ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.16" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 10.0 kilo ohm\n", + "Resistance R1 is 10.0 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.16\n", + "#Design the triangular wave generator of figure 7-23 so that fo=2 kHz and\n", + "#Vo(pp)=7V. The opamp is a 1458/772 and supply voltages =15,-15 V.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "fo=2*10**3 # Frequency of oscillation\n", + "vo=7 #Output voltage\n", + "Vsat=14 #Saturation voltage for opamp 1458\n", + "R3=40*10**3 #Assumption\n", + "C1=0.05*10**-6 #Assumption\n", + "\n", + "\n", + "#calculation\n", + "R2=(vo*R3)/(2*Vsat)\n", + "k=R3/(4*fo*R2) #Using fo=R3/(4*R1*C1*R2),k=R1*C1;\n", + "R1=k/C1\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3),\"kilo ohm\"\n", + "print \"Resistance R1 is\",round(R1/10**3),\"kilo ohm\"\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 7.17" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Terminal voltage Vc is 10.43 volts\n", + "Approximate Nominal freq fo is 26.09 kHz\n", + "Approximate Nominal freq fo1 is 41.67 kHz\n", + "Approximate Nominal freq fo2 is 8.33 kHz\n", + "Change in output freq delta_fo is 33.33 kHz\n" + ] + } + ], + "source": [ + "\n", + "#Example 7.17\n", + "#In the circuit of figure 7-25(c), V=12 V, R2=1.5 Kilo ohm, R1=R3=10 Kilo ohm\n", + "#and C1=0.001 uF.\n", + "#a)Determine the nominal frequency of all the output waveforms.\n", + "#b)Compute the modulation in the output frequencies if Vc is varied between 9.5 V\n", + "#and 11.5 V.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R2=1.5*10**3\n", + "R1=10*10**3\n", + "R3=10*10**3\n", + "C1=0.001*10**-6\n", + "V=12 #Supply voltage\n", + "Vc1=9.5\n", + "Vc2=11.5\n", + "\n", + "#calculation\n", + "Vc=R3*V/(R2+R3) #Using voltage divider rule\n", + "fo=2*(V-Vc)/(V*R1*C1)\n", + "fo1=2*(V-Vc1)/(V*R1*C1)\n", + "fo2=2*(V-Vc2)/(V*R1*C1)\n", + "delta_fo=fo1-fo2 #Change in output freq\n", + "\n", + "\n", + "#result\n", + "print \"Terminal voltage Vc is\",round(Vc,2),\"volts\"\n", + "print \"Approximate Nominal freq fo is\",round(fo/10**3,2),\"kHz\"\n", + "print \"Approximate Nominal freq fo1 is\",round(fo1/10**3,2),\"kHz\"\n", + "print \"Approximate Nominal freq fo2 is\",round(fo2/10**3,2),\"kHz\"\n", + "print \"Change in output freq delta_fo is\",round(delta_fo/10**3,2),\"kHz\"\n", + "\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter8.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter8.ipynb new file mode 100644 index 00000000..2cb88119 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter8.ipynb @@ -0,0 +1,399 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 8. Comparators and Converters" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.1" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f5548e4e5d0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "\n", + "#Example 8.1\n", + "#In the circuit of figure 8-4(a), R1=100 ohm,R2=56 kilo Ohm, Vin=V pp sine wave\n", + "#and the opamp is type 741 with supply voltages 15 V, -15 V.\n", + "#Determine the threshold voltages Vul and Vut and draw the output waveform.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "%matplotlib inline\n", + "import math\n", + "import array\n", + "import numpy as np\n", + "#Variable declaration\n", + "R1=100\n", + "R2=56*10**3\n", + "vin=1 #Input voltage in volt\n", + "pos_Vsat=14 #Positive saturation voltage in volt\n", + "neg_Vsat=-14 #Negative saturation voltage in volt\n", + "Vut=(R1/(R1+R2))*(pos_Vsat) #Upper threshold voltage\n", + "\n", + "\n", + "#calculation\n", + "Vut=(R1/(R1+R2))*(pos_Vsat) #Upper threshold voltage\n", + "Vlt=(R1/(R1+R2))*(neg_Vsat) #Lower threshold voltage\n", + "\n", + "t=np.arange(0,2*math.pi,0.1)\n", + "vut=0.5*np.sin(t)\n", + "subplot(211)\n", + "plt.plot(t,vut)\n", + "plt.ylabel('Vin')\n", + "plt.xlabel('t')\n", + "plt.title(r'$Input voltage$')\n", + "\n", + "import matplotlib.pyplot as plt\n", + "t1=math.asin(0.025/0.5)\n", + "t2=math.pi-math.asin(-0.025/0.5)\n", + "t3=2*math.pi\n", + "x=[0,t1,t2,t3]\n", + "y=[-14,14,-14,14]\n", + "plt.subplot(212)\n", + "plt.step(x,y) #Plotting square wave\n", + "plt.title('Output Waveform')\n", + "plt.xlabel('t')\n", + "plt.ylabel('Vo')\n", + "\n", + "#result\n", + "plt.show()\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.2" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "#Since zener diode is forward biased\n", + "Output voltage during positive half-cycle of the input is -0.7 V\n", + "Output voltage during negative half-cycle of the input is 5.1 V\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f55408e2f50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Example 8.2\n", + "#In the circuit of figure 8-7(a), Vin=500 mV peak 60-Hz sinewave, R=100 ohm.\n", + "#IN3826 zener with Vz=5.1 V and the supply voltages= 15 V, -15V.\n", + "#Determine the output voltage swing.Assume that the voltage drop across\n", + "#the forward biased zener=0.7V.\n", + "\n", + "%matplotlib inline\n", + "from __future__ import division #to perform decimal division\n", + "import numpy as np\n", + "import matplotlib.pyplot as plt\n", + "import math\n", + "import array\n", + "\n", + "#Variable declaration\n", + "vin=5*10**-3\n", + "R=100\n", + "Vd1=-0.7 # Output voltage during positive half-cycle of the input\n", + "Vd2=5.1 # Output voltage during negative half-cycle of the input\n", + "\n", + "#calculation\n", + "\n", + "\n", + "t=np.arange(0,2*math.pi,0.1)\n", + "vut=0.5*np.sin(t)\n", + "subplot(211)\n", + "plt.plot(t,vut)\n", + "plt.ylabel('Vin')\n", + "plt.xlabel('t')\n", + "plt.title(r'$Input voltage$')\n", + "\n", + "\n", + "\n", + "\n", + "t1=math.pi\n", + "t2=2*math.pi\n", + "x=[0,t1,t2]\n", + "y=[-0.7,-0.7,5.1]\n", + "subplot(2,1,2)\n", + "plt.step(x,y)\n", + "plt.title('Output Waveform')\n", + "plt.xlabel('t')\n", + "plt.ylabel('Vo')\n", + "\n", + "#result\n", + "print \"#Since zener diode is forward biased\"\n", + "print \"Output voltage during positive half-cycle of the input is\",Vd1,\"V\"\n", + "print \"Output voltage during negative half-cycle of the input is\",Vd2,\"V\" \n", + "plt.show()" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.3" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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7YnVjIJDpsOD8ycH/+TvMGOwW/B93Y/7/ycDNwKfr6GG88QXsPq7CHqIbsYd3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\n+PS6p2FJxz8G3IeFOK7MSyt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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f5522d5d710>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output freq Fo is 2000 Hz\n", + "Output freq Fo/2 is 1000 Hz\n" + ] + } + ], + "source": [ + "\n", + "#Example 8.3\n", + "#The V..F converter of figure 8-12 is initially adjusted for a 10 kHz full scale\n", + "#output frequency.Determine the output frequencies Fo and Fo/2 if the output\n", + "#signal Vin=2 V.\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "from matplotlib.pyplot import subplot\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "import array\n", + "\n", + "\n", + "\n", + "#Variable declaration\n", + "Vin=2 # Input voltage\n", + "Fo1=2*10**3 # Output freq Fo when Vin=2V\n", + "Fo2=1*10**3 # Output freq Fo/2 when Vin=2V\n", + "\n", + "#calculation\n", + "import array\n", + "v=array.array('i',(0 for i in range(0,50)))\n", + "\n", + "\n", + "count=1\n", + "for i in range(1,50): #for 5 cycles\n", + " if count<4:\n", + " v[i]=5\n", + " else:\n", + " v[i]=0\n", + " if count<10:\n", + " count=count+1\n", + " else:\n", + " count=1\n", + "\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(v)\n", + "plt.title('Output Waveform')\n", + "plt.xlabel('t(microsec)')\n", + "plt.ylabel('Pulse freq output,Fo(V)')\n", + "\n", + "import matplotlib.pyplot as plt\n", + "for i in range(1,50): #for 5 cycles\n", + " if count<10:\n", + " v[i]=5\n", + " else:\n", + " v[i]=0\n", + " if count<20:\n", + " count=count+1\n", + " else:\n", + " count=1\n", + "\n", + "plt.subplot(2,1,2)\n", + "plt.plot(v)\n", + "plt.title('Output Waveform')\n", + "plt.xlabel('t(microsec)')\n", + "plt.ylabel('Pulse freq output,Fo(V)')\n", + "plt.show()\n", + "\n", + "#result\n", + "print \"Output freq Fo is \",Fo1,\"Hz\"\n", + "print \"Output freq Fo/2 is\",Fo2,\"Hz\"" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage is 0.28 Volts\n" + ] + } + ], + "source": [ + "\n", + "#Example 8.4\n", + "#The F/V converterof figure 8-14(a) is initially adjusted for Vo=2.8V at Fin max\n", + "#of 10 kHz. Determine the output voltage Vo if fin= 1 kHz.\n", + "\n", + "#Variable decclaration\n", + "Vo=2.8 #At Finmax of 10kHz\n", + "\n", + "#Calculation\n", + "Vo1=Vo/10 #Output voltage at Fin=1kHz\n", + "\n", + "#Result\n", + "print \"Output voltage is\",Vo1,\"Volts\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 8.5" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f5522c75350>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "#Example 8.5\n", + "#In the circuit of figure 8-25(a), Vin=200 mV peak to peak sine wave at 100 Hz.\n", + "#Briefly describe the operation of the circuit and draw the output waveform.\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "from __future__ import division #to perform decimal division\n", + "import numpy as np\n", + "import math\n", + "import array\n", + "\n", + "\n", + "#Variable declaration\n", + "Vin=100*10**-3 # Input voltage\n", + "\n", + "\n", + "#calculation\n", + "\n", + "t=np.arange(0,math.pi,0.1) #time scale\n", + "v=Vin*np.sin(t)\n", + "\n", + "import matplotlib.pyplot as plt\n", + "plt.xlim(0,2*math.pi)\n", + "plt.ylim(0,0.1)\n", + "plt.plot(t,v)\n", + "plt.ylabel('Vin')\n", + "plt.xlabel('t')\n", + "plt.title(r'$Input voltage$')\n", + "\n", + "\n", + "#result\n", + "plt.show()" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter9.ipynb b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter9.ipynb new file mode 100644 index 00000000..117f377b --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/Chapter9.ipynb @@ -0,0 +1,813 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "# Chapter 9: Specialixed IC Applications" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.1" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 31.6 kilo ohm\n", + "Resistance R3 is 3.27 kilo ohm\n", + "Resistance R1 is Open\n", + "Resistance R4 is 25.15 kilo ohm\n", + "Resistance R5 is 25.15 kilo ohm\n", + "Resistance R6 is 1.8 kilo ohm\n", + "Resistance R7 is 9.0 kilo ohm\n", + "Resistance R8 is 1.5 kilo ohm\n" + ] + } + ], + "source": [ + "#Example 9.1\n", + "#The FLT-U2 is to be used as a second order inverting Butterworth low pass filter\n", + "#with a dc gain of 5,cutoff frequency of 2 kHz and Q=10. Determine the values\n", + "#of the external components.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "dc_gain=5\n", + "f1=2*10**3 # Cutoff freq in Hz\n", + "Q=10 # Figure of merit\n", + "\n", + "\n", + "#calculation\n", + "R2=(316*10**3)/10 #Resistance R2\n", + "R3=(100*10**3)/((3.16*Q)-1)\n", + "R4=(5.03*10**7)/f1\n", + "R5=R4\n", + "R6=1.8*10**3 # Assumption\n", + "R7=dc_gain*R6\n", + "R8=(R6*R7)/(R6+R7)\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R3 is\",round(R3/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R1 is Open\"\n", + "print \"Resistance R4 is\",round(R4/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R5 is\",round(R5/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R6 is\",round(R6/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R7 is\",round(R7/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R8 is\",round(R8/10**3,2),\"kilo ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.2" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 100.0 kilo ohm\n", + "Resistance R3 is 2.96 kilo ohm\n", + "Resistance R1 is Open ohm\n", + "Resistance R4 is 10.0 kilo ohm\n", + "Resistance R5 is 10.0 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.2\n", + "#Using the FLT-U2, design a second order inverting Butterworth bandpass filter\n", + "#with centre frequency f1=5 kHz and Q=10.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "f1=5*10**3 # Center freq in Hz\n", + "Q=10 # Figure of merit\n", + "R2=100*10**3 # Constant for band-pass filter\n", + "\n", + "\n", + "#calculation\n", + "R3=(100*10**3)/((3.48*Q)-1)\n", + "R4=(5.03*10**7)/f1\n", + "R5=R4\n", + "\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R3 is\",round(R3/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R1 is Open\",\"ohm\"\n", + "print \"Resistance R4 is\",round(R4/10**3),\"kilo ohm\"\n", + "print \"Resistance R5 is\",round(R5/10**3),\"kilo ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.3" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 100.0 kilo ohm\n", + "Resistance R3 is 2.96 kilo ohm\n", + "Resistance R1 is Open ohm\n", + "Resistance R4 is 10.06 kilo ohm\n", + "Resistance R5 is 10.06 kilo ohm\n", + "Resistance R6 is 10.0 kilo ohm\n", + "Resistance R7 is 10.0 kilo ohm\n", + "Resistance R8 is 10.0 kilo ohm\n", + "Resistance R9 is 3.33 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.3\n", + "#Using the FLT-U2, design a notch filter with 5 kHz notch out frequency and\n", + "#Q=10.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "f1=5*10**3 # Center freq in Hz\n", + "Q=10 # Figure of merit\n", + "R2=100*10**3 # Constant for band-pass filter\n", + "\n", + "\n", + "#calculation\n", + "R3=(100*10**3)/((3.48*Q)-1)\n", + "R4=(5.03*10**7)/f1\n", + "R5=R4\n", + "R6=10*10**3 #Assumption\n", + "R7=R6\n", + "R8=R6\n", + "R9=(R6*R7*R8)/(R6*R7+R6*R8+R7*R8) #Since R6||R7||R8\n", + "\n", + "\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R3 is\",round(R3/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R1 is Open\",\"ohm\"\n", + "print \"Resistance R4 is\",round(R4/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R5 is\",round(R5/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R6 is\",round(R6/10**3),\"kilo ohm\"\n", + "print \"Resistance R7 is\",round(R7/10**3),\"kilo ohm\"\n", + "print \"Resistance R8 is\",round(R8/10**3),\"kilo ohm\"\n", + "print \"Resistance R9 is\",round(R9/10**3,2),\"kilo ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.4" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2 is 20.0 kilo ohm\n", + "Resistance R3 is 14.14 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.4\n", + "#Using the MF5,design a second order Butterworth lowpass filter with a cutoff\n", + "#frequency of 500Hz and a passband gain of -2. Assume that a 5,-5 V power supply\n", + "#and a CMOS clock are used.\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "f1=500 #Cut-off freq in Hz\n", + "Holp=-2 #Passband gain\n", + "R1=10*10**3 #Assumption\n", + "Q=0.707 #Figure of merit Q is fixed for second order butterworth LPF\n", + "#calculation\n", + "\n", + "R2=-R1*Holp #Using Holp=-R2/R1;\n", + "R3=Q*R2 #Using Q=R3/R2\n", + "\n", + "#result\n", + "print \"Resistance R2 is\",round(R2/10**3),\"kilo ohm\"\n", + "print \"Resistance R3 is\",round(R3/10**3,2),\"kilo ohm\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.5" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Capacitance C is 1.0 micro Farad\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.5\n", + "#In the circuit of figure 9-16(a), Ra=10 Kilo ohm, the output pulse width\n", + "#tp=10 ms. Determine the value of C.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Ra=10*10**3 #Resistance in ohm\n", + "tp=10*10**-3 #Output pulse width\n", + "C=tp/(1.1*Ra)\n", + "\n", + "#calculation\n", + "C=tp/(1.1*Ra)\n", + "\n", + "#result\n", + "print \"Capacitance C is\",round(C*10**6),\"micro Farad\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance Ra is 54.55 kilo ohm\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.6\n", + "#The circuit of figure 9-16 (a) is to be used as a divide-by-2 network.\n", + "#The frequency of the input trigger signal is 2 kHz.If the value of C=0.01 uF\n", + "#What should be the value of Ra.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "f=2*10**3 #Freq of input trigger signal in Hz\n", + "C=0.01*10**-6\n", + "\n", + "#calculation\n", + "tp=1.2/f\n", + "Ra=tp/(1.1*C)\n", + "\n", + "#result\n", + "print \"Resistance Ra is\",round(Ra/10**3,2),\"kilo ohm\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Charging time of the capacitor is 0.42 ms\n", + "Discharging time of the capacitor is 0.27 ms\n", + "Freq of oscillation is 1.4 kHz\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.7\n", + "#In the astable multivibrator of figure 9-21(a), Ra=2.2 kilo ohm, Rb=3.9 kilo ohm\n", + "#and C=0.1 uF. Determine the pulse width tc, negative pulse width td and free\n", + "#running frequency fo.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Ra=2.2*10**3 # Resistance in ohm\n", + "Rb=3.9*10**3 # Resistance in ohm\n", + "C=0.1*10**-6 # capacitance in farad\n", + "\n", + "#calculation\n", + "tc=0.69*(Ra+Rb)*C # Charging time of the capacitor\n", + "td=0.69*Rb*C # Discharging time of the capacitor\n", + "T=tc+td\n", + "fo=1/T # Freq of oscillation\n", + "\n", + "\n", + "#result\n", + "print \"Charging time of the capacitor is\",round(tc*10**3,2),\"ms\"\n", + "print \"Discharging time of the capacitor is\",round(td*10**3,2),\"ms\"\n", + "print \"Freq of oscillation is\",round(fo/10**3,1),\"kHz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.8" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Freq of free running ramp generator is 5.16 kHz\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.8\n", + "#Referring to the circuit of figure 9-24(a),determine the frequency of the free-\n", + "#running ramp generator if R is set at 10 kHz.\n", + "#Assume that Vbe=Vd1=0.7 V.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R=10*10**3 #Resistance in ohm\n", + "Vcc=5 #Supply voltage in volt\n", + "Vbe=0.7 #Base to emitter voltage in volt\n", + "C=0.05*10**-6 #Capacitance in farad\n", + "\n", + "#calculation\n", + "Ic=(Vcc-Vbe)/R #Collector current in ampere\n", + "fo=(3*Ic)/(Vcc*C)\n", + "\n", + "#result\n", + "print \"Freq of free running ramp generator is\",round(fo/10**3,2),\"kHz\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.9" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Free running frequency of VCO is 2.5 kHz\n", + "Lock range frequency of VCO is 1.0 kHz\n", + "Capture range frequency of VCO is 66.49 Hz\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.9\n", + "#Referring to the circuit of figure 9-33(a),determine the free-running frequency\n", + "#fout, the lock range fl and the capture range fc.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "R1=12*10**3 # Resistance in ohm\n", + "V_plus=10 # Supply voltage in volt\n", + "V_minus=-10 # Supply voltage in volt\n", + "C1=0.01*10**-6 # Capacitance in farad\n", + "C2=10*10**-6 # Capacitance in farad\n", + "\n", + "#calculation\n", + "fout=1.2/(4*R1*C1)\n", + "V=V_plus-V_minus\n", + "fl=(8*fout)/V\n", + "fc=math.sqrt(fl/(2*math.pi*3.6*10**3*C2))\n", + "\n", + "#result\n", + "print \"Free running frequency of VCO is\",round(fout/10**3,1),\"kHz\"\n", + "print \"Lock range frequency of VCO is\",round(fl/10**3),\"kHz\"\n", + "print \"Capture range frequency of VCO is\",round(fc,2),\"Hz\"\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.10" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R is 20.0 ohm\n", + "Output voltage Vo is 17.0 Volt\n", + "Min input voltage Vin is 19.0 Volt\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.10\n", + "#Using the 7805C voltage regulator , design a current source that will deliver\n", + "#a 0.25 A current to a 48 ohm, 10 W load.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Vr=5 #Voltage in volt\n", + "Il=0.25 #Load current in ampere\n", + "Rl=48 #Load resistance in ohm\n", + "dropout_volt=2 #Constant for IC7805C\n", + "\n", + "#calculation\n", + "R=Vr/Il #Approximate result sice Iq is negligible in the eq. Il=(Vr/Il)+Iq where Iq is quiescent current\n", + "Vl=Rl*Il\n", + "Vo=Vr+Vl\n", + "Vin=Vo+dropout_volt\n", + "\n", + "#result\n", + "print \"Resistance R is\",R,\"ohm\"\n", + "print \"Output voltage Vo is\",Vo,\"Volt\"\n", + "print \"Min input voltage Vin is\",Vin,\"Volt\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.11" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Resistance R2_min is 0.71 kilo ohm\n", + "Resistance R2_max is 2.03 kilo ohm\n", + "Therefore resistance should be varied from R2_min to R2_max values\n", + "To do this we take R2 as 3kohm potentiometer\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.11\n", + "#Design an adjustable voltage regulator to satisfy the following specifications\n", + "#Output voltage Vo= 5 to 12 V\n", + "#Output current Io= 1 A.\n", + "#Voltage regulator is LM317.\n", + "\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Vo_min=5 #Min output voltage in volt\n", + "Vo_max=12 #Max output voltage in volt\n", + "Vref=1.25 #Reference voltage in volt\n", + "Iadj=100*10**-6 #Adjustment pin current in ampere\n", + "R1=240 #Assumption\n", + "C2=1*10**-6 #Added to the circuit to improve transient response\n", + "C3=1*10**-6 #Added to the circuit to obtain high ripple rejection ratios\n", + "\n", + "#calculation\n", + "R2_min=R1*(Vo_min-Vref)/(Vref+Iadj*R1) #Using Vo_min=Vref*(1+R2/R1)+Iadj*R2\n", + "R2_max=R1*(Vo_max-Vref)/(Vref+Iadj*R1) #Using Vo_max=Vref*(1+R2/R1)+Iadj*R2\n", + "\n", + "#result\n", + "print \"Resistance R2_min is\",round(R2_min/10**3,2),\"kilo ohm\"\n", + "print \"Resistance R2_max is\",round(R2_max/10**3,2),\"kilo ohm\"\n", + "print \"Therefore resistance should be varied from R2_min to R2_max values\"\n", + "print \"To do this we take R2 as 3kohm potentiometer\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.12" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sense current,Ipk is 1.0 A\n", + "Sense resistance,Rsc is 0.33 ohm\n", + "Constant K is 1.06\n", + "i.e, ton is K times of toff\n", + "OFF time period,toff is 24.27 us\n", + "ON time period,ton is 25.73 us\n", + "Inductance,L is 151.7 uH\n", + "Output capacitance,Co is 125.0 uF\n", + "Resistance R2 is 12.0 kilo ohm\n", + "Resistance R1 is 38.0 kilo ohm\n", + "efficiency is 81.0\n" + ] + } + ], + "source": [ + "\n", + "#Example 9.12\n", + "#Design a step down switching regulator according to the following\n", + "#specifications.\n", + "#Input voltage Vin= 12 V dc.\n", + "#Output voltage Vo= 5V at 500 m A maximum.\n", + "#Output ripple voltage Vripple= 50 mV or 1% of Vo\n", + "#Switching regulator :uA78S40.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Iomax=500*10**-3 # Max output current in ampere\n", + "Vo=5 # Output voltage in volt\n", + "Vd=1.25 # Voltage drop across the power diode in volt\n", + "Vin=12 # Input voltage in volt\n", + "Vs=1.1 # Output saturation voltage in volt\n", + "Vripple=50*10**-3 # Output ripple voltage in volt\n", + "Vref=1.245 # Reference voltage in volt\n", + "Vr2=1.2 # Voltage across resistance R2 in volt\n", + "\n", + "#calculation\n", + "Ipk=2*Iomax # Sense current in ampere\n", + "Rsc=0.33/Ipk # Sense resistance in ohm\n", + "K=(Vo+Vd)/(Vin-Vs-Vo) # K= ton/toff\n", + "f=20*10**3 # Assuming operating freq in Hz\n", + "T=1/f\n", + "toff=T/2.06 # Using ton+toff=T and substituting for ton\n", + "ton=1.06*toff\n", + "Ct=45*10**-5*toff # Oscillator timing capacitance in farad\n", + "L=((Vo+Vd)/Ipk)*toff # Inductance in henry\n", + "Co=Ipk*((ton+toff)/(8*Vripple)) # Output capacitance in farad\n", + "I2=0.1*10**-3 # Assuming the current through R2\n", + "R2=Vref/I2 # Resistance R2 in ohm\n", + "R2=12*10**3 # Taking approximate value\n", + "R1=(R2*(Vo-Vr2))/Vr2 # Using Vr2=(R1*Vo)/R1+R2, voltage divider rule\n", + "efficiency=((Vin-Vs+Vd)/Vin)*(Vo/(Vo+Vd))*100\n", + "\n", + "#result\n", + "print \"Sense current,Ipk is\",Ipk,\"A\"\n", + "print \"Sense resistance,Rsc is\",Rsc,\"ohm\"\n", + "print \"Constant K is\",round(K,2)\n", + "print \"i.e, ton is K times of toff\"\n", + "print \"OFF time period,toff is\",round(toff*10**6,2),\"us\"\n", + "print \"ON time period,ton is\",round(ton*10**6,2),\"us\"\n", + "print \"Inductance,L is\",round(L*10**6,2),\"uH\"\n", + "print \"Output capacitance,Co is\",round(Co*10**6,3),\"uF\"\n", + "print \"Resistance R2 is\",R2/10**3,\"kilo ohm\"\n", + "print \"Resistance R1 is\",R1/10**3,\"kilo ohm\"\n", + "print \"efficiency is\",efficiency\n" + ] + }, + { + "cell_type": "markdown", + "metadata": { + "collapsed": true + }, + "source": [ + "## Example 9.13" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Sense current,Ipk is 6 A\n", + "Sense resistance,Rsc is 0.055 ohm\n", + "Constant K 1.06\n", + "i.e, ton is K times of toff\n", + "OFF time period,toff is 24.27 us\n", + "ON time period,ton is 25.73 us\n", + "Oscillator timing capacitance,Ct is 10.9 nF\n", + "Inductance,L is = %.8f H 25.28 uH\n", + "Output capacitance,Co is = %.7f F 0.75 milli Farad\n" + ] + } + ], + "source": [ + "#Example 9.13\n", + "#Upgrade the switching regulator in Example 9-12 to provide +5V at 3A.\n", + "#Use the same specifications given in example 9-12,except the output ratings.\n", + "\n", + "from __future__ import division #to perform decimal division\n", + "import math\n", + "\n", + "\n", + "#Variable declaration\n", + "Iomax=3 #Max output current in ampere\n", + "Vo=5 #Output voltage in volt\n", + "Vd=1.25 #Voltage drop across the power diode in volt\n", + "Vin=12 #Input voltage in volt\n", + "Vs=1.1 #Output saturation voltage in volt\n", + "Vripple=50*10**-3 #Output ripple voltage in volt\n", + "Vref=1.245 #Reference voltage in volt\n", + "Vr2=1.2 #Voltage across resistance R2 in volt\n", + "\n", + "#calculation\n", + "Ipk=2*Iomax #Sense current in ampere\n", + "Rsc=0.33/Ipk #Sense resistance in ohm\n", + "K=(Vo+Vd)/(Vin-Vs-Vo) #K= ton/toff\n", + "f=20*10**3 #Assuming operating freq in Hz\n", + "T=1/f\n", + "toff=T/2.06 #Using ton+toff=T and substituting for ton\n", + "ton=1.06*toff\n", + "Ct=45*10**-5*toff #Oscillator timing capacitance in farad\n", + "L=((Vo+Vd)/Ipk)*toff #Inductance in henry\n", + "Co=Ipk*((ton+toff)/(8*Vripple)) #Output capacitance in farad\n", + "\n", + "#result\n", + "print \"Sense current,Ipk is\",Ipk,\"A\"\n", + "print \"Sense resistance,Rsc is\",Rsc,\"ohm\"\n", + "print \"Constant K\",round(K,2)\n", + "print \"i.e, ton is K times of toff\"\n", + "print \"OFF time period,toff is\",round(toff*10**6,2),\"us\"\n", + "print \"ON time period,ton is\",round(ton*10**6,2),\"us\"\n", + "print \"Oscillator timing capacitance,Ct is\",round(Ct*10**9,1),\"nF\"\n", + "print \"Inductance,L is = %.8f H\",round(L*10**6,2),\"uH\"\n", + "print \"Output capacitance,Co is = %.7f F\",round(Co*10**3,5),\"milli Farad\"\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.6" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/README.txt b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/README.txt new file mode 100644 index 00000000..cdf11029 --- /dev/null +++ b/OpAmps_And_Linear_Integrated_Circuits_by_Gayakwad/README.txt @@ -0,0 +1,10 @@ +Contributed By: Jumana - +Course: mtech +College/Institute/Organization: MPBtech. ECE, 6th SemVNIT Nagpur +Department/Designation: MPBtech. ECE, 6th SemVNIT Nagpur +Book Title: OpAmps And Linear Integrated Circuits +Author: Gayakwad +Publisher: PHI Learning Pvt. Ltd., New Delhi +Year of publication: 2004 +Isbn: 9788120320581 +Edition: 4
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a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_6.ipynb new file mode 100644 index 00000000..0f6f4355 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter10_6.ipynb @@ -0,0 +1,1279 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 10 : SINGLE STAGE TRANSISTOR AMPLIFIERS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.2 : page number 243-244" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of the emitter capacitor = 1.42 ЁЭЬЗF\n" + ] + } + ], + "source": [ + "from math import pi\n", + "#Variable declaration\n", + "f_min=2.0; #Minimum frequency of operation of amplifier, kHz\n", + "f_max=10.0; #Maximum frequency of operation of amplifier, kHz\n", + "RE=560.0; #Emitter resistor, тДж\n", + "\n", + "#Calculations\n", + "#X_CE(Emitter capacitor's capacitive reactance)\n", + "#X_CE=1/(2*pi*f_min*CE)=RE/10\n", + "#From the above equation.\n", + "CE=1/(2*pi*f_min*1000*(RE/10)); #Emitter capacitor, F,\n", + "\n", + "CE=CE*10**6; #Emitter capacitor, ЁЭЬЗF\n", + "\n", + "\n", + "#Results\n", + "print('The value of the emitter capacitor = %.2f ЁЭЬЗF'%(CE));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.5: Page number 252-253" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The operating point: VCE=8.55V and IC=2.15mA.\n", + "Maximum v_CE=9.62V and maximum i_C=19.25mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f895dcab908>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage in V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=10.0; #Resistor R1, kтДж\n", + "R2=5.0; #Resistor R2, kтДж\n", + "RC=1.0; #Collector resistor, kтДж\n", + "RE=2.0; #Emitter resistor, kтДж\n", + "RL=1.0; #Load resistor, kтДж\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#For d.c load line, from the equation: VCE=VCC-IC*(RC+RE),\n", + "#VCE is maximum when IC=0 and IC is maximum when VCE=0.\n", + "VCE_max=VCC; #Maximum collector-emitter voltage, V\n", + "IC_max=VCC/(RC+RE); #Maximum collector current, mA\n", + "\n", + "\n", + "#plot\n", + "VCE_plot=[i for i in range(0,(int)(VCC+1))]; #Plot variable for V_CE\n", + "IC_plot=[((VCC-i)/(RC+RE)) for i in (VCE_plot[:])]; #Plot variable for I_C\n", + "\n", + "plt.subplot(211)\n", + "plt.xlim(0,20)\n", + "plt.ylim(0,6)\n", + "plt.plot(VCE_plot,IC_plot);\n", + "plt.xlabel(\"VCE(V)\");\n", + "plt.ylabel(\"IC(mA)\");\n", + "plt.title(\"d.c load line\");\n", + "\n", + "\n", + "\n", + "#(ii)\n", + "#For operating point:\n", + "#Assuming VCC drops almost completely across R1 and R2,\n", + "V2=VCC*R2/(R1+R2); #Voltage across resistor R2, V\n", + "IE=(V2-VBE)/RE; #Emitter current, mA\n", + "IC=IE; #Collector current, mA\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage , V\n", + "\n", + "print(\"The operating point: VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n", + "\n", + "\n", + "#(iii)\n", + "#For a.c load line\n", + "RAC=(RC*RL)/(RC+RL); #a.c load, kтДж\n", + "VCE_ac_max=VCE+IC*RAC; #Maximum collector-emitter voltage, V\n", + "IC_ac_max=IC+VCE/RAC; #Maximum collector current, mA\n", + "print(\"Maximum v_CE=%.2fV and maximum i_C=%.2fmA\"%(VCE_ac_max,IC_ac_max));\n", + "\n", + "#plot\n", + "vCE_plot=[0,VCE_ac_max]; #Plot variable for V_CE\n", + "iC_plot=[IC_ac_max,0]; #Plot variable for I_C\n", + "\n", + "plt.subplot(212)\n", + "plt.xlim(0,10)\n", + "plt.ylim(0,20)\n", + "plt.plot(vCE_plot,iC_plot);\n", + "plt.xlabel(\"vCE(V)\");\n", + "plt.ylabel(\"iC(mA)\");\n", + "plt.title(\"a.c load line\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.6: Page number 253-254" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f895dbf5390>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as p\n", + "\n", + "#Variable declaration\n", + "RC=10; #Collector resistor, kтДж\n", + "RL=30; #Load resistor, kтДж\n", + "VCC=20; #Collector supply voltage, V\n", + "IC=1; #Collector current, mA\n", + "VCE=10; #Collector-emitter voltage, V\n", + "\n", + "\n", + "#Calculations\n", + "#For d.c load line:\n", + "#From the equation: VCE=VCC-IC*(RC+RE),\n", + "#When VCE=0, IC is maximum.\n", + "#Emitter resistor is neglected, assuming it as negligible\n", + "IC_max=VCC/RC; #Maximum collector current, mA\n", + "\n", + "#And, when IC=0, VCE is maximum\n", + "VCE_max=VCC; #Maximum collector-emitter voltage, V\n", + "\n", + "#plot\n", + "p.subplot(211)\n", + "p.xlim(0,20)\n", + "p.ylim(0,5)\n", + "VCE_plot=[0,VCE_max]; #Plot variable for V_CE\n", + "IC_plot=[IC_max,0]; #Plot variable for I_C\n", + "\n", + "p.plot(VCE_plot,IC_plot);\n", + "p.xlabel(\"VCE(V)\");\n", + "p.ylabel(\"IC(mA)\");\n", + "p.title(\"d.c load line\");\n", + "\n", + "\n", + "#For a.c load line:\n", + "RAC=(RC*RL)/(RC+RL); #a.c Load resistor, kтДж\n", + "\n", + "VCE_ac_max=VCE+IC*RAC; #Maximum collector-emitter voltage, V\n", + "IC_ac_max=IC+ VCE/RAC; #Maximum collector current, mA\n", + "\n", + "#plot\n", + "p.subplot(212)\n", + "p.xlim([0,25])\n", + "p.ylim([0,5])\n", + "vCE_plot=[0,VCE_ac_max]; #Plot variable for V_CE\n", + "iC_plot=[IC_ac_max,0]; #Plot variable for I_C\n", + "\n", + "p.plot(vCE_plot,iC_plot);\n", + "p.xlabel(\"vCE(V)\");\n", + "p.ylabel(\"iC(mA)\");\n", + "p.title(\"a.c load line\");\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.7 : Page number 254-255" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f898013a748>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as p\n", + "\n", + "#Variabe declaration\n", + "VCE_Q=8.0; #Q-point collector emitter voltage, V\n", + "IC_Q=1; #Q-point collector current, mA\n", + "ic_positive_peak=1.5; #Collector current at positive peak of signal, mA\n", + "ic_negative_peak=0.5; #Collector current at negative peak of signal, mA\n", + "vce_positive_peak=7; #Collector emitter voltage at positive peak of signal, V\n", + "vce_negative_peak=9; #Collector emitter voltage at negative peak of signal, V\n", + "\n", + "#Plot\n", + "vce_plot=[vce_positive_peak,vce_negative_peak]; #Plot variable of vce\n", + "ic_plot=[ic_positive_peak,ic_negative_peak]; #Plot variable of ic\n", + "\n", + "p.xlim(0,10)\n", + "p.ylim(0,2)\n", + "p.plot(vce_plot,ic_plot);\n", + "p.xlabel(\"vCE(V)\");\n", + "p.ylabel(\"iC(mA)\");\n", + "p.title(\"a.c load line\");\n", + "p.grid();\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.8 : Page number 256" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain= 24.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "RC=2.0; #Collector resistor, kтДж\n", + "Rin=1.0; #Input resistance, kтДж\n", + "beta=60.0; #Base current amplification factor\n", + "RL=0.5; #Load resistor, kтДж\n", + "\n", + "\n", + "#Calculation\n", + "RAC=(RC*RL)/(RC+RL); #a.c load resistor, kтДж\n", + "Av=beta*(RAC/Rin); #Voltage gain\n", + "\n", + "#Results\n", + "print(\"Voltage gain= %d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.9 : Page number 256" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage= 200mV.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "V_in=1.0; #Input voltage , mV\n", + "RC=10.0; #Collector resistor, kтДж\n", + "Rin=2.5; #Input resistance, kтДж\n", + "beta=100.0; #Base current amplification factor\n", + "RL=10.0; #Load resistor, kтДж\n", + "\n", + "#Calculations\n", + "RAC=(RC*RL)/(RC+RL); #Effective load, kтДж\n", + "Av=beta*(RAC/Rin); #Voltage gain\n", + "\n", + "V_out=V_in*Av; #Output voltage, V\n", + "\n", + "#Results\n", + "print(\"Output voltage= %dmV.\"%V_out);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.10 : Page number 256-257" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Beta= 100.\n", + "Input impedance=2 kтДж.\n", + "a.c load=3.3 kтДж.\n", + "Voltage gain= 165.\n", + "Power gain=16500.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "change_in_IB=10.0; #Change in base current, ЁЭЬЗA\n", + "change_in_IC=1.0; #Change in collector current, mA\n", + "change_in_VBE=0.02; #Change in Base-emitter voltage, V\n", + "RC=5.0; #Collector resistor, kтДж\n", + "RL=10.0; #Emitter resistor, kтДж\n", + "\n", + "#Calculations\n", + "#(i)\n", + "beta=(change_in_IC*1000)/change_in_IB; #Base current amplification factor\n", + "\n", + "#(ii)\n", + "Rin=(change_in_VBE/change_in_IB)*1000; #Input impedance, kтДж\n", + "\n", + "#(iii)\n", + "RAC=round((RC*RL)/(RC+RL),1); #a.c load, kтДж\n", + "\n", + "#(iv)\n", + "Av=beta*RAC/Rin; #Voltage gain\n", + "\n", + "#(v)\n", + "Ap=beta*Av; #Power gain\n", + "\n", + "\n", + "#Results\n", + "print(\"Beta= %d.\"%beta);\n", + "print(\"Input impedance=%d kтДж.\"%Rin);\n", + "print(\"a.c load=%.1f kтДж.\"%RAC);\n", + "print(\"Voltage gain= %d.\"%Av);\n", + "print(\"Power gain=%d.\"%Ap);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.11 : Page number 257" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=200mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "beta=50.0; #Base current amplification factor\n", + "RC=3.0; #Collector resistor,kтДж\n", + "RL=6.0; #Load resistor, kтДж\n", + "Rin=0.5; #Input impedance, kтДж\n", + "Vin=1; #Input voltage, mV\n", + "\n", + "#Calculation\n", + "RAC=(RC*RL)/(RC+RL); #a.c load, kтДж\n", + "Av=beta*RAC/Rin; #Voltage gain\n", + "Vout=Vin*Av; #Output voltage, V\n", + "\n", + "#Results\n", + "print(\"Output voltage=%dmV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.12 : Page number 257-258" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The circuit is not operating properly.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VT=6.0; #Collector potential, V\n", + "R1=1.0; #Resistor R1, kтДж\n", + "R2=2.0; #Resistor R2, kтДж\n", + "VB_found=4.0; #Measured base voltage, V\n", + "\n", + "#Calculations\n", + "VB=(VT*R1)/(R1+R2); #Theoretical base voltage, V\n", + "\n", + "if(VB_found==VB):\n", + " print(\"The circuit is operating properly.\");\n", + "else:\n", + " print(\"The circuit is not operating properly.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.13 : Page number 258-259" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "a.c emitter resistance= 38.46 тДж.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "R1=40.0; #Resistor R1, kтДж\n", + "R2=10.0; #Resistor R2, kтДж\n", + "RC=6.0; #Collector resistor, kтДж\n", + "RE=2.0; #Emitter resistor, kтДж\n", + "beta=80; #Base current amplification factor\n", + "VBE=0.7; #Base emitter voltage, V\n", + "\n", + "#Calculations\n", + "V2=(VCC*R2)/(R1+R2); #Voltage across resistor R2, V\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=VE/RE; #Emitter current, mA\n", + "re=25/IE; #a.c emitter resistance, тДж\n", + "\n", + "\n", + "#Results\n", + "print(\"a.c emitter resistance= %.2f тДж.\"%re);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.14 : Page number 262-263" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Voltage gain= 360.\n", + "(ii)Voltage gain= 5.37.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=150.0; #Resistor R1, kтДж\n", + "R2=20.0 #Resistor R2, kтДж\n", + "RC=12.0; #Collector resistor, kтДж\n", + "RE=2.2; #Emitter resistor, kтДж\n", + "\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),2); #Voltage across R2, V\n", + "VE=round(V2-VBE,2); #Voltage across emitter resistor, V\n", + "IE=round(VE/RE,2); #Emitter current, mA\n", + "re=round(25/IE,1); #a.c emitter resistance, тДж\n", + "\n", + "\n", + "#(i)\n", + "#CE(emitter capacitor) connected in the circuit:\n", + "Av=(RC*1000)/re; #Voltage gain for emitter capacitor connected.\n", + "\n", + "print(\"(i)Voltage gain= %d.\"%Av);\n", + "\n", + "#(ii)\n", + "#CE(emitter capacitor) removed from the circuit:\n", + "Av=(RC*1000)/(re+RE*1000); #Voltage gain for emitter capacitor removed.\n", + "\n", + "print(\"(ii)Voltage gain= %.2f.\"%Av);\n", + "\n", + "#Note: The answer in the text book has been approximated to 5.38 but it's actually coming 5.37.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.15 : Page number 263" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain= 120.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=6.0; #Collector resistor, kтДж\n", + "RL=12.0; #Load resistor, kтДж\n", + "re=33.3; #a.c emitter resistance, тДж\n", + "\n", + "\n", + "#Calculations\n", + "RAC=RC*RL/(RC+RL); #a.c effective load, kтДж\n", + "Av=RAC*1000/re; #Voltage gain\n", + "\n", + "#Result\n", + "print(\"Voltage gain= %d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.16 : Page number 263-264" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) a.c emitter resistance=250 тДж.\n", + "(ii) Voltage gain =80.\n", + "(iii) d.c voltage across input capacitor= 1V and emitter capacitor=0.3V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=9.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=240.0; #Resistor R1, kтДж\n", + "R2=30.0 #Resistor R2, kтДж\n", + "RC=20.0; #Collector resistor, kтДж\n", + "RE=3.0; #Emitter resistor, kтДж\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "V2=round(VCC*R2/(R1+R2),1); #Voltage across R2, V\n", + "VE=round(V2-VBE,1); #Voltage across emitter resistor, V\n", + "IE=round(VE/RE,1); #Emitter current, mA\n", + "re=25/IE; #a.c emitter resistance, тДж\n", + "\n", + "#(ii)\n", + "Av=RC*1000/re; #Voltage gain\n", + "\n", + "#(iii)\n", + "V_C_in=V2; #d.c voltage across input capacitor, V\n", + "V_C_E=VE; #d.c vooltage across emitter capacitor, V\n", + "\n", + "\n", + "\n", + "#Results\n", + "print(\"(i) a.c emitter resistance=%d тДж.\"%re);\n", + "print(\"(ii) Voltage gain =%d.\"%Av);\n", + "print(\"(iii) d.c voltage across input capacitor= %dV and emitter capacitor=%.1fV.\"%(V_C_in,V_C_E));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.17 : Page number 264-265" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) D.C bias levels: V2=3V, VE=2.3V, IE=2.3mA, IC=2.3mA, IB=0.023mA and VC=10.4V.\n", + "(ii) D.c voltage across: Cin=3V and CE=2.3V and CC=10.4V.\n", + "(iii) a.c emitter resistance=10.9тДж.\n", + "(iv) Voltage gain=61.2.\n", + "(v) VC>VE. Therefore, the transistor is in active state.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=40.0; #Resistor R1, kтДж\n", + "R2=10.0 #Resistor R2, kтДж\n", + "RC=2.0; #Collector resistor, kтДж\n", + "RE=1.0; #Emitter resistor, kтДж\n", + "RL=1.0; #Load resistor, kтДж\n", + "beta=100; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "#(i) D.C bias levels\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V\n", + "VE=round(V2-VBE,1); #Voltage across emitter resistor, V\n", + "IE=round(VE/RE,1); #Emitter current, mA\n", + "IC=IE; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VC=VCC-IC*RC; #Collector voltage, V\n", + "print(\"(i) D.C bias levels: V2=%dV, VE=%.1fV, IE=%.1fmA, IC=%.1fmA, IB=%.3fmA and VC=%.1fV.\"%(V2,VE,IE,IC,IB,VC));\n", + "\n", + "\n", + "#(ii)\n", + "Cin_V=V2; #Voltage across Cin capacitor, V\n", + "CE_V=VE; #Voltage across CE capacitor, V \n", + "CC_V=VC; #Voltage across CC capacitor, V\n", + "print(\"(ii) D.c voltage across: Cin=%dV and CE=%.1fV and CC=%.1fV.\"%(Cin_V,CE_V,CC_V));\n", + "\n", + "#(iii)\n", + "re=round(25/IE,1); #a.c emitter resistance, тДж\n", + "print(\"(iii) a.c emitter resistance=%.1fтДж.\"%re);\n", + "\n", + "\n", + "#(iv)\n", + "RAC=round(RC*RL/(RC+RL),3); #Total a.c collector resistance, kтДж\n", + "Av=RAC/(re/1000); #Voltage gain\n", + "print(\"(iv) Voltage gain=%.1f.\"%Av);\n", + "\n", + "#(v)\n", + "print(\"(v) VC>VE. Therefore, the transistor is in active state.\" );\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.18 : page number 265" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The power gain = 26400 and output power = 1.584W.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=132.0; #Voltage gain\n", + "beta=200.0; #Base current amplification factor\n", + "P_in=60.0; #Input power, ЁЭЬЗW\n", + "\n", + "\n", + "#Calculations\n", + "Ap=beta*Av; #Power gain\n", + "P_out=Ap*(P_in/10**6); #Output power, W\n", + "\n", + "\n", + "#Results\n", + "print(\"The power gain = %d and output power = %.3fW.\"%(Ap,P_out));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.19 : page number 265-266" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Current gain=49\n", + "(ii) Voltage gain=2.14\n", + "(iii) Power gain=105.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IB=200.0; #Base current, microampere\n", + "IE=10.0; #Emitter current, mA\n", + "R1=27.0; #Resistor R1, kilo ohm\n", + "R2=13.0 #Resistor R2, kilo ohm\n", + "RC=4.7; #Collector resistor, kilo ohm\n", + "RE=2.2; #Emitter resistor, kilo ohm\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "IC=IE-(IB/1000); #Collector current, mA\n", + "beta=IC/(IB/1000); #Current gain\n", + "\n", + "print(\"(i) Current gain=%d\"%beta);\n", + "\n", + "#(ii)\n", + "#a.c emitter resistance is neglected, voltage gain=(collector resistor)/(emitter resistor)\n", + "Av=RC/RE; #Voltage gain\n", + "\n", + "print(\"(ii) Voltage gain=%.2f\"%Av);\n", + "\n", + "#(iii)\n", + "Ap=round(beta*Av,0); #Power gain\n", + "\n", + "#Results\n", + "print(\"(iii) Power gain=%d.\"%Ap);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.20 : Page number 266-267" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance of the amplifier circuit= 3.46 kтДж.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=30.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base emitter voltage, V\n", + "R1=45.0; #Resistor R1, kтДж\n", + "R2=15.0 #Resistor R2, kтДж\n", + "RC=10.0; #Collector resistor,kтДж\n", + "RE=7.5; #Emitter resistor, kтДж\n", + "beta=200.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),1); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2; #Voltage across emitter resistor(base-emitter voltage is neglected), V\n", + "IE=VE/RE; #Emitter current, mA (OHM's LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm\n", + "Zin_base=(beta*re)/1000; #input impedance of transistor base,kтДж\n", + "R1_R2=(R1*R2)/(R1+R2); #Parallel resistance between R1 and R2, kтДж\n", + "Zin=((R1_R2)*Zin_base)/(R1_R2+Zin_base); #Input impedance of the amplifier circuit, kтДж\n", + "\n", + "\n", + "#Result\n", + "print(\"The input impedance of the amplifier circuit= %.2f kтДж.\"%Zin); \n", + "\n", + "#Note: The input impedance of the amplifier circuit is approximated as 3.45 kтДж in the text book, but actually it's 3.46 kтДж.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.21 : Page Number 268-269" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the swamped amplifier= 4.67.\n", + "Input impedance of transistor base of the swamped amplifier= 48.21 kтДж.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V.\n", + "RC=1.5; #Collector resistor, kтДж.\n", + "R1=18.0; #Resistor R1, kтДж.\n", + "R2=4.7; #Resistor R2, kтДж.\n", + "RE1=300.0; #Emitter resistor 1, тДж.\n", + "RE2=900.0; #Emitter resistor 2, тДж.\n", + "VBE=0.7; #Base-emitter voltage, V.\n", + "beta=150.0; #Base current amplification factor.\n", + "\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),1); #d.c voltage across R2, V. (Voltage divider rule)\n", + "VE=round(V2-VBE,1); #d.c voltage across RE, V.\n", + "IE=round((VE/(RE1+RE2))*1000,2); #d.c emitter current, mA.(OHM'S LAW)\n", + "re=round(25/IE,1); #a.c emitter resistance, тДж.\n", + "Av=RC*1000/(re+RE1); #Voltage gain\n", + "Zin_base=(beta*(re+RE1))/1000; #Input impedance of transistor base, kтДж.\n", + "\n", + "\n", + "#Results\n", + "print(\"The voltage gain of the swamped amplifier= %.2f.\"%Av);\n", + "print(\"Input impedance of transistor base of the swamped amplifier= %.2f kтДж.\"%Zin_base);\n", + "\n", + "#Note:In the textbook Av is approximated to 4.66and Zin_base to 48.22 kilo ohm, but the actual answers come as 4.67 and 48.21 kilo ohm.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.22 : Page number 269" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change from the original value= 6.42%(decrease)\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=1.5; #Collector resistor, kтДж.\n", + "RE1=300.0; #Emitter resistor 1, тДж.\n", + "re=21.5; #a.c emitter resistance, тДж.\n", + "\n", + "#Calculations\n", + "Av=round(RC*1000/(re+RE1),2); #Voltage gain.\n", + "Av_1=round(RC*1000/(2*re+RE1),2); #Voltage gain when re doubles.\n", + "change_in_gain=round(Av-Av_1,2); #Change in voltage gain.\n", + "change_percentage=change_in_gain*100/Av; #Change percentage\n", + "\n", + "\n", + "#Results\n", + "if(change_in_gain>0):\n", + " print(\"The percentage change from the original value= %.2f%%(decrease)\"%change_percentage);\n", + "else:\n", + " print(\"The percentage change from the original value= %.2f%%(increase)\"%change_percentage);\n", + "\n", + "\n", + "#Note: The percentage has been approximated in the text book as 6.22%, but the answer comes as 6.42%.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.23 : Page number 269-270" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) input impedance of transistor base for standard amplifier= 5 kilo ohm\n", + " input impedance of transistor base for swamped amplifier= 47 kilo ohm\n", + "(ii) input impedance for standard amplifier= 1.33 kilo ohm\n", + " input impedance for swamped amplifier= 1.74 kilo ohm\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base emitter voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=2.2; #Resistor R2, kilo ohm\n", + "RC=4.0; #Collector resistor, kilo ohm\n", + "RE=1.1; #Emitter resistor, kilo ohm\n", + "beta=200.0; #Base current amplification factor\n", + "RE1=210.0; #Emitter resistor 1 of swamped amplifier, ohm.\n", + "RE2=900.0; #Emitter resistor 2 of swamped amplifier, ohm.\n", + "\n", + "\n", + "#Calculations\n", + "V2=round(VCC*R2/(R1+R2),1); #d.c voltage across R2, V. (Voltage divider rule)\n", + "VE=round(V2-VBE,1); #d.c voltage across RE, V.\n", + "IE=(VE/RE); #d.c emitter current, mA.(OHM'S LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm.\n", + "\n", + "\n", + "#(i) Zin_base:\n", + "Zin_base_standard=(beta*re)/1000; #input impedance of transistor base for standard amplifier , kilo ohm.\n", + "Zin_base_swamped=(beta*(re+RE1))/1000; #input impedance of transistor base for swamped amplifier, kilo ohm.\n", + "\n", + "\n", + "#(ii) Zin:\n", + "#input impedance for standard amplifier circuit\n", + "Zin_standard=(((R1*R2)/(R1+R2))*Zin_base_standard)/(Zin_base_standard +((R1*R2)/(R1+R2))); #kilo ohm\n", + "\n", + "#input impedance for standard amplifier circuit\n", + "Zin_swamped=(((R1*R2)/(R1+R2))*Zin_base_swamped)/(Zin_base_swamped +((R1*R2)/(R1+R2))); #kilo ohm\n", + "\n", + "\n", + "#Results\n", + "print(\"(i) input impedance of transistor base for standard amplifier= %d kilo ohm\"%Zin_base_standard);\n", + "print(\" input impedance of transistor base for swamped amplifier= %d kilo ohm\"%Zin_base_swamped);\n", + "print(\"(ii) input impedance for standard amplifier= %.2f kilo ohm\"%Zin_standard);\n", + "print(\" input impedance for swamped amplifier= %.2f kilo ohm\"%Zin_swamped);\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.24 : Page number 270-271" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of standard amplifier=160.\n", + "The voltage gain of swamped amplifier=17.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=4.0; #Collector resistor, kilo ohm\n", + "re=25.0; #a.c emitter resistance, ohm (calculated in example 10.23)\n", + "RE_1=210.0; #Emitter resistor 1 of swamped amplifier,ohm\n", + "\n", + "#Calculation\n", + "Av_standard=(RC*1000)/re; #Voltage gain of standard common emitter amplifier\n", + "Av_swamped=(RC*1000)/(re+RE_1); #Voltage gain of swamped amplifier\n", + "\n", + "#Results\n", + "print(\"The voltage gain of standard amplifier=%d.\"%Av_standard);\n", + "print(\"The voltage gain of swamped amplifier=%d.\"%Av_swamped);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.26 : Page number 273-274" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required input signal voltage =2.5mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_0=1000.0; #Open circuit voltage gain\n", + "R_in=2.0; #Input resistance, kilo ohm\n", + "R_out=1.0; #Output resistance, ohm\n", + "RL=4; #Load resistor across the output, ohm\n", + "I_2=0.5; #Output signal current, A.\n", + "\n", + "\n", + "#Calculations\n", + "#Since A_0*(I_1*R_in) = I_2*(R_out+RL)\n", + "I_1=I_2*(R_out+RL)/(A_0*(R_in*1000)); #Input current, A\n", + "V_1=I_1*(R_in*1000); #Input signal voltage, V\n", + "V_1=V_1*1000; #Input signal voltage, mV\n", + "\n", + "print(\"The required input signal voltage =%.1fmV\"%V_1);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.27 : Page number 274" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The magnitude of output voltage = 4.9V\n", + "The power gain =98e-06.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_0=1000.0; #Open circuit voltage gain\n", + "R_in=7.0; #Input resistance, kilo ohm\n", + "R_out=15.0; #Output resistance, ohm\n", + "RL=35.0; #Load resistor across the output, ohm\n", + "R_s=3.0; #Internal resistance, kilo ohm\n", + "E_s=10.0; #Input signal voltage, mV.\n", + "\n", + "#Calculations\n", + "#(i)\n", + "I_1=E_s*(10**-3)/(R_s*1000+R_in*1000); #Input current, A\n", + "V_1=I_1*(R_in*1000); #Voltage across input resistance, V\n", + "\n", + "#Since, A_v=V_2/V_1 = A_0*RL/(R_out+RL)\n", + "A_v=A_0*RL/(R_out+RL); #Voltage gain\n", + "V_2=A_v*V_1; #Outout voltage, V\n", + "\n", + "\n", + "#(ii)\n", + "P_2=V_2**2/RL; #Output power, W\n", + "P_1=V_1**2/(R_in*1000); #Input power, W\n", + "A_p=round(P_2/P_1,-6); #Power gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The magnitude of output voltage = %.1fV\"%V_2);\n", + "print(\"The power gain =%de-06.\"%(A_p/10**6));\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 10.28 : Page number 274-275" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Necessary input signal voltage= 12.5mV\n", + "Input signal current =4.17 ╬╝A\n", + "Power gain = 9600.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_v=80.0; #Voltage gain\n", + "V_2=1.0; #Output voltage, V\n", + "A_i=120.0; #Current gain\n", + "RL=2; #Load resistor, kilo ohm\n", + "\n", + "#Calculation\n", + "V_1=(V_2/A_v)*1000; #Input signal voltage, mV\n", + "\n", + "#Since, A_i=A0*R_in/(R_out+RL) and A_v=A0*RL/(R_out+RL)\n", + "#So, A_v/A_i=RL/R_in\n", + "R_in=RL*A_i/A_v; #Input resistance, kilo ohm\n", + "I_1=V_1/R_in; #Input current, ╬╝A\n", + "A_p=A_i*A_v; #Power gain\n", + "\n", + "#Results\n", + "print(\"Necessary input signal voltage= %.1fmV\"%V_1);\n", + "print(\"Input signal current =%.2f ╬╝A\"%I_1);\n", + "print(\"Power gain = %d.\"%A_p);\n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_6.ipynb new file mode 100644 index 00000000..6e0fb200 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter11_6.ipynb @@ -0,0 +1,1025 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8e425c9c2dfbfee43b3a89e44b0fd7936ba869da73ac3c372e9b23848f1cded1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#CHAPTER 11 : MULTISTAGE TRANSISTOR AMPLIFIERS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 11.1 : Page number 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import log10\n", + "#Variable declaration\n", + "#(i)\n", + "A_v=30; #Voltage gain\n", + "\n", + "#(ii)\n", + "A_p=100; #Power gain\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "A_v_dB=20*log10(A_v); #Voltage gain, dB\n", + "A_p_dB=10*log10(A_p); #Power gain, dB\n", + "\n", + "#Results\n", + "print(\"(i) Voltage gain in dB=%.2fdB\"%A_v_dB);\n", + "print(\"(ii) Power gain in dB=%ddB\"%A_p_dB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Voltage gain in dB=29.54dB\n", + "(ii) Power gain in dB=20dB\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.2 : Page number 285-286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#(i)\n", + "A_p_dB=40.0; #Power gain in dB\n", + "A_p_b=A_p_dB/10; #Power gain in bel\n", + "A_p=10**A_p_b; #Power gain in number\n", + "\n", + "#Result\n", + "print(\"(i) Power gain in number=%d\"%A_p);\n", + "\n", + "#(ii)\n", + "A_p_dB=43.0; #Power gain in dB\n", + "A_p_b=A_p_dB/10; #Power gain in bel\n", + "A_p=round(10**A_p_b,-4); #Power gain in number\n", + "\n", + "#Result\n", + "print(\"(ii) Power gain in number=%d\"%A_p);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Power gain in number=10000\n", + "(ii) Power gain in number=20000\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.3 : Page number 286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import log10\n", + "\n", + "#Variable declaration\n", + "Av_1=100.0; #Voltage gain of stage 1\n", + "Av_2=200.0; #Voltage gain of stage 2\n", + "Av_3=400.0; #Voltage gain of stage 3\n", + "\n", + "#Calculations\n", + "Av_1_dB=20*log10(Av_1); #Voltage gain of stage 1, dB\n", + "Av_2_dB=20*log10(Av_2); #Voltage gain of stage 2, dB\n", + "Av_3_dB=20*log10(Av_3); #Voltage gain of stage 3, dB\n", + "\n", + "Av_T=Av_1_dB+Av_2_dB+Av_3_dB; #Total voltage gain\n", + "\n", + "#Result\n", + "print(\"The total voltage gain=%ddB\"%Av_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total voltage gain=138dB\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.4 : Page number 286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_p_absolute=30.0; #Absolute gain of each stage\n", + "number_of_stages=5.0; #number of stages\n", + "negative_feedback=10.0; #negative feedback, dB\n", + "\n", + "#Calculations\n", + "#(i)\n", + "A_p_dB=round(10*log10(A_p_absolute),2); #Power gain of one stage. dB\n", + "A_p_T=number_of_stages * A_p_dB; #Total power gain, dB\n", + "\n", + "#(ii)\n", + "A_p_resultant=A_p_T-negative_feedback; #Resultant power gain with negative feedback, dB\n", + "\n", + "#Results\n", + "print(\"The total power gain = %.2fdB.\"%A_p_T);\n", + "print(\"The resultant power gain with negative feedback = %.2fdB.\"%A_p_resultant);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total power gain = 73.85dB.\n", + "The resultant power gain with negative feedback = 63.85dB.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.5 : Page number 286\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "P_out_2kHz=1.5; #Output power at 2 kHz, W\n", + "P_out_20kHz=0.3; #Output power at 20 kHz, W\n", + "P_in=10.0; #Input power, mW\n", + "\n", + "#Calculations\n", + "A_p_dB_2kHz=10*log10(P_out_2kHz*1000/P_in); #dB power gain at 2 kHz\n", + "A_p_dB_20kHz=10*log10(P_out_20kHz*1000/P_in); #dB power gain at 20 kHz\n", + "Fall_in_gain=A_p_dB_2kHz-A_p_dB_20kHz; #Fall in gain from 2kHz to 20kHz\n", + "\n", + "#Results\n", + "print(\"The fall in gain from 2kHz to 20kHz=%.2fdB\"%Fall_in_gain);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The fall in gain from 2kHz to 20kHz=6.99dB\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.6 : Page number 287\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_v=15.0; #Voltage gain, dB\n", + "V_1=0.8; #Input signal voltage, V\n", + "\n", + "#Calculations\n", + "#Since, Av(in decibel)=20*log10(V_2/V_1),\n", + "V_2=V_1*(10**(A_v/20)); #Output voltage, V\n", + "\n", + "#Results\n", + "print(\"The output voltage= %.1fV.\"%V_2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage= 4.5V.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.7 : Page number 287\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A_0_dB=70.0; #Open circuit voltage gain, dB\n", + "A_v_dB=67.0; #Voltage gain, dB\n", + "R_out=1.5; #Output resistance, kilo ohm\n", + "\n", + "#Calculations\n", + "#Since, A_0_dB-A_v_dB=20*log10(A_0/A_v)\n", + "ratio_A0_Av=round(10**((A_0_dB-A_v_dB)/20),2); #Ratio of open-circuit voltage gain to normal voltage gain\n", + "\n", + "#Since, A_v/A_0 = RL/(R_out+RL)\n", + "RL=R_out/(ratio_A0_Av-1); #Load resistor, kilo ohm\n", + "\n", + "\n", + "#Results\n", + "print(\"The load resistance=%.2f kilo ohm.\"%RL);\n", + "\n", + "#Note: The value of load resistor is calculated to be 3.6585 kilo ohm and approximated to 3.66. But, in the text it has been approximated to 3.65.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load resistance=3.66 kilo ohm.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.8 : Page number 287\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=1.0; #Load resistance, kilo ohm\n", + "A_v=40.0; #Voltage gain, dB\n", + "V_in=10.0; #Input signal voltage, mV\n", + "\n", + "#Calcultaions\n", + "#(i)\n", + "#Since, A_v=20*log10(V_out/V_in)\n", + "V_out=V_in*(10**(A_v/20))/1000; #Output voltage, V\n", + "\n", + "#(ii)\n", + "P_L=(V_out**2/RL); #The load power, mW\n", + "\n", + "\n", + "#Results\n", + "print(\"(i)The output voltage is %dV.\"%V_out);\n", + "print(\"(ii)The load poweris %dmW.\"%P_L);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The output voltage is 1V.\n", + "(ii)The load poweris 1mW.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.9 : Page number 287-288\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import log10\n", + "\n", + "#Variable declaration\n", + "P_2=40.0; #Output power, W\n", + "R=10.0; #Resistance of speaker, ohm\n", + "\n", + "\n", + "#Calculations\n", + "#(i)\n", + "A_p_dB=25.0; #Power gain, dB\n", + "#Since, A_p_dB=10*log10(P_2/P_1)\n", + "P_1=(P_2/10**(A_p_dB/10))*1000; #Input power, mW\n", + "\n", + "\n", + "#(ii)\n", + "A_v_dB=40.0; #Voltage gain, dB\n", + "\n", + "#Since, P=(V**2)/R,\n", + "V_2=(P_2*R)**0.5; #Output voltage, V\n", + "\n", + "#Since, A_v_dB=20*log10(V_2/V_1)\n", + "V_1=(V_2/10**(A_v_dB/20))*1000; #Input voltage, mV\n", + "\n", + "\n", + "#Results\n", + "\n", + "print(\"(i)The input power=%.1fmW.\"%P_1);\n", + "print(\"(ii)The input voltage=%dmV.\"%V_1);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The input power=126.5mW.\n", + "(ii)The input voltage=200mV.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.10 : Page number 288\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A_v_max=2000.0; #Maximum voltage gain\n", + "f_max=2.0; #Frequency at which maximum voltage gain occurs,kHz\n", + "A_v=1414.0; #Voltage gain at 50 Hz and 10kHz\n", + "f1=50; #Lower frequency at which gain is 1414, Hz\n", + "f2=10; #Upper frequency at which gain is 1414, kHz\n", + "\n", + "#Calculation\n", + "#Since, bandwidth is the range of frequency over which gain is greater than or equal to 70.7% of maximum gain\n", + "if((A_v/A_v_max)*100 ==70.7): \n", + " print(\"(i)The bandwidth is from %dHz to %dkHz.\"%(f1,f2));\n", + " print(\"(ii)The lower cut-off frequency=%dHz.\"%f1);\n", + " print(\"(iii)The upper cut-off frequency=%dkHz.\"%f2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The bandwidth is from 50Hz to 10kHz.\n", + "(ii)The lower cut-off frequency=50Hz.\n", + "(iii)The upper cut-off frequency=10kHz.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.11 : Page number 291\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "A_v=60.0; #Voltage gain of single stage amplifier\n", + "R_C=500.0; #Collector load, ohm\n", + "R_in=1.0; #Input impedance, kilo ohm\n", + "\n", + "\n", + "#Calculation\n", + "#Since, there is no loading , second stage gain remains at A_v\n", + "#But, due to loading effect of input impedance of second stage, gain of first stage decreases\n", + "A_v_2=A_v; #Voltage gain of second stage\n", + "R_AC=round((R_C*R_in*1000)/(R_C+R_in*1000),0); #Effective load of first stage, ohm\n", + "A_v_1=A_v*R_AC/R_C; #Gain of first stage\n", + "A_v_T=A_v_1*A_v_2; #Total gain\n", + "\n", + "\n", + "#Results\n", + "print(\"The total gain=%d.\"%A_v_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total gain=2397.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.12 : Page number 291-292\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rin=1.0; #Input resistance, kilo ohm\n", + "beta=100.0; #base current amplification factor\n", + "RC=2.0; #Collector load, kilo ohm\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "R_AC=(RC*Rin)/(RC+Rin); #Effective load on first stage, kilo ohm\n", + "A_v_1=round(beta*(R_AC/Rin),0); #Voltage gain of first stage\n", + "\n", + "#(ii)\n", + "A_v_2=round(beta*RC/Rin,0); #Voltage gain of second stage\n", + "\n", + "#(iii)\n", + "A_v_T=A_v_1*A_v_2; #Total voltage gain\n", + "\n", + "\n", + "#Results\n", + "print(\"(i)The voltage gain of first stage =%d.\"%A_v_1);\n", + "print(\"(ii)The voltage gain of second stage =%d.\"%A_v_2);\n", + "print(\"(iii)The total voltage gain =%d.\"%A_v_T);\n", + "\n", + "#Note: The approximation inthe text for A_v_1=66.66 is taken as 66 but here it has been taken 67 and therefore the total voltage is 13400 instead of 13200.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The voltage gain of first stage =67.\n", + "(ii)The voltage gain of second stage =200.\n", + "(iii)The total voltage gain =13400.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.13 : Page number 292\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RC=10.0; #Collector load of single stage amplifier, kilo ohm\n", + "Rin=1.0; #Input resistance, kilo ohm\n", + "beta=100.0; #base current amplification factor\n", + "RL=100.0; #Load resistor, ohm\n", + "\n", + "\n", + "#Calculations\n", + "R_AC=round((RC*1000)*RL/(RC*1000+RL),-1); #Effective collector load,\n", + "A_v=beta*R_AC/(Rin*1000); #Voltage gain\n", + "\n", + "#Results\n", + "print(\"The voltage gain=%d.\"%A_v);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain=10.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.14 : Page number 292-293\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector suppply voltage, V\n", + "R1=10.0; #resistor R1, kilo ohm\n", + "R2=2.2; #resistor R2, kilo ohm\n", + "R3=10.0; #resistor R3, kilo ohm\n", + "R4=2.2; #resistor R4, kilo ohm\n", + "RC_1=3.6; #Collector resistor of first stage, kilo ohm\n", + "RC_2=4.0; #Collector resistor of second stage, kilo ohm\n", + "RE_1=900.0; #Emitter resistor of first stage, ohm\n", + "RE_2=1.0; #Emitter resistor of second stage, kilo ohm\n", + "\n", + "\n", + "#Calculations\n", + "#Biasing potential for the second stage is the voltage across R4 resistor,\n", + "#so, by voltage divider rule:\n", + "VB=VCC*R4/(R3+R4); #Biasing potential for second stage,(Voltage across R4), V\n", + "\n", + "print(\"The biasing voltage for the second stage=%.1fV.\"%VB);\n", + "\n", + "#If coupling capacitor C_c is replaced by a wire, RC_1 and R3 become parallel\n", + "Req=round((RC_1*R3)/(RC_1+R3),2); #Equivalent resistance of R3 parallel with RC_1, kilo ohm\n", + "VB=VCC*R4/(Req+R4); #Biasing voltage if coupling capacitor is replaced by a wire, V\n", + "\n", + "print(\"The biasing voltage after replacing coupling capacitor by wire=%.2fV.\"%VB);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The biasing voltage for the second stage=3.6V.\n", + "The biasing voltage after replacing coupling capacitor by wire=9.07V.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.15 : Page number 293-294\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return (r1*r2)/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "R1=22.0; #Resistor R1, kilo ohm\n", + "R2=3.3; #Resistor R2, kilo ohm\n", + "R3=5.0; #Resistor R3, kilo ohm\n", + "R4=1.0; #Resistor R4, kilo ohm\n", + "R5=15.0; #Resistor R5, kilo ohm\n", + "R6=2.5; #Resistor R6, kilo ohm\n", + "R7=5.0; #Resistor R7, kilo ohm\n", + "R8=1.0; #Resistor R8, kilo ohm\n", + "beta=200; #Base current amplification factor\n", + "RL=10.0; #Load resistor, kilo ohm\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculations\n", + "#for 2nd stage\n", + "V_R6=round(VCC*R6/(R5+R6),2); #Voltage across R6, V (voltage divider rule)\n", + "V_R8=round(V_R6-V_BE,2); #Voltage across R8, V\n", + "IE_2=round(V_R8/R8,2); #Emitter current through R8, mA (OHM's LAW)\n", + "re_2nd_stage=round(25/IE_2,1); #a.c emitter resistance for 2nd stage, ohm\n", + "\n", + "#For 1st stage\n", + "V_R2=round(VCC*R2/(R1+R2),2); #Voltage across R2, V (voltage divider rule)\n", + "V_R4=round(V_R2-V_BE,2); #Voltage across R4, V\n", + "IE_1=round(V_R4/R4,2); #Emitter current through R4, mA (OHM's LAW)\n", + "re_1st_stage=round(25/IE_1,1); #a.c emitter resistance for 1st stage, ohm\n", + "\n", + "#(i)\n", + "Zin_base_2nd_stage=round((beta*re_2nd_stage)/1000,2); #input resistance of transistor base of 2nd stage, kilo ohm\n", + "Zin=round(pr(pr(R5,R6),Zin_base_2nd_stage),2); #Input impedance of the 2nd stage, kilo ohm\n", + "R_AC_1st_stage=round(pr(R3,Zin),2); #Effective collector load for 1st stage, kilo ohm\n", + "A_v_1=round(R_AC_1st_stage*1000/re_1st_stage,0); #voltage gain of 1st stage\n", + "\n", + "#(ii)\n", + "R_AC_2nd_stage=round(pr(R7,RL),2); #Effective collector load for 2nd stage, kilo ohm\n", + "A_v_2=round(R_AC_2nd_stage*1000/re_2nd_stage,1); #voltage gain of 2nd stage\n", + "\n", + "#(iii)\n", + "A_v_overall=A_v_1*A_v_2; #overall voltage gain\n", + "\n", + "\n", + "#results\n", + "print(\"(i)The voltage gain of 1st stage=%.0f.\"%A_v_1);\n", + "print(\"(i)The voltage gain of 2nd stage=%.1f.\"%A_v_2);\n", + "print(\"(i)The overall voltage gain =%d.\"%A_v_overall);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The voltage gain of 1st stage=53.\n", + "(i)The voltage gain of 2nd stage=191.4.\n", + "(i)The overall voltage gain =10144.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.16 : Page number 297\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Primary_impedance=1000.0; #Primary impedance, ohm\n", + "Load_impedance=10.0; #Load impedance, ohm\n", + "\n", + "#Calculation\n", + "#since,for maximum power transfer primary impedance should be equal to output impedance\n", + "#and, impedance of secondary should be equal to load impedance\n", + "#therfore, primary_impedance/load_impedance=square of(primary to secondary turn ratio)\n", + "n=(Primary_impedance/Load_impedance)**0.5; #Primary to secondary turn ratio\n", + "\n", + "\n", + "#Result\n", + "print('The primary to secondary turn ratio for maximum power transfer=%d.'%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary to secondary turn ratio for maximum power transfer=10.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.17 : Page number 297\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=16.0; #Load resistor, ohm\n", + "R_p=10.0; #Output impedance of primary, kilo ohm\n", + "Vp=10.0; #Terminal voltage of the source, V\n", + "\n", + "#Calculation\n", + "#Since, for maximum power transfer, the impedance of the primary should be equal to output impedance of the source\n", + "n=(R_p*1000/RL)**0.5; #Primary to secondary turns ratio\n", + "\n", + "#Since, power in a transformer remains constant,\n", + "#ratio of primary to secondary voltageis equal to primary to secondary turns ratio\n", + "Vs=Vp/n; #Voltage across the external load, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The primary to secondary turns ratio=%d.\"%n);\n", + "print(\"The voltage across the external load=%.1fV.\"%Vs);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary to secondary turns ratio=25.\n", + "The voltage across the external load=0.4V.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.18 : Page number 297-298\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Rp=300.0; #D.C resistance of primary, ohm\n", + "RL=3.0; #Load resistance, ohm\n", + "R_out=3.0; #Ouput resistance of the transistor, kilo ohm \n", + "\n", + "#Calculation\n", + "#when no signal is applied, only Rp is seen to be the load.\n", + "#But, when a.c signal is applied, RL in secondary reflects as RL*(squre of turns ratio).\n", + "#Therefore, load is seen to be Rp in series with the reflected RL in primary.\n", + "#i.e, R_out=Rp+(n**2 * RL), where n is the turns ratio\n", + "n=((R_out*1000-Rp)/RL)**0.5; #turns ratio\n", + "\n", + "#Result\n", + "print(\"Turns ratio for maximum power transfer=%d.\"%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Turns ratio for maximum power transfer=30.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.19 : Page number 298" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "f=200.0; #Frequency, Hz\n", + "Z_out=10.0; #Output impedance of the transistor, kilo ohm\n", + "Z_in=2.5; #Input impedance of the next stage, kilo ohm\n", + "\n", + "#Calculation\n", + "#For perfect impedance matching,\n", + "#Z_out should be equal to primary impedance\n", + "#Z_out=2*pi*f*(primary inductance)\n", + "Lp=(Z_out*1000)/(2*pi*f); #Primary inductance, H\n", + "\n", + "#for the secondary side,\n", + "#Z_in should be equal to impedance of secondary\n", + "Ls=(Z_in*1000)/(2*pi*f); #Secondary inductance, H\n", + "\n", + "\n", + "#result\n", + "print(\"The primary inductance=%.0fH.\"%Lp);\n", + "print(\"The secondary inductance=%.0fH.\"%Ls);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary inductance=8H.\n", + "The secondary inductance=2H.\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.20 : Page number 299\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Lp=8.0; #Primary inductance, H\n", + "Ls=2.0; #Secondary inductance, H\n", + "K=10**-5; #Inductance to turns ratio, constant\n", + "\n", + "#Calculations\n", + "Np=(Lp/K)**0.5; #Primary turns\n", + "Ns=(Ls/K)**0.5; #Secondary turns\n", + "\n", + "#Result\n", + "print(\"The primary turns=%.0f.\"%Np);\n", + "print(\"The secondary turns=%.0f.\"%Ns);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The primary turns=894.\n", + "The secondary turns=447.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 11.21 : Page number 300-301\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "R1=100.0; #Resistor R1, kilo ohm\n", + "R2=22.0; #Resistor R2, kilo ohm\n", + "R3=22.0; #Resistor R3, kilo ohm\n", + "R4=4.7; #Resistor R4, kilo ohm\n", + "R5=10.0; #Resistor R5, kilo ohm\n", + "R6=10.0; #Resistor R6, kilo ohm\n", + "beta=125; #Base current amplification factor\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i) D.C voltages\n", + "#For 1st stage:\n", + "V_B1=VCC*R2/(R1+R2); #Voltage at the base of 1st transistor, V (Voltage across R2, using voltage divider rule)\n", + "V_E1=V_B1-V_BE; #Emitter voltage of the 1st transistor, V\n", + "I_E1=round(V_E1/R4,2); #Emitter current of 1st transistor, mA (OHM's LAW)\n", + "I_C1=I_E1; #Collector current of 1st transistor, mA(approximately equals to emitter current)\n", + "V_C1=VCC-I_C1*R3; #Collector voltage of 1st transistor, V\n", + "\n", + "#For 2nd stage:\n", + "V_B2=V_C1; #Voltage at the base of 2nd transistor, V (equals to collector voltage of 1st transistor)\n", + "V_E2=V_C1-V_BE; #Emitter voltage of the 2nd transistor, V\n", + "I_E2=V_E2/R6; #Emitter current of 2nd transistor, mA (OHM's LAW)\n", + "I_C2=I_E2; #Collector current 2nd transistor, mA(approximately equals to emitter current)\n", + "V_C2=VCC-I_C2*R5; #Collector voltage of 2nd transistor, V\n", + "\n", + "print(\"(i) D.C voltages\");\n", + "print(\"First stage: VB1=%.2fV , VE1=%.2fV and VC1=%.2fV\"%(V_B1,V_E1,V_C1));\n", + "print(\"First stage: VB2=%.2fV , VE2=%.2fV and VC2=%.2fV\"%(V_B2,V_E2,V_C2));\n", + "\n", + "#(ii)Voltage gain\n", + "#First stage\n", + "re_1=25/I_E1; #a.c emitter resistance of 1st transistor, ohm\n", + "re_2=25/I_E2; #a.c emitter resistance of 2nd transistor, ohm\n", + "Zin_2nd_stage=beta*re_2/1000; #Input impedance of 2nd stage, kilo ohm\n", + "R_AC=R3*Zin_2nd_stage/(R3+Zin_2nd_stage); #Total a.c collector load, kilo ohm\n", + "A_v1=round(R_AC*1000/re_1,0); #Voltage gain of first stage\n", + "\n", + "print(\"The voltage gain of first stage=%d.\"%A_v1);\n", + "\n", + "#Second stage\n", + "R_AC=R5; #Total a.c collector load for 2nd stage, kilo ohm(Due to no loading effect, equal to R5)\n", + "A_v2=round(R5*1000/re_2,0); #Voltage gain of 2nd stage\n", + "\n", + "print(\"The voltage gain of second stage=%d.\"%A_v2);\n", + "\n", + "A_vT=A_v1*A_v2; #Overall voltage gain\n", + "\n", + "print(\"Overall voltage gain=%d.\"%A_vT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) D.C voltages\n", + "First stage: VB1=2.16V , VE1=1.46V and VC1=5.18V\n", + "First stage: VB2=5.18V , VE2=4.48V and VC2=7.52V\n", + "The voltage gain of first stage=66.\n", + "The voltage gain of second stage=179.\n", + "Overall voltage gain=11814.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_6.ipynb new file mode 100644 index 00000000..05e3d9d8 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter12_6.ipynb @@ -0,0 +1,968 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:1e9209171152811793fc18d1ee8c80ddcef574d69421ec87eeaa8fb87a304f6d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 12: TRANSISTOR AUDIO POWER AMPLIFIERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.1 : Page number 308\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=2.2; #Resistor R2, kilo ohm\n", + "RC=3.6; #Collector resistor, kilo ohm\n", + "RE=1.1; #Emitter resistor, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "I1=VCC/(R1+R2); #Current through R1 and R2, mA (OHM's LAW)\n", + "V2=I1*R2; #Voltage across R2 resistor, V (OHM's LAW)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=VE/RE; #Emitter current, mA (OHM's LAW)\n", + "IC=IE; #Collector current, mA (approximately equal to emitter current)\n", + "I_T=I1+IC; #Total current drawn from the supply, mA\n", + "P_dc=VCC*I_T; #Total power drawn from the supply, mW\n", + "\n", + "\n", + "#Results\n", + "print(\"The total power drawn from the supply=%.1fmW.\"%P_dc);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total power drawn from the supply=18.2mW.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.2 : Page number 309\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_L=10.6; #Voltage across load, V.(from a.c voltmeter, therfore r.m.s value)\n", + "R_L=200.0; #Load resistance, ohm\n", + "\n", + "#Calculation\n", + "#Since, power =V**2/R,\n", + "P_O=(V_L**2/R_L)*1000; #A.C output power, mW\n", + "\n", + "#Result\n", + "print(\"The a.c output power = %.1fmW.\"%P_O);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c output power = 561.8mW.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.3 : Page number 309\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=100.0; #Load resistance, ohm\n", + "V_PP=18.0; #Peak-to-peak a.c voltage, V\n", + "\n", + "#Calculation\n", + "#Since, V(r.m.s)=(V(peak-to-peak)/2)/sqrt(2)\n", + "VL=V_PP/(2*(2**0.5)); #r.m.s value, V\n", + "\n", + "#Since, power=(square of voltage)/resistance\n", + "P_O_max=(VL**2/RL)*1000; #Maximum possible a.c load power, mW\n", + "\n", + "#Result\n", + "print(\"The maximum possible a.c load power=%dmW.\"%P_O_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum possible a.c load power=405mW.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.4 : Page number 310\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_battery=12.0; #Battery voltage, V\n", + "P_out=2.0; #Output power, W\n", + "\n", + "#Calculation\n", + "#Since, Power=Current*Voltage\n", + "IC=(P_out/V_battery)*1000; #Maximum collector current , mA\n", + "\n", + "#Result\n", + "print(\"The maximum collector current=%.1fmA.\"%IC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector current=166.7mA.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.5 : Page number 310\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_battery=12.0; #Battery voltage, V\n", + "RL=4.0; #Collector load, kilo ohm\n", + "\n", + "#Calculation\n", + "IC_max=V_battery/RL; #Maximum collector current, mA\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum collector current=%dmA.\"%IC_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector current=3mA.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.6 : Page number 310-311\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P=50.0; #Power supplied by power amplifier, W\n", + "R=8.0; #Resistance of speaker, ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, Power=Voltage _square/Resistance,\n", + "V=(P*R)**0.5; #a.c output voltage, V\n", + "\n", + "#(ii)\n", + "I=V/R; #a.c output current, A (OHM's LAW)\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The a.c output voltage=%dV.\"%V);\n", + "print(\"(ii) The a.c output current=%.1fA.\"%I);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The a.c output voltage=20V.\n", + "(ii) The a.c output current=2.5A.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.7 : Page number 315\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "ib_peak=10.0; #Base current(peak), mA\n", + "RB=1.0; #Base resistance, kilo ohm\n", + "RC=20.0; #Collector resistance, ohm\n", + "beta=25.0; #Base current amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "IB=round(VCC-VBE/RB,1); #Base current, mA (OHM's LAW)\n", + "IC=int(beta*IB); #Collector current, mA\n", + "VCE=VCC-(IC/1000)*RC; #Collector emitter voltage, V (KVL)\n", + "\n", + "#(i)\n", + "ic_peak=beta*ib_peak; #Collector current(peak), mA\n", + "P_o_ac=(ic_peak/1000)**2*RC/2; #Output power, W\n", + "\n", + "#(ii)\n", + "P_dc=VCC*IC/1000; #Input power, W\n", + "\n", + "#(iii)\n", + "collector_efficiency=(P_o_ac/P_dc)*100; #Collector efficiency of the amplifier circuit,\n", + "\n", + "#Result\n", + "print(\"(i) The output power=%.3fW.\"%P_o_ac);\n", + "print(\"(ii) The input power=%.1fW.\"%P_dc);\n", + "print(\"(iii) The collector efficiency=%.1f%%.\"%collector_efficiency);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output power=0.625W.\n", + "(ii) The input power=9.6W.\n", + "(iii) The collector efficiency=6.5%.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.8 : Page number 317\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_dc=10.0; #zero signal power dissipation, W\n", + "P_o=4.0; #a.c output power, W\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Collector_eff=(P_o/P_dc)*100; #collector efficiency\n", + "\n", + "#(ii)\n", + "#Zero signal power is the maximum power dissipation in a transistor, therefore,\n", + "Power_rating=P_dc; #Power rating of the transistor, W\n", + "\n", + "#Result\n", + "print(\"(i) The collector efficiency=%d%%.\"%Collector_eff);\n", + "print(\"(i) The power rating of the transistor=%dW.\"%Power_rating);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The collector efficiency=40%.\n", + "(i) The power rating of the transistor=10W.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.9 : Page number 317-318\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=100.0; #Secondary load, ohm\n", + "n=10.0; #Transformer turn ratio\n", + "IC=100.0; #Zero signal collector current, mA\n", + "\n", + "#Calculation\n", + "RL_reflected=n**2*RL; #Reflected load as seen by the primary of the transformer, ohm\n", + "P_o_ac_max=(IC/1000)**2*RL_reflected/2; #Maximum a.c power output, W \n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum a.c power output=%dW.\"%P_o_ac_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum a.c power output=50W.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.10 : Page number 318\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=5.0; #Collector supply voltage, V\n", + "IC=50.0; #Zero signal collector current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "P_o_max=VCC*IC/2; #Maximum a.c output power, mW\n", + "\n", + "#(ii)\n", + "P_dc=VCC*IC; #D.C input power, mW\n", + "#Since, maximum power is dissipated in the zero signal conditions\n", + "Power_rating=P_dc; #Power rating of transistor, mW\n", + "\n", + "#(iii)\n", + "Max_collector_eff=(P_o_max/P_dc)*100; #Maximum collector efficiency\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The maximum a.c output power=%dmW\"%P_o_max);\n", + "print(\"(ii) The power rating of the transistor=%dmW.\"%Power_rating);\n", + "print(\"(iii) The maximum collector efficiency =%d%%.\"%Max_collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The maximum a.c output power=125mW\n", + "(ii) The power rating of the transistor=250mW.\n", + "(iii) The maximum collector efficiency =50%.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.11 : Page number 318\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "ic_max=160.0; #Maximum a.c collector current, mA\n", + "ic_min=10.0; #Minimum a.c collector current, mA\n", + "vce_max=12.0; #Maximum collector-emitter voltage, V\n", + "vce_min=2.0; #Minimum collector-emitter voltage, V\n", + "\n", + "#Calculation\n", + "vce_pp=vce_max-vce_min; #peak to peak collector emitter voltage, V\n", + "ic_pp=ic_max-ic_min; #peak to peak collector current, V\n", + "P_o=(vce_pp/(2*sqrt(2)))*(ic_pp/(2*sqrt(2))); #a.c output power, mW\n", + "\n", + "\n", + "#Result\n", + "print(\"The a.c output power=%.1fmW.\"%P_o);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c output power=187.5mW.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.12 : Page number 319-320\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "\n", + "#Variable declaration\n", + "VCC=12.0; #Battey voltage, V\n", + "IC_max_change=100.0; #maximum collector current change, mA\n", + "RL=5.0; #Loudspeaker resistance, ohm\n", + "\n", + "#Calculation\n", + "VCE_max_change=VCC; #Maximum collector-emitter voltage change\n", + "#(i) Loud speaker directly connected in the collector\n", + "Vmax_speaker=(IC_max_change/1000)*RL; #Maximum voltage across the loudspeaker, V\n", + "P_speaker_directly_coupled=Vmax_speaker*IC_max_change; #Power developed in the loudspeaker,mW\n", + "\n", + "#(ii) Loudspeaker transformer coupled\n", + "Z_out=(VCE_max_change/IC_max_change)*1000; #Output impedance of transistor, ohm\n", + "\n", + "#For max power transfer, primary impedance should be Z_out\n", + "RL_reflected=Z_out; #Load resistance as seen by primary, ohm\n", + "n=sqrt(RL_reflected/RL); #Turns ratio of transformer\n", + "Vp=VCC; #Transformer primary voltage, V\n", + "Vs=Vp/n; #Transformer secondary voltage, V\n", + "IL=Vs/RL; #Load current, A\n", + "P_speaker_transformer_coupled=IL**2*RL*1000; #Power delivered to the speaker, mW\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The power transferred to the speaker when directly coupled=%dmW.\"%P_speaker_directly_coupled);\n", + "print(\"(ii) The power trasnferred to the speaker when transformer-coupled=%dmW.\"%P_speaker_transformer_coupled);\n", + "print(\" The turns ratio=%.1f.\"%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The power transferred to the speaker when directly coupled=50mW.\n", + "(ii) The power trasnferred to the speaker when transformer-coupled=1200mW.\n", + " The turns ratio=4.9.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.13 : Page number 320-321\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "beta=100.0; #Base current amplification factor\n", + "RL=81.6; #Load resistance, ohm\n", + "VCE_peak=30.0; #Peak value of collector voltage, V\n", + "IC_peak=35.0; #Peak value of collector current, mA\n", + "VCE_min=5.0; #Minimum value of collector voltage, V\n", + "IC_min=1.0; #Minimum value of collector current, mA\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IC_zero_signal=(IC_peak-IC_min)/2 +1; #Zero signal collector current, mA\n", + "\n", + "#(ii)\n", + "IB_zero_signal=IC_zero_signal/beta; #Zero signal base current, mA\n", + "\n", + "#(iii)\n", + "VCE_zero_signal=(VCE_peak-VCE_min)/2 +5; #Zero signal collector-emitter voltage, V\n", + "VCC=VCE_zero_signal; #Collector supply voltage,V (due to transformer coupling, aproximately equal to zero signal VCE)\n", + "P_dc=VCC*IC_zero_signal; #d.c input power, mW\n", + "VCE_ac=(VCE_peak-VCE_min)/(2*sqrt(2)); #a.c output voltage, V\n", + "IC_ac=(IC_peak-IC_min)/(2*sqrt(2)); #a.c output current, mA\n", + "P_ac=VCE_ac*IC_ac; #a.c output power, mW\n", + "\n", + "#(iv)\n", + "collector_eff=(P_ac/P_dc)*100; #Collector efficiency\n", + "\n", + "#(v)\n", + "#a.c resistance RL'=negative inverse of slope of the d.c load line\n", + "slope=(IC_peak-IC_min)/(VCE_min-VCE_peak); #Slope of he d.c load line, kilo mho\n", + "RL_ac=-(1/slope)*1000; #a.c resistance, ohm\n", + "n=sqrt(RL_ac/RL); #Transformer turn ratio\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The approximate value of zero signal collector current=%dmA.\"%IC_zero_signal);\n", + "print(\"(ii) The zero signal base current=%.2fmA.\"%IB_zero_signal);\n", + "print(\"(iii) The d.c input power= %dmW and a.c output power =%dmW.\"%(P_dc,P_ac));\n", + "print(\"(iv) The collector efficiency=%.1f%%.\"%collector_eff);\n", + "print(\"(v) The turn ratio of the transformer=%d.\"%n);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The approximate value of zero signal collector current=18mA.\n", + "(ii) The zero signal base current=0.18mA.\n", + "(iii) The d.c input power= 315mW and a.c output power =106mW.\n", + "(iv) The collector efficiency=33.7%.\n", + "(v) The turn ratio of the transformer=3.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.14 : Page number 321-322\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "RL=13.0; #Load resistance, ohm\n", + "RL_reflected=325.0; #Load resistance, when referred to primary, ohm\n", + "VCC=20.0; #Supply voltage, V\n", + "IC=58.0; #Quiscent value of collector current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "n=sqrt(RL_reflected/RL); #Transformer turn ratio\n", + "\n", + "#(ii)\n", + "P_ac=(((IC/1000)**2)*RL_reflected/2)*1000; #A.C output power, mW\n", + "\n", + "#(iii)\n", + "P_dc=VCC*IC; #d.c input power, mW\n", + "collector_eff=(P_ac/P_dc)*100; #Collector efficiency\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Transformer turn ratio=%d.\"%n);\n", + "print(\"(ii) The a.c output power=%dmW.\"%P_ac);\n", + "print(\"(iii) The collector efficiency=%d%%.\"%collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Transformer turn ratio=5.\n", + "(ii) The a.c output power=546mW.\n", + "(iii) The collector efficiency=47%.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.15 : Page number 323\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_total=4.0; #Total power dissipated by the power transistor, W\n", + "T_j_max=90.0; #Maximum junction temperature, degree celsius\n", + "theta=10.0; #Thermal resistance, degree celsius per watt\n", + "\n", + "#Calculation\n", + "#Since, Total power dissipation=half of(max. junc. temp. - ambient temp.)\n", + "T_amb=T_j_max-(P_total*theta); #Ambient temperature, degree celsius\n", + "\n", + "#Result\n", + "print(\"The ambient temperature=%d degree celsius.\"%T_amb);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The ambient temperature=50 degree celsius.\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.16 : Page number 323-324\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "theta=300.0; #Thermal resistance, degree celsius per watt\n", + "T_j_max=90.0; #Maximum junction temperature, degree celsius\n", + "T_amb=30.0; #Ambient temperature, degree celsius\n", + "\n", + "#Calculation\n", + "#(i) Without heat sink\n", + "P_total=((T_j_max-T_amb)/theta)*1000; #Maximum permissible power dissipation without sink, mW\n", + "\n", + "print(\"(i)The maximum permissible power dissipation without heat sink=%dmW.\"%P_total);\n", + "\n", + "#(ii) With heat sink\n", + "theta=60.0; #reduced thermal resistance, degree celsius per watt\n", + "P_total=((T_j_max-T_amb)/theta)*1000; #Maximum permissible power dissipation with heat sink, mW\n", + "\n", + "print(\"(ii)The maximum permissible power dissipation with heat sink=%dmW.\"%P_total);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i)The maximum permissible power dissipation without heat sink=200mW.\n", + "(ii)The maximum permissible power dissipation with heat sink=1000mW.\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.17 : Page number 324\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "theta=20.0; #Thermal resistance, degree celsius per watt\n", + "T_j_max=200.0; #Maximum junction temperature, degree celsius\n", + "T_amb=25.0; #Ambient temperature, degree celsius\n", + "VCE=4.0; #Collector-emitter voltage, V\n", + "\n", + "#Calculation\n", + "P_total=(T_j_max-T_amb)/theta; #Maximum permissible power dissipation, W\n", + "\n", + "#since, the max. power dissipation=VCE_max*IC_max,therefore\n", + "IC_max=P_total/VCE; #Maximum collector current, A\n", + "\n", + "print(\"The maximum collector current that the transistor can carry without destruction=%.2fA.\"%IC_max);\n", + "\n", + "#The ambient temperature rises\n", + "T_amb=75.0; #The risen ambibent temperature, degree celsius\n", + "P_total=(T_j_max-T_amb)/theta; #Maximum permissible power dissipation, W\n", + "IC_max=P_total/VCE; #Maximum collector current, A\n", + "\n", + "print(\"The maximum collector current for the risen ambient temperature=%.2fA.\"%IC_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector current that the transistor can carry without destruction=2.19A.\n", + "The maximum collector current for the risen ambient temperature=1.56A.\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.18 : Page number 328-329\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "VCC=12.0; #Supply voltage, V\n", + "RL=8.0; #Driving load, ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IC_sat=VCC/(2*RL); #Collector saturation current, A\n", + "P_o_max=round(VCC*IC_sat*0.25,2); #Maximum load power, W\n", + "\n", + "#(ii)\n", + "P_dc=round(VCC*IC_sat/round(pi,2),2); #d.c input power, W\n", + "\n", + "#(iii)\n", + "Collector_eff=(P_o_max/P_dc)*100; #Collector efficiency\n", + "\n", + "#Result\n", + "print(\"(i) The maximum load power =%.2fW.\"%P_o_max);\n", + "print(\"(ii) The d.c input power=%.2fW.\"%P_dc);\n", + "print(\"(iii) The collector efficiency=%.1f%%.\"%Collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The maximum load power =2.25W.\n", + "(ii) The d.c input power=2.87W.\n", + "(iii) The collector efficiency=78.4%.\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.19 : Page number 329\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_T=10.0; #Power rating of each transistor, W\n", + "max_eff=0.785; #Maximum collector effciency\n", + "\n", + "#Calculation\n", + "#Since, input power=max. a.c power + Power rating of transistor\n", + "#And, max. efficiency=max. a.c power/input d.c power\n", + "P_2T=2*P_T; #Total power dissipation by two transistors\n", + "P_o_max=(max_eff*P_2T)/(1-max_eff); #Maximum output a.c power, W\n", + "\n", + "#result\n", + "print(\"The maximum output power that can be obtained=%.2fW.\"%P_o_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum output power that can be obtained=73.02W.\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.20 : Page number 329\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "eff=60.0/100; #Efficiency of the amplifier\n", + "P_T=2.5; #Power dissipated by each transistor, W\n", + "\n", + "#Calculation\n", + "#Since, input power=max. a.c power + Power rating of transistor\n", + "#And, max. efficiency=max. a.c power/input d.c power\n", + "P_2T=2*P_T; #Total power dissipated by both transistors, W\n", + "P_ac=(eff*P_2T)/(1-eff); #Output a.c power, W\n", + "P_dc=P_ac+P_2T; #Input d.c power, W\n", + "\n", + "#Result\n", + "print(\"The a.c output power= %.1fW.\"%P_ac);\n", + "print(\"The d.c input power= %.1fW.\"%P_dc);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c output power= 7.5W.\n", + "The d.c input power= 12.5W.\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 12.21 : Page number 329-330\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "RL=10.0; #Load resistance, ohm\n", + "\n", + "#Calculation\n", + "IC_sat=(VCC/(2*RL))*1000; #Saturated collector current, mA\n", + "VCE_off=VCC/2; #Collector-emitter voltage in off state, V\n", + "\n", + "#Result\n", + "print(\"1st end point of a.c load line, IC(sat)=%dmA.\"%IC_sat);\n", + "print(\"2nd end point of a.c load line, VCE(off)=%dV.\"%VCE_off);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "1st end point of a.c load line, IC(sat)=500mA.\n", + "2nd end point of a.c load line, VCE(off)=5V.\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_6.ipynb new file mode 100644 index 00000000..ea59ae59 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter13_6.ipynb @@ -0,0 +1,1137 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 13: AMPLIFIERS WITH NEGATIVE FEEDBACK" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.1 : Page number 338" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the amplifier with negative feedback=97.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=3000.0; #Voltage gain without feedback\n", + "m_v=0.01; #Feedback fraction\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=Av/(1+Av*m_v); #Voltage gain of the amplifier with negative feedback\n", + "\n", + "#Result\n", + "print(\"The voltage gain of the amplifier with negative feedback=%.0f.\"%Avf);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.2 : Page number 339" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The fraction of output fedback to the input=1/20.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=140.0; #Voltage gain\n", + "Avf=17.5; #Voltage gain with negative feedback\n", + "\n", + "#Calculation\n", + "#Since, Avf=Av/(1+Av*mv), so,\n", + "mv=(Av-Avf)/(Av*Avf); #Fraction of output fedback to the input\n", + "\n", + "\n", + "#Result\n", + "print(\"The fraction of output fedback to the input=1/%.0f.\"%(1.0/mv));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.3 : Page number 339" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The fraction of output fedback to input=0.01.\n", + "(ii) The required amplifier gain for overall gain to be 75=300.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=100.0; #Voltage gain\n", + "Avf=50.0; #Voltage gain with negative feedback\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #The fraction of output fedback to input\n", + "\n", + "#(ii) Overall gain is to be 75:\n", + "Avf=75.0; #The required overall gain\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Av=Avf/(1-Avf*mv); #The required value of amplifier gain\n", + "\n", + "#result\n", + "print(\"(i) The fraction of output fedback to input=%.2f.\"%mv);\n", + "print(\"(ii) The required amplifier gain for overall gain to be 75=%d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.4 : Page number 339-340" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The voltage gain without feedback=40.\n", + "(ii) The feedback fraction = 1/40.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vout=10.0; #output voltage , V\n", + "Vin_f=0.5; #Input votage for amplifier with feedback, V\n", + "Vin=0.25; #Input votage for amplifier without feedback, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Av=Vout/Vin; #Voltage gain without negative feedback\n", + "\n", + "#(ii)\n", + "Avf=Vout/Vin_f; #Voltage gain with negative feedback\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #Feedback fraction\n", + "\n", + "#Result\n", + "print(\"(i) The voltage gain without feedback=%d.\"%Av);\n", + "print(\"(ii) The feedback fraction = 1/%d.\"%(1/mv));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.5 : Page number 340" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The percentage of reduction in stage gain without feedback=20%.\n", + "(ii) The percentage of reduction in net gain with feedback=11.2%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=50.0; #Gain without feedback\n", + "Avf=25.0; #Gain with negative feedback\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #Feedback fraction\n", + "\n", + "#(i)\n", + "#percentage of reduction without feedback\n", + "Av_reduced=40.0; #Reduced amplifier gain due to ageing\n", + "percentage_of_reduction=((Av-Av_reduced)/Av)*100; #Percentage of reduction in stage gain\n", + "\n", + "print(\"(i) The percentage of reduction in stage gain without feedback=%d%%.\"%percentage_of_reduction);\n", + "\n", + "#(ii)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf_reduced=round(Av_reduced/(1+mv*Av_reduced),1); #Reduced net gain with negative feedback \n", + "percentage_of_reduction_f=((Avf-Avf_reduced)/Avf)*100; #Percentage of reduction in net gain with feedback\n", + "\n", + "print(\"(ii) The percentage of reduction in net gain with feedback=%.1f%%\"%percentage_of_reduction_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.6 : Page number 340" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change in system gain=8.36%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=100.0; #Gain\n", + "mv=0.1; #feedback fraction\n", + "Av_fall=6.0; #fall in gain, dB\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=round(Av/(1+Av*mv),2); #Total system gain with feedback\n", + "\n", + "#Since, fall in gain=20*log10(Av/Av_1)\n", + "Av1=round(Av/10**(Av_fall/20),0); #New absolute voltage gain without feedback\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf_new=round(Av1/(1+Av1*mv),2); #New net system gain with feedback\n", + "\n", + "percentage_change=((Avf-Avf_new)/Avf)*100; #Percentage change in system gain\n", + "\n", + "#Result\n", + "print(\"The percentage change in system gain=%.2f%%\"%percentage_change);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.7 : Page number 341" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The feedback fraction=0.008.\n", + "The percentage fall in system gain=4.8%.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=500.0; #Voltage gain without feedback\n", + "Avf=100.0; #Voltage gain with negative feedback\n", + "Av_fall_percentage=20.0; #Gain fall percentage due to ageing\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #Feedback fraction\n", + "Av_reduced=((100-Av_fall_percentage)/100)*Av; #Reduced voltage gain\n", + "Avf_reduced=round(Av_reduced/(1+Av_reduced*mv),1); #Reduced total gain of the system\n", + "percentage_fall=((Avf-Avf_reduced)/Avf)*100; #Percentage of fall in total system gain\n", + "\n", + "#Result\n", + "print(\"The feedback fraction=%.3f.\"%mv);\n", + "print(\"The percentage fall in system gain=%.1f%%.\"%percentage_fall);\n", + "\n", + "#Note: The percentage gain is calculated in the text as 4.7% due to approximation of Avf to 95.3 whose actual approximation will be (95.238)~95.2. So, the percentage fall calculated here is 4.8%\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.8 : Page number 341" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain with feedback=31622.\n", + "The feedback fraction=2.16e-05.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "Av=100000.0; #Open loop voltage gain\n", + "f_dB=10.0; #Negative feedback, dB\n", + "\n", + "#Calculation\n", + "Av_dB=20*log10(Av); #dB voltage gain without feedback, dB\n", + "Avf_dB=Av_dB-f_dB; #dB voltage gain with feedback, dB\n", + "Avf=10**(Avf_dB/20); #Voltage gain with feedback\n", + "\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "mv=(Av-Avf)/(Av*Avf); #feedback fraction\n", + "\n", + "#Result\n", + "print(\"The voltage gain with feedback=%d.\"%Avf);\n", + "print(\"The feedback fraction=%.2e.\"%mv);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.9 : Page number 341-342" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain with feedback=47.4.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ao=1000.0; #Open circuit voltage gain\n", + "Rout=100.0; #Output resistance, ohm\n", + "RL=900.0; #Resistive load, ohm\n", + "mv=1/50; #feedback fraction\n", + "\n", + "#Calculation\n", + "#Since, Av=Ao*RL/(Rout+RL)\n", + "Av=Ao*RL/(Rout+RL); #Voltage gain without feedback\n", + "Avf=Av/(1+Av*mv); #Voltage gain with feedback\n", + "\n", + "#Result\n", + "print(\"The voltage gain with feedback=%.1f.\"%Avf);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.10 : Page number 342" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "100=Av/(1+Av*mv) ------Eq. 1\n", + "99=0.8*Av/(1+0.8*Av*mv) ------Eq. 2\n", + "99 + 79.2*Av*mv=0.8Av ------Eq. 3 from Eq. 2\n", + "79.2 + 79.2*Av*mv=0.792Av ------Eq. 4 from Eq. 1\n", + "Subtracting Eq.4 from Eq.3\n", + "19.8 = 0.008*Av\n", + "Av=2475.\n", + "mv=0.0096.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Avf=100.0; #Voltage gain with feedback\n", + "vary_f=1; #Vary percentage in voltage gain with feedback\n", + "vary_wf=20; #Vary percentage in voltage gain without feedback\n", + "\n", + "#Calculation\n", + "#Avf=Av/(1+Av*mv)\n", + "print(\"%d=Av/(1+Av*mv) ------Eq. 1\"%Avf); #Equation 1\n", + "\n", + "#considering variation in gains\n", + "Avf_vary=Avf*(1- vary_f/100.0); #Gain with feedback, considering variation\n", + "print(\"%d=%.1f*Av/(1+%.1f*Av*mv) ------Eq. 2\"%(Avf_vary,(1-vary_wf/100.0),(1-vary_wf/100.0))); #Equation 2\n", + "\n", + "#Solving the above two equations\n", + "print(\"%d + %.1f*Av*mv=%.1fAv ------Eq. 3 from Eq. 2\"%(Avf_vary,Avf_vary*(1-vary_wf/100.0),(1-vary_wf/100.0))); #Equation 3\n", + "\n", + "#multiplying Eq. 1 with (Avf_vary*(1-vary_wf/100.0))/100=0.792\n", + "print(\"%.1f + %.1f*Av*mv=%.3fAv ------Eq. 4 from Eq. 1\"%(Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf*Avf_vary*(1-vary_wf/100.0)/100.0,Avf_vary*(1-vary_wf/100.0)/100.0)); #Equation 4\n", + "\n", + "print(\"Subtracting Eq.4 from Eq.3\" );\n", + "print(\"%.1f = %.3f*Av\"%(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0,(1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0));\n", + "Av=(Avf_vary-Avf*Avf_vary*(1-vary_wf/100.0)/100.0)/((1-vary_wf/100.0)-Avf_vary*(1-vary_wf/100.0)/100.0);\n", + "print(\"Av=%.0f.\"%Av);\n", + "mv=(Av-Avf)/(Av*Avf);\n", + "print(\"mv=%.4f.\"%mv);\n", + " " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.11 : Page number 345" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Feedback fraction=0.1.\n", + "(ii) Voltage gain with feedback=10.\n", + "(iii) Output voltage=10mV.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=10000.0; #Volage gain without feedback\n", + "R1=2.0; #Resistor R1, kilo ohm\n", + "R2=18.0; #Resistor R2, kilo ohm\n", + "Vin=1.0; #input voltage, mV\n", + "\n", + "#Calculation\n", + "#(i)\n", + "mv=R1/(R1+R2); #feedback fraction\n", + "\n", + "#(ii)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=round(Av/(1+Av*mv),0); #Voltage gain with feedback\n", + "\n", + "#(iii)\n", + "Vout=Avf*Vin; #Output voltage, mV\n", + "\n", + "#Result\n", + "print(\"(i) Feedback fraction=%.1f.\"%mv);\n", + "print(\"(ii) Voltage gain with feedback=%d.\"%Avf);\n", + "print(\"(iii) Output voltage=%dmV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.12 : Page number 345-346" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Feedback fraction=0.1.\n", + "(ii) The voltage gain with feedback=10.\n", + "(iii) Increased input impedance due to negative feedback=10 mega ohm\n", + "(iv) Decreased output impedance due to negative feedback=0.1 ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=10000.0; #Volateg gain without feedback\n", + "Zin=10.0; #Input impedance, kilo ohm\n", + "Zout=100.0; #Output impedance, ohm\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=90.0; #Resistor R2, kilo ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "mv=R1/(R1+R2); #Feedback fraction\n", + "\n", + "#(ii)\n", + "#Since, Gain_with_feedback= Gain_without_feedback/(1+Gain_without_feedback*feedback_fraction),\n", + "Avf=round(Av/(1+Av*mv),0); #Voltage gain with feedback\n", + "\n", + "#(iii)\n", + "Zin_feedback=((1+Av*mv)*Zin)/1000; #Increased input impedance due to negative feedback, mega ohm\n", + "\n", + "#(iv)\n", + "Zout_feedback=Zout/(1+Av*mv); #Decreased output impedance due to negative feedback, ohm\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Feedback fraction=%.1f.\"%mv);\n", + "print(\"(ii) The voltage gain with feedback=%d.\"%Avf);\n", + "print(\"(iii) Increased input impedance due to negative feedback=%.0f mega ohm\"%Zin_feedback);\n", + "print(\"(iv) Decreased output impedance due to negative feedback=%.1f ohm.\"%Zout_feedback);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.13 : Page number 346" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Distortion with negative feedback=0.312%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=150.0; #Voltage gain\n", + "D=5/100.0; #Distortion\n", + "mv=10/100.0; #Feedback fraction\n", + "\n", + "#Calculation\n", + "Dvf=round((D/(1+Av*mv))*100,3); #Distortion with negative feedback\n", + "\n", + "\n", + "#Result\n", + "print(\"Distortion with negative feedback=%.3f%%\"%Dvf);\n", + "\n", + "#Note: In the text, value of Dvf=0.3125% has been approximated to 0.313%. But, here the approximation is done to 0.312%\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.14 : Page number 346" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The new lower cut-off frequency=136.4Hz\n", + "The new upper cut-off frequency=5.52MHz\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Av=1000.0; #Voltage gain\n", + "f1=1.5; #Lower cut-off frequency, kHz\n", + "f2=501.5; #Upper cut-off frequency, kHz\n", + "mv=1/100.0; #Feedbcack fraction\n", + "\n", + "#Calculation\n", + "f1_f=(f1/(1+mv*Av))*1000; #New lower cut-off frequency, Hz\n", + "f2_f=(f2*(1+mv*Av))/1000; #New upper cut-off frequency, MHz\n", + "\n", + "\n", + "#Result\n", + "print(\"The new lower cut-off frequency=%.1fHz\"%f1_f);\n", + "print(\"The new upper cut-off frequency=%.2fMHz\"%f2_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.15 : Page number 348" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The effective current gain=58.82.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=200.0; #Current gain without feedback\n", + "mi=0.012; #Current attenuation\n", + "\n", + "#Calculation\n", + "Aif=Ai/(1+Ai*mi);\n", + "\n", + "#Result\n", + "print(\"The effective current gain=%.2f.\"%Aif);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.16 : Page number 349" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance when negative feedback is applied=3.26 kilo ohm\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=240.0; #Current gain\n", + "Zin=15.0; #Input impedance without feedback, kilo ohm\n", + "mi=0.015; #Current feedback fraction\n", + "\n", + "#Calculations\n", + "Zin_f=Zin/(1+mi*Ai); #Input impedance with feedback, kilo ohm\n", + "\n", + "\n", + "#Result\n", + "print(\"The input impedance when negative feedback is applied=%.2f kilo ohm\"%Zin_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.17 : Page number 349" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output impedance with negative feedback=9kilo ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=200.0; #Current gain without feedback\n", + "Zout=3.0; #Output impedance without feedback, kilo ohm\n", + "mi=0.01; #current feedback fraction\n", + "\n", + "#Calculation\n", + "Zout_f=Zout*(1+mi*Ai); #Output impedance with negative feedback, kilo ohm\n", + "\n", + "#Result\n", + "print(\"The output impedance with negative feedback=%dkilo ohm.\"%Zout_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.18 : Page number 349" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Bandwidth when negative feedback is applied=1400kHz.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Ai=250.0; #Current gain without feedback\n", + "BW=400.0; #Bandwidth, kHz\n", + "mi=0.01; #current feedback fraction\n", + "\n", + "#Calculation\n", + "BW_f=BW*(1+mi*Ai); #Bandwidth when negative feedback is applied, kHz\n", + "\n", + "#Result\n", + "print(\"Bandwidth when negative feedback is applied=%dkHz.\"%BW_f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.19 : Page number 350-351" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of VE=9.72V and IE=10.68mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f84ace55f98>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as p\n", + "\n", + "#Variable declaration\n", + "VCC=18.0; #Supply voltage, V\n", + "R1=16.0; #Resistor R1, kilo ohm\n", + "R2=22.0; #Resistor R2, kilo ohm\n", + "RE=910.0; #Emitter resistor, ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculations\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=(VE/RE)*1000; #Emitter current, mA (OHM's LAW)\n", + "\n", + "#D.C load line\n", + "IC_sat=(VCC/RE)*1000; #Collector saturation current, mA\n", + "VCE_off=VCC; #Collector-emitter voltage in off state, V\n", + "\n", + "#Result\n", + "print(\"Value of VE=%.2fV and IE=%.2fmA\"%(VE,IE));\n", + "\n", + "#Plotting\n", + "VCE_plot=[0,VCE_off]; #Plotting variable for VCE\n", + "IC_plot=[IC_sat,0]; #Plotting variable for IC\n", + "p.plot(VCE_plot,IC_plot);\n", + "p.xlim(0,20)\n", + "p.ylim(0,25)\n", + "p.xlabel('VCE(V)');\n", + "p.ylabel('IC(mA)');\n", + "p.title('d.c load line');\n", + "p.grid();\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.20 : Page number 352" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the emitter follower circuit=0.994.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=10.0; #Resistor R2, kilo ohm\n", + "RE=5.0; #Emitter resistance, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=(VE/RE); #Emitter current, mA (OHM's LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm\n", + "Av=RE*1000/(re+RE*1000); #Voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The voltage gain of the emitter follower circuit=%.3f.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.21 : Page number 352-353" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain=0.988\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RE=5.0; #Emitter resistance, kilo ohm\n", + "re=29.1; #a.c emitter resistance, ohm\n", + "RL=5.0; #Load resistance, kilo ohm\n", + "\n", + "#Calculation\n", + "RE_ac=(RE*RL)/(RE+RL); #New effective value of emitter resistance, kilo ohm\n", + "Av=RE_ac*1000/(re+RE_ac*1000); #Voltage gain\n", + "\n", + "#Result\n", + "print(\"The voltage gain=%.3f\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.22 : Page number 354" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance of the emitter follower =4.96 kilo ohm\n", + "The approximate value of the input impedance=5 kilo ohm\n" + ] + } + ], + "source": [ + "def pr(r1,r2): #Function for calculating parallel resistance\n", + " return (r1*r2)/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "R1=10.0; #Resistor R1, kilo ohm\n", + "R2=10.0; #Resistor R2, kilo ohm\n", + "RE=4.3; #Emitter resistor, kilo ohm\n", + "RL=10.0; #Load resistance, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=200.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "VE=V2-VBE; #Emitter voltage, V\n", + "IE=(VE/RE); #Emitter current, mA (OHM's LAW)\n", + "re=25/IE; #a.c emitter resistance, ohm\n", + "RE_eff=pr(RE,RL); #Effective external emitter resistance, kilo ohm\n", + "Zin_base=beta*(re/1000+RE_eff); #Input impedance of the base of the transistor, kilo ohm\n", + "Zin=pr(pr(R1,R2),Zin_base); #Input impedance of emitter follower, kilo ohm\n", + "#Approximate value of input impedance taken as parallel resistance of R1 and R2 and ignoring Zin_base due to its relatively large value\n", + "Zin_approx=pr(R1,R2); #Approximate input impedance, kilo ohm\n", + "\n", + "#Result\n", + "print(\"The input impedance of the emitter follower =%.2f kilo ohm\"%Zin);\n", + "print(\"The approximate value of the input impedance=%d kilo ohm\"%Zin_approx);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.23 : Page number 355" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output impedance=22.3 ohm\n" + ] + } + ], + "source": [ + "def pr(r1,r2): #Function for calculating parallel resistance\n", + " return (r1*r2)/(r1+r2);\n", + "\n", + "\n", + "#Variable declaration\n", + "re=20.0; #a.c emitter resistance, ohm\n", + "R1=3.0; #Resistor R1, kilo ohm\n", + "R2=4.7; #Resistor R2, kilo ohm\n", + "RS=600.0; #Source resistance, kilo ohm\n", + "beta=200.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "Rin_ac=pr(pr(R1,R2)*1000,RS); #Input a.c resistance, ohm\n", + "Zout=re + Rin_ac/beta; #Output impedance, ohm\n", + "\n", + "#Result\n", + "print(\"The output impedance=%.1f ohm\"%Zout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.24 : Page number 358" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) d.c value of current in RE=1.09mA\n", + "(ii) Input impedance=16.17 mega ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Supply voltage, V\n", + "R1=120.0; #Resistor R1, kilo ohm\n", + "R2=120.0; #Resistor R2, kilo ohm\n", + "RE=3.3; #Emitter resistor, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta_1=70.0; #Base current amplification factor of 1st transistor\n", + "beta_2=70.0; #Base current amplification factor of 2nd transistor\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V2=VCC*R2/(R1+R2); #Voltage across R2, V (Voltage divider rule)\n", + "IE_2=(V2-2*VBE)/RE; #Emitter current, mA (OHM's LAW)\n", + "\n", + "#(ii)\n", + "Zin=(beta_1*beta_2*RE)/1000; #Input impedance, mega ohm\n", + "\n", + "#Result\n", + "print(\"(i) d.c value of current in RE=%.2fmA\"%IE_2);\n", + "print(\"(ii) Input impedance=%.2f mega ohm.\"%Zin);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 13.25 : Page number 358-359" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) D.C Bias levels: \n", + " VB1= 4V, VE1=3.3V, VB2=3.3V, VE2=2.6V, IE2=1.3mA and IE1=0.013mA.\n", + "(ii) A.C Analysis: \n", + " re1=1923 ohm and re2=19.23 ohm \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Supply voltage, V\n", + "R1=20.0; #Resistor R1, kilo ohm\n", + "R2=10.0; #Resistor R2, kilo ohm\n", + "RC=4.0; #Collector resistor, kilo ohm\n", + "RE=2.0; #Emitter resistor, kilo ohm\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100.0; #Base current amplification factor of 1st transistor\n", + "\n", + "#Calculation\n", + "#(i) D.C Bias levels\n", + "VB1=VCC*R2/(R1+R2); #Base voltage of 1st transistor, V (Voltage divider rule)\n", + "VE1=VB1-VBE; #Emitter voltage of 1st transistor, V\n", + "VB2=VE1; #Base voltage of 2nd transistor, V\n", + "VE2=VB2-VBE; #Emitter voltage of 2nd transistor, V\n", + "IE2=VE2/RE; #Emitter current of 2nd transistor, mA (OHM' LAW)\n", + "IE1=IE2/beta; #Emitter current of 1st transistor, mA (IE~IC=beta*IB, here IB2=IE1)\n", + "\n", + "#(ii) A.C analysis\n", + "re1=25/IE1; #a.c emitter resistance of 1st transistor\n", + "re2=25/IE2; #a.c emitter resistance of 2nd transistor\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) D.C Bias levels: \\n VB1= %dV, VE1=%.1fV, VB2=%.1fV, VE2=%.1fV, IE2=%.1fmA and IE1=%.3fmA.\"%(VB1,VE1,VB2,VE2,IE2,IE1));\n", + "print(\"(ii) A.C Analysis: \\n re1=%d ohm and re2=%.2f ohm \"%(re1,re2));\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_6.ipynb new file mode 100644 index 00000000..5e35882c --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter14_6.ipynb @@ -0,0 +1,502 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:ec0d27209d08b0b95750f66ce9ee21af5ef586e23d0bf0ea218aa20a4ce63e43" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#CHAPTER 14: SINUSOIDAL OSCILLATORS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.1 : Page number 371-372\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "L1=58.6; #Inductance, micro henry\n", + "C1=300.0; #Capacitance, pF\n", + "\n", + "#Calculation\n", + "f=(1/(2*round(pi,2)*sqrt(L1*10**-6*C1*10**-12)))/1000; #Frequency of oscillation, kHz\n", + "\n", + "\n", + "#Result\n", + "print(\"frequency of oscillation=%dkHz\"%f);\n", + "\n", + "\n", + "#Note : The frequency has been calculated in the text as 1199kHz but here the answer gets approximated to 1200kHz.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "frequency of oscillation=1200kHz\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.2 : Page number 372\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "L1=1.0; #Inductance , mH\n", + "f=1.0; #frequency of oscillation, GHz\n", + "\n", + "#Calculation\n", + "#Since, f=1/(2*pi*sqrt(L1*C1)),\n", + "C1=(1/(L1*10**-3*(f*10**12*2*pi)**2))*10**12; #Capacitance, pF\n", + "\n", + "\n", + "#Result\n", + "print(\"The Capacitance of the capacitor of the LC oscillator=%.2epF\"%C1);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Capacitance of the capacitor of the LC oscillator=2.53e-11pF\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.3 : Page number 373-374\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "C1=0.001; #Capacitor C1, microfarad\n", + "C2=0.01; #Capacitor C2, microfarad\n", + "L=15.0; #Inductance, microhenry\n", + "\n", + "#Calculation\n", + "CT=C1*C2/(C1+C2); #Total capacitance\n", + "\n", + "#(i) Operating frequency\n", + "f=(1/(2*pi*sqrt(CT*10**-6*L*10**-6)))/1000; #Operating frequency, kHz\n", + "\n", + "#(ii) Feedback fraction\n", + "mv=C1/C2; #Feedback fraction\n", + "\n", + "#Result\n", + "print(\"(i) The operating frequency=%dkHz\"%f);\n", + "print(\"(ii) The feedback fraction=%.1f\"%mv);\n", + "\n", + "#Note : The operating frequency is calculated in the text as 1361kHz but here it has been approximated to 1362kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The operating frequency=1362kHz\n", + "(ii) The feedback fraction=0.1\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.4 : Page number 374: Page number\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "mv=0.25; #Feedback fraction\n", + "L=1.0; #Inductance, mH\n", + "f=1.0; #Operating frequeny, MHz\n", + "\n", + "#Calculation\n", + "#Since, f=1/(2*pi*sqrt(L*C))\n", + "CT=round((1/(L*10**-3*(2*pi*f*10**6)**2))*10**12,1); #Total capacitance, pF\n", + "\n", + "#Since, mv=C1/C2 and CT=C1*C2/(C1+C2) or CT=C2/(1+ (C2/C1)),\n", + "#From the above equations, substituting value of mv and calculaing value of C2,\n", + "C2=CT*(1+(1/mv)); #Capacitance of C2 capactior, pF\n", + "C1=mv*C2; #Capacitance of C1 capacitor, pF\n", + "\n", + "#Result\n", + "print(\"C1=%.1fpF and C2=%.1fpF\"%(C1,C2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "C1=31.6pF and C2=126.5pF\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.5 : Page number 375-376\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable decalaration\n", + "L1=1000.0; #Inductance of L1 inductor, microhenry\n", + "L2=100.0; #Inductance of L2 inductor, microhenry\n", + "M=20.0; #Mutual inductance, microhenry\n", + "C=20.0; #Capacitance, pF\n", + "\n", + "#Calculation\n", + "LT=L1+L2+2*M; #Total inductance, microhenry\n", + "\n", + "#(i) Operating frequency\n", + "f=(1/(2*pi*sqrt(LT*10**-6*C*10**-12)))/1000; #Operating frequency, kHz\n", + "\n", + "#(ii)\n", + "mv=L2/L1; #feedback fraction\n", + "\n", + "#Result\n", + "print(\"(i) The operating frequency=%dkHz.\"%f);\n", + "print(\"(ii) The feedback fraction=%.1f.\"%mv);\n", + "\n", + "#Note : The operating frequecy has been calculated in the text as 1052kHz but here it gets approximated to 1054kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The operating frequency=1054kHz.\n", + "(ii) The feedback fraction=0.1.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.6 : Page number 376\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=1.0; #Capacitance, pF\n", + "f=1.0; #Frequency, MHz\n", + "mv=0.2; #Feedback frequency\n", + "\n", + "\n", + "#Calculation\n", + "LT=(1/(C*10**-12*(2*pi*f*10**6)**2))*1000; #Total inductance, mH\n", + "\n", + "#Since, mv=L2/L1 or L2=mv*L1 and L1+L2=LT or L1(1+mv)=LT,\n", + "L1=LT/(1+mv); #Inductance of L1 inductor, mH\n", + "L2=L1*mv; #inductance of L2 inductor, mH\n", + "\n", + "#Result\n", + "print(\"L1=%.1fmH and L2=%.2fmH.\"%(L1,L2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "L1=21.1mH and L2=4.22mH.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.7 : Page number 378\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "R1=1.0; #Resistor R1, mega ohm\n", + "R2=R1; #Resistor R2, mega ohm\n", + "R3=R1; #Resistor R3, mega ohm\n", + "C1=68.0; #Capacitor C1, pF\n", + "C2=C1; #Capacitor C2, pF\n", + "C3=C1; #Capacitor C3, pF\n", + "\n", + "\n", + "#Calculation\n", + "R=R1*10**6; #Resistance of the resistors of phase shift circuit, ohm\n", + "C=C1*10**-12; #Capacitance of the capacitors of phase shift circuit, F\n", + "fo=1/(2*pi*R*C*sqrt(6)); #Frequency of oscillation, Hz\n", + "\n", + "#Result\n", + "print(\"The frequency of oscillation=%dHz\"%fo);\n", + "\n", + "#Note: The frequency of oscillation had been calculated in the text as 954Hz, but here it gets approximated to 955 HZ.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of oscillation=955Hz\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.8 : Page number 378\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "C=5.0; #Capacitance of the capacitors of phase shift circuit, pF\n", + "fo=800.0; #Required frequency of oscillation, kHz\n", + "\n", + "#Calculation\n", + "#Since, fo=1/(2*pi*R*C*sqrt(6))\n", + "R=(1/(2*pi*C*10**-12*fo*10**3*sqrt(6)))/1000; #Resistance of the resistors of phase shift circuit, kilo ohm\n", + "\n", + "#Result\n", + "print(\"R=%.1f kilo ohm.\"%R);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "R=16.2 kilo ohm.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.9 : Page number 380\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "#Resistance of R1 and R2 resistors of the R-C bridge circuit\n", + "R1=220.0; #kilo ohm \n", + "R2=220.0; #kilo ohm\n", + "\n", + "#Capacitance of C1 and C2 the capacitors of the R-C bridge circuit\n", + "C1=250.0; #pF\n", + "C2=250.0; #pF\n", + "\n", + "#Calculation\n", + "#Since, R1=R2 and C1=C2, R1=R2 is taken as R and C1=C2 is taken as C\n", + "#And, f=1/(2*pi*sqrt(R1*R2*C1*C2))is transformed to f=1/(2*pi*R*C).\n", + "R=R1*10**3; #kilo ohm\n", + "C=C1*10**-12; #pF\n", + "f=1/(2*pi*R*C); #Frequency of oscillation, Hz\n", + "\n", + "\n", + "#Result\n", + "print(\"The frequency of oscillation=%dHz.\"%f);\n", + "\n", + "\n", + "#Note : The frequency of oscillation is calculated in the text as 2892Hz but here it gets approximated to 2893 Hz.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of oscillation=2893Hz.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 14.11 : Page number 384\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "#a.c equivalent values of the crystal:\n", + "L=1.0; #Inductance , H\n", + "C=0.01; #Capacitance , pF\n", + "R=1000.0; #Resistance , ohm\n", + "Cm=20.0; #Mounting capacitance, pF\n", + "\n", + "#Calculation\n", + "fs=(1/(2*round(pi,2)*sqrt(L*C*10**-12)))/1000; #Series resonant frrequency, kHz\n", + "CT=(C*Cm/(C+Cm)); #Total capacitance, pF\n", + "fp=(1/(2*round(pi,2)*sqrt(L*CT*10**-12)))/1000; #Prallel resonant frequency, kHz\n", + "\n", + "#Result\n", + "print(\"fs=%.0fkHz and fp=%.0fkHz.\"%(fs,fp));\n", + "\n", + "#Note: fs and fp are calculated in the text as 1589kHz and 1590kHz, but here it gets approximated to 1592kHz and 1593kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "fs=1592kHz and fp=1593kHz.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_6.ipynb new file mode 100644 index 00000000..e649cc91 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter15_6.ipynb @@ -0,0 +1,482 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:034eec32676d4e7abdedfb3bf68426d81a2d1483fc668bcbfdb5be18cec2e406" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 15: TRANSISTOR TUNED AMPLIFIERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.1 : Page number 394" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=250.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "L=1.25*10**-3; #Inductor of the parallel resonant circuit, H\n", + "R=10.0; #Resistor of the parallel resonant circuit, ohm\n", + "\n", + "#Calculation\n", + "#(i) Resonant frequency\n", + "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000; #Resonant frequecy, kHz\n", + "\n", + "#(ii) Impedance of the circuit at resonance\n", + "Zr=(L/(C*R))/1000; #Impedance of the circuit at resonance, kilo ohm\n", + "\n", + "#(iii) Quality factor of the circuit\n", + "Q=2*pi*(fr*10**3)*L/R; #Quality factor of the circuit\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%.1fkHz.\"%fr);\n", + "print(\"(ii) The impedance of the circuit at resonance=%d kilo ohm.\"%Zr);\n", + "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=284.7kHz.\n", + "(ii) The impedance of the circuit at resonance=500 kilo ohm.\n", + "(iii) The quality factor of the circuit=223.6.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.2 : Page number 394-395\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=100.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "L=100.0*10**-6; #Inductor of the parallel resonant circuit, H\n", + "R=10.0; #Resistor of the parallel resonant circuit, ohm\n", + "V=10.0; #Supply voltage, V\n", + "\n", + "#Calculation\n", + "#(i) Resonant frequency\n", + "fr=((1/(2*pi))*sqrt((1/(L*C))-(R/L)**2))/1000; #Resonant frequecy, kHz\n", + "\n", + "#(ii) Impedance of the circuit at resonance\n", + "Zr=(L/(C*R))/10**6; #Impedance of the circuit at resonance, mega ohm\n", + "\n", + "I=V/Zr; #Line current at resonance, microampere\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%.2fkHz.\"%fr);\n", + "print(\"(ii) The impedance of the circuit at resonance=%.1f mega ohm.\"%Zr);\n", + "print(\"The line current at resonance=%d micro ampere.\"%I);\n", + "\n", + "#Note : The resonant frequency in the text has been calculated as 1592.28 kHz, but here it gets approximated to 1591.47 kHz.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=1591.47kHz.\n", + "(ii) The impedance of the circuit at resonance=0.1 mega ohm.\n", + "The line current at resonance=100 micro ampere.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.3 : Page number 395\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=250.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "Zr=500.0*10**3; #Dynamic impedance, ohm\n", + "R=10.0; #Resistance of the coil, ohm\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since,Zr=L/CR,\n", + "L=(Zr*C*R)*10**3; #Inductance of the coil, mH\n", + "\n", + "#(ii) Resonant frequency\n", + "fr=((1/(2*pi))*sqrt((1/(L*10**-3*C))-(R/(L*10**-3))**2))/1000; #Resonant frequecy, kHz\n", + "\n", + "#(iii) Quality factor of the circuit\n", + "Q=2*pi*(fr*10**3)*(L*10**-3)/R; #Quality factor of the circuit\n", + "\n", + "#Result\n", + "print(\"(i) The inductance of the coil=%.2fmH.\"%L);\n", + "print(\"(ii) The resonant frequency=%.1fkHz.\"%fr);\n", + "print(\"(iii) The quality factor of the circuit=%.1f.\"%Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The inductance of the coil=1.25mH.\n", + "(ii) The resonant frequency=284.7kHz.\n", + "(iii) The quality factor of the circuit=223.6.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.4 : Page number 397\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Q=60.0; #Quality factor of the tuned amplifier\n", + "fr=1200.0; #Resonant frequency, kHz\n", + "\n", + "#Calculation\n", + "#(i)\n", + "BW=fr/Q; #Bandwidth, kHz\n", + "\n", + "#(ii)\n", + "f1=fr-(BW/2); #Lower cut-off frequency, kHz\n", + "f2=fr+(BW/2); #Upper cut-off frequency, kHz\n", + "\n", + "#Result\n", + "print(\"(i) The bandwidth=%dkHz\"%BW);\n", + "print(\"(ii) The lower and upper cut-off frequencies are=%dkHz and %dkHz.\"%(f1,f2));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The bandwidth=20kHz\n", + "(ii) The lower and upper cut-off frequencies are=1190kHz and 1210kHz.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.5 : Page number 397\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fr=2.0; #Resonant frequency, MHz\n", + "BW=50.0; #Bandwidth, kHz\n", + "\n", + "#Calculation\n", + "#Since, bandwidth=resonant_frequency/quality_factor\n", + "Q=(fr*10**6)/(BW*10**3); #Quality factor\n", + "\n", + "#Result\n", + "print(\"The quality factor=%d\"%Q);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The quality factor=40\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.7 : Page number 400\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=0.1*10**-6; #Capacitor of parallel resonant circuit, F\n", + "L=33.0*10**-3; #Inductor of the parallel resonant circuit, H\n", + "R=25.0; #Resistor of the parallel resonant circuit, ohm\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "fr=(1/(2*pi*sqrt(L*C)))/1000; #Resonant frequency, kHz\n", + "\n", + "#(ii)\n", + "XL=2*pi*(fr*10**3)*L; #Inductive reactance, ohm\n", + "Q=round(XL/R,0); #Quality factor\n", + "\n", + "#(iii)\n", + "BW=(fr*10**3)/Q; #Bandwidth\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%.2fkHz\"%fr);\n", + "print(\"(ii) The quality factor= %d.\"%Q);\n", + "print(\"(iii) The bandwidth=%dHz.\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=2.77kHz\n", + "(ii) The quality factor= 23.\n", + "(iii) The bandwidth=120Hz.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.8 : Page number 401-402\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "BW_dt=200.0; #Bandwidth, kHz\n", + "fr=10.0; #Operating frequency, MHz\n", + "\n", + "#Calculation\n", + "#Since, BW_dt=k*fr (i.e.,co-efficient_of_coupling * operating_frequency)\n", + "k=BW_dt/(fr*10**3); #co-efficient of coupling\n", + "\n", + "#Result\n", + "print(\"The co-efficient of coupling=%.2f.\"%k);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The co-efficient of coupling=0.02.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.9 : Page number 405\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "C=500.0*10**-12; #Capacitor of parallel resonant circuit, F\n", + "L=50.7*10**-6; #Inductor of the parallel resonant circuit, H\n", + "R=10.0; #Resistor of the parallel resonant circuit, ohm\n", + "RL=1.0; #Load resistance, mega ohm\n", + "\n", + "#Calculation\n", + "#(i)\n", + "fr=round((1/(2*pi*sqrt(L*C)))/1000); #Resonant frequency, Hz\n", + "\n", + "#(ii)\n", + "R_dc=R; #d.c load, ohm\n", + "XL=2*pi*(fr*1000)*L; #Inductive reactance, ohm\n", + "Q_coil=round(XL/R,1); #Quality factor\n", + "R_P=(Q_coil*XL)/1000 ; #Equivalent parallel resistance, kilo ohm\n", + "R_AC=(R_P*RL*10**3)/(R_P+RL*10**3); #A.C load,kilo ohm\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The resonant frequency=%dkHz\"%fr);\n", + "print(\"(ii) d.c load=%d ohm and a.c load=%d kilo ohm.\"%(R_dc,R_AC));\n", + "\n", + "#Note: In the text resonant frequency has been wrongly calculated to 106kHz but its actual value is approximately 1000kHz\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The resonant frequency=1000kHz\n", + "(ii) d.c load=10 ohm and a.c load=10 kilo ohm.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.10 : Page number 406-407\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "RL=50.0; #Load resistance, ohm\n", + "n=5; #Turns ratio of the transformer\n", + "VCC=50.0; #Supply voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "R_ac=n**2*RL; #A.C load, ohm\n", + "\n", + "#(ii)\n", + "P_o_max=VCC**2/(2*R_ac); #Maximum load power, W\n", + "\n", + "#Result\n", + "print(\"(i) The a.c load=%d ohm\"%R_ac);\n", + "print(\"(ii) Maximum load power=%dW\"%P_o_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The a.c load=1250 ohm\n", + "(ii) Maximum load power=1W\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 15.11 : Page number 407\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "P_D=4.0; #Maximum power dissipation, mW\n", + "P_o_max=1.0; #Maximum load power, W\n", + "\n", + "\n", + "#Calculation\n", + "max_collector_eff=(P_o_max/(P_o_max+(P_D/1000)))*100; #Maximum collector efficiency\n", + "\n", + "#Result\n", + "print(\"The maximum collector efficiency=%.1f%%\"%max_collector_eff);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum collector efficiency=99.6%\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_6.ipynb new file mode 100644 index 00000000..b2262b8f --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter16_6.ipynb @@ -0,0 +1,936 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d4ffda068787fb0974622fa8de40f7d54b5df2a00735e870e01cb9df45be78f9" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 16 : MODULATION AND DEMODULATION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.2 : Page number 416-417\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variabledeclaration\n", + "V_pp_max=16.0; #Maximum peak-to-peak voltage of an AM wave, mV\n", + "V_pp_min=4.0; #Minimum peak-to-peak voltage of an AM wave, mV\n", + "\n", + "#Calculation\n", + "Vmax=V_pp_max/2; #Maximum voltage of AM wave, mV\n", + "Vmin=V_pp_min/2; #Minimum voltage of AM wave, mV\n", + "m=(Vmax-Vmin)/(Vmax+Vmin); #Modulation factor.\n", + "\n", + "#Result\n", + "print(\"The modulation factor=%.1f.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulation factor=0.6.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.3 : Page number 417\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Es=50.0; #signalvoltage amplitude, V\n", + "Ec=100.0; #Carrier voltage amplitude, V\n", + "\n", + "\n", + "#Calculation\n", + "m=Es/Ec; #Modulation factor\n", + "\n", + "#Result\n", + "print(\"Modulation factor=%.1f.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Modulation factor=0.5.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.4 : Page number 419\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fc=2500.0; #Carrier frequency, kHz\n", + "f1=50.0; #Lower frequency of the audio signal, Hz\n", + "f2=15000.0; #Upper frequency of the audio signal, Hz\n", + "\n", + "#Calculation\n", + "fl_usb=fc+(f1/1000); #Lower frequency of upper sideband, kHz\n", + "fu_usb=fc+(f2/1000); #Upper frequency of upper sideband, kHz\n", + "\n", + "fu_lsb=fc-(f1/1000); #Lower frequency of upper sideband, kHz\n", + "fl_lsb=fc-(f2/1000); #Upper frequency of upper sideband, kHz\n", + "\n", + "#Since, f1=50Hz is negligible with respect to f2=15000Hz,\n", + "BW=(fc+(f2/1000))-(fc-(f2/1000)); #Bandwidth, kHz\n", + "\n", + "#Result\n", + "print(\"The upper sideband=%.2fkHz to %dkHz.\"%(fl_usb,fu_usb));\n", + "print(\"The lower sideband=%dkHz to %.2fkHz.\"%(fl_lsb,fu_lsb));\n", + "print(\"The bandwidth=%dkHz\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The upper sideband=2500.05kHz to 2515kHz.\n", + "The lower sideband=2485kHz to 2499.95kHz.\n", + "The bandwidth=30kHz\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.5 : Page number 420\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "EC=5.0; #Carrier amplitude, V\n", + "m=0.6; #modulation factor\n", + "ws=6280.0; #angular frequency of signal, radians/s\n", + "wc=211*10**4; #angular frequency of carrier, radians/s\n", + "\n", + "#Calculation\n", + "fs=(ws/(2*pi))/1000; #Signal frequency, kHz\n", + "fc=(wc/(2*pi))/1000; #Carrier frequency, kHz\n", + "\n", + "#(i)\n", + "Max_amp=EC+m*EC; #Maximum amplitude of AM wave, V\n", + "Min_amp=EC-m*EC; #Minimum amplitude of AM wave, V\n", + "\n", + "#(ii)\n", + "frequency_components=[fc-fs,fc,fc+fs]; #frequency components, kHz\n", + "amplitudes=[m*EC/2,EC,m*EC/2]; #Corresponding amplitudes, V\n", + "\n", + "#Result\n", + "print(\"(i) The maximum and minimum amplitudes of AM wave=%dV and %dV.\"%(Max_amp,Min_amp));\n", + "print(\"(ii) The frequency components of the AM wave=%.0f,%.0f,%.0f.\"%(frequency_components[0],frequency_components[1],frequency_components[2]));\n", + "print(\" The corresponding amplitudes are =%.1fV, %dV, %.1fV.\"%(amplitudes[0],amplitudes[1],amplitudes[2]));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The maximum and minimum amplitudes of AM wave=8V and 2V.\n", + "(ii) The frequency components of the AM wave=335,336,337.\n", + " The corresponding amplitudes are =1.5V, 5V, 1.5V.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.6 : Page number 420-421\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fc=1000.0; #Carrier frequency, kHz\n", + "fs=5.0; #Signal frequency, kHz\n", + "m=0.5; #Modulation factor\n", + "EC=100.0; #Amplitude of the carrier, V\n", + "\n", + "#Calculation\n", + "f_lsb=fc-fs; #Lower sideband frequency,kHz\n", + "f_usb=fc+fs; #Upper sideband frequency, kHz\n", + "Amplitude=m*EC/2; #Amplitude of each sideband, V\n", + "\n", + "#Result\n", + "print(\"The lower and upper sideband frequencies are=%dkHz and %dkHz.\"%(f_lsb,f_usb));\n", + "print(\"The amplitude of each sideband =%dV\"%Amplitude);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lower and upper sideband frequencies are=995kHz and 1005kHz.\n", + "The amplitude of each sideband =25V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.7 : Page number 421\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "EC=10.0; #Carrier amplitude, V\n", + "ES=6.0; #Signal amplitude, V\n", + "fc=10.0; #Carrier frequency, MHz\n", + "fs=5/1000.0; #Signal frequency. MHz\n", + "\n", + "#Calculation\n", + "#(i)\n", + "m=ES/EC; #Modulation factor\n", + "\n", + "#(ii)\n", + "f_lsb=fc-fs; #Lower sideband frequency,MHz\n", + "f_usb=fc+fs; #Upper sideband frequency, MHz\n", + "\n", + "#(iii)\n", + "Amplitude=m*EC/2; #Amplitude of each sideband, V\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The modulation factor=%.1f.\"%m);\n", + "print(\"(ii) The lower and upper sideband frequencies are=%.3fMHz and %.3fMHz.\"%(f_lsb,f_usb));\n", + "print(\"(iii) The amplitude of each sideband =%dV\"%Amplitude);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The modulation factor=0.6.\n", + "(ii) The lower and upper sideband frequencies are=9.995MHz and 10.005MHz.\n", + "(iii) The amplitude of each sideband =3V\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.8 : Page number 423\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Pc=500.0; #Carrier power, W\n", + "m=1.0; #Modulation factor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Ps=(1/2.0)*m**2*Pc; #Sideband power, W\n", + "\n", + "#(ii)\n", + "PT=Pc+Ps; #Power of AM wave, W\n", + "\n", + "#Result\n", + "print(\"(i) The power in sidebands=%dW\"%Ps);\n", + "print(\"(ii) The power of AM wave=%dW\"%PT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The power in sidebands=250W\n", + "(ii) The power of AM wave=750W\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Exmaple 16.9 : Page number 423\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Pc=50.0; #Power of carrier, kW\n", + "\n", + "#Calculation\n", + "#(i)\n", + "m=80/100.0; #Modulation factor\n", + "Ps=(1/2.0)*m**2*Pc; #Sideband Power, kW\n", + "print(\"(i) The sideband power for 80%% modulation=%dkW.\"%Ps);\n", + "\n", + "#(ii)\n", + "m=10/100.0; #Modulation factor\n", + "Ps=(1/2.0)*m**2*Pc; #Sideband Power, kW\n", + "print(\"(ii) The sideband power for 10%% modulation=%.2fkW.\"%Ps);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The sideband power for 80% modulation=16kW.\n", + "(ii) The sideband power for 10% modulation=0.25kW.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.10 : Page number 423-424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Pc=40.0; #Carrier power, kW\n", + "m=100/100.0; #Modulation index\n", + "amplifier_eff=72/100.0; #Efficiency of modulated RF amplifier\n", + "\n", + "\n", + "#Calculation\n", + "#(i)Carrier power remains same after modulation\n", + "\n", + "#(ii)\n", + "Ps=(1/2.0)*(m**2)*Pc; #Sideband power\n", + "P_audio=Ps/amplifier_eff; #Required audio power, kW\n", + "\n", + "#Result\n", + "print(\"(i) The carrier power=%dkW.\"%Pc);\n", + "print(\"(ii) The required audio power=%.1fkW.\"%P_audio);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The carrier power=40kW.\n", + "(ii) The required audio power=27.8kW.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.11 : Page number 424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fs=1.0; #Signal frequency, kHz\n", + "fc=500.0; #Carrier frequency, kHz\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "sideband_f=[fc-fs,fc+fs]; #Sideband frequencies, kHz\n", + "\n", + "#(ii)\n", + "BW=(fc+fs)-(fc-fs); #Bandwidth required, kHz\n", + "\n", + "#Result\n", + "print(\"(i) The sideband frequencies=%dkHz and %dkHz.\"%(sideband_f[0],sideband_f[1]));\n", + "print(\"(ii) The bandwidth required=%dkHz\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The sideband frequencies=499kHz and 501kHz.\n", + "(ii) The bandwidth required=2kHz\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.12 : Page number 424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "IC=8.0; #Antenna current due to carrier,A\n", + "m=40/100.0; #Modulation index\n", + "\n", + "#Calculation\n", + "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n", + "#that implies, (PT/Pc)=1+(m\u00b2/2),\n", + "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n", + "IT=IC*sqrt(1+(m**2/2.0)); #Total current, A\n", + "\n", + "\n", + "#Result\n", + "print(\"The total antenna current=%.2fA.\"%IT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total antenna current=8.31A.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.13 : Page number 424\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "IC=8.0; #Antenna current when only carrier is sent, A\n", + "IT=8.93; #Total antenna current, A\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n", + "#that implies, (PT/Pc)=1+(m\u00b2/2),\n", + "#So, square_of(Total_current/Carrier_current)=(IT/IC)\u00b2=1+(m\u00b2/2).\n", + "m=sqrt((((IT/IC)**2)-1)*2)*100; #The %age of modulation\n", + "\n", + "#Result\n", + "print(\"The %%age of modulation=%.1f%%.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The %age of modulation=70.1%.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.14 : Page number 425\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "Vc=100.0; #Carrier voltage, V\n", + "V_T=110.0; #The total voltage after modulation, V\n", + "\n", + "#Calculation\n", + "#Since, Ps=(1/2)*m\u00b2*Pc and PT=Pc+Ps (Total_power=carrier_power+signal_power)\n", + "#that implies, (PT/Pc)=1+(m\u00b2/2),\n", + "#So, square_of(Total_voltage/Carrier_voltage)=(V_T/Vc)\u00b2=1+(m\u00b2/2).\n", + "m=sqrt((((V_T/Vc)**2)-1)*2); #The %age of modulation\n", + "\n", + "#Result\n", + "print(\"The modulation index =%.3f.\"%m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulation index =0.648.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.15 : Page number 425-426\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vc=5.0; #Carrier voltage, V\n", + "V_lsb=2.5; #Lower sideband component, V\n", + "V_usb=2.5; #Upper sideband component, V\n", + "R=2.0; #Resistor driven by AM wave, k\u03a9\n", + "\n", + "#Calculation\n", + "#Since, power=(r.m.s_voltage)\u00b2/resistance\n", + "#(i)\n", + "Pc=round((0.707*Vc)**2/R,2); #Carrier power mW\n", + "\n", + "#(ii)\n", + "P_lower=round((0.707*V_lsb)**2/R,3); #Power delivered by lower sideband, mW\n", + "\n", + "#(iii)\n", + "P_upper=round((0.707*V_usb)**2/R,3); #Power delivered by upper sideband, mW\n", + "\n", + "P_T=round(Pc+P_lower+P_upper,3); #Total power delivered by the AM wave, mW\n", + "\n", + "#Result\n", + "print(\"(i) The carrier power=%.2fmW\"%Pc);\n", + "print(\"(ii) The power delivered by lower sideband=%.3fmW\"%P_lower);\n", + "print(\"(iii) The power delivered by upper sideband=%.3fmW\"%P_upper);\n", + "print(\"The total power delivered by the AM wave=%.3fmW\"%P_T);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The carrier power=6.25mW\n", + "(ii) The power delivered by lower sideband=1.562mW\n", + "(iii) The power delivered by upper sideband=1.562mW\n", + "The total power delivered by the AM wave=9.374mW\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.16 : Page number 428\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "wc=6e08; #Carrier angular frequency, rad/s\n", + "ws=1250.0; #Signal angular frequency, rad/s\n", + "mf=5; #Modulation index\n", + "Ec=12.0; #Carrier amplitude, V\n", + "R=10.0; #Resistor, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "fc=wc/(2*pi); #Carrier frequency, Hz\n", + "\n", + "#(ii)\n", + "fs=ws/(2*pi); #Signal frequency, Hz\n", + "\n", + "#(iv)\n", + "delta_f=mf*fs; #Maximum frequency deviation, Hz\n", + "\n", + "#(v)\n", + "P=(Ec/sqrt(2))**2/R; #Power dissipated, W\n", + "\n", + "#Result\n", + "print(\"(i) The carrier frequency=%.1fe06 Hz.\"%(fc/10**6));\n", + "print(\"(ii) The signal frequency=%.0f Hz.\"%fs);\n", + "print(\"(iii) The modulation index=%d.\"%mf);\n", + "print(\"(iv) The maximum frequency deviation=%.0fHz.\"%delta_f);\n", + "print(\"(v) The power dissipated=%.1fW.\"%P);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The carrier frequency=95.5e06 Hz.\n", + "(ii) The signal frequency=199 Hz.\n", + "(iii) The modulation index=5.\n", + "(iv) The maximum frequency deviation=995Hz.\n", + "(v) The power dissipated=7.2W.\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.17 : Page number 428-429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "fc=25.0; #Carrier frequency, MHz\n", + "fs=400.0; #Signal frequency, Hz\n", + "Ec=4.0; #Carrier amplitude, V\n", + "delta_f=10.0; #Maximum frequency deviation, kHz\n", + "\n", + "#Calculation\n", + "wc=2*pi*fc*10**6; #Carrier angular frequency, rad/s\n", + "ws=2*pi*fs; #Signal angular frequency, rad/s\n", + "mf=delta_f*1000/fs; #Modulation index\n", + "\n", + "\n", + "#Result\n", + "print(\"e=%dcos(%.2et + %dsin%dt)\"%(Ec,wc,mf,ws));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "e=4cos(1.57e+08t + 25sin2513t)\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.18 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "delta_f=50.0; #Maximum frequency deviation, kHz\n", + "fs=5.0; #Modulating frequency, kHz\n", + "\n", + "#Calculation\n", + "mf=delta_f/fs; #Modulation index\n", + "\n", + "#Result\n", + "print(\"The modulation index=%d\"%mf);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The modulation index=10\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.19 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fc=1000.0; #Carrier frequency, kHz\n", + "fs=15.0; #Modulating frequency, kHz\n", + "\n", + "#Calculation\n", + "first_3_usb_f=[fc+fs,fc+2*fs,fc+3*fs]; #First three upper sideband frequncies, kHz\n", + "first_3_lsb_f=[fc-fs,fc-2*fs,fc-3*fs]; #First three lowerr sideband frequncies, kHz\n", + "\n", + "\n", + "#Result\n", + "print(\"The first three upper sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_usb_f[0],first_3_usb_f[1],first_3_usb_f[2]));\n", + "print(\"The first three lower sideband frequencies=%dkHz ,%dkHz and %dkHz.\"%(first_3_lsb_f[0],first_3_lsb_f[1],first_3_lsb_f[2]));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The first three upper sideband frequencies=1015kHz ,1030kHz and 1045kHz.\n", + "The first three lower sideband frequencies=985kHz ,970kHz and 955kHz.\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.20 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fs=15.0; #Modulating frequency, kHz\n", + "delta_f=75.0; #Maximum frequency deviation, kHz\n", + "\n", + "#Calculation\n", + "BW=2*(delta_f+fs); #Bandwidth, kHz\n", + "\n", + "#Result\n", + "print(\"The bandwidth of the FM signal=%dkHz.\"%BW);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The bandwidth of the FM signal=180kHz.\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.21 : Page number 429\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "k=75.0; #Frequency deviation constant, kHz/V\n", + "Es=2.0; #Amplitude of signal, V\n", + "\n", + "\n", + "#Calculation\n", + "delta_f=k*Es; #Maximum frequency deviation, kHz\n", + "\n", + "#Result\n", + "print(\"The maximum frequency deviation=%dkHz.\"%delta_f);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum frequency deviation=150kHz.\n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 16.22 : Page number 429-430\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "fs1=500.0; #First audio frequency, Hz\n", + "fs2=200.0; #Second audio frequency (decreased), Hz\n", + "Es=2.4; #AF voltage, V\n", + "delta_f1=4.8; #Frequency deviation,kHz\n", + "\n", + "#Calculation\n", + "k=delta_f1/Es; #Frequency deviation constant, kHz/V\n", + "Es=7.2; #AF voltage, V (increased)\n", + "delta_f2=k*Es; #2nd frequency deviation, kHz\n", + "Es=10.0; #AF voltage, V (increased)\n", + "delta_f3=k*Es; #3rd frequency deviation, kHz\n", + "\n", + "mf1=delta_f1/(fs1/1000); #Modulation index in 1st case\n", + "mf2=delta_f2/(fs1/1000); #Modulation index in 2nd case\n", + "mf3=delta_f3/(fs2/1000); #Modulation index in 3rd case\n", + "\n", + "#Result\n", + "print(\"The frequency deviation in second case=%.1fkHz.\"%delta_f2);\n", + "print(\"The frequency deviation in third case=%dkHz.\"%delta_f3);\n", + "print(\"The modulation index in 1st case=%.1f\"%mf1);\n", + "print(\"The modulation index in 2nd case=%.1f\"%mf2);\n", + "print(\"The modulation index in 3rd case=%d\"%mf3);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency deviation in second case=14.4kHz.\n", + "The frequency deviation in third case=20kHz.\n", + "The modulation index in 1st case=9.6\n", + "The modulation index in 2nd case=28.8\n", + "The modulation index in 3rd case=100\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_6.ipynb new file mode 100644 index 00000000..2c9a49a5 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter17_6.ipynb @@ -0,0 +1,877 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:33d9a7285654630dd24f2d6229210244e78961e0605c11041ed1b2f130cb19a1" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 17 : REGULATED D.C POWER SUPPLY" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.1 : Page number 444\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_NL=400.0; #Output voltage with no-load, V\n", + "V_FL=300.0; #Output voltage with full-load, V\n", + "\n", + "\n", + "#Calculation\n", + "percentage_voltage_regulation=((V_NL-V_FL)/V_FL)*100; #Percentage of voltage regulation\n", + "\n", + "#Result\n", + "print(\"The percentage of voltage regulation=%.2f%%.\"%percentage_voltage_regulation);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage of voltage regulation=33.33%.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.2 : Page number 444\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_regulation=1.0; #%age voltage regulation\n", + "V_NL=30.0; #Output voltage with no-load,V\n", + "\n", + "#Calculation\n", + "#Since, %age_of_voltage_regulation=((V_NL-V_FL)/V_FL)*100\n", + "V_FL=V_NL/(1+(V_regulation/100)); #Output voltage with full-load, V\n", + "\n", + "#Result\n", + "print(\"The full-load voltage=%.1fV.\"%V_FL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The full-load voltage=29.7V.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.3 : Page number 445\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_NL_A=30.0; #Output voltage of supply A with no-load, V\n", + "V_FL_A=25.0; #Output voltage of supply A with full-load, V\n", + "V_NL_B=30.0; #Output voltage of supply B with no-load, V\n", + "V_FL_B=29.0; #Output voltage of supply B with full-load, V\n", + "\n", + "\n", + "#Calculation\n", + "V_regulation_A=((V_NL_A-V_FL_A)/V_FL_A)*100; #%age of voltage regulation in power supply A\n", + "V_regulation_B=((V_NL_B-V_FL_B)/V_FL_B)*100; #%age of voltage regulation in power supply B\n", + "\n", + "#Result\n", + "if(V_regulation_A<V_regulation_B):\n", + " print(\"Power supply A is better than B.\");\n", + "else :\n", + " print(\"Power supply B is better than A.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power supply B is better than A.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.4 : Page number 445\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_NL=500.0; #Output voltage with no-load, V\n", + "V_FL=300.0; #Output voltage with full-load, V\n", + "I_FL=120.0; #Output current with full-load, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Regulation=((V_NL-V_FL)/V_FL)*100; #Voltage regulation percentage\n", + "\n", + "#(ii)\n", + "RL_min=V_FL/I_FL; #Minimum load resistance, k\u03a9\n", + "\n", + "#Result\n", + "print(\"(i) The voltage regulation=%.1f%%.\"%Regulation);\n", + "print(\"(ii)The minimum load resistance=%.1fk\u03a9.\"%RL_min);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The voltage regulation=66.7%.\n", + "(ii)The minimum load resistance=2.5k\u03a9.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.5 : Page number 445\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VL_1=10.5; #Initial output voltage with load, V\n", + "VL_2=10.0; #Decreased output voltage with additional load, V\n", + "IL_1=1.0; #Initial load current, A\n", + "IL_added=1.0; #Added load current, A\n", + "\n", + "#Calculation\n", + "delta_VL=VL_1-VL_2; #Change in output voltage, V\n", + "delta_IL=IL_added; #Change in load current, A\n", + "\n", + "#(i)\n", + "Zo=delta_VL/delta_IL; #Output impedance of power supply, \u03a9 (OHM's LAW)\n", + "\n", + "#(ii)\n", + "#Since, Output_impedance=change_in_output_voltage/change_in_output_current\n", + "#Zo=(V_NL-VL_1)/delta_IL,\n", + "delta_IL=IL_1; #Change in load current, A\n", + "V_NL=VL_1+(delta_IL*Zo); #Output voltage with no load, V\n", + "\n", + "#Result\n", + "print(\"(i) The output impedance=%.1f\u03a9.\"%Zo);\n", + "print(\"(ii) The output voltage with no-load=%dV.\"%V_NL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output impedance=0.5\u03a9.\n", + "(ii) The output voltage with no-load=11V.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.6 : Page number 446\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Zo=0.01; #Output impedance, \u03a9\n", + "IL_max=1.0; #Maximum output current, A\n", + "IL_min=0.5 #Minimum output current, A\n", + "f=10.0; #Frequency, kHz\n", + "\n", + "#Calculation\n", + "#Since, Zo=delta_VL/delta_IL\n", + "delta_IL=IL_max-IL_min; #Maximum change in output current, A\n", + "delta_VL=(Zo*delta_IL)*1000; #Fluctuations in output voltage, mV\n", + "\n", + "#Result\n", + "print(\"The output voltage will have %dmV peak-to-peak fluctuation at a rate of %dkHz.\"%(delta_VL,f));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage will have 5mV peak-to-peak fluctuation at a rate of 10kHz.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.7 : Page number 446\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "delta_Vout=10.0; #Change in output voltage, \u03bcV\n", + "delta_Vin=5.0; #Change in input voltage, V\n", + "\n", + "#Calculation\n", + "Line_regulation=delta_Vout/delta_Vin; #Line regulation, \u03bcV/V\n", + "\n", + "#Result\n", + "print(\"The line regulation of the voltage regulator=%d\u03bcV/V.\"%Line_regulation);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The line regulation of the voltage regulator=2\u03bcV/V.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.8 : Page number 449-450\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vin=24.0; #Input voltage, V\n", + "Vz=12.0; #Zener voltage, V\n", + "Rs=160.0; #Series resistance, \u03a9\n", + "RL_max=float('inf'); #Maximum load resistance, \u03a9\n", + "RL_min=200.0; #Minimum load resistance, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout=Vz; #Output voltage,(equal to zener regulated voltage), V\n", + "Is=((Vin-Vout)/Rs)*1000; #Current through series resistance, mA\n", + "\n", + "#(ii)\n", + "IL_min=Vout/RL_max; #Minimum load current, A\n", + "IL_max=(Vout/RL_min)*1000; #Maximum load current, mA\n", + "\n", + "#(iii)\n", + "IZ_min=Is-IL_max; #Minimum zener current, mA\n", + "IZ_max=Is-IL_min; #Maximum zener current, mA\n", + "\n", + "#Result\n", + "print(\"(i) The current through the series resistance=%dmA\"%Is);\n", + "print(\"(ii) The minimum and maximum load currents are=%dA and %dmA\"%(IL_min,IL_max));\n", + "print(\"(iii) The minimum and maximum zener currents are=%dmA and %dmA\"%(IZ_min,IZ_max));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The current through the series resistance=75mA\n", + "(ii) The minimum and maximum load currents are=0A and 60mA\n", + "(iii) The minimum and maximum zener currents are=15mA and 75mA\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.9 : Page number 450\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VZ=15.0; #Zener voltage, V\n", + "Vin_min=22.0 #Minimum input voltage, V\n", + "Vin_max=40.0 #Maximum input voltage, V\n", + "Vout=VZ; #Regulated output voltage, V\n", + "IL_max=100.0; #Maximum load current, mA\n", + "IL_min=20.0; #Minimum load current, mA\n", + "\n", + "#Calculation\n", + "RS_max=(Vin_min-Vout)/(IL_max/1000); #Maximum value of series resistance, \u03a9 (OHM'S lAW)\n", + "\n", + "#Result\n", + "print(\"The maximum load resistance to hold the voltage constant=%d\u03a9.\"%RS_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum load resistance to hold the voltage constant=70\u03a9.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.10 : Page number 450-451\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=3.3; #Zener voltage, V\n", + "Iz_min=3.0; #Minimum zener current, mA\n", + "Iz_max=100.0; #Maximum zener current, mA\n", + "RL_max=2.0; #Maximum load resistance, k\u03a9\n", + "RL_min=500.0; #Minimum load resistance, \u03a9\n", + "Vin=20.0; #Input voltage, V\n", + "\n", + "#Calculation\n", + "Rs_min=(Vin-Vz)/(Iz_max/1000); #Minimum series resistance required, \u03a9\n", + "\n", + "#Result\n", + "print(\"The minimum series resistance required to limit the zener current=%.0f\u03a9.\"%Rs_min);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum series resistance required to limit the zener current=167\u03a9.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.11 : Page number 451\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=3.3; #Zener voltage, V\n", + "Iz_min=3.0; #Minimum zener current, mA\n", + "Iz_max=100.0; #Maximum zener current, mA\n", + "RL_max=2.0; #Maximum load resistance, k\u03a9\n", + "RL_min=500.0; #Minimum load resistance, \u03a9\n", + "Vin=20.0; #Input voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "IL_max=(Vz/RL_min)*1000; #Maximum load current, mA\n", + "Rs_max=((Vin-Vz)/(IL_max+Iz_min))*1000; #Maximum series resistance, \u03a9\n", + "\n", + "#Result\n", + "print(\"The maximum allowable value of series resistance=%d\u03a9.\"%Rs_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable value of series resistance=1739\u03a9.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.12 : Page number 452\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=10.0; #Zener voltage, V\n", + "beta=100.0; #Base current amplification factor\n", + "RL=1000.0; #Load resistance, \u03a9\n", + "VBE=0.5; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "Vout=Vz-VBE; #Output voltage, V\n", + "IL=(Vout/RL)*1000; #Load current, mA\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.1fV.\"%Vout);\n", + "print(\"The load current=%.1fmA\"%IL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage=9.5V.\n", + "The load current=9.5mA\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.3 : Page number 452\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "IC=1.0; #Required current(collector current), A\n", + "Vout=6.0; #Constant output voltage, V\n", + "Vin=10.0; #Supply voltage, V\n", + "beta=50.0; #Base current amplification factor\n", + "VBE=0.5; #Base-emitter voltage, V\n", + "Iz=10.0; #Minimum zener current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IB=(IC/beta)*1000; #Base current, mA\n", + "\n", + "#Since, Vout=Vz-VBE;\n", + "Vz=Vout+VBE; #Zener breakdown voltage, V\n", + "\n", + "#(ii)\n", + "V_Rs=Vin-Vz; #Voltage across series resistance Rs, V\n", + "Rs=(V_Rs/(IB+Iz))*1000; #Series resistance, \u03a9\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The zener breakdown voltage=%.1fV\"%Vz);\n", + "print(\"(ii)The series resistance=%.0f\u03a9.\"%Rs);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The zener breakdown voltage=6.5V\n", + "(ii)The series resistance=117\u03a9.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.14: Page number 452-453\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=12.0; #Zener voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "Vin=20.0; #Input voltage, V\n", + "RS=220.0; #Series resistance, \u03a9\n", + "RL=1.0; #Load resistance, k\u03a9\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout=Vz-VBE; #Output voltage, V\n", + "\n", + "#(ii)\n", + "V_RS=Vin-Vz; #Voltage across series resistance, RS, V\n", + "IR=(V_RS/RS)*1000; #Current through series resistance, mA\n", + "IL=Vout/RL; #Load current, mA\n", + "\n", + "#Since, IL is emitter current and emitter current is approx. equal to collector current,\n", + "IC=IL; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "Iz=IR-IB; #Zener current, mA\n", + "\n", + "#Result\n", + "print(\"(i) The output voltage=%.1fV.\"%Vout);\n", + "print(\"(ii) The zener current=%dmA\"%Iz);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output voltage=11.3V.\n", + "(ii) The zener current=36mA\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.15 : Page number 453-454\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "IL_min=0; #Minimum load current, A\n", + "IL_max=1.0; #Maximum load current, A\n", + "Vin_min=12.0; #Minimum input voltage, V\n", + "Vin_max=18.0; #Maximum input voltage, V\n", + "Iz_min=1.0; #Minimum zener current, mA\n", + "Vz=8.5; #Zener voltage, V\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculation\n", + "IB_max=(IL_max/beta)*1000; #Maximum base current, mA\n", + "I_RS=Iz_min+IB_max; #Current through the series resistance, mA\n", + "RS=((Vin_min-Vz)/I_RS)*1000; #Series resistance, \u03a9\n", + "\n", + "#(ii)\n", + "V_RS_max=Vin_max-Vz; #Maximum voltage across series resistance, V\n", + "P_max_RS=ceil((V_RS_max**2/RS)*1000)/1000; #Maximum power dissipation in series resistance RS, W\n", + "\n", + "#(iii)\n", + "I_RS_max=V_RS_max/floor(RS); #Maximum current through series resistance,mA\n", + "Iz_max=I_RS_max; #Maximum zener current, mA\n", + "P_z_max=Vz*Iz_max; #Maximum power dissipated in zener diode, W\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The series resistance=%d\u03a9.\"%RS);\n", + "print(\"(ii) The maximum power dissipated in series resistance=%.3fW.\"%P_max_RS);\n", + "print(\"(iii)The maximum power dissipated in zener diode=%.3fW.\"%P_z_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The series resistance=166\u03a9.\n", + "(ii) The maximum power dissipated in series resistance=0.542W.\n", + "(iii)The maximum power dissipated in zener diode=0.486W.\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.16 : Page number 456\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=2.0; #Resistor R1, k\u03a9\n", + "R2=1.0; #Resistor R2, k\u03a9\n", + "Vz=6.0; #Zener voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "m=R2/(R1+R2); #Feedback fraction\n", + "A_CL=1/m; #Closed-loop voltage gain\n", + "Vout=A_CL*(Vz+VBE); #Regulated output voltage, V\n", + "\n", + "#Result\n", + "print(\"The regulated output voltage=%.1fV\"%Vout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulated output voltage=20.1V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.17 : Page number 456\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=30.0; #Resistor R1, k\u03a9\n", + "R2=10.0; #Resistor R2, k\u03a9\n", + "\n", + "#Calculation\n", + "m=R2/(R1+R2); #Feedback fraction\n", + "A_CL=1/m; #Closed-loop voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The closed-loop voltage gain=%d.\"%A_CL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The closed-loop voltage gain=4.\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.18 : Page number 457\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vin=22.0; #Input voltage, V\n", + "Rs=130.0; #Series resistance, \u03a9\n", + "Vz=8.3; #Zener voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "RL=100.0; #Load resistance, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout=Vz+VBE; #Output voltage, V\n", + "\n", + "#(ii)\n", + "IL=(Vout/RL)*1000; #Load current, mA (OHM's LAW)\n", + "IS=((Vin-Vout)/Rs)*1000; #Current through series resistance, mA (OHM's LAW)\n", + "IC=IS-IL; #Collector current, mA\n", + "\n", + "#Result\n", + "print(\"(i) The regulated output voltage=%dV\"%Vout);\n", + "print(\"(ii) Various currents for the shunt regulator are: IL=%dmA , IS=%dmA and IC=%dmA\"%(IL,IS,IC));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The regulated output voltage=9V\n", + "(ii) Various currents for the shunt regulator are: IL=90mA , IS=100mA and IC=10mA\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.20 : Page number 463\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=240.0; #Resistor R1 of the regulator, \u03a9\n", + "R2=2.4; #Variable resistance R2 of the regulator, k\u03a9\n", + "\n", + "#Calculation\n", + "Vout=1.25*(R2*1000/R1 + 1); #Regulated output voltage, V\n", + "\n", + "#Result\n", + "print(\"The regulated output voltage=%.2fV.\"%Vout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulated output voltage=13.75V.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 17.21 : Page number 463\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vout_adj=8.0; #Output voltage (adjusted), V\n", + "Vd=40.0; #Input/output differential rating, V\n", + "\n", + "#Calculation\n", + "Vin_max=Vout_adj+Vd; #Maximum allowable input voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum allowable input voltage=%dV.\"%Vin_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable input voltage=48V.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_6.ipynb new file mode 100644 index 00000000..9e4eeafa --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter18_6.ipynb @@ -0,0 +1,793 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 18 : SOLID-STATE SWITCHING CIRCUITS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.1 : Page number 472" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Input voltage required to saturate the transistor switch=5.4V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10; #Supply voltage, V\n", + "RC=1.0; #Collector resistor, kΩ\n", + "RB=47.0; #Base resistor, kΩ\n", + "beta=100.0; #Base current amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "IC_sat=VCC/RC; #Collector saturation current, mA\n", + "IB=IC_sat/beta; #Base current, mA\n", + "V=IB*RB+VBE; #Input voltage, V\n", + "\n", + "#Result\n", + "print(\"Input voltage required to saturate the transistor switch=%.1fV.\"%V);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.2 : Page number 475" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The collector emitter voltage at cut-off=9.99V.\n", + "(ii) The collector emitter voltage at saturation=0.7V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10; #Supply voltage, V\n", + "RC=1.0; #Collector resistor, kΩ\n", + "ICBO=10.0; #Collector leakage current, μA\n", + "V_knee=0.7; #Knee voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IC=ICBO; #Collector current, μA\n", + "VCE=VCC-(ICBO/1000)*RC; #Collector-emitter voltage, V\n", + "\n", + "print(\"(i) The collector emitter voltage at cut-off=%.2fV.\"%VCE);\n", + "\n", + "#(ii)\n", + "#Since, saturation current=IC_sat=(VCC-V_knee)/RC; \n", + "VCE=V_knee; #Collector-emitter voltage, V\n", + "\n", + "print(\"(ii) The collector emitter voltage at saturation=%.1fV.\"%VCE);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.3 : Page number 475-476" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Minimum β=19.4.\n", + "(ii) The transistor will not be saturated.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10; #Supply voltage, V\n", + "RC=1; #Collector resistor, kΩ\n", + "VBB=2; #Supply voltage to base, V\n", + "RB=2.7; #Base resistor, kΩ\n", + "V_knee=0.7; #Knee voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IB=round((VBB-VBE)/RB,2); #Base current, mA\n", + "Ic_sat=(VCC-V_knee)/RC; #Collector saturation current, mA\n", + "beta_min=Ic_sat/IB; #Minimum value of base current amplification factor\n", + "print(\"(i) Minimum β=%.1f.\"%beta_min);\n", + "\n", + "#(ii)\n", + "VBB=1; #Supply voltage to base(changed), V\n", + "beta=50; #Base current amplification factor\n", + "IB=(VBB-VBE)/RB; #Base current, mA\n", + "IC=beta*IB; #Collector current,mA\n", + "\n", + "if(IC<Ic_sat):\n", + " print(\"(ii) The transistor will not be saturated.\");\n", + "else:\n", + " print(\"(ii) The transistor will be saturated.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.4 : Page number 480" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Time period of the square wave=0.14 m sec.\n", + "Time frequency of the square wave=7 kHz.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R2=10; #Resistor R2, kΩ\n", + "R3=10; #Resistor R3, kΩ\n", + "C1=0.01; #Capacitor of 1st transistor, μF\n", + "C2=0.01; #Capacitor of 2nd transistor, μF\n", + "\n", + "#Calculation\n", + "R=R2*1000; #Resistance, Ω\n", + "C=C1*10**-6; #Capacitance, F\n", + "T=round((1.4*R*C)*1000,2); #Time period,m sec\n", + "f=1/(T*10**-3); #Frequency, Hz\n", + "f=f/1000; #Frequency, kHz\n", + "\n", + "#Result\n", + "print(\"Time period of the square wave=%.2f m sec.\"%T);\n", + "print(\"Time frequency of the square wave=%d kHz.\"%f);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.6 : Page number 485" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage=0.55V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=10; #Resistance in differentiating circuit, kΩ\n", + "C=2.2; #Capacitance in differentiating circuit, μF\n", + "d_ei=10; #Change in input voltage, V\n", + "dt=0.4; #Time in which change occurs, s\n", + "\n", + "#Calculation\n", + "eo=R*1000*C*10**-6*d_ei/dt\n", + "\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.2fV.\"%eo);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.7 : Page number 489" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The peak output voltage=11.3V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vin_peak=12; #Peak value of input voltage, V\n", + "V_D=0.7; #Forward bias voltage of diode, V\n", + "\n", + "#Calculation\n", + "Vout_peak=Vin_peak-V_D; #Peak value of output voltage, V\n", + "\n", + "#Result\n", + "print(\"The peak output voltage=%.1fV.\"%Vout_peak);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.8 : Page number 489" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The peak output voltage=8V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vin_peak=10; #Peak value of input voltage, V\n", + "R=1; #Input resistor, kΩ\n", + "RL=4; #Load resistor, kΩ\n", + "\n", + "#Calculation\n", + "Vout_peak=(Vin_peak*RL)/(R+RL); #Peak output voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The peak output voltage=%dV.\"%Vout_peak);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.9 : Page number 490" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The diode will be forward biased for the negative half-cycle of input signal.\n", + "The output voltage=-0.7V.\n", + "The voltage across R=-9.3V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Vin=-10; #Input voltage, V\n", + "V_D=0.7; #Forward bias voltage of the diode, V\n", + "R=1; #Resistance, kΩ\n", + "\n", + "\n", + "print(\"The diode will be forward biased for the negative half-cycle of input signal.\");\n", + "Vout=-V_D; #Output voltage, V\n", + "V_R=Vin-(-V_D); #Voltage across resistor R, V\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.1fV.\"%Vout);\n", + "print(\"The voltage across R=%.1fV.\"%V_R);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.10 : Page number 490-491" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "During the positive half cycle, the diode is foward biased and can be replaced by battery of 0.7V.\n", + "Therefore, Vout=0.7V.\n", + "During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\n", + "Therefore, Vout_peak=-8.33V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_F=0.7; #Forward bias voltage of diode, V\n", + "R=200.0; #Input resistor of the circuit, Ω\n", + "RL=1.0; #Load resistor, kΩ\n", + "Vin_peak=10.0; #Peak input voltage, V\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "#Positive half-cycle:\n", + "print(\"During the positive half cycle, the diode is foward biased and can be replaced by battery of %.1fV.\"%V_F);\n", + "print(\"Therefore, Vout=%.1fV.\"%V_F);\n", + "\n", + "#Negative half-cycle:\n", + "print(\"During the negative half cycle, the diode is reverse biased and hence behaves as an open circuit.\");\n", + "Vout_peak=RL*(-Vin_peak)/(R/1000+RL);\n", + "print(\"Therefore, Vout_peak=%.2fV.\"%Vout_peak);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.12 : Page number 491" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa23cd37e80>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "from math import sin\n", + "from math import pi\n", + "\n", + "V_biasing=10.0; #Biasing voltage, V\n", + "vin=[30*sin(t/10.0) for t in range(0,(int)(2*pi*10))] #input voltage waveform, V\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(vin);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in vin[:]:\n", + " if(v-V_biasing)>0 : #Diode is forward biased.\n", + " vout.append(v-V_biasing);\n", + " else: #Diode is reverse biased.\n", + " vout.append(0);\n", + "\n", + "plt.subplot(212) \n", + "plt.plot(vout);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.13 : Page number 492" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa23dcfd668>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "Vin=[]; #Input voltage waveform, V\n", + "t1=50; #Assumed time interval, s\n", + "t2=100; #Assumed time interval, s\n", + "V_biasing=10; #Biasing voltage, V\n", + "for t in range(0,151): #time interval from 0s to 151s\n", + " if(t<=t1): \n", + " Vin.append(15); #Value of input voltage for time 0 to t1 seconds \n", + " elif(t<=t2 and t>t1):\n", + " Vin.append(-30); #Value of input voltage for time t1 to t2 seconds\n", + " else :\n", + " Vin.append(15); #Value of input voltage after t2 seconds\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(Vin);\n", + "plt.xlim([0,160])\n", + "plt.ylim([-35,20])\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=0):\n", + " vout.append(0); #Diode reverse biased\n", + " else:\n", + " vout.append(v-V_biasing); #Diode forward biased\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout);\n", + "plt.xlim(0,160)\n", + "plt.ylim(-35,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.14 : Page number 492-493" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa23c928d30>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "Vin=[]; #Input voltage waveform, V\n", + "t1=50; #Assumed time interval, s\n", + "t2=100; #Assumed time interval, s\n", + "V_biasing=5; #Biasing voltage, V\n", + "for t in range(0,151): #time interval from 0s to 151s\n", + " if(t<=t1): \n", + " Vin.append(10); #Value of input voltage for time 0 to t1 seconds \n", + " elif(t<=t2 and t>t1):\n", + " Vin.append(-10); #Value of input voltage for time t1 to t2 seconds\n", + " else :\n", + " Vin.append(0);\n", + "\n", + "plt.subplot(211) \n", + "plt.plot(Vin);\n", + "plt.xlim(0,101)\n", + "plt.ylim(-20,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=0):\n", + " vout.append(v); #Diode reverse biased\n", + " else:\n", + " vout.append(v-V_biasing); #Diode forward biased\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout);\n", + "plt.xlim(0,101)\n", + "plt.ylim(-20,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.15 : Page number 493" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa23cd370b8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "Vin=[]; #Input voltage waveform, V\n", + "t1=50; #Assumed time interval, s\n", + "t2=100; #Assumed time interval, s\n", + "V_D1=0.6; #Forward Biasing voltage of the 1st diode, V\n", + "V_D2=0.6; #Forward Biasing voltage of the 2nd diode, V\n", + "for t in range(0,151): #time interval from 0s to 151s\n", + " if(t<=t1): \n", + " Vin.append(10); #Value of input voltage for time 0 to t1 seconds \n", + " elif(t<=t2 and t>t1):\n", + " Vin.append(-10); #Value of input voltage for time t1 to t2 seconds\n", + " else :\n", + " Vin.append(0);\n", + "\n", + "plt.subplot(211);\n", + "plt.plot(Vin);\n", + "plt.xlim(0,110)\n", + "plt.ylim(-20,20)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=0):\n", + " vout.append(-V_D1); #Diode D1 forward biased, \n", + " else:\n", + " vout.append(V_D2); #Diode D2 forward biased\n", + "\n", + "plt.subplot(212) \n", + "plt.plot(vout);\n", + "plt.xlim(0,110)\n", + "plt.ylim(-1,1)\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.16 : Page number 493-494" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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XvVPSHdHt44HfAi8Rupx2kSTCbMYxwFaSviO0mpB0lKQPJC2Nxpp2zbju59Hr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\nLRcZYxRRpdkixwEfVEmUpVi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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa23cac80f0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "from math import sin\n", + "from math import pi\n", + "\n", + "VZ=20; #Assumed zener voltage, V\n", + "VF=0.7; #Assumed forward biasing voltage of the zener diode, V\n", + "Vin=[]; #Input voltage waveform, V\n", + "for t in range(0,(int)(2*pi*10)): #time interval from 0s to 151s\n", + " Vin.append(30*sin(t/10.0));\n", + "\n", + "plt.subplot(211)\n", + "plt.plot(Vin);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=-VF):\n", + " vout.append(-VF); #Zener diode forward biased, \n", + " elif(v>=VZ):\n", + " vout.append(VZ); #Input voltage exceeds zener voltage\n", + " else:\n", + " vout.append(v); #Zener diode reverse biased\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout);\n", + "plt.xlim([0,80])\n", + "plt.ylim([-1,40])\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 18.17 : Page number 494" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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2Bx7Fu5x2lSR8NuM4oJWkz/BWE5J6SnpN0tJkrGm3gud9O3m9V4AvJDVMzp2d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\n2Gwz+OlP4dprfRe+EEL1VbR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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fa23c83bc50>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "from math import sin\n", + "from math import pi\n", + "\n", + "VZ1=20; #Assumed zener voltage, V\n", + "VF1=0.7; #Assumed forward biasing voltage of the zener diode, V\n", + "VZ2=20; #Assumed zener voltage, V\n", + "VF2=0.7; #Assumed forward biasing voltage of the zener diode, V\n", + "Vin=[]; #Input voltage waveform, V\n", + "for t in range(0,(int)(2*pi*10)): #time interval from 0s to 151s\n", + " Vin.append(30*sin(t/10.0));\n", + " \n", + "plt.subplot(211)\n", + "plt.plot(Vin);\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vin(V)');\n", + "plt.title('Input waveform');\n", + "\n", + "\n", + "vout=[]; #Output voltage waveform, V\n", + "for v in Vin[:]: #Loop iterating input voltage \n", + " if(v<=-(VZ1+VF2)):\n", + " vout.append(-(VZ1+VF2)); #Zener diode forward biased, \n", + " elif(v>=VZ2+VF1):\n", + " vout.append(VZ2+VF1); #Input voltage exceeds zener voltage\n", + " else:\n", + " vout.append(v); #Zener diode reverse biased\n", + "plt.subplot(212)\n", + "plt.plot(vout); \n", + "plt.xlim([0,80])\n", + "plt.ylim([-40,40])\n", + "plt.xlabel('t-->');\n", + "plt.ylabel('Vout(V)');\n", + "plt.title('Output waveform');\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_6.ipynb new file mode 100644 index 00000000..f86a8046 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter19_6.ipynb @@ -0,0 +1,1660 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 19 : FIELD EFFECT TRANSISTORS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.1 : Page number 515" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "ID=12[1 + VGS/5]²mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=12.0; #Shorted gate drain current, mA\n", + "V_GS_off=-5.0; #Gate-source cut-off voltage, V\n", + "\n", + "#Result\n", + "print(\"ID=%d[1 + VGS/%d]²mA.\"%(I_DSS,abs(V_GS_off)));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.2 : Page number 515" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain current=6.12mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=32.0; #Shorted gate drain current, mA\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "V_GS=-4.5; #Gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current mA\n", + "\n", + "#Result\n", + "print(\"The drain current=%.2fmA.\"%I_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.3 : Page number 515" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) VGS=-1.76V.\n", + "(ii) VP=6V\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "I_DSS=10.0; #Shorted gate drain current, mA\n", + "V_GS_off=-6.0; #Gate-source cut-off voltage, V\n", + "I_D=5.0; #Drain current mA\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, I_D=I_DSS*[1 - (V_GS/V_GS_off)]²\n", + "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS)); #Gate-source voltage, V\n", + "\n", + "#(ii)\n", + "V_P=-V_GS_off; #Pinch-off voltage, V \n", + "\n", + "#Result\n", + "print(\"(i) VGS=%.2fV.\"%V_GS);\n", + "print(\"(ii) VP=%dV\"%V_P);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.4 : Page number 515-516" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The minimum value of VDD required =10.72V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS_off=-4.0; #Gate-source cut-off voltage, V\n", + "I_DSS=12.0; #Shorted gate drain current, mA\n", + "R_D=560.0; #Drain resistor, Ω\n", + "\n", + "#Calculation\n", + "V_P=-V_GS_off; #Pinch-off voltage, V\n", + "V_DS=V_P; #Minimum drain-source voltage for JFET to be in constant current region, V\n", + "I_D=I_DSS; #Maximum drain current, mA (V_GS=0)\n", + "V_RD=(I_D/1000)*R_D; #Voltage across drain resistor, V (OHM's LAW)\n", + "V_DD=V_DS+V_RD; #Minimum value of supply voltage to drain, V\n", + "\n", + "#Result\n", + "print(\"The minimum value of VDD required =%.2fV.\"%V_DD);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.5 : Page number 516" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain current=1.33mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=3.0; #Shorted gate drain current, mA\n", + "V_GS_off=-6.0; #Gate-source cut-off voltage, V\n", + "V_GS=-2.0; #Gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current mA\n", + "\n", + "#Result\n", + "print(\"The drain current=%.2fmA.\"%I_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.6 : Page number 516" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "p-channel JFET requires a positive gate-to-source voltage to pass drain current.\n", + "More positive voltage, the less the drain current. \n", + "Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VGS_off=4; #Gate-source cut-off voltage, V\n", + "VGS=6; #Gate source voltage, V\n", + "\n", + "print(\"p-channel JFET requires a positive gate-to-source voltage to pass drain current.\");\n", + "print(\"More positive voltage, the less the drain current. \");\n", + "print(\"Any further increase in VGS keeps the JFET cut-off. Therefore, ID=0A.\");" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.7 : Page number 517-518" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The gate to source resistance=15000MΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS=15.0; #Gate-source voltage, V\n", + "I_G=1e-03; #Gate current, μA\n", + "\n", + "#Calculation\n", + "R_GS=(V_GS/(I_G*10**-6))/10**6; #Gate to source resistance, MΩ (OHM's LAW)\n", + "\n", + "#Result\n", + "print(\"The gate to source resistance=%dMΩ.\"%R_GS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.8 : Page number 518" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Transconductance=3000 μ mho\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "V_GS_max=-3.1; #Maximum gate to source voltage, V\n", + "V_GS_min=-3.0; #Minimum gate to source voltage, V\n", + "I_D_max=1.3; #Maximum drain current, mA\n", + "I_D_min=1.0; #Minimum drain current, mA\n", + "\n", + "\n", + "#Calculation\n", + "delta_V_GS=abs(V_GS_max-V_GS_min); #Change in gate to source voltage, V\n", + "delta_I_D=I_D_max-I_D_min; #Change in drain current, mA\n", + "g_fs=(delta_I_D/delta_V_GS)*1000; #Transconductance, μ mho\n", + "\n", + "\n", + "#Result\n", + "print(\"Transconductance=%.0f μ mho\"%g_fs);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.9 : Page number 518" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VGS= 0V 0V -0.2V\n", + "VDS= 7V 15V 15V\n", + "ID = 10V 10.25V 9.65V\n", + "(i) The a.c drain resistance=32kΩ.\n", + "(ii) The transconductance=3000 μ mho.\n", + "(iii) The amplification factor=96.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS=[0,0,-0.2]; #Readings of Gate-source voltage, V\n", + "V_DS=[7,15,15]; #Readings of Drain-source voltage, V\n", + "ID=[10,10.25,9.65]; #Readings of drain current, mA\n", + "\n", + "\n", + "#Displaying the readings:\n", + "print(\"VGS= %dV %dV %.1fV\"%(V_GS[0],V_GS[1],V_GS[2]));\n", + "print(\"VDS= %dV %dV %dV\"%(V_DS[0],V_DS[1],V_DS[2]));\n", + "print(\"ID = %dV %.2fV %.2fV\"%(ID[0],ID[1],ID[2]));\n", + "\n", + "#Calculations\n", + "#(i)\n", + "#V_GS constant at 0V,\n", + "delta_VDS=V_DS[1]-V_DS[0]; #Change in drain-source voltage, V\n", + "delta_ID=ID[1]-ID[0]; #Change in drain current, mA\n", + "rd=delta_VDS/delta_ID; #a.c drain resistance, kΩ\n", + "\n", + "#(ii)\n", + "#V_DS constant at 15V,\n", + "delta_VGS=V_GS[2]-V_GS[1]; #Change in gate-source voltage, V\n", + "delta_ID=ID[2]-ID[1]; #Change in drain current, mA\n", + "g_fs=round((delta_ID/delta_VGS)*1000,); #Transconductance, μ mho\n", + "\n", + "#(iii)\n", + "amplification_factor=rd*1000*g_fs*10**-6; #Amplification factor\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The a.c drain resistance=%dkΩ.\"%rd);\n", + "print(\"(ii) The transconductance=%d μ mho.\"%g_fs);\n", + "print(\"(iii) The amplification factor=%d.\"%amplification_factor );\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.10 : Page number 519" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The transconductance=2500 μS.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "g_mo=4000.0; #Maximum transconductance, μS\n", + "V_GS=-3.0; #Gate to source voltage, V\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "\n", + "#Calculation\n", + "g_m=g_mo*(1-(V_GS/V_GS_off)); #Transconductance, μS\n", + "\n", + "#Result\n", + "print(\"The transconductance=%d μS.\"%g_m);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.11 : Page number 519" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The transconductance=1667 μS.\n", + "The drain current=333 μA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "g_mo=5000.0; #Maximum transconductance, μS\n", + "V_GS=-4.0; #Gate to source voltage, V\n", + "V_GS_off=-6.0; #Gate-source cut-off voltage, V\n", + "I_DSS=3.0; #Shorted-gate drain current, mA\n", + "\n", + "#Calculation\n", + "g_m=g_mo*(1-(V_GS/V_GS_off)); #Transconductance, μS\n", + "I_D=(I_DSS*(1-(V_GS/V_GS_off))**2)*1000; #Drain current μA\n", + "\n", + "\n", + "#Result\n", + "print(\"The transconductance=%.0f μS.\"%g_m);\n", + "print(\"The drain current=%d μA.\"%I_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.12 : Page number 520" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The gate-source voltage=-5V.\n", + "The drain current=2.25mA.\n", + "The drain-source voltage=5.05V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "I_DSS=16.0; #Shorted-gate drain current, mA\n", + "R_D=2.2; #Drain resistor, kΩ\n", + "R_G=1.0; #Gate resistor, MΩ\n", + "V_DD=10.0; #Drain supply voltage, V\n", + "V_GG=-5.0; #Gate supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_GS=V_GG; #Gate-source voltage, V\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current μA\n", + "V_DS=V_DD-I_D*R_D; #Drain-source voltage, V (Kirchhoff's voltage law)\n", + "\n", + "#Result\n", + "print(\"The gate-source voltage=%dV.\"%V_GS);\n", + "print(\"The drain current=%.2fmA.\"%I_D);\n", + "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.13 : Page number 521" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain-source voltage=7.65V.\n", + "The gate-source voltage=-2.35V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_D=5.0; #Drain current mA\n", + "V_DD=15.0; #Drain supply voltage, V\n", + "V_G=0; #Gate voltage, V\n", + "R_D=1.0; #Drain resistor, kΩ\n", + "R_S=470.0; #Source resistor, Ω\n", + "\n", + "\n", + "#Calculation\n", + "V_S=(I_D/1000)*R_S; #Source voltage, V (OHM's LAW)\n", + "V_D=V_DD-I_D*R_D; #Drain voltage, V (Kirchhoff's voltage law)\n", + "V_DS=V_D-V_S; #Drain-source voltage, V\n", + "V_GS=V_G-V_S; #Gate-source voltage, V\n", + "\n", + "#Result\n", + "print(\"The drain-source voltage=%.2fV.\"%V_DS);\n", + "print(\"The gate-source voltage=%.2fV.\"%V_GS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.14 : Page number 521" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The required source resistor=800 Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_GS=-5.0; #Gate-source voltage, V\n", + "I_D=6.25; #Drain current mA\n", + "\n", + "\n", + "#Calculation\n", + "R_S=abs(V_GS/(I_D/1000)); #Required source resistor, Ω (OHM's LAW)\n", + "\n", + "#Result\n", + "print(\"The required source resistor=%d Ω.\"%R_S);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.15 : Page number : 521" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The source resistance=450Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=25.0; #Shorted gate drain current, mA\n", + "V_GS_off=15.0; #Gate-source cut-off voltage, V\n", + "V_GS=5.0; #Gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_D=I_DSS*(1-(V_GS/V_GS_off))**2; #Drain current mA\n", + "R_S=V_GS/(I_D/1000); #Required source resistor, Ω (OHM's LAW)\n", + "\n", + "\n", + "#Result\n", + "print(\"The source resistance=%.0fΩ.\"%R_S);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.16 : Page number 522" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + " RS=313 Ω and RD=800 Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_DSS=15.0; #Shorted gate drain current, mA\n", + "V_GS_off=-8.0; #Gate-source cut-off voltage, V\n", + "V_DD=12.0; #Drain supply voltage,V\n", + "V_D=V_DD/2; #Drain voltage(half of V_DD), V\n", + "\n", + "#Calculation\n", + "I_D=I_DSS/2; #Drain current(approximately half of I_DSS), mA\n", + "V_GS=V_GS_off/3.4; #Gate-source voltage, V\n", + "R_S=abs(V_GS/(I_D/1000)); #Source resistor, Ω (OHM's LAW)\n", + "#Since,V_D=V_DD-I_D*R_D; \n", + "R_D=(V_DD-V_D)/(I_D/1000); #Drain resistor, Ω (OHM's LAW)\n", + "\n", + "#Result\n", + "print(\" RS=%d Ω and RD=%d Ω.\"%(R_S,R_D));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.17 : Page number 522" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The source resistance=0.6 kΩ\n", + "The drain resistance=6 kΩ.\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "I_DSS=5.0; #Shorted gate drain current, mA\n", + "V_GS_off=-2.0; #Gate-source cut-off voltage, V\n", + "V_DS=10.0; #Drain-source voltage,V\n", + "I_D=1.5; #Drain current, mA\n", + "V_DD=20.0; #Drain supply voltage,V\n", + "V_G=0; #Gate voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Drain current, I_D=I_DSS*(1-(V_GS/V_GS_off))**2; \n", + "V_GS=V_GS_off*(1-sqrt(I_D/I_DSS)); #Gate-source voltage, V\n", + "\n", + "#Since, V_GS=V_G-V_S,\n", + "V_S=V_G-V_GS; #Source voltage, V\n", + "\n", + "R_S=V_S/I_D; #Source resistor, kΩ\n", + "\n", + "#Since, V_DD=I_D*R_D +V_DS+ I_D*R_S,\n", + "R_D=(V_DD-I_D*R_S-V_DS)/I_D; #Drain resistor, kΩ\n", + "\n", + "#Calculation\n", + "print(\"The source resistance=%.1f kΩ\"%R_S);\n", + "print(\"The drain resistance=%d kΩ.\"%R_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.18 : Page number 522-523" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The drain-source voltage=17V.\n", + "The gate-source voltage=-0.5V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_DD=30.0; #Drain supply voltage, V\n", + "R_D=5.0; #Drain resistor, kΩ\n", + "I_D=2.5; #Drain current, mA\n", + "R_S=200.0; #Source resistor, Ω\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V_DS=V_DD-I_D*(R_D+(R_S/1000)); #Drain-source voltage, V\n", + "\n", + "#(ii)\n", + "V_GS=-(I_D/1000)*R_S; #Gate-source voltage, V\n", + "\n", + "#Result\n", + "print(\"The drain-source voltage=%dV.\"%V_DS);\n", + "print(\"The gate-source voltage=%.1fV.\"%V_GS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.19 : Page number 523" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "drain voltage of 1st stage=12.37V.\n", + "Source voltage of 1st stage=1.46V.\n", + "drain voltage of 2nd stage=11.7V.\n", + "Source voltage of 2nd stage=2.01V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ID_1=2.15; #First stage drain current, mA\n", + "ID_2=9.15; #Second stage drain current, mA\n", + "VDD=30; #Drain supply voltage, V\n", + "RS_1=0.68; #Source resistance of 1st stage, kΩ\n", + "RS_2=0.22; #Source resistance of 2nd stage, kΩ\n", + "RD_1=8.2; #Drain resistor of 1st stage, kΩ\n", + "RD_2=2; #Drain resistor of 2nd stage, kΩ\n", + "\n", + "#Calculation\n", + "V_RD1=ID_1*RD_1; #Voltage drop across 8.2kΩ\n", + "VD_1=VDD-V_RD1; #Drain voltage of 1st stage, V\n", + "VS_1=ID_1*RS_1; #D.C potential of source of first stage, V\n", + "V_RD2=ID_2*RD_2; #Voltage drop across 2kΩ\n", + "VD_2=VDD-V_RD2; #Drain voltage of 2nd stage, V\n", + "VS_2=ID_2*RS_2; #D.C potential of source of 2nd stage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"drain voltage of 1st stage=%.2fV.\"%VD_1);\n", + "print(\"Source voltage of 1st stage=%.2fV.\"%VS_1);\n", + "print(\"drain voltage of 2nd stage=%.1fV.\"%VD_2);\n", + "print(\"Source voltage of 2nd stage=%.2fV.\"%VS_2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.20 : Page number 524" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=1.52mA.\n", + "Gate-source voltage=-1.2V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=12; #Drain supply voltage, V\n", + "VD=7; #Drain voltage, V\n", + "R1=6.8; #Resistor R1, MΩ\n", + "R2=1; #Resistor R2, MΩ\n", + "RS=1.8; #Source resistance, kΩ\n", + "RD=3.3; #Drain resistor, kΩ\n", + "\n", + "#Calculation\n", + "ID=(VDD-VD)/RD; #Second stage drain current, mA\n", + "VS=ID*RS; #Source voltage, V\n", + "VG=VDD*R2/(R1+R2); #Drain voltage, V\n", + "VGS=VG-VS; #Drain-source voltage, V\n", + "\n", + "#Calculation\n", + "print(\"Drain current=%.2fmA.\"%ID);\n", + "print(\"Gate-source voltage=%.1fV.\"%VGS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.21 : Page number 524-525" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Source resistor, RS=5kΩ.\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "VDD=30; #Drain supply voltage, V\n", + "ID=2.5; #Drain current, mA\n", + "VDS=8; #Drain-source voltage, V\n", + "VGS_off=-5; #Gate-source cutoff voltage, V\n", + "R1=1; #Resistor R1, MΩ\n", + "R2=500; #Resistor R2, kΩ\n", + "IDSS=10; #Shorted gate drain current, mA\n", + "\n", + "#Calculation\n", + "#ID=IDSS*square_of(1-(VGS/VGS_off))\n", + "VGS=VGS_off*(1-sqrt(ID/IDSS)); #Gate-source voltage, V\n", + "V2=VDD*R2/(R1*1000+R2); #Voltage across R2, V\n", + "\n", + "\n", + "#V2=VGS+ID*RS\n", + "RS=(V2-VGS)/ID; #Source resistor, kΩ\n", + "\n", + "#Result\n", + "print(\"Source resistor, RS=%dkΩ.\"%RS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.22 : Page number 528-529" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f70cf070f28>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline \n", + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VDD=20.0; #Drain supply voltage, V\n", + "RS=50.0; #Source resistor, Ω\n", + "RD=150.0; #Drain resistor, Ω\n", + "\n", + "#Calculation\n", + "VDS_max=VDD; #Maximum drain source voltage, V\n", + "ID_max=(VDD/(RD+RS))*1000; #Maximum drain current, mA\n", + "\n", + "\n", + "#plot\n", + "x=[i for i in range(0,(int)(VDS_max+1))]; #Plot variable for V_DS\n", + "y=[(i/(RD+RS))*1000 for i in reversed(x[:])]; #Plot variable for ID\n", + "\n", + "\n", + "plt.plot(x,y);\n", + "plt.xlabel(\"VDS(V)\");\n", + "plt.ylabel(\"ID(mA)\");\n", + "plt.title(\"d.c load line\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.23 : Page number 529" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f70cf06ab38>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline \n", + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VDD=20; #Drain supply voltage, V\n", + "RD=500.0; #Drain resistor, Ω\n", + "\n", + "#Calculation\n", + "VDS_max=VDD; #Maximum drain source voltage, v \n", + "ID_max=(VDD/RD)*1000; #Maximum drain current, mA\n", + "\n", + "#Plot\n", + "x=[i for i in range(0,(int)(VDS_max+1))]; #Plot variable for V_DS\n", + "y=[(i/RD)*1000 for i in reversed(x[:])]; #Plot variable for ID\n", + "\n", + "plt.plot(x,y);\n", + "plt.xlabel(\"VDS(V)\");\n", + "plt.ylabel(\"ID(mA)\");\n", + "plt.title(\"d.c load line\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.24 : Page number 530-531" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain=4.8.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=20; #Drain supply voltage, V\n", + "RD=12.0; #Drain resistor, kΩ\n", + "RL=8.0; #Load resistor, kΩ\n", + "RG=1.0; #Gate resistor, MΩ\n", + "gm=1.0; #transconductance, mA/V\n", + "\n", + "#Calculation\n", + "gm=gm*10**-3; #transconductance, mho\n", + "RAC=(RD*RL)/(RD+RL); #Total a.c load, kΩ\n", + "Av=gm*RAC*1000; #Voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"Voltage gain=%.1f.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.25 : Page number 531" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain=30.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "gm=3000; #transconductance, μmho\n", + "RD=10; #Drain resistance, kΩ\n", + "\n", + "#Calculation\n", + "Av=gm*10**-6*RD*1000; #Voltage gain\n", + "\n", + "#Result\n", + "print(\"Voltage gain=%d.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.26 : Page number 531" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=257mV(r.m.s).\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IDSS=8; #Shorted gate drain current, mA\n", + "VGS_off=-10; #Gate-source cut-off voltage, V\n", + "ID=1.9; #Drain current, mA\n", + "RD=3.3; #Drain resistance, kΩ\n", + "RS=2.7; #Source resistor, kΩ\n", + "vin=100; #Input voltage, mV\n", + "\n", + "#Calculation\n", + "VGS=-ID*RS; #Gate-source voltage, V\n", + "gmo=2*IDSS*10**-3/abs(VGS_off); #Maximum transconductance, S\n", + "gm=gmo*(1-(VGS/VGS_off)); #Transconductance, S\n", + "Av=gm*RD*1000; #Voltage gain\n", + "vout=Av*vin; #Output voltage, mA\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dmV(r.m.s).\"%vout);\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.27 : Page number 531-532" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=151mV(r.m.s).\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RL=4.7; #Load resistor, Ω\n", + "RD=3.3; #Drain resistance, kΩ\n", + "gm=779*10**-6; #Transconductance, S\n", + "vin=100; #Input voltage, mV\n", + "\n", + "\n", + "#Calculation\n", + "RAC=RD*RL/(RD+RL); #Total a.c drain resistance, kΩ\n", + "Av=gm*RAC*1000; #Voltage gain\n", + "vout=Av*vin; #Output voltage, mA\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dmV(r.m.s).\"%vout);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.28 : Page number 532-533" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Voltage gain=1.85.\n", + "Voltage gain, if RS resistor is bypassed=6.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RD=1.5; #Drain resistance, kΩ\n", + "gm=4; #Transconductance, mS\n", + "RS=560; #Source resistance, Ω\n", + "\n", + "#Calculation\n", + "Av=gm*10**-3*RD*1000/(1+gm*10**-3*RS);\n", + "print(\"Voltage gain=%.2f.\"%Av);\n", + "\n", + "#If RS is bypassed by a capacitor\n", + "Av=gm*10**-3*RD*1000;\n", + "print(\"Voltage gain, if RS resistor is bypassed=%d.\"%Av);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.29 : Page number 533" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Voltage gain with RS bypassed=4.155.\n", + "(ii) Voltage gain with RS unbypassed=1.35.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "from math import sqrt\n", + "\n", + "IDSS=10; #Shorted gate drain current, mA\n", + "VGS_off=-3.5; #Gate-source cut-off voltage, V\n", + "RD=1.5; #Drain resistance, kΩ\n", + "RS=750; #Source resistance, Ω\n", + "\n", + "\n", + "#Calculation\n", + "#From d.c biasing\n", + "ID=2.3; #Drain current, mA\n", + "VGS=round(VGS_off*(1-sqrt(ID/IDSS)),1); #Gate-source voltage, V\n", + "gm=round(round((2*IDSS/abs(VGS_off)),1)*round((1-(VGS/VGS_off)),3),2); #Transconductance, mS\n", + "\n", + "\n", + "#(i)\n", + "Av=gm*RD; #Voltage gain with RS resistor bypassed\n", + "print(\"(i) Voltage gain with RS bypassed=%.3f.\"%Av);\n", + "\n", + "#(ii)\n", + "Av=Av/(1+gm*(RS/1000.0));\n", + "print(\"(ii) Voltage gain with RS unbypassed=%.2f.\"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.30 : Page number 539-540" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) n-channel D-MOSFET\n", + "(ii) Drain current=3.91mA\n", + "(iii) Drain current=18.9mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IDSS=10.0; #Shorted gate drain current, mA\n", + "VGS_off=-8.0; #Gate-source cut-off voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "if(VGS_off<0):\n", + " print(\"(i) n-channel D-MOSFET\");\n", + "else:\n", + " print(\"(i) p-channel D-MOSFET\");\n", + " \n", + "\n", + "#(ii)\n", + "VGS=-3.0; #Gate-source voltage, V\n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"(ii) Drain current=%.2fmA\"%ID);\n", + "\n", + "#(iii)\n", + "VGS=3.0; #Gate-source voltage, V\n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"(iii) Drain current=%.1fmA\"%ID);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.31 : Page number 540" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Point 1: VGS=0V and ID=1mA.\n", + "Point 2: VGS=-6V and ID=0mA.\n", + "Point 3: VGS=-3V and ID=0.25mA.\n", + "Point 4: VGS=-1V and ID=0.694mA.\n", + "Point 5: VGS=1V and ID=1.36mA.\n", + "Point 6: VGS=3V and ID=2.25mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "IDSS=1.0; #Shorted gate drain current, mA\n", + "VGS_off=-6.0; #Gate-source cut-off voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#Point 1\n", + "VGS=0; #Gate source voltage, V \n", + "ID=IDSS; #Drain current, mA\n", + "print(\"Point 1: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n", + "\n", + "#Point 2\n", + "VGS=VGS_off; #Gate source voltage, V \n", + "ID=0; #Drain current, mA\n", + "print(\"Point 2: VGS=%dV and ID=%dmA.\"%(VGS,ID));\n", + "\n", + "#locating more points by changing VG values\n", + "VGS=-3; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 3: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n", + "\n", + "VGS=-1; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 4: VGS=%dV and ID=%.3fmA.\"%(VGS,ID));\n", + "\n", + "VGS=1; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 5: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));\n", + "\n", + "VGS=3; #Gate source voltage, V \n", + "ID=IDSS*(1-(VGS/VGS_off))**2; #Drain current mA\n", + "print(\"Point 6: VGS=%dV and ID=%.2fmA.\"%(VGS,ID));" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.32 : Page number 541" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain source voltage=10.6V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=18; #Drain supply voltage, V\n", + "RD=620.0; #Drain resistor, Ω\n", + "IDSS=12.0; #Shorted gate drain current, mA\n", + "VGS_off=-8.0; #Gate-source cut-off voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "ID=IDSS; #Drain current, mA\n", + "VDS=VDD-IDSS*(RD/1000); #Drain source voltage, V\n", + "\n", + "#Result\n", + "print(\"Drain source voltage=%.1fV.\"%VDS);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.33 : Page number 542" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Drain source voltage=7.56V.\n", + "(ii) Output voltage=922mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=15; #Drain supply voltage\n", + "RD=620.0; #Drain resistor, Ω\n", + "RL=8.2; #Load resistor, kΩ\n", + "vin=500.0; #Input voltage, V\n", + "IDSS=12.0; #Shorted gate drain current, mA\n", + "gm=3.2; #Transconductance, mS\n", + "\n", + "#Calculation\n", + "#(i)\n", + "VDS=VDD-IDSS*(RD/1000.0); #Drain source voltage, V\n", + "\n", + "#(ii)\n", + "RAC=RD*RL*1000/(RD+RL*1000); #Total a.c drain resistace, Ω\n", + "vout=(gm/1000.0)*RAC*vin; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"(i) Drain source voltage=%.2fV.\"%VDS);\n", + "print(\"(ii) Output voltage=%dmV\"%vout);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.34 : Page number 545" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=98.7mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ID_on=500.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=1.0; #Threshold value of gate-source voltage, V\n", + "VGS=5; #Gate-source voltage, V\n", + "\n", + "#Calculation\n", + "K=round(ID_on/(VGS_on-VGS_th)**2,2); #Constant for a E-MOSFET, mA/V²\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "\n", + "#Result\n", + "print(\"Drain current=%.1fmA\"%ID);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.35 : Page number 545" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "K=0.061e-03A/V².\n", + "For VGS=5V, Drain current=0.244mA\n", + "For VGS=8V, Drain current=1.525mA\n", + "For VGS=10V, Drain current=2mA\n", + "For VGS=12V, Drain current=4.94mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ID_on=3.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=3.0; #Threshold value of gate-source voltage, V\n", + "\n", + "#Calculation\n", + "K=round((ID_on/(VGS_on-VGS_th)**2),3); #Constant for a E-MOSFET, mA/V²\n", + "print(\"K=%.3fe-03A/V².\"%K);\n", + "\n", + "#Determining different points for plotting\n", + "VGS=5; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=5V, Drain current=%.3fmA\"%ID);\n", + "VGS=8; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=8V, Drain current=%.3fmA\"%ID);\n", + "VGS=10; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=10V, Drain current=%.dmA\"%ID);\n", + "VGS=12; #Gate-source voltage, V\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "print(\"For VGS=12V, Drain current=%.2fmA\"%ID);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.36 : Page number 546-547" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain-source voltage=10.8V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=24.0; #Drain supply voltage, V\n", + "RD=470.0; #Drain resistor, Ω\n", + "R1=100.0; #Resistor R1, kΩ\n", + "R2=15.0; #Resistor R2, kΩ\n", + "ID_on=500.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=1.0; #Threshold value of gate-source voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "VGS=VDD*R2/(R1+R2); #Gate-source voltage, V (Voltage divider rule)\n", + "K=round((ID_on/(VGS_on-VGS_th)**2),2); #Constant for a E-MOSFET, mA/V²\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "VDS=VDD-(ID/1000)*RD; #Drain-source voltage, V\n", + "\n", + "#Result\n", + "print(\"Drain-source voltage=%.1fV.\"%VDS);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.37 : Page number 547" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=10mA.\n", + "Drain-source voltage=10V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=20.0; #Drain supply voltage, V\n", + "RD=1.0; #Drain resistor, kΩ\n", + "RG=5.0; #Gate resistor , MΩ\n", + "ID_on=10.0; #Drain current for MOSFET ON, mA\n", + "\n", + "#Calculation\n", + "#since, VGS=VDS\n", + "ID=ID_on; #Drain current, mA\n", + "VDS=VDD-ID*RD; #Drain-source voltage, V\n", + "\n", + "#Result\n", + "print(\"Drain current=%dmA.\"%ID);\n", + "print(\"Drain-source voltage=%dV.\"%VDS);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 19.38 : Page number 547-548" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Drain current=1.69mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VDD=10.0; #Drain supply voltage, V\n", + "RD=3.0; #Drain resistor, kΩ\n", + "R1=1.0; #Resistor R1, MΩ\n", + "R2=1.0; #Resistor R2, MΩ\n", + "ID_on=10.0; #Drain current for MOSFET ON, mA\n", + "VGS_on=10.0; #Gate-source voltage for MOSFET ON, V\n", + "VGS_th=1.5; #Threshold value of gate-source voltage, V\n", + "\n", + "#Calculation\n", + "K=round((ID_on/(VGS_on-VGS_th)**2),3); #Constant for a E-MOSFET, mA/V²\n", + "VGS=VDD*R2/(R1+R2); #Gate-source voltage, V (Voltage divider rule)\n", + "ID=K*(VGS-VGS_th)**2; #Drain current, mA\n", + "\n", + "#Result\n", + "print(\"Drain current=%.2fmA.\"%ID);\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_6.ipynb new file mode 100644 index 00000000..49519941 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter1_6.ipynb @@ -0,0 +1,646 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:0ac98582dd0b2497034e459e869a2a3bd28001d0d4c4b37a61a8ed5d05f228e3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 1: INTRODUCTION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 1.1: Page Number 8" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "Eg=24.0; #Generated voltage in V\n", + "Ri=0.01; #Internal Resistance in \u03a9\n", + "P=100; #Power supplied in watts\n", + "\n", + "#Calculations\n", + "# (i)\n", + "I=P/Eg; #Load current in A\n", + "V_Ri=I*Ri; #Voltage drop in internal resistance\n", + "\n", + "# (ii)\n", + "V=Eg-(I*Ri); #Terminal Voltage\n", + "\n", + "#Results\n", + "print (\"The voltage drop in internal resistance is %.4f V\"%V_Ri);\n", + "print (\"The terminal voltage is %.2f V\"%V);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage drop in internal resistance is 0.0417 V\n", + "The terminal voltage is 23.96 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 1.2: Page number 10" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Eg=500.0; #Generated voltage in V\n", + "Ri=1000.0; #Internal Resistance in \u03a9\n", + "\n", + "\n", + "#Calculations\n", + "# (i)\n", + "RL=10; #Load resistance of case 1 in \u03a9 \n", + "I= Eg/(RL+Ri); #Load current in A\n", + "\n", + "print(\"The load current for RL=10\u03a9 is %.3f A\"%I);\n", + "\n", + "# (ii)\n", + "RL=50; #Load resistance of case 2 in \u03a9 \n", + "I= Eg/(RL+Ri); #Load current in A\n", + "\n", + "print(\"The load current for RL=50\u03a9 is %.3f A\"%I);\n", + "\n", + "# (iii)\n", + "RL=100; #Load resistance of case 3 in \u03a9 \n", + "I= Eg/(RL+Ri); #Load current in A\n", + "I=round(I,3);\n", + "\n", + "print(\"The load current for RL=100\u03a9 is %.3f A\"%I);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load current for RL=10\u03a9 is 0.495 A\n", + "The load current for RL=50\u03a9 is 0.476 A\n", + "The load current for RL=100\u03a9 is 0.455 A\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3: Page Number 11" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E=10.0; #voltage of voltage source in V\n", + "Ri=10.0; #Internal Resistance of the voltage source in \u03a9\n", + "\n", + "#Calculation\n", + "Isc=E/Ri; #short circuit current in A\n", + "I=Isc; #Current value of current source in A\n", + "R=Ri; #Internal Resistence of the current source in \u03a9\n", + "\n", + "#Results\n", + "print(\"The current value of the current source= %d A\"%Isc);\n", + "print(\"The internal resistance of the current source =%d \u03a9 \"%R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current value of the current source= 1 A\n", + "The internal resistance of the current source =10 \u03a9 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "EXAMPLE 1.4: Page number 11-12" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I=6.0; # current value of current source in mA\n", + "Ri=2000.0; #Internal Resistance of the current source in \u03a9\n", + "\n", + "#Calcultion\n", + "V=(I/1000)*Ri; #Voltage of voltage source in V\n", + "R=Ri; #Internal resistance of voltage source in \u03a9\n", + "\n", + "#Results\n", + "print(\"The voltage of voltage source is %d V\"%V);\n", + "print(\"The internal resistance of the voltage source is %d \u03a9\"%R);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage of voltage source is 12 V\n", + "The internal resistance of the voltage source is 2000 \u03a9\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5: Page number 13\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "E=200.0; #Generated voltage in V\n", + "Ri=100.0; #Internal Resistance of generator in \u03a9\n", + "\n", + "#Calculations\n", + "#(i)\n", + "RL=100; #Load resistance for 1st case in \u03a9\n", + "I=E/(RL+Ri); #Load current in 1st case A\n", + "P=(I*I)*RL; #Power delivered to load of 2nd case in watts\n", + "Pt=(I*I)*(Ri+RL); #Total power generated in watts\n", + "\n", + "print(\"Power delivered for RL=100\u03a9 is %d watts\"%P);\n", + "print(\"Total power generated for RL=100\u03a9 is %d watts\"%Pt);\n", + "\n", + "\n", + "#(ii)\n", + "RL=300; #Load resistance for 2nd case in \u03a9\n", + "I=E/(RL+Ri); #Load current in 2nd case in A\n", + "P=(I*I)*RL; #Power delivered to load of 2nd case in watts\n", + "Pt=(I*I)*(Ri+RL); #Total power generated in watts\n", + "\n", + "print(\"Power delivered for RL=300\u03a9 is %d watts\"%P);\n", + "print(\"Total power generated for RL=300\u03a9 is %d watts\"%Pt);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power delivered for RL=100\u03a9 is 100 watts\n", + "Total power generated for RL=100\u03a9 is 200 watts\n", + "Power delivered for RL=300\u03a9 is 75 watts\n", + "Total power generated for RL=300\u03a9 is 100 watts\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6: Page number 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=12.0; #Output from amplifier in V\n", + "R_out_eq=15; #Equivalent resistance in \u03a9\n", + "\n", + "#Calculations\n", + "RL=R_out_eq; #Load resistance in \u03a9\n", + "Rt=RL+R_out_eq; #Total resistance in \u03a9\n", + "I=V/Rt; #Circuit current in A\n", + "PL=pow(I,2)*RL; #Power delivered to load in W\n", + "\n", + "#Results\n", + "print(\"Load resistance required is = %d \u03a9\"%RL);\n", + "print(\"Power delivered to load = %.1f W\"%PL);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load resistance required is = 15 \u03a9\n", + "Power delivered to load = 2.4 W\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7, Page number 14" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=50.0; #voltage from ac generator in V\n", + "R=100.0; #Resistance of internal impedance in \u03a9\n", + "XL=50.0; #inductive reactance of internal impedance in \u03a9\n", + "\n", + "#Calculation\n", + "Zi=100+(50j); #Internal impedance in complex form (\u03a9)\n", + "ZL=conjugate(Zi); #Load impedance (conjugate of internal impedance ) in \u03a9\n", + "Zt=Zi+ZL; #Total impedance in \u03a9\n", + "I=real(V/Zt); #Circuit current in A\n", + "\n", + "Max_Power=pow(I,2)*R; #Maximum power transferred to the load in watts\n", + "\n", + "\n", + "#Results\n", + "print (\"Load impedance %d %dj \u03a9\"%(real(ZL),imag(ZL)));\n", + "print(\"Maximum power transferred to the load =%.2f W\"%Max_Power);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Load impedance 100 -50j \u03a9\n", + "Maximum power transferred to the load =6.25 W\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8: Page number 16" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Function for calculating parallel resistance\n", + "def pR(R1,R2):\n", + " return((R1*R2)/(R1+R2));\n", + "\n", + "\n", + "#Variable declaration\n", + "E=100.0; #Source voltage in V\n", + "R1=10.0; #Resistance of resistor 1 in \u03a9\n", + "R2=20.0; #Resistance of resistor 2 in \u03a9\n", + "R3=12.0; #Resistance of resistor 3 in \u03a9\n", + "R4=8.0; #Resistance of resistor 4 in \u03a9\n", + "RL=100.0; #Resistance of load in \u03a9\n", + "\n", + "#Calculation\n", + "Req=R1+pR(R3+R4,R2); #Equivalent resistance after removing RL ,in \u03a9\n", + "I=E/Req; #Total circuit current in A\n", + "I8=I*R2/(R2+R3+R4);\n", + "\n", + "#Thevenin's equivalent circuit's parameters\n", + "E0=I8*R4; #Thevenin voltage V\n", + "R0=pR(pR(R1,R2)+R3,R4); #Thevenin resistance \n", + "I_RL=E0/(R0+RL); #Load current in A \n", + "\n", + "#Result \n", + "print (\"Current through load = %.2f A.\"%I_RL);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current through load = 0.19 A.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9: Page number 17" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating parallel resistance\n", + "def pR(R1,R2):\n", + " return((R1*R2)/(R1+R2));\n", + "\n", + "\n", + "#Variable declaration\n", + "V=20.0; #Voltage source in V\n", + "R1=1000.0; #resistance of resistor 1 in \u03a9\n", + "R2=1000.0; #resistance of resistor 2 in \u03a9\n", + "R3=1000.0; #resistance of resistor 3 in \u03a9\n", + "\n", + "#calculation\n", + "#parameter for Thevenin's equivalent circuit\n", + "E0=(V*R3)/(R1+R3); #thevenin voltage in V\n", + "R0=pR(R1,R3)+R2; #Thevenins resistance in \u03a9\n", + "\n", + "#result\n", + "print(\"The thevenin voltage = %d V\"%E0);\n", + "print(\"The thevenin resistance = %d \u03a9\"%R0);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The thevenin voltage = 10 V\n", + "The thevenin resistance = 1500 \u03a9\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10: Page number 18" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=120.0; #Supply voltage in V\n", + "R1=40.0; #Resistor 1's resistance in \u03a9\n", + "R2=20.0; #Resistor 2's resistance in \u03a9\n", + "R3=60.0; #Resistor 3's resistance in \u03a9\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, Thevenin's voltage and resistance are calculated\n", + "E0=(V*R2)/(R1+R2); #Thevenin voltage (voltage across the load resistance RL, after removing RL)in V\n", + "R0=(R1*R2)/(R1+R2) + R3; #Thevenin's resistance (Resistance between the terminals of load RL, with RL removed and source voltage shorted)in \u03a9 \n", + "RL=R0; #Value of load resistance to be connected for maximum power transfer in \u03a9\n", + "Pmax=pow(E0,2)/(4*RL); #Maximum power transferred to load in watts\n", + "\n", + "#Results\n", + "print(\"The value of load resistance RL to which maximum power will be transferred = %.2f \u03a9.\"%RL);\n", + "print(\"The maximum power transferred to load =%.2f W.\"%Pmax);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of load resistance RL to which maximum power will be transferred = 73.33 \u03a9.\n", + "The maximum power transferred to load =5.45 W.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11: Page number 18-19-20" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=80.0; #Supply voltage in V\n", + "R1=100.0; #Resistor 1's resistance in \u03a9\n", + "R2=100.0; #Resistor 2's resistance in \u03a9\n", + "R3=30.0; #Resistor 3's resistance in \u03a9\n", + "R4=80.0; #Resistor 4's resistance in \u03a9\n", + "R5=20.0; #Resistor 5's resistance in \u03a9\n", + "R6=60.0; #Resistor 6's resistance in \u03a9\n", + "R7=20.0; #Resistor 7's resistance in \u03a9\n", + "R8=50.0; #Resistor 8's resistance in \u03a9\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem,\n", + "E0=(V*R2)/(R1+R2); #Thevenin's voltage for the circuit containing V, R1, R2 in V.\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance for R1, R2 in \u03a9.\n", + "\n", + "#Using Thevenin's theorem again on E0, R0 and rest of the circuit resistors.\n", + "E0_1=(E0*R4)/(R0+R3+R4); #Thevenin's voltage for the cicruit containing E0, R0, R3, R4 in V\n", + "R0_1=((R0+R3)*R4)/(R0+R3+R4); #Thevenin's resistance of R0,R3,R4 (R0 and R3 in series and both in parallel with R4), in \u03a9 \n", + "\n", + "#Using Thevenin's theorem again on E0_1, R0_1, and rest of the circuit resistors.\n", + "E0_2=(E0_1*R6)/(R0_1+R5+R6); #Thevenin's voltage for the circuit containing E0_1, R0_1, R5, R6 in V\n", + "R0_2=((R0_1+R5)*R6)/(R0_1+R5+R6); #Thevenin's resistance of R0_1,R5,R6 (R0 and R3 in series and both in parallel with R4), in \u03a9\n", + "\n", + "\n", + "I_50=E0_2/(R0_2+R7+R8); #Current through the 50 \u03a9 resistor in A\n", + "\n", + "\n", + "#Results\n", + "print(\"The current through the 50 \u03a9 resistor =%.1f A.\"%I_50);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through the 50 \u03a9 resistor =0.1 A.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12: Page number 22\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "from math import floor\n", + "#Variable declaration\n", + "V=40.0; #Voltage supply in V\n", + "R1=4.0; #Resistor 1's resistance in \u03a9\n", + "R2=6.0; #Resistor 2's resistance in \u03a9\n", + "R3=5.0; #Resistor 3's resistance in \u03a9\n", + "R4=8.0; #Resistor 4's resistance in \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#Using Norton's theorem,\n", + "#calculating Norton current by removing the load resistance R4 and short circuiting those two terminals of the circuit\n", + "R=R1 + (R2*R3)/(R2+R3); #Load on source after removing R4 resistor, in \u03a9\n", + "I=V/R; #Source current in A\n", + "\n", + "#Using current dividing rule ,calculating the short circuit current.\n", + "I_N=(I*R2)/(R2+R3); #Norton's equivalent current or the short circuit current in A\n", + "\n", + "R_N=R3 + (R1*R2)/(R1+R2); #Norton's equivalent resistance in \u03a9\n", + "\n", + "I_8=(I_N*R_N)/(R_N+R4); #Current through the 8 \u03a9 resistance in A\n", + "\n", + " \n", + "\n", + "#Results\n", + "print(\"The current through the 8\u03a9 resistance =%.2f A.\"%I_8);\n", + "\n", + "#Note: The answer in the book is 1.55 A, but in the above code the approximate value is obtained, i.e not 1.55A but 1.56A\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through the 8\u03a9 resistance =1.56 A.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13 :Page number 23\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V1=30.0; #Voltage source 1, V\n", + "V2=18.0; #Voltage source 2, V\n", + "R1=20.0; #1st resistor, \u03a9\n", + "R2=10.0; #2nd resistor, \u03a9\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "#Finding Thevenin's Equivalent circuit\n", + "I=(V1-V2)/(R1+R2); #Current in the circuit, A\n", + "\n", + "#Applying Kirchhoff's voltage law to 1st loop of the circuit,\n", + "#V1-I*R1-E0=0, where E0 is the voltage across the points X-Y.\n", + "E0=V1-I*R1; #Thevenin's voltage source, V\n", + "\n", + "R0=R1*R2/(R1+R2); #Thevenin's resistance, \u03a9\n", + "\n", + "#Finding Norton's equivalent circuit\n", + "IN=E0/R0; #Norton's equivalent current source, A\n", + "RN=R0; #Norton's equivanlent resistance, \u03a9\n", + "\n", + "#Result\n", + "print(\"IN=%.1fA and RN=%.2f \u03a9\"%(IN,RN));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "IN=3.3A and RN=6.67 \u03a9\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_6.ipynb new file mode 100644 index 00000000..cad31534 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter20_6.ipynb @@ -0,0 +1,677 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:369e36634d005b832372dcae6796c76b979f32b499d8baadce951517f2201533" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 20 : SILICON CONTROLLED RECTIFIERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.2 : Page number 559\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I=50.0; #Surge current, A\n", + "t=12.0; #Time for which surge current lasts, ms\n", + "circuit_fusing_rating_max=90; #Maximum circuit fusing rating, A\u00b2s\n", + "\n", + "#Calculation\n", + "circuit_fusing_rating=I**2*(t*10**-3); #Circuit fusing rating, A\u00b2s\n", + "\n", + "#Result\n", + "if(circuit_fusing_rating<circuit_fusing_rating_max):\n", + " print(\"The device will not be destroyed.\");\n", + "else:\n", + " print(\"The device will be destroyed.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The device will not be destroyed.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.3 : Page number 559\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I2_t_rating=50.0; #circuit fuse rating, A\u00b2s\n", + "Is=100.0; #Surge current, A\n", + "\n", + "#Calculation\n", + "t_max=(I2_t_rating/Is**2)*1000; #Maximum allowable duration, ms\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum allowable duration =%dms\"%t_max);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum allowable duration =5ms\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.4 : Page number 559\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "R=220.0; #Gate resistor, \u03a9\n", + "I_G=7.0; #Gate current, mA\n", + "V_GK=0.7; #Junction voltage, V\n", + "\n", + "#Calculation\n", + "V_in=V_GK+(I_G/1000)*R; #Input voltage, V (Kirchhoff's voltage law)\n", + "\n", + "#Result\n", + "print(\"The required input voltage=%.2fV.\"%V_in);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required input voltage=2.24V.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.5 : Page number 564\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "from math import cos\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "I_G=1.0; #Gate current, mA\n", + "V_m=200.0; #Peak value of input sinusoidal voltage, V\n", + "v=100.0; #Forward breakdown voltage of SCR, V\n", + "R_L=100.0; #Load resistance, \u03a9\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#v=Vm*sin(theta)\n", + "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n", + "theta=asin(v/V_m); #angle for input voltage = forward breakdown voltage, rad\n", + "theta=theta*180/pi; #angle for input voltage = forward breakdown voltage, degrees\n", + "\n", + "alpha=round(theta,0); #Firing angle, degrees\n", + "\n", + "#(ii)\n", + "phi=180-alpha; #Conduction angle, degrees\n", + "\n", + "#(iii)\n", + "V_avg=(V_m/(2*pi))*(1+cos(alpha*pi/180)); #Average voltage, V\n", + "I_avg=V_avg/R_L; #Average current, A\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n", + "print(\"(ii) The conduction angle=%.0f\u00b0\"%phi);\n", + "print(\"(iii) The average current=%.4fA \"%I_avg);\n", + "\n", + "#Note: In the text book has approximated the average current to 0.5925A but in the code it gets approximated to 0.5940A.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The firing agle=30\u00b0\n", + "(ii) The conduction angle=150\u00b0\n", + "(iii) The average current=0.5940A \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.6 : Page number 564\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "from math import cos\n", + "from math import pi\n", + "from math import floor\n", + "\n", + "\n", + "#Variable declaration\n", + "I_G=1.0; #Gate current, mA\n", + "V_m=400.0; #Peak value of input sinusoidal voltage, V\n", + "v=150.0; #Forward breakdown voltage of SCR, V\n", + "R_L=200.0; #Load resistance, \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#v=Vm*sin(theta)\n", + "#Finding angle theta, for input voltage (v)= (V_f)forward_breakdown_voltage\n", + "theta=asin(v/V_m); #angle for input voltage = forward breakdown voltage, rad\n", + "theta=theta*180/pi; #angle for input voltage = forward breakdown voltage, degrees\n", + "\n", + "alpha=theta; #Firing angle, degrees\n", + "\n", + "#(ii)\n", + "V_av=floor((V_m/(2*pi))*(1+cos(alpha*pi/180))*10)/10; #Average voltage, V\n", + "\n", + "#(iii)\n", + "I_av=V_av/R_L; #Average current, A\n", + "\n", + "#(iv)\n", + "P_out=V_av*I_av; #Output power, W\n", + "\n", + "#Result\n", + "print(\"(i) The firing agle=%d\u00b0\"%alpha);\n", + "print(\"(ii) The average output voltage=%.1f V\"%V_av);\n", + "print(\"(iii) The average current=%.3fA \"%I_av);\n", + "print(\"(iv) The output power=%.2f W\"%P_out);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The firing agle=22\u00b0\n", + "(ii) The average output voltage=122.6 V\n", + "(iii) The average current=0.613A \n", + "(iv) The output power=75.15 W\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.7 : Page number 564-565\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import asin\n", + "\n", + "#Variable declaration\n", + "v=180.0; #Forward breakdown voltage, V\n", + "V_m=240.0; #Peak value of input voltage, V\n", + "w=314.0; #Angular frequency of input ,rad/s\n", + "\n", + "\n", + "#Calculation\n", + "#v=Vm*sin(w*t)\n", + "#So, t=asin(v/Vm)/w\n", + "t=(asin(v/V_m)/w)*1000; #Time for which SCR remains off, ms\n", + "\n", + "#Result\n", + "print(\"The SCR remains off for %.1f ms.\"%t);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The SCR remains off for 2.7 ms.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.8 : Page number 565\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import cos\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "I_dc=1.0; #d.c load current, A\n", + "alpha=30.0; #Firing angle, \u00b0\n", + "\n", + "\n", + "#Calculation\n", + "I_av=I_dc; #Average current(= d.c current), A\n", + "\n", + "#Since, Iav=(Vm/(2*pi*RL))*(1+cos(alpha)) and Im=Vm/RL\n", + "I_m=floor((2*pi*I_av/(1+cos(alpha*pi/180)))*100)/100; #Peak-load current, A\n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"Peak-load current=%.2f A.\"%I_m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak-load current=3.36 A.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.9: Page number 565\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "from math import sin\n", + "\n", + "#Variable declaration\n", + "alpha=60.0; #Firing angle, \u00b0\n", + "P=100.0; #Power rating of tungsten lamp, W\n", + "V=110.0; #Voltage rating of tungsten lamp, V\n", + "V_ac=110.0; #a.c supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_m=V_ac*sqrt(2); #Peak value of input voltage, V\n", + "\n", + "alpha=alpha*pi/180; #firing angle, rad\n", + "\n", + "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n", + "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "E_rms=round(V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi))); #r.m.s voltage developed in the lamp, V\n", + "\n", + "\n", + "RL=V**2/P; #Load resistance, \u03a9\n", + "\n", + "I_rms=E_rms/RL; #r.m.s current developed in the lamp, A\n", + "\n", + "#Result\n", + "print(\"The r.m.s voltage developed in the lamp=%d V.\"%E_rms);\n", + "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The r.m.s voltage developed in the lamp=70 V.\n", + "The r.m.s current developed in the lamp=0.58 A.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.10 : Page number 567\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import cos\n", + "from math import pi\n", + "\n", + "#Variable declaration\n", + "RL=100.0; #Load resistance, \u03a9\n", + "V_m=200.0; #Peak a.c voltage, V\n", + "alpha=60; #firing angle, \u00b0\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V_av=(V_m/pi)*(1+cos(alpha*pi/180)); #D.C output voltage, V\n", + "\n", + "#(ii)\n", + "I_av=V_av/RL; #Load current, A\n", + "\n", + "#Result\n", + "print(\"(i) d.c output voltage=%.1f V.\"%V_av);\n", + "print(\"(ii) Load current=%.3f A\"%I_av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) d.c output voltage=95.5 V.\n", + "(ii) Load current=0.955 A\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.11: Page number 567\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import pi\n", + "from math import sin\n", + "\n", + "#Variable declaration\n", + "alpha=60.0; #Firing angle, \u00b0\n", + "P=100.0; #Power rating of tungsten lamp, W\n", + "V=110.0; #Voltage rating of tungsten lamp, V\n", + "V_ac=110.0; #a.c supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_m=round(V_ac*sqrt(2)); #Peak value of input voltage, V\n", + "\n", + "alpha=alpha*pi/180; #firing angle, rad\n", + "\n", + "#Since, E_rms\u00b2=(1/2*pi) \u222b V_m\u00b2sin\u00b2(theta) d(theta), limits: alpha to pi\n", + "# E_rms\u00b2=Vm\u00b2*((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "# E_rms=Vm*sqrt((2*(pi-alpha) + sin(2*alpha))/(8*pi)),\n", + "E_rms=V_m*sqrt((2*(pi-alpha) + sin(2*alpha))/(4*pi)); #r.m.s voltage developed in the lamp, V\n", + "\n", + "RL=V**2/P; #Load resistance, \u03a9\n", + "\n", + "I_rms=E_rms/RL; #r.m.s current developed in the lamp, A\n", + "\n", + "#Result\n", + "print(\"The r.m.s voltage developed in the lamp=%.1f V.\"%E_rms);\n", + "print(\"The r.m.s current developed in the lamp=%.2f A.\"%I_rms);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The r.m.s voltage developed in the lamp=98.9 V.\n", + "The r.m.s current developed in the lamp=0.82 A.\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.12 : Page number 572\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=15; #Suuply voltage, V\n", + "V_T=0.7; #Gate trigger voltage, V\n", + "I_T=7.0; #Gate trigger current, mA\n", + "I_H=6.0; #Holding current. mA\n", + "R_Vin=1; #Resistance at Vin, k\u03a9\n", + "R_VCC=100; #Resistance at Vcc, \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "#(i) when SCR is off, there is no current, therefore no voltage drop across the resistor\n", + "V_out=VCC; #Output voltage, when SCR is off, V\n", + "\n", + "#(ii)\n", + "V_in=V_T+I_T*R_Vin; #Input voltage required to trigger the SCR, V\n", + "\n", + "#(iii)\n", + "#Since, I_H=(Vcc-VT)/R_Vin;\n", + "VCC_SCR_open=(I_H/1000)*R_VCC+V_T; #Decreased value of supply voltage at which SCR opens, V\n", + "\n", + "#Result\n", + "print(\"(i) The output voltage when SCR is off=%dV.\"%V_out);\n", + "print(\"(ii) The input voltage required to trigger the SCR=%.1f V.\"%V_in);\n", + "print(\"(iii) The decreased supply voltage at which SCR opens=%.1f V.\"%VCC_SCR_open);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The output voltage when SCR is off=15V.\n", + "(ii) The input voltage required to trigger the SCR=7.7 V.\n", + "(iii) The decreased supply voltage at which SCR opens=1.3 V.\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.13 : Page number 572-573\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=5.6; #zener voltage, V\n", + "V_T=0.7; #Trigger voltage of SCR, V\n", + "\n", + "#Calculation\n", + "VCC=Vz+V_T; #Required supply voltage to turn on the crowbar, V\n", + "\n", + "#Result\n", + "print(\"The required supply voltage to turn on the crowbar=%.1fV.\"%VCC);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required supply voltage to turn on the crowbar=6.3V.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.14 : Page number 573\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=12; #Zener breakdown voltage, V\n", + "V_T=1.5; #Trigger voltage, V\n", + "tolerance_z=10.0; #Tolerance of zener diode, %\n", + "\n", + "\n", + "#Calculation\n", + "Vz_max=Vz*(1+tolerance_z/100); #Maximum value of zener breakdown, V\n", + "Vz_min=Vz*(1-tolerance_z/100); #Minimum value of zener breakdown, V\n", + "V_crowbar=Vz_max+V_T; #Maximum value of supply voltage for crowbarring, V\n", + "\n", + "#Result\n", + "print(\"The maximum value of supply voltage for crowbarring=%.1fV\"%V_crowbar);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum value of supply voltage for crowbarring=14.7V\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 20.15 : Page number 573\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=25.0; #Supply voltage, V\n", + "\n", + "#Calculation\n", + "#When brights light is on, LASCR conducts and thus gets short circuited to ground, hence,\n", + "V_out=0; #Output voltage, V\n", + "\n", + "print(\"Output voltage when bright light is on=%dV\"%V_out);\n", + "\n", + "\n", + "#When brights light is off, LASCR stops conducting and thus no current through resistor, hence,\n", + "V_out=VCC; #Output voltage, V\n", + "print(\"Output voltage when bright light is off=%dV\"%V_out);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output voltage when bright light is on=0V\n", + "Output voltage when bright light is off=25V\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_6.ipynb new file mode 100644 index 00000000..acca0cfa --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter21_6.ipynb @@ -0,0 +1,467 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:412bf04e25192c77f9fa9664d995cc0ae6446a81f631fb5e0e755ebfa36436bf" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 21 : POWER ELECTRONICS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.3: Page number 585\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_GT=2; #Gate triggering voltage, V\n", + "V_F=0.7; #Forward voltage for diode D1\n", + "\n", + "#Calculation\n", + "#(i)Triggering only by a positive gate voltage,\n", + "#A diode is connected at the gatewith the n-side connected to thegate of the device,\n", + "V_A=V_F+V_GT; #Required voltage to trigger the device, V\n", + "\n", + "print(\"The required voltage to trigger the device only by positive voltage=%.1fV.\"%V_A);\n", + "\n", + "#(ii)\n", + "print(\"In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required voltage to trigger the device only by positive voltage=2.7V.\n", + "In order to trigger the triac only by negative voltage, the direction of diode D1 is reversed.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.4 : Page number 585-586\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=50.0; #Resitor, \u03a9\n", + "V=50.0; #Supply voltage, V\n", + "V_drop=1.0; #Drop across the triac in conduction, V\n", + "\n", + "#Calculation\n", + "#(i) Ideal triac\n", + "#Since the triac is ideal, voltage drop across it is zero,\n", + "I=V/R; #Current through the 50 \u03a9 resistor, A\n", + "\n", + "print(\"(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=%dA.\"%I);\n", + "\n", + "#(ii) Triac has a drop of 1V\n", + "I=(V-V_drop)/R; #Current through the 50 \u03a9 resistor, A\n", + "\n", + "print(\"(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=%.2fA.\"%I);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The cuurent through the 50 \u03a9 resistor when the triac is ideal=1A.\n", + "(ii) The current through the 50 \u03a9 resistor when the triac has a drop of 1V=0.98A.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.5 : Page number 588-589\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_GT=2; #Gate triggering voltage, V\n", + "V_BO=20; #Breakover voltage,V\n", + "\n", + "#Calculation\n", + "print(\"The triggering level is raised by using a diac.\");\n", + "V_A=V_BO+V_GT; #Gate trigger signal, V\n", + "\n", + "#Result\n", + "print(\"In order to turn on the triac, the gate trigger signal=%dV.\"%V_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The triggering level is raised by using a diac.\n", + "In order to turn on the triac, the gate trigger signal=22V.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.6 : Page number 589\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BO=30; #Breakover voltage of diac, V\n", + "V_GT=1; #Trigger voltage of the triac, V\n", + "I_T=10; #Trigger current, mA\n", + "\n", + "\n", + "#Calculation\n", + "V_A=V_BO+V_GT; #Voltage required for triggering the triac, V\n", + "\n", + "#Result\n", + "print(\"The minimum capacitor voltage that will trigger the triac=%d V.\"%V_A);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum capacitor voltage that will trigger the triac=31 V.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.7 : Page number 593\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Variable declaration\n", + "eta=0.6; #Intrinsic stand-off ratio for UJT\n", + "R_BB=10; #Inter-base resistance, k\u03a9\n", + "\n", + "#Calculation\n", + "#Since, RBB=RB1+RB2 and eta=RB1/(RB1+RB2),\n", + "#eta=RB1/RBB.\n", + "R_B1=eta*R_BB; #Resistance of the bar between B1 and emitter junction, k\u03a9\n", + "R_B2=R_BB-R_B1; #Resistance of the bar between B2 and emitter junction, k\u03a9 \n", + "\n", + "#Result\n", + "print(\"Resistance of the bar between B1 and emitter junction=%d k\u03a9.\"%R_B1);\n", + "print(\"Resistance of the bar between B2 and emitter junction=%d k\u03a9.\"%R_B2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance of the bar between B1 and emitter junction=6 k\u03a9.\n", + "Resistance of the bar between B2 and emitter junction=4 k\u03a9.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.8 : Page number 593\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BB=10; #Interbase voltage, V\n", + "eta=0.65; #Intrinsic stand-off ratio for UJT\n", + "V_D=0.7; #Voltage drop in the pn junction, V\n", + "\n", + "#Calculation\n", + "V_stand_off=eta*V_BB; #Stand off voltage, V\n", + "V_P=V_stand_off+V_D; #Peak-point voltage, V\n", + "\n", + "#Result\n", + "print(\"Stand off voltage=%.1f V.\"%V_stand_off);\n", + "print(\"Peak-point voltage=%.1f V.\"%V_P);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Stand off voltage=6.5 V.\n", + "Peak-point voltage=7.2 V.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.9 : Page number 593\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_BB=25; #Interbase voltage, V\n", + "eta_max=0.86; #Maximum intrinsic stand-off ratio for UJT\n", + "eta_min=0.74; #Minimum intrinsic stand-off ratio for UJT\n", + "V_D=0.7; #Voltage drop in the pn junction, V\n", + "\n", + "#Calculation\n", + "V_P_max=eta_max*V_BB+V_D; #Maximum peak-point, V\n", + "V_P_min=eta_min*V_BB+V_D; #Minimum peak-point, V\n", + "\n", + "#Result\n", + "print(\"Maximum peak-point voltage=%.1fV\"%V_P_max);\n", + "print(\"Minimum peak-point voltage=%.1fV\"%V_P_min);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Maximum peak-point voltage=22.2V\n", + "Minimum peak-point voltage=19.2V\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.10 : Page number 593-594\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "eta=0.65; #Intrinsic stand-off ratio for UJT\n", + "R_BB=7.0; #Inter-base resistance, k\u03a9\n", + "R1=100.0; #Resistor R1, \u03a9\n", + "R2=400.0; #Resistor R2, \u03a9\n", + "V_S=12.0; #Source voltage, V\n", + "V_D=0.7; #Voltage drop in the pn junction, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, eta=RB1/RBB,\n", + "R_B1=eta*R_BB; #Resistance of the bar between B1 and emitter junction, k\u03a9\n", + "R_B2=R_BB-R_B1; #Resistance of the bar between B2 and emitter junction, k\u03a9 \n", + "\n", + "print(\"(i) Resistance of the bar between B1 and emitter junction=%.2f k\u03a9.\"%R_B1);\n", + "print(\" Resistance of the bar between B2 and emitter junction=%.2f k\u03a9.\"%R_B2);\n", + "\n", + "#(ii)\n", + "V_B2_B1=V_S*R_BB/(R_BB + (R1/1000) + (R2/1000)); #Voltage across B2-B1, V (voltage divider rule)\n", + "V_P=eta*V_B2_B1+V_D; #Peak-point voltage, V\n", + "\n", + "print(\"(ii) The voltage across the base B2-B1=%.1fV.\"%V_B2_B1);\n", + "print(\" Peak-point voltage=%.2fV\"%V_P);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Resistance of the bar between B1 and emitter junction=4.55 k\u03a9.\n", + " Resistance of the bar between B2 and emitter junction=2.45 k\u03a9.\n", + "(ii) The voltage across the base B2-B1=11.2V.\n", + " Peak-point voltage=7.98V\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.11 : Page number 596\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "#Variable declaration\n", + "RE_initial=5; #Initial value of emitter resistor, k\u03a9\n", + "RE_adjusted=10; #Adjusted value of emitter resistor, k\u03a9\n", + "C=0.2; #Capacitance, \u03bcF\n", + "eta=0.54; #intrinsic stand-off ratio\n", + "\n", + "#Calculation\n", + "#(i)\n", + "t=round((RE_initial*1000*C*10**-6*log(1/(1-eta)))*1000,2); #Time period, ms\n", + "f=(1/t)*1000; #frequency, Hz\n", + "\n", + "print(\"Frequency for 5k\u03a9 setting=%dHz.\"%f);\n", + "\n", + "#(i)\n", + "t=round((RE_adjusted*1000*C*10**-6*log(1/(1-eta)))*1000,2); #Time period, ms\n", + "f=(1/t)*1000; #frequency, Hz\n", + "\n", + "print(\"Frequency for 10k\u03a9 setting=%dHz.\"%f);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Frequency for 5k\u03a9 setting=1282Hz.\n", + "Frequency for 10k\u03a9 setting=645Hz.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 21.12 : Page number 596-597\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "#Variable declaration\n", + "V_S=12; #Supply voltage, V\n", + "R_BB=5; #Interbase resistance, k\u03a9\n", + "R_1=50; #Resistor R1, k\u03a9\n", + "R_2=0.1; #Resistor R2, k\u03a9\n", + "C=0.1; #Capacitance, \u03bcF\n", + "eta=0.6; #intrinsic stand-off ratio\n", + "V_D=0.7; #Voltage drop across pn junction, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "#Since, \u03b7=R_B1/R_BB,\n", + "R_B1=eta*R_BB; #Resitance between base B1 and emitter junction, k\u03a9\n", + "\n", + "#Since, R_BB=R_B1+R_B2\n", + "R_B2=R_BB-R_B1; #Resitance between base B2 and emitter junction, k\u03a9\n", + "\n", + "#(ii)\n", + "V_RB1_R2=V_S*(R_B1+R_2)/(R_BB+R_2); #Voltage drop across R_B1 and R_2 resistors, V\n", + "V_P=V_D+V_RB1_R2; #Peak-point voltage, V\n", + "\n", + "#(iii)\n", + "t=round((R_1*1000*C*10**-6*log(1/(1-eta)))*1000,2); #Time period, ms\n", + "f=(1/t)*1000; #frequency, Hz\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) R_B1=%dk\u03a9 and R_B2=%dk\u03a9\"%(R_B1,R_B2));\n", + "print(\"(ii) The peak-point voltage to turn on the UJT=%.0fV.\"%V_P);\n", + "print(\"(iii) Frequency of oscillations=%dHz.\"%f);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) R_B1=3k\u03a9 and R_B2=2k\u03a9\n", + "(ii) The peak-point voltage to turn on the UJT=8V.\n", + "(iii) Frequency of oscillations=218Hz.\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_6.ipynb new file mode 100644 index 00000000..5f13ea0e --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter22_6.ipynb @@ -0,0 +1,668 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a688629536ad6915939234eacc1ed3eaaf36e0aaa88de5df5b6a309da4d2c64d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 22: ELECTRONIC INSTRUMENTS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.1 : Page number 606\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "I_g=1; #Full scale deflection current, mA\n", + "\n", + "#Calculation\n", + "MS=1/(I_g/1000.0); #Multimeter sensitivity, \u03a9 per volt\n", + "\n", + "#Result\n", + "print(\"The multimeter sensitivity=%d \u03a9 per volt.\"%MS);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The multimeter sensitivity=1000 \u03a9 per volt.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.2 : Page number 606-607\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "meter_sensitivity=1000.0; #Meter sensitivity, \u03a9 per volt\n", + "V_full_scale=50.0; #Full scale volts\n", + "R=50000.0; #Resistance to be measured, \u03a9\n", + "\n", + "#Calculation\n", + "meter_resistance=V_full_scale*meter_sensitivity; #Meter resistance, \u03a9\n", + "R_p=R*meter_resistance/(R+meter_resistance); #Parallel resistance, \u03a9\n", + "\n", + "#Result\n", + "print(\"When the meter is used to measure the voltage across the resistance %d\u03a9, total resistance =%d\u03a9.\"%(R,R_p));\n", + "print(\"\u2234 Meter will give highly incorrect reading.\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "When the meter is used to measure the voltage across the resistance 50000\u03a9, total resistance =25000\u03a9.\n", + "\u2234 Meter will give highly incorrect reading.\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.3 : Page number 607\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "meter_sensitivity=4.0; #Meter sensitivity, k\u03a9/V\n", + "R_1=10.0; #Resistance across which voltage is to be measured, k\u03a9\n", + "R_2=10.0; #Resistance, k\u03a9\n", + "range_max=10.0; #Maximum range of the meter, V\n", + "range_min=0; #Minimum range of the meter, V\n", + "V=20.0; #Battery voltage, V\n", + "\n", + "#Calculation\n", + "R_meter=meter_sensitivity*range_max; #Resistance of the meter, k\u03a9\n", + "R_T=(R_meter*R_1)/(R_1+R_meter) + R_2; #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T,2); #Circuit current, mA\n", + "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter)); #Voltage read by multimeter, V\n", + "\n", + "\n", + "#Result\n", + "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage read by multimeter=8.88V.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.4 : Page number 607-608\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "meter_sensitivity=20.0; #Meter sensitivity, k\u03a9/V\n", + "R_1=10.0; #Resistance across which voltage is to be measured, k\u03a9\n", + "R_2=10.0; #Resistance, k\u03a9\n", + "range_max=10.0; #Maximum range of the meter, V\n", + "range_min=0; #Minimum range of the meter, V\n", + "V=20.0; #Battery voltage, V\n", + "\n", + "#Calculation\n", + "R_meter=meter_sensitivity*range_max; #Resistance of the meter, k\u03a9\n", + "R_T=round((R_meter*R_1)/(R_1+R_meter) + R_2,1); #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T,2); #Circuit current, mA\n", + "V_multimeter=I_circuit*((R_meter*R_1)/(R_1+R_meter)); #Voltage read by multimeter, V\n", + "\n", + "\n", + "#Result\n", + "print(\"Voltage read by multimeter=%.2fV.\"%V_multimeter);\n", + "\n", + "\n", + "#Note: The circuit current=1.0256mA, has been approximated in the text as 1.04mA. But, in the code 1.03 mA has been used. Therefore, the final answer is obtained as 9.81V and not 9.88V.\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage read by multimeter=9.81V.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.5 : Page number 608-609\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "#Variable declaration\n", + "R_A=20.0; #Resistance after point A, k\u03a9\n", + "R_B=20.0; #Resistance after point B, k\u03a9\n", + "R_C=30.0; #Resistance after point C, k\u03a9\n", + "R_D=30.0; #Resistance after point D, k\u03a9\n", + "R_meter=60.0; #Resistance of the meter, k\u03a9\n", + "V=100.0; #Battery voltage, V\n", + "\n", + "#Calculation\n", + "#(i) When meter is not connected:\n", + "R_T=R_A+R_B+R_C+R_D; #Total circuit resistance, k\u03a9\n", + "I_circuit=V/R_T; #Circuit current, mA\n", + "V_A=V; #Voltage at point A, V\n", + "V_B=V-(I_circuit*R_A); #Voltage at point B, V\n", + "V_C=V-(I_circuit*(R_A+R_B)); #Voltage at point C, V\n", + "V_D=V-(I_circuit*(R_T-R_D)); #Voltage at point D, V\n", + "\n", + "print(\"(i) When meter is not connected:\");\n", + "print(\" Voltage at point A=%dV.\"%V_A);\n", + "print(\" Voltage at point B=%dV.\"%V_B);\n", + "print(\" Voltage at point C=%dV.\"%V_C);\n", + "print(\" Voltage at point D=%dV.\"%V_D);\n", + "\n", + "\n", + "#(ii) When meter is connected:\n", + "#(a) Since, point A is directly connected to the source, voltage at point A is equal to source voltage.\n", + "V_A=V; #Voltage at point A, V\n", + "\n", + "#(b)\n", + "R_T_B=R_A + round((R_T-R_A)*R_meter/(R_meter + (R_T-R_A)),2); #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T_B,2); #Circuit current, mA\n", + "V_B=I_circuit*(R_T-R_A)*R_meter/(R_meter + (R_T-R_A)); #Voltage at point B, V\n", + "\n", + "\n", + "#(c)\n", + "R_T_C=(R_A+R_B) + (R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)); #Total circuit resistance, k\u03a9\n", + "I_circuit=V/R_T_C; #Circuit current, mA\n", + "V_C=floor((I_circuit*(R_T-R_A-R_B)*R_meter/(R_meter + (R_T-R_A-R_B)))*10)/10; #Voltage at point C, V\n", + "\n", + "\n", + "\n", + "#(c)\n", + "R_T_D=(R_T-R_D) + R_D*R_meter/(R_meter + R_D); #Total circuit resistance, k\u03a9\n", + "I_circuit=round(V/R_T_D,2); #Circuit current, mA\n", + "V_D=I_circuit*(R_D*R_meter)/(R_meter + R_D); #Voltage at point D, V\n", + "\n", + "\n", + "print(\"(ii) When meter is connected:\");\n", + "print(\" Voltage at point A=%dV.\"%V_A);\n", + "print(\" Voltage at point B=%dV.\"%V_B);\n", + "print(\" Voltage at point C=%.1fV.\"%V_C);\n", + "print(\" Voltage at point D=%.1fV.\"%V_D);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) When meter is not connected:\n", + " Voltage at point A=100V.\n", + " Voltage at point B=80V.\n", + " Voltage at point C=60V.\n", + " Voltage at point D=30V.\n", + "(ii) When meter is connected:\n", + " Voltage at point A=100V.\n", + " Voltage at point B=63V.\n", + " Voltage at point C=42.8V.\n", + " Voltage at point D=22.2V.\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.6 : Page number 614\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=12.0; #Supply voltage, V\n", + "R_m=1.0; #Meter resistance, k\u03a9\n", + "I_m_fsd=2.0; #Full scale deflection of meter current, mA\n", + "beta=80.0; #Base current amplification factor\n", + "E=5.0; #Voltage to be measured, V\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_E=E-V_BE; #Emitter voltage, V\n", + "\n", + "#(i)\n", + "#I_m_fsd=V_E/(R_s+R_m), (OHM's LAW)\n", + "R_s=((V_E/I_m_fsd)-R_m)*1000; #Multiplier resistor, \u03a9\n", + "\n", + "#(ii)\n", + "IB=I_m_fsd/beta; \t\t\t\t#Base current, mA\n", + "R_i=E/IB; \t\t\t#Input resistance of voltmeter, k\u03a9\n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The multiplier resistor=%d\u03a9.\"%R_s);\n", + "print(\"(ii) The voltmeter input resistance=%dk\u03a9\"%R_i);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The multiplier resistor=1150\u03a9.\n", + "(ii) The voltmeter input resistance=200k\u03a9\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.7 : Page number 614\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20; #Supply voltage, V\n", + "Rs_Rm=9.3; #Sum of multipier resistance and meter resistance, k\u03a9\n", + "I_m=1; #Meter current, mA\n", + "beta=100; #Base current amplification factor\n", + "E=10; #Voltage to be measured, V\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "V_E=E-V_BE; #Emitter voltage, V\n", + "I_m=V_E/Rs_Rm; #Meter current, mA\n", + "\n", + "#(ii)\n", + "I_B=I_m/beta; #Base current, mA\n", + "R_i_T=(E/I_B)/1000; #Input resistance of voltmeter, with transistor, M\u03a9\n", + "R_i_WT=Rs_Rm; #Input resistance of voltmeter, without transistor, k\u03a9\n", + "\n", + "#Result\n", + "print(\"(i) The meter current=%dmA\"%I_m);\n", + "print(\"(ii) The input resistance of voltmeter with transistor=%dM\u03a9.\"%R_i_T);\n", + "print(\" The input resistance of voltmeter without transistor=%.1fk\u03a9.\"%R_i_WT);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The meter current=1mA\n", + "(ii) The input resistance of voltmeter with transistor=1M\u03a9.\n", + " The input resistance of voltmeter without transistor=9.3k\u03a9.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.8 : Page number 614-615\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCC=20; #Supply voltage, V\n", + "Rs_Rm=9.3; #Sum of multipier resistance and meter resistance, k\u03a9\n", + "I_m=1; #Meter current, mA\n", + "beta=100; #Base current amplification factor\n", + "E=5; #Voltage to be measured, V\n", + "V_BE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "I_m=(E-V_BE)/Rs_Rm; #Meter current, mA\n", + "\n", + "#Result\n", + "print(\"The meter current=%.2fmA\"%I_m);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The meter current=0.46mA\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.9 : Page number 616\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "I_m_fsd=100.0; #Full scale deflection of meter current, \u03bcA\n", + "R_m=1.0; #Meter resistance, k\u03a9\n", + "V_rms=100.0; #r.m.s voltage to be measured, V\n", + "V_F=0.7; #Forward voltage drop of rectifier diode, V \n", + "\n", + "#Calculation\n", + "V_m=round(sqrt(2)*V_rms,1); #Peak value of applied voltage, V\n", + "V_rectifier_drop=2*V_F; #Total rectifier drop, V\n", + "I_peak=round(I_m_fsd/0.637,2); #Peak f.s.d current, \u03bcA\n", + "R_s=floor(((((V_m-V_rectifier_drop)/(I_peak*10**-6))-(R_m*1000))/1000)*10)/10; #Multiplier resistance, k\u03a9 (OHM's LAW)\n", + "\n", + "\n", + "#Result\n", + "print(\"The multiplier resistance=%.1fk\u03a9.\"%R_s);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The multiplier resistance=890.7k\u03a9.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.10 : Page number 616\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "I_av=75; #Full scale deflection of meter current, \u03bcA\n", + "R_s=708; #Multiplier resistor, k\u03a9\n", + "R_m=900; #Meter coil resistor, \u03a9\n", + "\n", + "#Calculation\n", + "I_peak=I_av*10**-6/0.637; #Peak f.s.d meter current, A\n", + "R_T=R_s*1000+R_m; #Total circuit resistance, \u03a9\n", + "\n", + "#I_peak=(Vm-V_drop)/R_T; (OHM's LAW)\n", + "#And, Vm=sqrt(2)*Vrms\n", + "V_rms=(I_peak*R_T+(2*0.7))/sqrt(2) ; #applied r.m.s voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"The applied r.m.s voltage=%dV\"%V_rms);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The applied r.m.s voltage=60V\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.11 : Page number 618\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deflection_sensitivity=0.01; #Deflection sensitivity, mm/V\n", + "V=400; #Applied voltage, V\n", + "\n", + "#Calculation\n", + "spot_shift=V*deflection_sensitivity; #Spot shift produced, mm\n", + "\n", + "\n", + "#Result\n", + "print(\"The shift produced in the spot=%dmm.\"%spot_shift);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The shift produced in the spot=4mm.\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.12 : Page number 618-619\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deflection_sensitivity=0.03; #Deflection sensitivity, mm/V\n", + "spot_shift=3; #Spot shift produced, mm\n", + "\n", + "\n", + "#Calculation\n", + "#Since, spot_shift=Applied_Voltage*deflection_sensitivity,\n", + "V=spot_shift/deflection_sensitivity; #Applied voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"Applied voltage=%dV.\"%V);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Applied voltage=100V.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.13 : Page number 622\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "deflection=2; #Deflection produced by applied voltage, cm\n", + "V=200; #Applied voltage, V\n", + "deflection_by_another_voltage=3; #Deflection by another voltage, cm\n", + "\n", + "#Calculation\n", + "deflection_sensitivity=V/deflection; #deflection sensitivity, V/cm\n", + "V_unknown=deflection_sensitivity*deflection_by_another_voltage; #Unknown voltage, V\n", + "\n", + "#Result\n", + "print(\"The unknown voltage=%dV.\"%V_unknown);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The unknown voltage=300V.\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 22.14 : Page number 622\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "f_H=1000; #Frequency applied to horizontal plates, Hz\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Loops_H=1; #Number of loops cut by horizontal line\n", + "Loops_V=1; #Number of loops cut by vertical line\n", + "f_V=f_H*(Loops_H/Loops_V); #Unknown frequency, Hz\n", + "\n", + "print(\"(i) Unknown frequency=%dHz.\"%f_V);\n", + "\n", + "#(ii)\n", + "Loops_H=2; #Number of loops cut by horizontal line\n", + "Loops_V=1; #Number of loops cut by vertical line\n", + "f_V=f_H*(Loops_H/Loops_V); #Unknown frequency, Hz\n", + "\n", + "print(\"(ii) Unknown frequency=%dHz.\"%f_V);\n", + "\n", + "#(iii)\n", + "Loops_H=6; #Number of loops cut by horizontal line\n", + "Loops_V=1; #Number of loops cut by vertical line\n", + "f_V=f_H*(Loops_H/Loops_V); #Unknown frequency, Hz\n", + "\n", + "print(\"(iii) Unknown frequency=%dHz.\"%f_V);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Unknown frequency=1000Hz.\n", + "(ii) Unknown frequency=2000Hz.\n", + "(iii) Unknown frequency=6000Hz.\n" + ] + } + ], + "prompt_number": 21 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_6.ipynb new file mode 100644 index 00000000..19741354 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter23_6.ipynb @@ -0,0 +1,133 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:87bd5c8d9448f5bb2e75909f89934a6fb2b64e65e6ea37b085ce080a58026071" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 23 : INTEGRATED CIRCUITS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23.1: Page number 637" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=240; #Adjusted resistance of R2 resistor of LM317 voltage regulator, in kilo ohm\n", + "R2=2.4; #Fixed value of R1 resistor of LM317 voltage regulator, in ohm\n", + "\n", + "#Calculations\n", + "#Output voltage of LM317 voltage regulator IC = 1.25(R2/R1 +1)\n", + "Vout=1.25*((R2*1000)/R1 + 1); #Regulated d.c output voltage for the circuit in V\n", + "\n", + "#Results\n", + "print(\"The regulated d.c output voltage = %.2fV\"%Vout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulated d.c output voltage = 13.75V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23.2 : Page number 638" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R=1.2; #Value of resistance of monostable multivibrator in kilo ohm\n", + "C=0.1; #Value of capacitance of monostable multivibrator in microfarad\n", + "\n", + "#Calculations\n", + "T=1.1*(R*1000)*C; #Time for which the circuit is ON, in microseconds\n", + "\n", + "#Results\n", + "print(\"Time for which the circuit is ON = %d microseconds.\"%T); \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Time for which the circuit is ON = 132 microseconds.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 23.3 : Page number 639\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=3.0; #Resistance of R1 resistor of 555 timer circuit in kilo ohm\n", + "R2=2.7; #Resistance of R2 resistor of 555 timer circuit in kilo ohm\n", + "C=0.033; #Capacitance of the capacitor of 555 timer circuit in microfarad\n", + "\n", + "#Calculations\n", + "f=1.44/(((R1*1000) + 2*(R2*1000))*(C*pow(10,-6))); #Frequency of the circuit in Hz\n", + "f=f/1000; #Frequency of the circuit in kHz\n", + "\n", + "#Results\n", + "print(\"The frequency of the circuit = %.2fkHz\"%f);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The frequency of the circuit = 5.19kHz\n" + ] + } + ], + "prompt_number": 3 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_6.ipynb new file mode 100644 index 00000000..e63c17a6 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter24_6.ipynb @@ -0,0 +1,604 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:947f358cf49d029c94d008f72a340051744678cf2e36ecc199100b78d31fcba5" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 24 : HYBRID PARAMETERS" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.1 : Page number 644-645\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=10.0; #1st resistor, \u03a9\n", + "R2=5.0; #2nd resistor, \u03a9\n", + "\n", + "\n", + "#Calculation\n", + "print(\"To find h11 and h21, output terminals are shorted.\");\n", + "h11=R1; #Input impedance with output shorted, \u03a9\n", + "\n", + "print(\"h11=%d\u03a9.\"%h11);\n", + "\n", + "print(\"Output current flowing into the box= input current flowing out of the box.\");\n", + "print(\"i2=-i1\"); #Output current flowing into the box= input current flowing out of the box.\n", + "print(\"h21=i2/i1 = -i1/i1= -1.\"); #Current gain with output shorted.\n", + "\n", + "\n", + "print(\"For finding h22 and h12, voltage source is connected at the output\");\n", + "#As, there will be no current through 10k\u03a9 resistor due to open circuited input,\n", + "print(\"v1=v2\"); #Output voltage is equal to input voltage(equal to voltage drop across 5k\u03a9 resistor)\n", + "print(\"h12=v1/v2 = v2/v2 = 1\"); #Voltage feedback ratio with input terminals open\n", + "\n", + "h22=1/R2; #Output admittance, mho\n", + "print(\"h22=%.1f mho\"%h22);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "To find h11 and h21, output terminals are shorted.\n", + "h11=10\u03a9.\n", + "Output current flowing into the box= input current flowing out of the box.\n", + "i2=-i1\n", + "h21=i2/i1 = -i1/i1= -1.\n", + "For finding h22 and h12, voltage source is connected at the output\n", + "v1=v2\n", + "h12=v1/v2 = v2/v2 = 1\n", + "h22=0.2 mho\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.2 : Page number 645-646\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=4.0; #1st resistor(at the input side), \u03a9\n", + "R2=4.0; #2nd resistor(at the middle), \u03a9\n", + "R3=4.0; #3rd resistor(at the output side), \u03a9\n", + "\n", + "#Calculation\n", + "print(\"To find h11 and h21, output terminals are shorted.\");\n", + "h11=R1 + (R2*R3/(R2+R3)); #Input impedance with output shorted, \u03a9\n", + "print(\"h11=%d\u03a9.\"%h11);\n", + "\n", + "#As the input current gets divided in half due to R2=R3.\n", + "print(\"Output current flowing into the box=negative of half of input current flowing out of the box.\");\n", + "print(\"i2=-i1/2 = -0.5i1\"); \n", + "print(\"h21=i2/i1 = -0.5i1/i1= -0.5.\"); #Current gain with output shorted.\n", + "\n", + "print(\"For finding h22 and h12, voltage source is connected at the output\");\n", + "#As, there will be no current through the 1st 4k\u03a9 resistor due to open circuited input,\n", + "#Voltage gets equally divided across R2 and R3 resistor\n", + "print(\"v1=v2/2 = 0.5v2\"); #Input voltage is equal to half of input voltage\n", + "print(\"h12=v1/v2 = 0.5v2/v2 = 0.5\"); #Voltage feedback ratio with input terminals open\n", + "\n", + "h22=1/(R2+R3); #Output admittance, mho\n", + "print(\"h22=%.3f mho\"%h22);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "To find h11 and h21, output terminals are shorted.\n", + "h11=6\u03a9.\n", + "Output current flowing into the box=negative of half of input current flowing out of the box.\n", + "i2=-i1/2 = -0.5i1\n", + "h21=i2/i1 = -0.5i1/i1= -0.5.\n", + "For finding h22 and h12, voltage source is connected at the output\n", + "v1=v2/2 = 0.5v2\n", + "h12=v1/v2 = 0.5v2/v2 = 0.5\n", + "h22=0.125 mho\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.3 ; Page number 649-650\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R1=10.0; #Resistor at the input side, \u03a9\n", + "R2=5.0; #Resistor at the middle, \u03a9\n", + "rL=5.0; #Load resistor, \u03a9\n", + "\n", + "#h-parameter values from 24.1\n", + "h11=10.0; #Input impedance with output shorted, \u03a9\n", + "h21=-1.0; #Current gain with output shorted\n", + "h12=1.0; #Voltage feedback ratio with input terminal open\n", + "h22=0.2; #Output admittance, mho\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Zin=h11-(h12*h21/(h22+(1/rL))); #Input impedance, \u03a9\n", + "\n", + "#(ii)\n", + "Av=-h21/(Zin*(h22+(1/rL))); #voltage gain,\n", + "\n", + "#Result\n", + "print(\"(i) The input impedance=%.1f\u03a9.\"%Zin );\n", + "print(\"(ii) The voltage gain=1/%d.\"%(1/Av));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) The input impedance=12.5\u03a9.\n", + "(ii) The voltage gain=1/5.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.4 : Page number 652-653\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "VCE=10.0; #Collector-emitter voltage, V\n", + "IC=1.0; #Collector current, mA\n", + "rL=600.0; #a.c load seen by the transistor,\u03a9\n", + "\n", + "#h-parameters\n", + "hie=2000.0; #Input impedance with output shorted, \u03a9\n", + "hoe=10**-4; #Output impedance, mho\n", + "hre=10**-3; #Voltage feedback ratio with input terminal open\n", + "hfe=50.0; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Zin=hie - (hre*hfe/(hoe+(1/rL))); #Input impedance, \u03a9\n", + "print(\"Input impedance=%.0f \u03a9. \\n As second term in the expression of Zin is small compared to first, Zin~hie=%d\u03a9.\"%(Zin,hie));\n", + "\n", + "#(ii)\n", + "Ai=hfe/(1+hoe*rL); #Current gain\n", + "print(\"Current gain=%d\"%Ai);\n", + "print(\"if hoe*rL<<1, then Ai~hfe=%d.\"%hfe);\n", + "\n", + "#(iii)\n", + "Av=-hfe/(Zin*(hoe+(1/rL))); #Voltage gain\n", + "print(\"Voltage gain=%.1f\"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input impedance=1972 \u03a9. \n", + " As second term in the expression of Zin is small compared to first, Zin~hie=2000\u03a9.\n", + "Current gain=47\n", + "if hoe*rL<<1, then Ai~hfe=50.\n", + "Voltage gain=-14.4\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.5 : Page number 653\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import ceil\n", + "\n", + "#Variable declaration\n", + "VCE=5.0; #Collector-emitter voltage, V\n", + "IC=1.0; #Collector current, mA\n", + "rL=2.0; #a.c load seen by the transistor,\u03a9\n", + "\n", + "\n", + "#h-parameters\n", + "hie=1700.0; #Input impedance with output shorted, \u03a9\n", + "hoe=6*10**-6; #Output impedance, mho\n", + "hre=1.3*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=38.0; #Current gain with output shorted\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Zin=hie - (hre*hfe/(hoe+(1/(rL*1000)))); #Input impedance, \u03a9\n", + "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n", + "\n", + "#(ii)\n", + "Ai=ceil((hfe/round((1+hoe*rL*1000),3))*10)/10; #Current gain\n", + "print(\"Current gain=%.1f\"%Ai);\n", + "\n", + "#(iii)\n", + "Av=-hfe/(Zin*(hoe+(1/(rL*1000)))); #Voltage gain\n", + "print(\"Voltage gain=%.1f\"%abs(Av));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input impedance=1690 \u03a9.\n", + "Current gain=37.6\n", + "Voltage gain=44.4\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.6 : Page number 653-654\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return r1*r2/(r1+r2);\n", + "\n", + "\n", + "#Variable declaration\n", + "RC=10.0; #Collector resistance, k\u03a9\n", + "RL=30.0; #Load resistance, k\u03a9\n", + "R1=80.0; #Resistor R1, k\u03a9\n", + "R2=40.0; #Resistor R2, k\u03a9\n", + "\n", + "#h-parameters\n", + "hie=1500.0; #Input impedance with output shorted, \u03a9\n", + "hoe=5*10**-5; #Output impedance, mho\n", + "hre=4*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=50.0; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "rL=((RC*RL)/(RC+RL))*1000; #a.c load as seen by resistance, \u03a9\n", + "\n", + "#(i)\n", + "Zin=round(hie - (hre*hfe/(hoe+(1/rL))),-1); #Input impedance, \u03a9\n", + "print(\"Input impedance=%.0f \u03a9.\"%Zin);\n", + "\n", + "#Input impedance of stage=input impedance || bias resistors\n", + "Zin_stage=round(pr(pr(R1,R2)*1000,Zin),-1); #\u03a9\n", + "print(\"Input impedance of the stage=%.0f \u03a9.\"%Zin_stage);\n", + "\n", + "#(ii)\n", + "Av=-hfe/(Zin*(hoe+(1/rL))); #Voltage gain\n", + "print(\"Voltage gain=%d\"%Av);\n", + "print(\"The negative sign represents phase reversal.\");\n", + "\n", + "\n", + "#(iii)\n", + "Zout=(1/(hoe-(hfe*hre/hie)))/1000; #Output impedance of transistor, k\u03a9\n", + "Zout_stage=pr(Zout,pr(RL,RC)); #Output impedance of the stage, k\u03a9\n", + "print(\"Output impedance=%.2f k\u03a9.\"%Zout);\n", + "print(\"Output impedance of the stage=%.2f k\u03a9.\"%Zout_stage);\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input impedance=1390 \u03a9.\n", + "Input impedance of the stage=1320 \u03a9.\n", + "Voltage gain=-196\n", + "The negative sign represents phase reversal.\n", + "Output impedance=27.27 k\u03a9.\n", + "Output impedance of the stage=5.88 k\u03a9.\n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.7 : Page number 654\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return r1*r2/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "RC=4.7; #Collector resistance, k\u03a9\n", + "RL=10.0; #Load resistance, k\u03a9\n", + "R1=33.0; #Resistor R1, k\u03a9\n", + "R2=10.0; #Resistor R2, k\u03a9\n", + "\n", + "#h-parameters\n", + "hie=1; #Input impedance with output shorted, k\u03a9\n", + "hoe=25; #Output impedance, \u03bcS\n", + "hre=2.5*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=50; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "rL=(RC*RL)/(RC+RL); #a.c load as seen by resistance, k\u03a9\n", + "\n", + "Ai=hfe/(1+hoe*10**-6*rL*1000); #Current gain\n", + "print(\"Current gain=%.1f\"%Ai);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current gain=46.3\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.8 : Page number 654-655\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "R_S=100.0; #Series resistance, \u03a9 \n", + "\n", + "#h-parameters\n", + "hie=1.0; #Input impedance with output shorted, k\u03a9\n", + "hoe=25.0; #Output impedance, \u03bcS\n", + "hre=2.5*10**-4; #Voltage feedback ratio with input terminal open\n", + "hfe=50.0; #Current gain with output shorted\n", + "\n", + "\n", + "#Calculation\n", + "Zout=(1/(hoe*10**-6-(hfe*hre/(hie*1000+R_S))))/1000; #Output impedance of transistor, k\u03a9\n", + "print(\"Output impedance=%.1f k\u03a9.\"%Zout);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Output impedance=73.3 k\u03a9.\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.9 : Page number 656\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import floor\n", + "\n", + "#Function for calculating parallel resistance\n", + "def pr(r1,r2):\n", + " return r1*r2/(r1+r2);\n", + "\n", + "#Variable declaration\n", + "RC=12.0; #Collector resistance, k\u03a9\n", + "RL=15.0; #Load resistance, k\u03a9\n", + "R1=50.0; #Resistor R1, k\u03a9\n", + "R2=5.0; #Resistor R2, k\u03a9\n", + "hie=1.94; #Input impedance with output shorted, k\u03a9\n", + "hfe=71.0; #Current gain with output shorted\n", + "\n", + "\n", + "\n", + "#Calculation\n", + "rL=(RC*RL)/(RC+RL); #a.c load as seen by resistance, \u03a9\n", + "\n", + "#(i)\n", + "Zin_base=hie; #Transistor input impedance, k\u03a9\n", + "Zin_circuit=floor(pr(Zin_base,pr(R1,R2))*100)/100; #Circuit input impedance, k\u03a9\n", + "print(\"Circuit input impedance=%.2fk\u03a9\"%Zin_circuit);\n", + "\n", + "\n", + "#(ii)\n", + "Av=hfe*rL/hie; #Voltage gain\n", + "print(\"Voltage gain=%.0f\"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Circuit input impedance=1.35k\u03a9\n", + "Voltage gain=244\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.10 : Page number 656\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "hie_min=600; #Minimum input impedance with output shorted, \u03a9\n", + "hfe_min=110; #Minimum current gain with output shorted\n", + "hie_max=800; #Maximum input impedance with output shorted, \u03a9\n", + "hfe_max=140; #Maximum current gain with output shorted\n", + "rL=460; #a.c collector load, \u03a9\n", + "\n", + "#Calculation\n", + "hie=round(sqrt(hie_min*hie_max)); #Input impedance with output shorted, \u03a9\n", + "hfe=round(sqrt(hfe_min*hfe_max)); #Current gain with output shorted\n", + "Av=hfe*rL/hie; #Voltage gain\n", + "\n", + "#Result\n", + "print(\"Voltage gain=%.1f\"%Av);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage gain=82.3\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 24.11 : Page number 658-659\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#(a)Variable declaration\n", + "Ib=10; #Base current, \u03bcA\n", + "Ic=1; #Collector current, mA\n", + "Vbe=10; #Base-emitter voltage, mV\n", + "\n", + "#Calculation\n", + "hie=Vbe*10**-3/(Ib*10**-6); #Input impedance with output shorted, \u03a9\n", + "hfe=Ic*10**-3/(Ib*10**-6); #Current gain with output shorted\n", + "\n", + "#(b) Variable declaration\n", + "Vbe=0.65; #Base-emitter voltage, mV\n", + "Ic=60; #Collector current, \u03bcA\n", + "Vce=1; #Collector-emitter voltage, V\n", + "\n", + "#Calculation\n", + "hre=Vbe*10**-3/Vce; #Voltage feedback ratio with input terminal open\n", + "hoe=Ic/Vce; #Output impedance, \u03bcmho\n", + "\n", + "\n", + "#Result\n", + "print(\"hie=%d\u03a9\"%hie);\n", + "print(\"hfe=%d\"%hfe);\n", + "print(\"hre=%.2fe\u201303\"%(hre*1000));\n", + "print(\"hoe=%d\u03bcmho\"%hoe);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "hie=1000\u03a9\n", + "hfe=100\n", + "hre=0.65e\u201303\n", + "hoe=60\u03bcmho\n" + ] + } + ], + "prompt_number": 20 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_6.ipynb new file mode 100644 index 00000000..69b193da --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter25_6.ipynb @@ -0,0 +1,2511 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 25 : OPERATIONAL AMPLIFIERS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.1: Page number 664" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage of the differential amplifier = 10V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A=100.0; #Open-circuit voltage gain of differential amplifier\n", + "V1=3.25; #Input voltage to terminal 1 in V\n", + "V2=3.15; #Input voltage to terminal 2 in V\n", + "\n", + "#Calculations\n", + "V0=A*(V1-V2); #Output voltage in V\n", + "\n", + "#Results\n", + "print(\"The output voltage of the differential amplifier = %dV\"%V0);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.2: Page number 672" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio = 10000.\n", + "The common mode rejection ratio in decibels= 80dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_DM=2000.0; #Differential mode voltage gain\n", + "A_CM=0.2; #Common mode voltage gain\n", + "\n", + "#Calculations\n", + "CMRR=A_DM/A_CM; #Common mode rejection ratio\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in dB\n", + "\n", + "\n", + "#Results\n", + "print(\"The common mode rejection ratio = %d.\"%CMRR);\n", + "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.3: Page number 672" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The common mode rejection ratio in decibels= 46dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "VD_in=10.0; #Differential mode input in mV\n", + "VD_out=1.0; #Output for differential mode input in V\n", + "VC_in=10.0; #Common mode input in mV\n", + "VC_out=5.0; #Output for common mode input in mV\\\n", + "\n", + "#Calculations\n", + "A_DM=(VD_out*1000)/VD_in; #Differntial mode voltage gain\n", + "A_CM=VC_out/VC_in; #Common mode voltage gain\n", + "CMRR=A_DM/A_CM; #Common mode rejection ratio\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in dB\n", + "\n", + "#Results\n", + "print(\"The common mode rejection ratio in decibels= %ddB.\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.4: Page number 672" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage =7.5V\n", + "Noise on output = 4.7x10^-6V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_DM=150.0; #Differential mode voltage gain\n", + "CMRR_dB=90.0; #Common mode rejection ratio\n", + "V1=100.0; #Input voltage for terminal 1 in mV\n", + "V2=50.0; #Input voltage for terminal 2 in mV\n", + "V_noise=1.0; #Voltage of noise signal in mV\n", + "\n", + "#Calculation\n", + "\n", + "#Case(i)\n", + "V_out=A_DM*(V1-V2)/1000.0; #Output voltage for differntial mode input, in V\n", + "\n", + "#Since CMRR_dB=20*log10(differential mode gain/common mode gain),\n", + "A_CM=A_DM/pow(10,(CMRR_dB/20)); #Common mode gain\n", + "V_OUT_noise=A_CM*(V_noise/1000); #Noise on output in V\n", + "\n", + "\n", + "#Results\n", + "print(\"Output voltage =%.1fV\"%V_out);\n", + "print(\"Noise on output = %.1fx10^-6V\"%(V_OUT_noise*pow(10,6)));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.5 : Page number 672-673" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common mode gain =0.083\n", + "Common mode rejection ratio in decibels=89.5dB\n", + "r.m.s output signal =1.25V\n", + "r.m.s interfernce output voltage = 83mV\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "A_DM=2500.0; #Differential mode voltage gain\n", + "CMRR=30000.0; #Common mode rejection ratio\n", + "Input_signal=500.0; #Single ended input r.m.s signal in microvolts\n", + "Interference=1.0; #Interference signal, in V\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "A_CM=A_DM/CMRR; #Common mode gain\n", + "\n", + "#(ii)\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in decibels\n", + "\n", + "#(iii)\n", + "V_out=A_DM*(Input_signal/pow(10,6)-0); #r.m.s output signal in V\n", + "\n", + "#(iv)\n", + "Interference_out=A_CM*Interference; #r.m.s interference output in V\n", + "Interference_out=Interference_out*1000; #r.m.s interference output in mV\n", + "\n", + "\n", + "#Results\n", + "print(\"Common mode gain =%.3f\"%A_CM);\n", + "print(\"Common mode rejection ratio in decibels=%.1fdB\"%CMRR_dB);\n", + "print(\"r.m.s output signal =%.2fV\"%V_out);\n", + "print(\"r.m.s interfernce output voltage = %dmV\"%Interference_out);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.6 : Page number 674-675" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VE=-0.7V\n", + "IE=0.452mA\n", + "IE1=0.226mA\n", + "IE2=0.226mA\n", + "IC1=0.226mA\n", + "IC2=0.226mA\n", + "IB1=2.26μA\n", + "IB2=2.26μA\n", + "VC1=12V\n", + "VC2=9.7V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RB=10; #Base resistor, kΩ\n", + "RC2=10; #Collector resistor, kΩ\n", + "RE=25; #Emitter resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100; #Base amplification factor\n", + "\n", + "#Calculation\n", + "VE=-VBE; #Emitter voltage, V (Ignoring the base current)\n", + "IE=(VEE-VBE)/RE; #Tail current, mA\n", + "IE1=IE/2; #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "IC1=IE1; #Collector current(= emitter current) of 1st transistor, mA\n", + "IC2=IC1; #Collector current of 2nd transistor, mA\n", + "IB1=(IC1/beta)*1000; #Base current of 1st transistor, μA\n", + "IB2=IB1; #Base current of 2nd transistor, μA\n", + "VC1=VCC; #Collector voltage of 1st transistor, V\n", + "VC2=VCC-IC2*RC2; #Collector voltage of 2nd transistor, V\n", + "\n", + "#Result\n", + "print(\"VE=%.1fV\"%VE);\n", + "print(\"IE=%.3fmA\"%IE);\n", + "print(\"IE1=%.3fmA\"%IE1);\n", + "print(\"IE2=%.3fmA\"%IE2);\n", + "print(\"IC1=%.3fmA\"%IC1);\n", + "print(\"IC2=%.3fmA\"%IC2);\n", + "print(\"IB1=%.2fμA\"%IB1);\n", + "print(\"IB2=%.2fμA\"%IB2);\n", + "print(\"VC1=%dV\"%VC1);\n", + "print(\"VC2=%.1fV\"%VC2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.7 : Page number 675" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output voltage=7.85V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15; #Collector supply voltage, V\n", + "VEE=15; #Emitter supply voltage, V\n", + "RB=33; #Base resistor, kΩ\n", + "RC=15; #Collector resistor, kΩ\n", + "RE=15; #Emitter resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "IE_tail=(VEE-VBE)/RE; #Tail current, mA\n", + "IE=round(IE_tail/2,3); #Emitter current in each transistor, mA\n", + "IC=IE; #Collector current(=emitter current), mA\n", + "Vout=VCC-IC*RC; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"The output voltage=%.2fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.8 : Page number 675" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) IB1=5.56μA\n", + " IB2=4.55μA\n", + "(ii) VB1=-0.183V\n", + " VB2=-0.15V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VEE=15.0; #Emitter supply voltage, V\n", + "RB=33.0; #Base resistor, kΩ\n", + "RC=15.0; #Collector resistor, kΩ\n", + "RE=15.0; #Emitter resistor, kΩ\n", + "VBE=0; #Base-emitter voltage, V\n", + "beta_dc_l=90.0; #base current amplification factor for left transistor\n", + "beta_dc_r=110.0; #base current amplification factor for right transistor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IE_tail=(VEE-VBE)/RE; #Tail current, mA\n", + "IE=IE_tail/2; #Emitter current in each transistor, mA\n", + "IB1=(IE/beta_dc_l)*1000; #Base current of 1st transistor, μA\n", + "IB2=(IE/beta_dc_r)*1000; #Base current of 2nd transistor, μA\n", + "\n", + "#(ii)\n", + "VB1=-IB1/1000*RB; #Base voltage of 1st transistor, V\n", + "VB2=-IB2/1000*RB; #Base voltage of 1st transistor, V\n", + "\n", + "#Result\n", + "print(\"(i) IB1=%.2fμA\"%IB1);\n", + "print(\" IB2=%.2fμA\"%IB2);\n", + "print(\"(ii) VB1=%.3fV\"%VB1);\n", + "print(\" VB2=%.2fV\"%VB2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.9 : Page number 675-676" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VE=-0.7V\n", + "Emitter current in each transistor=0.5mA.\n", + "IC1~IE1=0.5mA and IC2~IE2=0.5mA\n", + "VC1=VC2=10V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VEE=15.0; #Emitter supply voltage, V\n", + "RB=10.0; #Base resistor, kΩ\n", + "RC1=10.0; #Collector resistor of 1st transistor, kΩ\n", + "RC2=10.0; #Collector resistor of 2nd transistor, kΩ\n", + "IE=1.0; #Tail current, mA\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "VE=-VBE; #Emitter voltage, V (Ignoring the base current)\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "IC1=IE1; #Collector current(= emitter current) of 1st transistor, mA\n", + "IC2=IE2; #Collector current of 2nd transistor, mA\n", + "VC1=VCC-IC1*RC1; #Collector voltage of 1st transistor, V\n", + "VC2=VCC-IC2*RC2; #Collector voltage of 2nd transistor, V\n", + "\n", + "\n", + "#Result\n", + "print(\"VE=%.1fV\"%VE);\n", + "print(\"Emitter current in each transistor=%.1fmA.\"%(IE/2.0));\n", + "print(\"IC1~IE1=%.1fmA and IC2~IE2=%.1fmA\"%(IE1,IE2));\n", + "print(\"VC1=VC2=%dV.\"%VC2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.10 : Page number 676-677" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VE=0.7V\n", + "Tail current=0.452mA.\n", + "Emitter current in each transistor=0.226mA.\n", + "IC1~IE1=0.226mA and IC2~IE2=0.226mA\n", + "VC1=-12V\n", + "VC2=-9.74V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "VEE=12.0; #Emitter supply voltage, V\n", + "RC2=10.0; #Collector resistor of 2nd transistor, kΩ\n", + "RE=25.0; #Emitter current, kΩ\n", + "VBE=-0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "VE=-VBE; #Emitter voltage, V (Ignoring the base current)\n", + "IE=(VCC-VE)/RE; #Tail current, mA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "IC1=IE1; #Collector current(= emitter current) of 1st transistor, mA\n", + "IC2=IE2; #Collector current of 2nd transistor, mA\n", + "VC1=-VEE; #Collector voltage of 1st transistor, V\n", + "VC2=-VEE+IC2*RC2; #Collector voltage of 2nd transistor, V\n", + "\n", + "\n", + "#Result\n", + "print(\"VE=%.1fV\"%VE);\n", + "print(\"Tail current=%.3fmA.\"%IE);\n", + "print(\"Emitter current in each transistor=%.3fmA.\"%(IE/2.0));\n", + "print(\"IC1~IE1=%.3fmA and IC2~IE2=%.3fmA\"%(IC1,IC2));\n", + "print(\"VC1=%dV\"%VC1);\n", + "print(\"VC2=%.2fV\"%VC2);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.11 : Page number 679" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The input offset current=15.1nA\n", + "(ii) The input bias current=75.8nA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15; #Collector supply voltage, V\n", + "VEE=15; #Emitter supply voltage, V\n", + "RB=1; #Base resistor, MΩ\n", + "RC2=1; #Collector resistor, MΩ\n", + "RE=1; #Emitter resistor, MΩ\n", + "VBE=0; #Base-emitter voltage, V (Neglected)\n", + "beta_dc_l=90.0; #base current amplification factor for left transistor\n", + "beta_dc_r=110.0; #base current amplification factor for right transistor\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "IE=(VEE-VBE)/RE; #Tail current, μA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, μA\n", + "IE2=IE1; #Emitter current of 2nd transistor, μA\n", + "IB1=round((IE1/beta_dc_l)*1000,1); #Base current of 1st transistor, nA\n", + "IB2=round((IE2/beta_dc_r)*1000,1); #Base current of 2nd transistor, nA\n", + "I_in_offset=IB1-IB2; #Input offset current, nA\n", + "\n", + "#(ii)\n", + "I_in_bias=(IB1+IB2)/2; #Input bias current, nA\n", + "\n", + "#Result\n", + "print(\"(i) The input offset current=%.1fnA\"%I_in_offset);\n", + "print(\"(ii) The input bias current=%.1fnA\"%I_in_bias);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.12 : Page number 679" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The two base currents are: IB1=90nA and IB2=70nA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_in_offset=20; #Input offset current, nA\n", + "I_in_bias=80; #Input bias current, nA\n", + "\n", + "#Calculation\n", + "IB1=I_in_bias+I_in_offset/2; #Base current in 1st transistor, nA\n", + "IB2=I_in_bias-I_in_offset/2; #Base current in 2nd transistor, nA\n", + "\n", + "\n", + "#Result\n", + "print(\"The two base currents are: IB1=%dnA and IB2=%dnA.\"%(IB1,IB2));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.13 : Page number 679-680" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input offset voltage=2mV.\n", + "The output offset voltage=0.3V.\n" + ] + } + ], + "source": [ + "#Variable declration\n", + "I_in_offset=20; #Input offset current, nA\n", + "I_in_bias=80; #Input bias current, nA\n", + "A=150; #Voltage gain\n", + "RB=100; #Base resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "V_io=(I_in_offset*10**-9*RB*1000)*1000; #Input offset voltage, mV\n", + "V_out_offset=(A*V_io)/1000; #Output offset voltage, V\n", + "\n", + "#Result\n", + "print(\"The input offset voltage=%dmV.\"%V_io);\n", + "print(\"The output offset voltage=%.1fV.\"%V_out_offset);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.14 : Page number 682" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Output voltage=0.15V.\n", + "(ii) Output voltage=-0.15V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15; #Collector supply voltage, V\n", + "VEE=15; #Emitter supply voltage, V\n", + "RE=1; #Emitter resistor, MΩ\n", + "RC=1; #Collector resistor, MΩ\n", + "\n", + "\n", + "#Calculation\n", + "IE=VEE/RE; #Tail current, μA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, μA\n", + "IE2=IE1; #Emitter current of 2nd transistor, μA\n", + "re=25/IE1; #a.c emitter resistance, kΩ\n", + "A_DM=RC/(2.0*re); #Differential voltage gain,\n", + "\n", + "#(i)\n", + "vin=1; #Input voltage, V\n", + "Vout=A_DM*vin; #Output voltage, V\n", + "\n", + "print(\"(i) Output voltage=%.2fV.\"%Vout);\n", + "\n", + "#(ii)\n", + "vin=-1; #Input voltage, V\n", + "Vout=A_DM*vin; #Output voltage, V;\n", + "print(\"(ii) Output voltage=%.2fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.15 : Page number 682-683" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The input impedance=194kΩ.\n", + "(ii) The differential voltage gain=136.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RE=100; #Emitter resistor, kΩ\n", + "RC1=120; #Collector resistor of 1st transistor, kΩ\n", + "RC2=120; #Collector resistor of 2nd transistor, kΩ\n", + "beta=220; #Base amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calcualtion\n", + "IE=((VEE-VBE)/RE)*1000; #Tail current, μA\n", + "IE1=IE/2.0; #Emitter current of 1st transistor, μA\n", + "IE2=IE1; #Emitter current of 2nd transistor, μA\n", + "re=(25/IE1)*1000; #a.c emitter resistance, Ω\n", + "Zin=2*beta*re/1000; #Input impedance, kΩ\n", + "A_DM=RC1*1000/(2.0*re); #Differential voltage gain,\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) The input impedance=%dkΩ.\"%Zin);\n", + "print(\"(ii) The differential voltage gain=%.0f.\"%A_DM);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.16: Page number 683-684" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Differential voltage gain=56.6.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RE=200; #Emitter resistor, kΩ\n", + "RC=100; #Collector resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "#Calculation\n", + "IE=round((VEE-VBE)/RE,4); #Tail current, mA\n", + "IE1=round(IE/2,4); #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "re=round(25/IE1,1); #a.c emitter resistance, Ω\n", + "A_DM=RC*1000/(2*re); #Differential voltage gain,\n", + "\n", + "#Result\n", + "print(\"Differential voltage gain=%.1f.\"%A_DM);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.17 : Page number 685-686" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common mode rejection ratio=666.7.\n", + "Common mode rejection ratio in decibel=56.48dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "v1=0.5; #Voltage in terminal 1, mV\n", + "v2=-0.5; #Voltage in terminal 2, mV\n", + "vo=8.0; #Output voltage, V\n", + "vo_cm=12.0; #Common mode output, mV\n", + "\n", + "#Calculation\n", + "vin=v1-v2; #Differential input, mV\n", + "A_DM=vo/(vin/1000.0); #Differential mode gain,\n", + "vin_cm=1; #Common mode input, mV\n", + "A_CM=vo_cm/vin_cm; #Common mode gain\n", + "CMRR=A_DM/A_CM; #Common mode rejection ratio\n", + "CMRR_dB=20*log10(CMRR); #Common mode rejection ratio in dB\n", + "\n", + "#Result\n", + "print(\"Common mode rejection ratio=%.1f.\"%CMRR)\n", + "print(\"Common mode rejection ratio in decibel=%.2fdB\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.18 : Page number 686" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Common mode voltage gain=6.32.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_DM=200000; #Differential mode gain\n", + "CMRR_dB=90; #Common mode rejection ratio, dB\n", + "\n", + "#Calculation\n", + "CMRR=10**(CMRR_dB/20.0); #Common mode rejection ratio\n", + "A_CM=A_DM/CMRR; #Common mode gain\n", + "\n", + "\n", + "\n", + "#Result\n", + "print(\"Common mode voltage gain=%.2f.\"%A_CM);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.19 : Page number 686" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The Common mode gain=0.0081\n", + "(ii) The common mode rejection ratio=81.8dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "vin_cm=3.2; #Common input voltage, V\n", + "vout=26; #Output voltage, V\n", + "A_DM=100; #Open-circuit voltage gain\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CM=vout*10**-3/vin_cm; #Common mode gain\n", + "\n", + "#(ii)\n", + "CMRR_dB=20*log10(A_DM/A_CM); #Common mode rejection ratio, dB\n", + "\n", + "#Result\n", + "print(\"(i) The Common mode gain=%.4f\"%A_CM);\n", + "print(\"(ii) The common mode rejection ratio=%.1fdB.\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.20 : Page number 686-687" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Common mode gain=0.25\n", + "(ii)Common mode rejection ratio=47.09dB\n" + ] + } + ], + "source": [ + "from math import log10\n", + "from math import floor\n", + "\n", + "#Variable declaration\n", + "VCC=12; #Collector supply voltage, V\n", + "VEE=12; #Emitter supply voltage, V\n", + "RE=200.0; #Emitter resistor, kΩ\n", + "RC=100.0; #Collector resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CM=round(RC/(2*RE),2); #Common mode voltage gain\n", + "\n", + "#(ii)\n", + "IE=round((VEE-VBE)/RE,4); #Tail current, mA\n", + "IE1=round(IE/2,4); #Emitter current of 1st transistor, mA\n", + "IE2=IE1; #Emitter current of 2nd transistor, mA\n", + "re=round(25/IE1,1); #a.c emitter resistance, Ω\n", + "A_DM=RC*1000/(2*re); #Differential voltage gain,\n", + "CMRR_dB=floor(20*log10(A_DM/A_CM)*100)/100; #Common mode rejection ratio, dB\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Common mode gain=%.2f\"%A_CM);\n", + "print(\"(ii)Common mode rejection ratio=%.2fdB\"%CMRR_dB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.21 : Page number 691" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "f2=30kHz\n", + "ACL=75 or 37.5dB.\n" + ] + } + ], + "source": [ + "from math import log10\n", + "\n", + "#Variable declaration\n", + "ACL=500; #closed loop gain\n", + "f_unity=15; #frequency with cloased-loop unity gain, MHz\n", + "\n", + "\n", + "#Calculation\n", + "f2=f_unity*1000/500 #Upper frequency of bandwidth,kHz\n", + "BW=f2-0; #Bandwidth, kHz\n", + "A_CL=f_unity*1000/200; #Maximum value of A_CL when f2=200kHz\n", + "A_CL_dB=20*log10(A_CL); #Maximum value of A_CL in decibel\n", + "\n", + "\n", + "#Result\n", + "print(\"f2=%dkHz\"%f2);\n", + "print(\"ACL=%d or %.1fdB.\"%(A_CL,A_CL_dB));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.22 : Page number 691-692" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Operating Bandwidth=1.5MHz.\n", + "(ii) Operating Bandwidth=150kHz.\n", + "(iii) Operating Bandwidth=15kHz.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "GBW=1.5; #Gain-bandwidth, MHz\n", + "\n", + "#Calculation\n", + "#(i) For A_CL=1;\n", + "A_CL=1; #Closed loop gain\n", + "BW=GBW/A_CL; #Bandwidth, MHz\n", + "\n", + "print(\"(i) Operating Bandwidth=%.1fMHz.\"%BW);\n", + "\n", + "#(ii) For A_CL=10;\n", + "A_CL=10; #Closed loop gain\n", + "BW=(GBW/A_CL)*1000; #Bandwidth, kHz\n", + "\n", + "print(\"(ii) Operating Bandwidth=%dkHz.\"%BW);\n", + "\n", + "#(iii) For A_CL=100;\n", + "A_CL=100; #Closed loop gain\n", + "BW=(GBW/A_CL)*1000; #Bandwidth, kHz\n", + "\n", + "print(\"(iii) Operating Bandwidth=%dkHz.\"%BW);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.23 : Page number 692" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum operating frequency=9.95kHz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "V_supply=10; #Supply voltage, V\n", + "\n", + "\n", + "#Calculation\n", + "V_sat=V_supply-2; #Saturation voltage, V\n", + "V_pk=V_sat; #Maximum peak-output voltage, V\n", + "f_max=((slew_rate*10**6)/(2*pi*V_pk))/1000; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"Maximum operating frequency=%.2fkHz.\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.24 : Page number 692" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum operating frequency=796kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "V_pk=100.0; #Peak-output voltage, mV\n", + "\n", + "\n", + "#Calculation\n", + "V_pk=V_pk/1000.0; #Peak-output voltage, V\n", + "f_max=(slew_rate*10**6/(2*pi*V_pk))/1000.0; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"Maximum operating frequency=%.0fkHz\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.25 : Page number 695-696" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Feedback resistor=220kΩ\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_CL=-100; #Closed-loop voltage gain\n", + "Ri=2.2; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "#Since, A_CL=-(Rf/Ri)\n", + "Rf=-A_CL*Ri; #Feedback resistor, kΩ\n", + "\n", + "#Result\n", + "print(\"Feedback resistor=%dkΩ\"%Rf);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.26 : Page number 696" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-0.25V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "vin=2.5; #Input voltage, mV\n", + "Rf=200; #Feedback resistor, kΩ\n", + "Ri=2; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "vout=A_CL*vin/1000; #Output voltage,V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.2fV\"%vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.27 : Page number 696" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain=-1\n", + "Therefore, output will have same amplitude but 180° phase inversion.\n" + ] + } + ], + "source": [ + "#Varaiable declaration\n", + "Rf=1.0; #Feedback resistor, kΩ\n", + "Ri=1.0; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "\n", + "#Result\n", + "print(\"Closed-loop voltage gain=%d\"%A_CL);\n", + "print(\"Therefore, output will have same amplitude but 180° phase inversion.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.28 : Page number 696-697" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Closed-loop voltage gain=-40\n", + "Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=40; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "\n", + "#Calculation\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"Closed-loop voltage gain=%d\"%A_CL);\n", + "print(\"Supply voltage=±15V, saturation voltage=±13V. Since gain=-40, op-Amp will be driven to saturation.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.29 : Page number 697" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) A_CL=-10.\n", + "(ii) Zi=10kΩ\n", + "(iii) Maximum operating frequency=15.9kHz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=10; #Input resistor, kΩ\n", + "Vpp=1; #Input peak-peak voltage, V\n", + "slew_rate=0.5; #Slew rate, V//μs\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CL=-(Rf/Ri); #Closed-loop voltage gain\n", + "\n", + "#(ii)\n", + "Zi=Ri; #Input impedance(~ Input resistor), kΩ\n", + "\n", + "#(iii)\n", + "Vout=A_CL*Vpp; #Peak-to-peak voltage, V\n", + "Vpk=Vout/2; #Peak output voltage, V\n", + "f_max=(slew_rate*10**6/(2*pi*abs(Vpk)))/1000; #Maximum operating frequency, kHz\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) A_CL=%d.\"%A_CL);\n", + "print(\"(ii) Zi=%dkΩ\"%Zi);\n", + "print(\"(iii) Maximum operating frequency=%.1fkHz.\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.30 : Page number 697" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Rf=20kΩ and Ri=5kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A_CL=-4; #Closed loop voltage gain\n", + "R=[1.0,5.0,10.0,20.0]; #List of available resistors, kΩ\n", + "\n", + "#Calculation\n", + "for i in R[:]:\n", + " for j in R[:]:\n", + " if -(i/j)==A_CL :\n", + " print(\"Rf=%dkΩ and Ri=%dkΩ.\"%(i,j));\n", + " break;\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.31 : Page number 697-698" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Closed loop voltage gain=-100.\n", + "(ii) Closed loop voltage gain=-50.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "R_source=0; #Source resistor, kΩ\n", + "A_CL=-Rf/(R_source+Ri); #Closed-loop voltage gain\n", + "\n", + "print(\"(i) Closed loop voltage gain=%d.\"%A_CL);\n", + "\n", + "#(ii)\n", + "R_source=1; #Source resistor, kΩ\n", + "A_CL=-Rf/(R_source+Ri); #Closed-loop voltage gain\n", + "\n", + "print(\"(ii) Closed loop voltage gain=%d.\"%A_CL);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.32 : Page number 699-700" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=12.12mV\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=240; #Feedback resistor, kΩ\n", + "Ri=2.4; #Input resistor, kΩ\n", + "Vin=120; #Input voltage, μV\n", + "\n", + "\n", + "#Calculation\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "Vout=(A_CL*Vin)/1000; #Output voltage, mV\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.2fmV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.33 : Page number 700" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Output voltage=11V\n", + "(ii) Output voltage=-11V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=10; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "#(i)\n", + "Vin=1; #Input voltage, V\n", + "Vout=A_CL*Vin; #Output voltage, V\n", + "\n", + "print(\"(i) Output voltage=%dV\"%Vout);\n", + "\n", + "\n", + "#(ii)\n", + "Vin=-1; #Input voltage, V\n", + "Vout=A_CL*Vin; #Output voltage, V\n", + "\n", + "print(\"(ii) Output voltage=%dV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.34 : Page number 700" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Peak to peak output voltage=12V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=5; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "Vin_max=1; #Maximum input voltage, V\n", + "Vin_min=-1; #Minimum input voltage, V\n", + "\n", + "#Calculation\n", + "V_inpp=Vin_max-Vin_min; #Peak-peak input voltage, V\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "Vout_pp=A_CL*V_inpp; #Peak-peak output voltage, V\n", + "\n", + "#Result\n", + "print(\"Peak to peak output voltage=%dV\"%Vout_pp);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.35 : Page number 700-701" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Closed-loop voltage gain=11\n", + "(ii) Maximum operating frequency=14.47kHz\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=10; #Input resistor, kΩ\n", + "Vpp=1; #Input peak-peak voltage, V\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "\n", + "#Calculation\n", + "#(i)\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "\n", + "#(ii)\n", + "Vout_pp=A_CL*Vpp; #Peak-peak output voltage, V\n", + "Vpk=Vout_pp/2.0; #Peak output voltage, V\n", + "f_max=((slew_rate*10**6)/(2*pi*Vpk))/1000.0; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"(i) Closed-loop voltage gain=%d\"%A_CL);\n", + "\n", + "print(\"(ii) Maximum operating frequency=%.2fkHz\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.36 : Page number 701" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Bandwidth=44.3kHz.\n", + "(ii) Bandwidth=63.8kHz.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=220; #Feedback resistor, kΩ\n", + "Ri=3.3; #Input resistor, kΩ\n", + "unity_gain_BW=3; #Unity gain bandwidth, MHz\n", + "\n", + "#Calculation\n", + "#(i) For non-inverting amplifier\n", + "A_CL=1+(Rf/Ri); #Closed loop voltage gain\n", + "BW=unity_gain_BW*1000.0/A_CL; #Bandwidth, kHz\n", + "\n", + "print(\"(i) Bandwidth=%.1fkHz.\"%BW);\n", + "\n", + "#(ii) For inverting amplifier\n", + "Rf=47; #Feedback resistor, kΩ\n", + "Ri=1; #Input resistor, kΩ\n", + "A_CL=-(Rf/Ri); #Closed loop voltage gain\n", + "BW=unity_gain_BW*1000.0/abs(A_CL); #Bandwidth, kHz\n", + "\n", + "print(\"(ii) Bandwidth=%.1fkHz.\"%BW);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.37 : Page number 701-702" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) For voltage follower A_CL=1.\n", + "(ii) The maximum output frequency=26.53kHz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#(i)\n", + "A_CL=1; #Closed loop voltage gain for voltage follower\n", + "print(\"(i) For voltage follower A_CL=1.\");\n", + "\n", + "\n", + "#(ii)\n", + "slew_rate=0.5; #Slew rate, V/μs\n", + "V_inpp=6; #peak-peak input voltage, V\n", + "Vout=A_CL*V_inpp; #Peak-peak output voltage, V\n", + "Vpk=Vout/2; #Peak output voltage, V\n", + "\n", + "f_max=(slew_rate*10**6/(2*pi*Vpk))/1000; #Maximum operating frequency, kHz\n", + "\n", + "#Result\n", + "print(\"(ii) The maximum output frequency=%.2fkHz.\"%f_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.38 : Page number 702" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=1.78V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=470.0; #Feedback resistor, kΩ\n", + "R1=4.3; #Input resistor of 1st op-Amp, kΩ\n", + "R2=33.0; #Input resistor of 2nd op-Amp, kΩ\n", + "R3=33.0; #Input resistor of 3rd op-Amp, kΩ\n", + "Vin=80.0; #Input voltage, μV.\n", + "\n", + "#Calculation\n", + "A1=1+Rf/R1; #Gain of first op-Amp\n", + "A2=-round(Rf/R2,1); #Gain of second op-Amp\n", + "A3=-round(Rf/R3,1); #Gain of third op-Amp\n", + "A=A1*A2*A3; #Overall gain\n", + "Vout=A*Vin*10**-6; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.2fV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.39 : Page number 702-703" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1=30kΩ, R2=15kΩ and R3=10kΩ.\n", + "Output voltage=0.729V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "A1=10; #Voltage gain of 1st op-Amp\n", + "A2=-18; #Voltage gain of 2nd op-Amp\n", + "A3=-27; #Voltage gain of 3rd op-Amp\n", + "Rf=270; #Feedback resistor, kΩ\n", + "Vin=150; #Input voltage, μV \n", + "\n", + "\n", + "#Calculation\n", + "R1=Rf/(A1-1); #Input resistor of 1st op-Amp, kΩ\n", + "R2=-Rf/A2; #Input resistr of 2nd op-Amp, kΩ\n", + "R3=-Rf/A3; #Input resistor of 3rd op-Amp, kΩ\n", + "\n", + "A=A1*A2*A3; #overall gain,\n", + "Vout=Vin*10**-6*A; #Output voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n", + "print(\"Output voltage=%.3fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.40 : Page number 703-704" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1=50kΩ, R2=25kΩ and R3=10kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=500; #Feedback resistor, kΩ\n", + "A1=-10; #Gain of 1st op-Amp\n", + "A2=-20; #Gain of 2nd op-Amp\n", + "A3=-50; #Gain of 3rd op-Amp\n", + "\n", + "#Calculation\n", + "R1=-Rf/A1; #Input resistor of 1st op-Amp, kΩ\n", + "R2=-Rf/A2; #Input resistor of 2nd op-Amp, kΩ\n", + "R3=-Rf/A3; #Input resistor of 3rd op-Amp, kΩ\n", + "\n", + "\n", + "#Result\n", + "print(\"R1=%dkΩ, R2=%dkΩ and R3=%dkΩ.\"%(R1,R2,R3));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.41 : Page number 705" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The input impedance=17202MΩ and output impedance=8.7e-03Ω.\n", + "(ii) The closed loop voltage gain=23.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Zin=2.0; #Input impedance of op-Amp, MΩ\n", + "Zout=75.0; #Output impedance of op-Amp, Ω\n", + "A_OL=200000.0; #Open-loop voltage gain\n", + "Rf=220.0; #Feedback resistor, kΩ\n", + "Ri=10.0; #Input resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "mv=round(Ri/(Ri+Rf),3); #Feedback fraction\n", + "Zin_NI=Zin*(1+(A_OL*mv)); #Input impedance, MΩ\n", + "Zout_NI=Zout/(1+A_OL*mv); #Output impedance, Ω\n", + "\n", + "#(ii)\n", + "A_CL=1+Rf/Ri; #Closed loop voltage gain\n", + "\n", + "#Result\n", + "print(\"(i) The input impedance=%dMΩ and output impedance=%.1eΩ.\"%(Zin_NI,Zout_NI));\n", + "print(\"(ii) The closed loop voltage gain=%d.\"%A_CL);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.42 : Page number 705-706" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance=400002MΩ and output impedance=0.38e-03Ω.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "#For voltage follower,\n", + "mv=1.0; #Feedback fraction\n", + "A_OL=200000.0; #Open-loop voltage gain\n", + "Zin=2.0; #Input impedance of op-Amp, MΩ\n", + "Zout=75.0; #Output impedance of op-Amp, Ω\n", + "\n", + "\n", + "#Calculation\n", + "Zin_VF=Zin*(1+(A_OL*mv)); #Input impedance, MΩ\n", + "Zout_VF=round(round(Zout/(1+A_OL*mv),6),5); #Output impedance, Ω\n", + "\n", + "#Result\n", + "print(\"The input impedance=%dMΩ and output impedance=%.2fe-03Ω.\"%(Zin_VF,Zout_VF*1000));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.43 : Page number 706" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The input impedance=1kΩ and output impedance=50Ω.\n", + "Closed-loop voltage gain=-100\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "Ri=1.0; #Input resistor, kΩ\n", + "Zin=4; #Input impedance of op-Amp, MΩ\n", + "Zout=50; #Output impedance of op-Amp, Ω\n", + "\n", + "\n", + "#Calculation\n", + "Zin_I=Ri; #Input impedance, kΩ\n", + "Zout_I=Zout; #Output impedance, Ω\n", + "A_CL=-(Rf/Ri); #Closed loop voltage gain\n", + "\n", + "\n", + "#Result\n", + "print(\"The input impedance=%dkΩ and output impedance=%dΩ.\"%(Zin_I,Zout_I));\n", + "print(\"Closed-loop voltage gain=%d\"%A_CL);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.44 : Page number 709" + ] + }, + { + "cell_type": "code", + "execution_count": 45, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-12V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=10; #Feedback resistor, kΩ\n", + "Ri=10; #Input resistor, kΩ\n", + "V1=3; #Input voltage 1st, V\n", + "V2=1; #Input voltage 2nd, V\n", + "V3=8; #Input voltage 3rd, V\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Rf=Ri, Vout=-(Rf/Ri)*(V1+V2+V3)= -(V1+V2+V3);\n", + "Vout=-(V1+V2+V3); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.45 : Page number 709" + ] + }, + { + "cell_type": "code", + "execution_count": 46, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-7V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=10; #Feedback resistor, kΩ\n", + "R1=1; #Input resistor for input 1, kΩ\n", + "R2=1; #Input resistor for input 2, kΩ\n", + "V1=0.2; #Input voltage 1st, V\n", + "V2=0.5; #Input voltage 2nd, V\n", + "\n", + "\n", + "#Calculation\n", + "R=R1; #Input resistor(=R1 or R2), kΩ\n", + "Vout=-(Rf/R)*(V1+V2); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%dV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.46 : Page number 709-710" + ] + }, + { + "cell_type": "code", + "execution_count": 47, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=-2.5V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=1; #Feedback resistor, kΩ\n", + "Ri=10.0; #Input resistor, kΩ\n", + "V1=10; #Input voltage 1st, V\n", + "V2=8.0; #Input voltage 2nd, V\n", + "V3=7.0; #Input voltage 3rd, V\n", + "\n", + "\n", + "#Calculation\n", + "#Since, Vout=-(Rf/Ri)*(V1+V2+V3);\n", + "Vout=-(Rf/Ri)*(V1+V2+V3); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.1fV.\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.47 : Page number 710" + ] + }, + { + "cell_type": "code", + "execution_count": 48, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Output voltage=2.5V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V1=0.6; #Input voltage to 1st input resistor, V\n", + "V2=-1.4; #Input voltage to 2nd input resistor, V\n", + "Rf=200; #Feedback resistor, kΩ\n", + "R1=400; #Input resistor 1, kΩ\n", + "R2=100.0; #Input resistor 2, kΩ\n", + "\n", + "#Calculation\n", + "Vout=-Rf*(V1/R1 +V2/R2); #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Output voltage=%.1fV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.48 : Page number 710-711" + ] + }, + { + "cell_type": "code", + "execution_count": 49, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The output voltage=-12.5V\n", + "(ii) The output voltage=-7.5V\n", + "(iii) The output voltage=-17.5V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=1.0; #Feedback resistor, kΩ\n", + "R1=1.0; #Input resistor 1, kΩ\n", + "R2=2.0; #Input resistor 2, kΩ\n", + "R3=4.0; #Input resistor 3, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "Rf_R1=Rf/R1; #Ratio of feedback resistor and 1st input resistor\n", + "Rf_R2=Rf/R2; #Ratio of feedback resistor and 2nd input resistor\n", + "Rf_R3=Rf/R3; #Ratio of feedback resistor and 3rd input resistor\n", + "\n", + "#(i) First input combination\n", + "V1=10; #Input voltage to 1st input resistor, V\n", + "V2=0; #Input voltage to 2nd input resistor, V\n", + "V3=10; #Input voltage to 3rd input resistor, V\n", + "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3); #Output voltage, V\n", + "print(\"(i) The output voltage=%.1fV\"%Vout);\n", + "\n", + "#(i) First input combination\n", + "V1=0; #Input voltage to 1st input resistor, V\n", + "V2=10; #Input voltage to 2nd input resistor, V\n", + "V3=10; #Input voltage to 3rd input resistor, V\n", + "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3); #Output voltage, V\n", + "print(\"(ii) The output voltage=%.1fV\"%Vout);\n", + "\n", + "\n", + "#(i) First input combination\n", + "V1=10; #Input voltage to 1st input resistor, V\n", + "V2=10; #Input voltage to 2nd input resistor, V\n", + "V3=10; #Input voltage to 3rd input resistor, V\n", + "Vout=-(V1*Rf_R1 +V2*Rf_R2 +V3*Rf_R3); #Output voltage, V\n", + "print(\"(iii) The output voltage=%.1fV\"%Vout);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.49 : Page number 711" + ] + }, + { + "cell_type": "code", + "execution_count": 50, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vout=-[0.5sin(1000t)+0.33sin(3000t)]V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Rf=330; #Feedback resistor, kΩ\n", + "R1=33.0; #Input resistor 1, kΩ\n", + "R2=10.0; #Input resistor 2, kΩ\n", + "V1_m=50; #Peak voltage of 1st input, mV\n", + "V2_m=10; #Peak voltage of 2nd input, mV\n", + "\n", + "#Calculation\n", + "#Since, Vout=-((Rf/R1)*V1 + (Rf/R2)*V2)\n", + "print(\"Vout=-[%.1fsin(1000t)+%.2fsin(3000t)]V\"%((V1_m/1000.0)*(Rf/R1),(V2_m/1000.0)*(Rf/R2)));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.50 : Page number 715" + ] + }, + { + "cell_type": "code", + "execution_count": 51, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo=-1*(1/RC)∫vi dt.\n", + "=>Vo=-1*(1/1)∫vi dt\n", + "=>Vo=∫vi dt\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=100; #Input resistor, kΩ\n", + "C=10; #Feedback capacitor, μF\n", + "\n", + "#Calculation\n", + "RC=R*10**3*C*10**-6; #product of input resistance and feedback capacitance, s\n", + "\n", + "\n", + "#Result\n", + "print(\"Vo=-1*(1/RC)∫vi dt.\");\n", + "print(\"=>Vo=-1*(1/%d)∫vi dt\"%RC);\n", + "print(\"=>Vo=∫vi dt\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.51 : Page number 715-716" + ] + }, + { + "cell_type": "code", + "execution_count": 52, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The critical frequency=159Hz.\n" + ] + } + ], + "source": [ + "from math import pi\n", + "\n", + "#Variable declaration\n", + "Rf=100; #Feedback resistor, kΩ\n", + "C=0.01; #Feedback capacitor, μF\n", + "\n", + "\n", + "#Calculation\n", + "fc=1/(2*pi*Rf*1000*C*10**-6); #Crictical frequency, Hz\n", + "\n", + "#Result\n", + "print(\"The critical frequency=%dHz.\"%fc);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.52 : Page number 716" + ] + }, + { + "cell_type": "code", + "execution_count": 53, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Vout=-1*(1/RC)∫vi dt.\n", + " ΔVout/dt = -vin/RC = -50mV/μs.\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f8b042c6b70>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "R=10.0; #Input resistor, kΩ\n", + "C=0.01; #Feedback capacitor, μF\n", + "vin=5; #Input voltage, V\n", + "\n", + "#Calculation\n", + "#(i)\n", + "Vout_change_rate=-vin/(R*C); #Rate of change of output voltage, V/μs \n", + "print(\"(i) Vout=-1*(1/RC)∫vi dt.\");\n", + "print(\" ΔVout/dt = -vin/RC = %dmV/μs.\"%Vout_change_rate);\n", + "\n", + "#(ii) Plotting the output waveform\n", + "vin_plot=[]; #Plotting variable for input waveform, V\n", + "dt=100; #time between edges, μs\n", + "for i in range(0,3*dt+1):\n", + " if i<dt or i>2*dt :\n", + " vin_plot.append(0);\n", + " else:\n", + " vin_plot.append(5); \n", + "\n", + "plt.subplot(211);\n", + "plt.plot(vin_plot);\n", + "plt.xlim([0,300])\n", + "plt.ylim([-5,10])\n", + "plt.xlabel(\"t(microsecond)\");\n", + "plt.ylabel(\"Vin(V)\");\n", + "plt.title(\"Input waveform\");\n", + "\n", + " \n", + "vout_plot=[]; #Plotting variable for output waveform, V\n", + "t=[i for i in range(0,301)]; #Time scale, μs\n", + "for i in t[:] :\n", + " if i<dt:\n", + " vout_plot.append(0);\n", + " elif i>2*dt:\n", + " vout_plot.append((Vout_change_rate/1000.0)*dt);\n", + " else :\n", + " vout_plot.append((-vin_plot[i]/(R*C))/1000*(i-dt));\n", + "\n", + "plt.subplot(212)\n", + "plt.plot(vout_plot);\n", + "plt.xlim([0,300])\n", + "plt.ylim([-5,5]);\n", + "plt.xlabel('t(microsecond)');\n", + "plt.ylabel(\"Vout(V)\");\n", + "plt.title(\"output waveform\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.53 : Page number 716-717" + ] + }, + { + "cell_type": "code", + "execution_count": 54, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vout=-1*(1/RC)∫vi dt.\n", + "Vout=-5*t volts\n", + "Time required=2.6seconds.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_supply=15; #Supply voltage, V\n", + "R=10; #Input resistor, kΩ\n", + "C=0.2; #Feedback capacitor, μF\n", + "vin=10; #Input voltage, mV\n", + "\n", + "\n", + "#Calculation\n", + "Vs=-V_supply+2; #Saturation voltage, V\n", + "print(\"Vout=-1*(1/RC)∫vi dt.\");\n", + "print(\"Vout=%d*t volts\"%(-vin/(R*C)));\n", + "t=Vs/(-vin/(R*C)); #Time required, seconds\n", + "print(\"Time required=%.1fseconds.\"%t);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.54 : Page number 717-718" + ] + }, + { + "cell_type": "code", + "execution_count": 55, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo=-5V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=1; #Feedback resistor, kΩ\n", + "C=0.1; #Input capacitor, μF\n", + "Vin_change=5; #Change in input voltage, V\n", + "t=0.1; #Time taken for change in input voltage, ms\n", + "\n", + "#Calcualtion\n", + "dvi_dt=Vin_change/(t/1000); #Rate of change of input voltage, V/s\n", + "RC=R*1000*C*10**-6; #Product of feedback resistance and input capacitance, s\n", + "#Since, Vo=-R*C*(dvi/dt);\n", + "Vo=-RC*dvi_dt; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Vo=%dV.\"%Vo);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.55 : Page number 718" + ] + }, + { + "cell_type": "code", + "execution_count": 56, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vo=-0.55V.\n", + "The output voltage stays constant at -0.55V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=10; #Feedback resistor, kΩ\n", + "C=2.2; #Input capacitor, μF\n", + "Vin_change=10; #Change in input voltage, V\n", + "t=0.4; #Time taken for change in input voltage, s\n", + "\n", + "#Calcualtin\n", + "dvi_dt=Vin_change/t; #Rate of change of input voltage, V/s\n", + "RC=R*1000*C*10**-6; #Product of feedback resistance and input capacitance, s\n", + "#Since, Vo=-R*C*(dvi/dt);\n", + "Vo=-RC*dvi_dt; #Output voltage, V\n", + "\n", + "#Result\n", + "print(\"Vo=%.2fV.\"%Vo);\n", + "print(\"The output voltage stays constant at %.2fV.\"%Vo);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 25.56 : Page number 718-719" + ] + }, + { + "cell_type": "code", + "execution_count": 57, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "vo=-1*(dvi/dt).\n", + "vo=-5V.\n", + "Therefore, between 0 to 0.2s, the output voltage is constant at -5V.\n", + "For t>0.2s, the input is constant so that output voltage is zero.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R=100; #Feedback resistor, kΩ\n", + "C=10; #Input capacitor, μF\n", + "Vin_change=1; #Change in input voltage, V\n", + "t=0.2; #Time taken for change in input voltage, s\n", + "\n", + "\n", + "#Calculation\n", + "RC=R*1000*C*10**-6; #Product of feedback resistor and input capacitance, s\n", + "#(i)\n", + "print(\"vo=-%d*(dvi/dt).\"%RC);\n", + "\n", + "#(ii)\n", + "dvi_dt=Vin_change/t; #Rate of change of input voltage, V\n", + "vo=-dvi_dt; #Output voltage, V\n", + "print(\"vo=%dV.\"%vo);\n", + "\n", + "print(\"Therefore, between 0 to 0.2s, the output voltage is constant at %dV.\"%vo);\n", + "print(\"For t>0.2s, the input is constant so that output voltage is zero.\");\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": true + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_6.ipynb new file mode 100644 index 00000000..670a61b0 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter26_6.ipynb @@ -0,0 +1,472 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:a1845801144904256bc26f3ca2e0294eb55dcabb139a523d403624121bc6876a" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#CHAPTER 26 : DIGITAL ELECTRONICS" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.1 : Page 732" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "d=37; #Given decimal number\n", + "\n", + "#Calculation\n", + "b=int(bin(d)[2:]); #Equivalent Octal number \n", + "\n", + "#Result\n", + "print(\"The equivalent binary number=%s.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equivalent binary number=100101.\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.2 : Page number 733" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "d=23; #Given decimal number\n", + "\n", + "#Calculation\n", + "b=int(bin(d)[2:]); #Equivalent Octal number\n", + "\n", + "#Result\n", + "print(\"The equivalent binary number=%d.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equivalent binary number=10111.\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.3 : Page number 733" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "b=0b110001; #Given binary number\n", + "\n", + "#Calculation\n", + "d=int(b); #Equivalent decimal number\n", + "\n", + "#Result\n", + "print(\"Equivalent decimal number=%d.\"%d);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent decimal number=49.\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.4 : Page number 735" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "d1=76; #Given decimal number\n", + "d2=255; #Given decimal number\n", + "d3=372; #Given decimal number\n", + "\n", + "#Calculation\n", + "o1=int(oct(d1)[1:]); #Equivalent octal number\n", + "o2=int(oct(d2)[1:]); #Equivalent octal number\n", + "o3=int(oct(d3)[1:]); #Equivalent octal number\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Equivalent octal number=%d.\"%o1);\n", + "print(\"(ii) Equivalent octal number=%d.\"%o2);\n", + "print(\"(iii) Equivalent octal number=%d.\"%o3);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Equivalent octal number=114.\n", + "(ii) Equivalent octal number=377.\n", + "(iii) Equivalent octal number=564.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.5 : Page number 735" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "o=24.6; #Given octal number\n", + "\n", + "#Calculation\n", + "o_f=o%1; #Floating part of octal number\n", + "o_i=(int)(o-(o%1)); #Integer part of octal number\n", + "d=int(str(o_i),8); #Equivalent decimal number\n", + "\n", + "s=str(o_f); #String value of floating part \n", + "i=2\n", + "while(i<len(s)):\n", + " d=d+int(s[i])*8**-(i-1);\n", + " i+=1;\n", + "#Result\n", + "print(\"Equivalent decimal number=%.2f.\"%d);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent decimal number=20.75.\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.6 : Page number 735" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "d=177; #Given decimal number\n", + "\n", + "#Calculation\n", + "o=oct(d)[1:]; #Equivalent octal number\n", + "\n", + "b=\"\";\n", + "for i in o:\n", + " bo=bin(int(i))[2:]; #Binary of individual octal digit\n", + " b=b+((\"0\" if len(bo)==2 else (\"00\" if len(bo)==1 else\"\")) +bo); #Equivalent binary number\n", + " \n", + "#Result\n", + "print(\"Equivalent octal number=%s.\"%o);\n", + "print(\"Equivalent binary number=%s.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent octal number=261.\n", + "Equivalent binary number=010110001.\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.7 : Page number 737" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given \n", + "d=541; #Given decimal number\n", + "\n", + "#Calculation\n", + "h=hex(d)[2:]; #Equivalent hexadecimal number\n", + "\n", + "#Result\n", + "print(\"Equivalent hexadecimal number=%s.\"%h);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent hexadecimal number=21d.\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.8 : Page number 737" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "hex_to_dec={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,'b':11,'c':12,'d':13,'e':14,'f':15};\n", + " \n", + "#Given \n", + "d=378; #Given decimal number\n", + "\n", + "#Calculation\n", + "h=hex(d)[2:]; #Equivalent Hexadecimal number\n", + "\n", + "\n", + "b=\"\";\n", + "for i in h:\n", + " bh=bin(hex_to_dec[i])[2:]; #Binary of individual hexadecimaldigit\n", + " b=b+((\"0\" if len(bh)==3 else (\"00\" if len(bh)==2 else (\"000\" if len(bh)==1 else \"\")))+bh); #Equivalent binary number\n", + "\n", + "#Result\n", + "print(\"Equivalent hexadeciaml number=%s.\"%h);\n", + "print(\"Equivalent binary number=%s.\"%b);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent hexadeciaml number=17a.\n", + "Equivalent binary number=000101111010.\n" + ] + } + ], + "prompt_number": 56 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 26.9 : Page number 737" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "h=0xB2F; #Given hexadecimal number\n", + "\n", + "#Calculation\n", + "o=oct(h)[1:]; #Equivalent octal number\n", + "\n", + "#Result\n", + "print(\"Equivalent octal number=%s.\"%o);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Equivalent octal number=5457.\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.10 : Page number 738" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Given\n", + "BCD=\"0100 0000 0010\" #Given BCD string\n", + "BCD_split=BCD.split(\" \"); #Splitting th binary string into individual BCD \n", + "d=0;\n", + "for i in range(len(BCD_split),0,-1):\n", + " d+=int(BCD_split[len(BCD_split)-i],2)*10**(i-1);\n", + "\n", + "#Result\n", + "print(\"The equivalent decimal =%d.\"%d);\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The equivalent decimal =402.\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.11 : Page number 745" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print(\"Boolean Expression obtained from the circuit: \\n Y'=A+B \\n Y=((A+B).A)\");\n", + "print(\"Truth Table:\");\n", + "print(\"a\\tb\\tY'=A+B\\t Y=Y'.A\");\n", + "for b in range(0,2):\n", + " for a in range(0,2):\n", + " Y_dash=1 if a or b else 0;\n", + " Y=1 if Y_dash and a else 0;\n", + " print(\"%d\\t%d\\t%d\\t %d\"%(a,b,Y_dash,Y));" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boolean Expression obtained from the circuit: \n", + " Y'=A+B \n", + " Y=((A+B).A)\n", + "Truth Table:\n", + "a\tb\tY'=A+B\t Y=Y'.A\n", + "0\t0\t0\t 0\n", + "1\t0\t1\t 1\n", + "0\t1\t1\t 0\n", + "1\t1\t1\t 1\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "##Example 26.12 : Page number 745-746" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "print(\"Boolean Expression obtained from the circuit: \\n Y'=A'.B \\n Y=Y'+B'\");\n", + "print(\"Truth Table:\");\n", + "print(\"A\\tB\\tA'\\tY'=A'.B\\t B'\\tY=Y'+B'\");\n", + "for b in range(0,2):\n", + " for a in range(0,2):\n", + " a_dash=1 if not a else 0;\n", + " b_dash=1 if not b else 0;\n", + " Y_dash=1 if a_dash and b else 0;\n", + " Y=1 if Y_dash or b_dash else 0;\n", + " print(\"%d\\t%d\\t%d\\t%d\\t %d\\t%d\"%(a,b,a_dash,Y_dash,b_dash,Y));" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Boolean Expression obtained from the circuit: \n", + " Y'=A'.B \n", + " Y=Y'+B'\n", + "Truth Table:\n", + "A\tB\tA'\tY'=A'.B\t B'\tY=Y'+B'\n", + "0\t0\t1\t0\t 1\t1\n", + "1\t0\t0\t0\t 1\t1\n", + "0\t1\t1\t1\t 0\t1\n", + "1\t1\t0\t0\t 0\t0\n" + ] + } + ], + "prompt_number": 30 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_6.ipynb new file mode 100644 index 00000000..06a555a6 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter2_6.ipynb @@ -0,0 +1,125 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:c0ff4f67576afe73a11c06eedd0a50709b7f5831737f83db1cd640098e3e9740" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 2 : ELECTRONIC EMISSION" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.1: Page number 31\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "\n", + "from math import exp\n", + "from math import pi\n", + "\n", + "l=5.0; #length of tungsten filament in cm\n", + "d=0.01; #diameter of the filament in cm\n", + "T=2500.0; #operating temperature in K\n", + "A=60.2*pow(10,4); #constant, depending upon the type of thermionic emitter, in amp/m\u00b2/K\u00b2\n", + "phi=4.517; #work function of emitter in eV\n", + "\n", + "\n", + "#Calculation\n", + "b=round(11600*phi,-1); #constant for a metal, in K\n", + "Js=round(A*T*T*exp(-b/T),-2); #Emission current density in amp/m\u00b2\n", + "a=pi*(d/100)*(l/100); #Surface area of the cathode in m\u00b2\n", + "E_I=Js*a; #Emission current in A\n", + "\n", + "#Result\n", + "print(\"emission current =%.3f A\"%E_I);" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "emission current =0.047 A\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 2.2:Page number 31\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import log\n", + "\n", + "#Variable declaration\n", + "Js=0.1; #Emission current density in amp/cm\u00b2\n", + "A=60.2; #Constant depending upon the type of thermionic emitter, in amp/cm\u00b2/K\u00b2\n", + "T=1900.0; #Absolute temperature in K\n", + "\n", + "\n", + "#calculations\n", + "#Calculating b according to the formula Js=A*T\u00b2*exp(-b/T) for emission current density\n", + "b=-T*(log(Js/(A*T*T))); #constant for emitter, in K\n", + "phi= round(b/11600,2); # work function in eV\n", + "\n", + "print (\"Work function of the tungsten wire = %.2f eV\"%phi);\n", + "\n", + "if(phi==4.52):\n", + "\tprint(\"Given sample is pure Tungsten\");\n", + "elif(phi!=4.52 and phi>=2.63 and phi<=4.52):\n", + "\tprint (\"The sample is not pure Tungsten\");\n", + " \n", + "#Note : In the text book, the work function has been approximated to 3.56eV, but in the code it calculates as 3.52eV\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Work function of the tungsten wire = 3.52 eV\n", + "The sample is not pure Tungsten\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_6.ipynb new file mode 100644 index 00000000..a6008ee1 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter6_6.ipynb @@ -0,0 +1,1624 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:3452607f2168b562d941493f83083042eaa5a2d316715f9d9f089ff03d73fdb8" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 6: SEMICONDUCTOR DIODE" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, Page number 81" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration \n", + "Vf =20; #Peak Input Voltage in V\n", + "rf=10; #Forward Resistance in ohms\n", + "RL=500.0; #Load Resistance in ohms\n", + "V0=0.7; #Potential Barrier Voltage of the diodes in V\n", + "\n", + "#Calculation\n", + "#(1)\n", + "If_peak=(Vf-V0)/(rf+RL); #Peak current through the diode in A\n", + "If_peak=If_peak*1000; #Peak current through the diode in mA\n", + "#(2)\n", + "V_out_peak =If_peak * RL/1000 ; #Peak output voltage in V\n", + "\n", + "#For an Ideal diode\n", + "If_peak_ideal=Vf/RL; #Peak current through the ideal diode in A\n", + "If_peak_ideal=If_peak_ideal*1000; #Peak current through the ideal diode in mA\n", + "\n", + "V_out_peak_ideal=If_peak_ideal * RL/1000; # Peak output voltage in case of the ideal diode in V\n", + "\n", + "#Result\n", + "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n", + "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n", + "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n", + "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) Peak current through the diode = 37.8 mA \n", + "(ii) Peak output voltage = 18.9 V\n", + "(iii) Peak current through the ideal diode = 40 mA \n", + "(iv) Peak output voltage in case of the ideal diode = 20 V\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3, Page number 82" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V =10.0; #Battery voltage in V\n", + "R1=50.0; #Resistor 1's resistance in ohms\n", + "R2=5.0; #Resistor 2's resistance in ohms\n", + "\n", + "#Calculation\n", + "#Using Thevenin's Theorem to find current in the diode\n", + "E0=(R2/(R1+R2))*V; #Thevenin's Voltage in V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's Resistance in ohms\n", + "\n", + "I0=E0/R0; #Current through the diode in A\n", + "I0=I0*1000; #Current through the diode in mA\n", + "\n", + "#Result\n", + "print 'Current through the diode = %d mA '%Io;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current through the diode = 200 mA \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page number 82-83 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V =10.0; #Battery voltage in V\n", + "R0=48.0; #Resistance of the resistor in ohms\n", + "Rd=1.0; #Forward resistance of the diodes in ohms\n", + "Vd=0.7; #Potential barrier of the diodes in V\n", + "#Calculation\n", + "V_net=V-Vd-Vd; #Net voltage in the circuit in V\n", + "R_net=R0+Rd+Rd #Net resistance of the circuit in ohms\n", + "I_net=V_net/R_net; #Net current in the circuit in A\n", + "I_net=I_net*1000; #Net current in mA\n", + "\n", + "#Result\n", + "print 'Net current in the circuit = %d mA '%I_net;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Net current in the circuit = 172 mA \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5, Page number 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "E1=24; #Voltage of first source in V\n", + "E2=4; #Voltage of second source in V\n", + "V0=0.7; #Potential barrier of diodes in V\n", + "R=2000; #Resistance of the given resistor in ohms\n", + "Rd=0; #Forward resistance of the diodes in ohms\n", + "\n", + "#Calculation\n", + "I=(E1-E2-V0)/(R+Rd); #Current in the circuit in A\n", + "I=I*1000; #Current in the circuit in mA \n", + "\n", + "#Result\n", + "print 'Current in the circuit = %.2f mA '%I;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in the circuit = 9.65 mA \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6, Page number 83-84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V=20; #Voltage of source in V\n", + "V0=0.3; #Potential barrier of Germanium diode in V\n", + "V0_Si=0.7; #Potetial barrier of Silicon diode in V \n", + "\n", + "#Calculation\n", + "#As only Ge diode is turned on due to less potential barrier,\n", + "VA=V-V0; #Voltage VA acroos resistor of 3k ohms\n", + "\n", + "#Result\n", + "print 'Voltage VA = %.1f mA '%VA;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage VA = 19.7 mA \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7, Page number 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V=10; #Voltage of source in V\n", + "V0=0.7; #Potetial barrier of Silicon diode in V \n", + "# Resistance of all resistors in ohms\n", + "R1=2000;\n", + "R2=2000;\n", + "R3=2000;\n", + "\n", + "#Calculation\n", + "Id=(V-V0)/(R2+2*R3); #Current through the diodes in A\n", + "VQ=2*Id*R3; #Voltage VQ across the grounded 2k ohm resistor in V\n", + "Id=Id*1000; #Current through the diodes in mA\n", + "\n", + "#Result\n", + "print 'Voltage VQ = %.1f V '%VQ;\n", + "print 'Current through the diodes, Id = %.2f mA '%Id;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Voltage VQ = 6.2 V \n", + "Current through the diodes, Id = 1.55 mA \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8, Page number 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "V=15; #Voltage of source in V\n", + "V0=0.7; #Potetial barrier of Silicon diode in V \n", + "R=500 # Resistance of all resistors in ohms\n", + "\n", + "#Calculation\n", + "I1=(V-V0)/R; #total current in the circuit in A\n", + "Id1=I1/2; #current in first diode in A\n", + "Id1=Id1*1000; #current in first diode in mA\n", + "Id2=Id1 #current in second diode in mA\n", + "\n", + "#Result\n", + "print ('Current in first diode = %.1f mA'%Id1);\n", + "print ('Current in second diode = %.1f mA'%Id2);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current in first diode = 14.3 mA\n", + "Current in second diode = 14.3 mA\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9, Page number 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "E=20; #Voltage of source in V\n", + "V0_d1=0.7; #Potetial barrier of first Silicon diode in V\n", + "V0_d2=0.7; #Potetial barrier of second Silicon diode in V\n", + "R1=5600; # Resistance of first resistor in ohms\n", + "R2=3300; # Resistance of second resistor in ohms\n", + "\n", + "#Calculation\n", + "I2=V0_d2/R2; #Current I2 through resistor R2 in A\n", + "I2=round((I2*1000),3); #Current I2 through resistor R2 in mA\n", + "I1=(E-V0_d1-V0_d2)/R1; #Current I1 through resistor R1 in A\n", + "I1=round((I1*1000),2); #Current I1 through resistor R1 in mA\n", + "I3=I1-I2; #Current I3 through diode D2 in mA\n", + "\n", + "#Result\n", + "print 'Current I1= %.2f mA'%I1;\n", + "print 'Current I1= %.3f mA'%I2;\n", + "print 'Current I1= %.3f mA'%I3;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current I1= 3.32 mA\n", + "Current I1= 0.212 mA\n", + "Current I1= 3.108 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10, Page number 85-86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable Declaration\n", + "E=10.0; #Voltage of source in V\n", + "V0=0.7; #Potetial barrier of Silicon diode in V\n", + "R1=2000; # Resistance of first resistor in ohms\n", + "R2=8000; # Resistance of second resistor in ohms\n", + "R3=4000; #Resistance of third resistor in ohms\n", + "R4=6000; #Resistance of fourth resistor in ohms\n", + "\n", + "#Calculation\n", + "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n", + "V1=(E/(R1+R2))*R2; #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n", + "V2=(E/(R3+R4))*R4; #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n", + "\n", + "#Result\n", + "if((V1-V2)>=V0):\n", + " print 'Our assumption was wrong and, the diode is forward biased';\n", + "else:\n", + " print 'The diode is reverse biased';\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Our assumption was wrong and, the diode is forward biased\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11, Page number 86" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=2; #Supply voltage in V\n", + "V0=0.7; #Potential barrier voltage of the diode in V \n", + "R1=4000.0; #Resistance of first resistor in \u03a9\n", + "R2=1000.0; ##Resistance of second resistor in \u03a9\n", + "\n", + "#Calculation\n", + "#Assuming the diode to be in ON state\n", + "I1=((V-V0)/R1)*1000; #Current through resistor R1, in mA\n", + "I2=(V0/R2)*1000; #Current through resistor R2, in mA\n", + "ID=I1-I2; #Diode current, in mA\n", + "\n", + "if(ID<0):\n", + " #Since the diode current is negative, the diode must be OFF \n", + " ID=0; #True value of diode current, mA\n", + " \n", + "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n", + "VD=V*R2/(R1 +R2); #Voltage across the diode, in V\n", + "\n", + "#Result\n", + "print 'ID =%d mA'%ID;\n", + "print 'VD =%.1f V'%VD;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "ID =0 mA\n", + "VD =0.4 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12, Page number 89-90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "AC_Input_Power=100.0; #Input AC Power in watts\n", + "AC_Output_Power=40.0; #Output AC Power in watts\n", + "Accepted_Power=50.0; #Power accepted by the half-wave rectifier in watt\n", + "\n", + "#Calculation\n", + "R_eff=(AC_Output_Power/AC_Input_Power)*100; #Rectification efficiency of the half-wave rectifier\n", + "Unused_power=AC_Input_Power-Accepted_Power; #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n", + "Power_dissipated=Accepted_Power-AC_Output_Power; #Power dissipated by the diode watt\n", + "\n", + "#Result\n", + "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n", + "\n", + "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The rectification efficiency of the half-wave rectifier= 40% \n", + "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13, Page number 90" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "#Variable declaration\n", + "Vrms=230.0; #AC supply RMS voltage in V\n", + "Turns_Ratio=10/1; #turn ratio of the transformer \n", + "\n", + "#Calculation\n", + "Vpm=sqrt(2)*Vrms; #Maximum primary voltage in V\n", + "Vsm=Vpm/Turns_Ratio; #Maximum secondary voltage in V\n", + "#Case 1\n", + "Vdc=Vsm/(round(pi,2)); #Output D.C voltage, which is the average voltage in V\n", + "Vdc=round(Vdc,2);\n", + "#Case 2\n", + "PIV=Vsm; #Peak Inverse Voltage in V\n", + "\n", + "#Result\n", + "print 'The output d.c voltage= %.2f V'%Vdc;\n", + "print 'The peak inverse voltage= %.2f V'%PIV;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output d.c voltage= 10.36 V\n", + "The peak inverse voltage= 32.53 V\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14, Page number 90-91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Variable declaration\n", + "rf=20.0; #Internal resistance of the crystal diode in ohms\n", + "Vm=50.0; #Maximum applied voltage in V\n", + "RL=800.0; #Load Resistance in ohms\n", + "\n", + "#Calculation\n", + "# 1\n", + "Im=Vm/(rf+RL); #Maximum current in A\n", + "Im=Im*1000; #Maximum current in \n", + "Im=round(Im,0);\n", + "Idc=Im/pi; #Average voltage in mA\n", + "Idc=round(Idc,1);\n", + "Irms=Im/2; #RMS value of the current in mA\n", + "Irms=round(Irms,1)\n", + "\n", + "# 2\n", + "AC_Input_Power=pow(Irms/1000,2)*(rf+RL); #Input a.c power in watt\n", + "\n", + "DC_Output_Power=pow(Idc/1000,2)*RL; #Output d.c power in watt\n", + "\n", + "# 3\n", + "DC_Output_Voltage=(Idc/1000)*RL; #Output d.c voltage in V\n", + "\n", + "# 4\n", + "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100; # Efficiency of rectification of the half-wave rectifier\n", + "\n", + "#Result\n", + "print ' i:';\n", + "print ' Im = %d mA'%Im;\n", + "print ' Idc = %.1f mA'%Idc;\n", + "print ' Irms = %.1f mA'%Irms;\n", + "print ' ii: ';\n", + "print ' a.c input power= %.3f watt'%AC_Input_Power;\n", + "print ' d.c output power= %.3f watt'%DC_Output_Power;\n", + "print ' iii: ';\n", + "print ' d.c output voltage = %.2f volts'%DC_Output_Voltage;\n", + "print ' iv: '\n", + "print ' Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n", + "\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " i:\n", + " Im = 61 mA\n", + " Idc = 19.4 mA\n", + " Irms = 30.5 mA\n", + " ii: \n", + " a.c input power= 0.763 watt\n", + " d.c output power= 0.301 watt\n", + " iii: \n", + " d.c output voltage = 15.52 volts\n", + " iv: \n", + " Efficiency of rectification = 39.5%\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15, Page number 91" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Variable declaration\n", + "Vdc=50.0; #Output d.c voltage in V\n", + "rf=25; #Diode resistance in ohm\n", + "RL=800; #Load resistance in ohm\n", + "\n", + "\n", + "#Calculation\n", + "Vm=(pi*(rf+RL)*Vdc)/RL; #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge of Vdc from the half-wave rectifier, in V\n", + "Vm=round(Vm,0); \n", + "#Result\n", + "print 'The a.c voltage required should have maximum value of = %d V' %Vm;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The a.c voltage required should have maximum value of = 162 V\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16, Page number 95" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "from math import pi\n", + "#Variable declaration\n", + "rf=20; #Internal resistance of the diodes in ohm\n", + "Vrms=50; #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n", + "RL=980; #Load resistance in ohm\n", + "\n", + "#Calculation\n", + "Vm=Vrms*sqrt(2); #Maximum a.c voltage in V\n", + "Im=Vm/(rf+RL); #Maximum load current in A\n", + "Im=Im*1000; #Maximum load current in mA\n", + " \n", + "# 1:\n", + "Idc=2*Im/pi; #Mean load current\n", + "\n", + "# 2:\n", + "Irms=Im/sqrt(2); #RMS value of load current in A\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print' The mean load current= %d mA'%Idc;\n", + "print 'ii:';\n", + "print ' The r.m.s value of the load current = %d mA'%Irms; " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " The mean load current= 45 mA\n", + "ii:\n", + " The r.m.s value of the load current = 50 mA\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17, Page number 95-96" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "#Variable declaration\n", + "RL=100; #Load resistance in ohm \n", + "rf=0; #Internal resistance of the diodes in ohm\n", + "Turns_ratio=5/1; #Primary to secondary turns ratio of transformer \n", + "P_Vrms=230; #R.M.S value of voltage in primary winding in V\n", + "S_Vrms=P_Vrms/Turns_ratio; #R.M.S value of voltage in secondary winding in V\n", + "S_Vm=S_Vrms*sqrt(2); #Maximum voltage across secondary winding in V\n", + "Vm=S_Vm/2; #Maximum voltage across half seconfdary winding in V\n", + "\n", + "\n", + "#Calculation\n", + "# 1:\n", + "Idc=2*Vm/(pi*RL); #Average current in A\n", + "Vdc=Idc*RL; #d.c output voltage in V\n", + "\n", + "# 2:\n", + "PIV=S_Vm; #Peak Invers Voltage(= Maximum secondary voltage) in V\n", + "\n", + "# 3:\n", + "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL); #a.c input power in watt\n", + "Pdc=(pow(Idc,2)*RL); #d.c output power in watt\n", + "R_eff=(Pdc/Pac)*100; #Rectification efficiency\n", + "R_eff=round(R_eff,1);\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' The d.c output voltage= %.1f V'%Vdc;\n", + "print 'ii:';\n", + "print ' The peak inverse voltage= %d V'%PIV;\n", + "print 'iii:';\n", + "print ' Rectification efficiency= %.1f%%'%R_eff;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " The d.c output voltage= 20.7 V\n", + "ii:\n", + " The peak inverse voltage= 65 V\n", + "iii:\n", + " Rectification efficiency= 81.1%\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%." + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18, Page number 96-97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt \n", + "#Variable declaration\n", + "fin=50; #frequency of input ac source in Hz\n", + "RL=200; #Load resistance in ohm\n", + "Turns_ratio=4/1; #Transformers turns ratio, primary to secondary.\n", + "P_Vrms=230.0; #R.M.S value of voltage in primary winding in V\n", + "S_Vrms=P_Vrms/Turns_ratio #R.M.S value of voltage in secondary winding in V\n", + "Vm=S_Vrms*sqrt(2); #Maximum voltage across secondary winding in V\n", + "\n", + "#Calculation\n", + "# 1:\n", + "Idc=2*Vm/(pi*RL); # Average current in A\n", + "Vdc=Idc*RL; #Output d.c voltage in V\n", + "Vdc=round(Vdc,0);\n", + "# 2:\n", + "PIV= Vm; #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n", + "\n", + "# 3:\n", + "fout=2*fin; #Output frequency in Hz\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' The d.c output voltage = %d V' %Vdc;\n", + "print 'ii:';\n", + "print ' The peak inverse voltage = %.1f V'%PIV;\n", + "print 'iii:';\n", + "print ' The output frequency = %d Hz'%fout;\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " The d.c output voltage = 52 V\n", + "ii:\n", + " The peak inverse voltage = 81.3 V\n", + "iii:\n", + " The output frequency = 100 Hz\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.19, Page number 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "RL=100.0; #Load Resistance in ohm\n", + "Turns_ratio=5/1; #Primary to secondary turns ratio of the transformer\n", + "Vin=230.0; #R.M.S value of input voltage in V\n", + "fin=50; #Input frequency in Hz\n", + "\n", + "#Calculation\n", + "Vs_rms=Vin/Turns_ratio; #R.M.S value of the voltage in secondary winding, in v\n", + "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across secondary, in V\n", + "\n", + "# (i)\n", + "#Case i: Centre-tap circuit\n", + "Vm=Vs_max/2; #Maximum voltage across half secondary winding, in V \n", + "Vdc=2*Vm*RL/(pi*RL); #DC output voltage, in V \n", + "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n", + "\n", + "#Case ii:\n", + "Vm=Vs_max; #Maximum voltage across secondary, in V\n", + "Vdc=2*Vm*RL/(pi*RL); #DC output voltage, in V \n", + "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n", + "\n", + "# ii:\n", + "#Case i: Centre-tap circuit\n", + "Turns_ratio=5/1; #Turns ratio of the transformer\n", + "Vs_rms=Vin/Turns_ratio; #R.M.S value of the secondary voltage in V\n", + "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across the secondary in V\n", + "Vm=Vs_max/2; #Maximum voltage across half of the secondary in V\n", + "PIV=2*Vm; #Peak Inverse Voltage in V\n", + "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n", + "\n", + "#Case ii: Bridge circuit\n", + "Turns_ratio=10/1; #Turns ratio of the transformer\n", + "Vs_rms=Vin/Turns_ratio; #R.M.S value of the secondary voltage in V\n", + "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across the secondary in V\n", + "PIV=Vm; #Peak Inverse Voltage in V\n", + "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The d.c output voltage for the centre-tap circuit = 20.7 V\n", + "The d.c output voltage for the bridge circuit = 41.4 V\n", + "PIV in case of centre-tap circuit = 65 V\n", + "PIV in case of bridge circuit = 32.5 V\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.20, Page number 98" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\n", + "#Variable declaration\n", + "rf=1; #forward resistance of diodes of the rectifier in ohm\n", + "RL=480; #Load resistance in ohm\n", + "Vrms=240.0; #a.c supply voltage in V\n", + "Vm=Vrms*sqrt(2); #Maximum a.c voltage in V \n", + "\n", + "#Calculation\n", + "# 1:\n", + "Rt=2*rf+RL; #Total circuit resistance at any instance in ohm\n", + "Im=Vm/Rt; #Maximum load current in A\n", + "Idc=2*Im/pi; #Mean load current in A\n", + "\n", + "# 2:\n", + "Irms=Im/2; #R.M.S value of current in A\n", + "P=pow(Irms,2)*rf; #Power dissipated in each diode in watt\n", + "\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' Mean load current = %.2f A'%Idc;\n", + "print 'ii:';\n", + "print ' Power dissipated in each diode= %.3f W'%P;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " Mean load current = 0.45 A\n", + "ii:\n", + " Power dissipated in each diode= 0.124 W\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W." + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.21, Page number 98-99" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt,pi\n", + "#Variable declaration\n", + "RL=12000; #Load resistance in ohm\n", + "V0=0.7; #Potential barrier voltage of diodes in V\n", + "Vrms=12; #R.M.S value of input a.c voltage in V\n", + "Vs_pk=Vrms*sqrt(2); #Peak secondary voltage in V\n", + "\n", + "#Calculation\n", + "# 1:\n", + "Vout_pk=Vs_pk-(2*V0); #Peak output voltage in V\n", + "Vav=2*Vout_pk/pi; #Average output voltage in V\n", + "Vav=round(Vav,2);\n", + "\n", + "# 2:\n", + "Iav=Vav/RL; #Average output current in A\n", + "Iav=Iav*pow(10,6); #Average output current in \u03bcA\n", + "\n", + "\n", + "#Result\n", + "print 'i:';\n", + "print ' Average output voltage=%.2f V'%Vav;\n", + "print 'ii:';\n", + "print ' Average output current=%.1f \u03bcA'%Iav;\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i:\n", + " Average output voltage=9.91 V\n", + "ii:\n", + " Average output current=825.8 \u03bcA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.22, Page number 102" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vdc_A=10; #Supply voltage of A in V\n", + "Vdc_B=25; #Supply voltage of B in V\n", + "Vac_rms_a=0.5; #Ripples in power supply A in V\n", + "Vac_rms_b=0.001; #Ripples in power supply B in V\n", + "\n", + "#Calculation\n", + "#For power supply A\n", + "ripple_factor_A=Vac_rms_a/Vdc_A; #Ripple factor of power supply A\n", + "\n", + "#For power supply B\n", + "ripple_factor_B=Vac_rms_b/Vdc_B; #Ripple factor of power supply B\n", + "\n", + "#Result\n", + "if(ripple_factor_A<ripple_factor_B):\n", + " print 'Power supply A is better';\n", + "else :\n", + " print 'Power supply B is better';" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Power supply B is better\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.23, Page number 105-106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "#Variable declaration\n", + "RL=2200; #Load resistance in ohm\n", + "C=50*pow(10,-6); #Capacitance of the capacitor used in filter circuit in F\n", + "V0=0.7; #Potential barrier voltage of the diodes of the rectifier in V\n", + "Vrms=115.0; #R.M.S value of input a.c voltage in V \n", + "fin=60; #Frequency of input a.c voltage in Hz\n", + "Turns_ratio=10/1; #Primary to secondary, turns ratio of the transformer \n", + "\n", + "#Calculation\n", + "Vp_prim=Vrms*sqrt(2); #Peak primary voltage in V\n", + "Vp_sec=Vp_prim/Turns_ratio; #Peak secondary voltage in V\n", + "Vp_in= Vp_sec - 2*V0; #Peak full wave rectified voltage at the filter input in V\n", + "f=2*fin; #Output frequency in Hz\n", + "Vdc=Vp_in*(1-(1/(2*f*RL*C))); #Output d.c voltage in V\n", + "\n", + "#Result\n", + "print 'The output d.c voltage is = %.1f V'%Vdc;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output d.c voltage is = 14.3 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.24, Page number 106" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "#Variable declaration\n", + "R=25; #d.c resistance of the choke in ohm\n", + "RL=750; #Load resistance in ohm\n", + "Vm=25.7; #Maximum value of the pulsating output from the rectifier in V\n", + "\n", + "#Calculation\n", + "V_dc=2*Vm/pi; #d.c component of the pulsating output in V\n", + "V_dc=round(V_dc,1);\n", + "V_dc_out=(V_dc*RL)/(R+RL); #Output d.c voltage in V\n", + "V_dc_out=round(V_dc_out,1);\n", + "\n", + "#Result\n", + "print ' The output d.c voltage accross the load resistance is = %.1f V'%V_dc_out;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " The output d.c voltage accross the load resistance is = 15.9 V\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.25, Page number 113-114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei=120.0; #Input Voltage in V\n", + "Vz=50.0; #Zener Voltage in V\n", + "R=5000.0; #Resistance of the series resistor in ohm\n", + "RL=10000.0; #Load resistance in ohm\n", + "\n", + "#Calculation\n", + "V=Ei*RL/(R+RL); #Voltage across the open circuit if the zener diode is removed\n", + "if(V>Vz):\n", + " #Zener diode is in ON state\n", + " # i:\n", + " Output_voltage=Vz; #Voltage across load resistance, in V\n", + " #ii:\n", + " Voltage_R=Ei-Vz; #Voltage across the series resistance R, in V\n", + " #iii:\n", + " IL=Vz/RL; #Load current through RL in A\n", + " IL=IL*1000; #Load current through RL in mA\n", + " I=Voltage_R/R; #Current through the series resistance in A\n", + " I=I*1000; #Current through the series resistance in mA\n", + " Iz=I-IL; #Applying Kirchhoff's first law, Zener current in mA\n", + " \n", + " #Result\n", + " print 'i) The output voltage across the load resistance RL = %d V'%Output_voltage;\n", + " print 'ii) The voltage drop across the series resistance R = %d V'%Voltage_R;\n", + " print 'iii) The current through the zener diode = %d mA'%Iz;\n", + " " + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) The output voltage across the load resistance RL = 50 V\n", + "ii) The voltage drop across the series resistance R = 70 V\n", + "iii) The current through the zener diode = 9 mA\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.26, Page number 114-115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Max_V=120.0; #Maximum input voltage in V\n", + "Min_V=80.0; #Minimum input voltage in V\n", + "R=5000.0; #Series resistance in ohm\n", + "RL=10000.0; #Load resistance in ohm\n", + "Vz=50.0; #Zener voltage in V\n", + "\n", + "\n", + "#Calculation\n", + "#Case i: Maximum zener current\n", + "#Zener current will be maximum when the input voltage is maximum\n", + "V_R_max=Max_V-Vz; #Voltage across series resistance R, in V\n", + "I_max=V_R_max/R; #Current through series resistance R, in A\n", + "I_max=I_max*1000; #Current through series resistance R, in mA\n", + "IL_max=Vz/RL; #Load current in A\n", + "IL_max=IL_max*1000; #Load current in mA\n", + "Iz_max=I_max-IL_max; #Applying Kirchhoff's first law, Zener current in mA;\n", + "\n", + "#Case ii: Minimum zener current\n", + "#The zener will conduct minimum current when the input voltage is minimum\n", + "V_R_min=Min_V-Vz; #Voltage across series resistance R, in V\n", + "I_min=V_R_min/R; #Current through series resistance R, in A\n", + "I_min=I_min*1000; #Current through series resistance R, in mA\n", + "IL_min=Vz/RL; #Load current in A\n", + "IL_min=IL_min*1000; #Load current in mA\n", + "Iz_min=I_min-IL_min; #Applying Kirchhoff's first law, Zener current in mA\n", + "\n", + "#Result\n", + "print 'Case i: ';\n", + "print 'Maximum zener current = %d mA'%Iz_max;\n", + "print 'Case ii: ';\n", + "print 'Minimum zener current = %d mA'%Iz_min;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Case i: \n", + "Maximum zener current = 9 mA\n", + "Case ii: \n", + "Minimum zener current = 1 mA\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.27, Page number 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei=12; #Input voltage in V\n", + "Vz=7.2; #Zener voltage in V\n", + "E0=Vz; #Voltage to be maintained across the load in V\n", + "IL_max=0.1; #Maximum load current in A\n", + "IL_min=0.012; #Minimum load current in A\n", + "Iz_min=0.01; #Minimum zener current in A\n", + "\n", + "#Calculation\n", + "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n", + "R=(Ei-E0)/(Iz_min+IL_max); #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n", + "\n", + "#Result\n", + "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.28, Page number 115" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei_min=22; #Minimum input voltage in V\n", + "Ei_max=28; #Maximum input voltage in V\n", + "Vz=18; #Zener voltage in V\n", + "E0=Vz; #Constant voltage maintained across the load resistance in V\n", + "Iz_min=0.2; #Minimum zener current in A\n", + "Iz_max=2; #Maximum zener current in A\n", + "RL=18; #Load resistance in \u03a9\n", + "\n", + "#Calculation\n", + "IL=Vz/RL; #Constant value of load current in A\n", + "#When the input voltage is minimum, the zener current will be minimum\n", + "R=(Ei_min-E0)/(Iz_min+IL) #The value of series resistance so that the voltage E0 across RL remains constant\n", + "\n", + "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.29, Page number 116 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=10 #Zener voltage in V\n", + "Ei_min=13; #Minimum input voltage in V\n", + "Ei_max=16; #Maximum input voltage in V\n", + "Iz_min=0.015; #Minimum zener current in A\n", + "IL_min=0.01; #Minimum load current in A \n", + "IL_max=0.085; #Maximum load curremt in A\n", + "E0=Vz; #Constant voltage to be maintained in V \n", + "\n", + "#Calculation\n", + "#The zener current will be minimum when the input voltage will be minimum and at that time the load current will be maximum\n", + "R=(Ei_min-E0)/(Iz_min+IL_max); #The value of series resistance R to maintain a constant voltage across load\n", + "\n", + "\n", + "#Result\n", + "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.30, Page number 116" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Iz=0.2; #Current rating of each zener in A\n", + "Vz=15; #Voltage rating of each zener in V\n", + "Ei=45; #Input voltage in V\n", + "\n", + "#Calculation\n", + "# i: Regulated output voltage across the two zener diodes \n", + "E0=2*Vz; # V\n", + "\n", + "# ii: Value of series resistance \n", + "R=(Ei-E0)/Iz; # \u03a9\n", + "\n", + "#Result\n", + "print 'i) The regulated output voltage = %d V'%E0;\n", + "print 'ii) The value of the series resistance = %d \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "i) The regulated output voltage = 30 V\n", + "ii) The value of the series resistance = 75 \u03a9\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.31, Page number 116-117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Vz=10; #Voltage rating of each zener in V\n", + "Iz=1; #Current rating of each zener in A\n", + "Ei=45; #Input unregulated voltage in V\n", + "\n", + "#Calculation\n", + "#Regulated output voltage across the three zener diodes\n", + "E0=3*Vz; # V\n", + "\n", + "#Value of series resistance to obtain a 30V regulated output voltage\n", + "R=(Ei-E0)/Iz; # \u03a9\n", + "\n", + "#Result\n", + "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.32, Page number 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#variable declaration\n", + "RL=2000.0; #Load resistance in \u03a9\n", + "R=200.0; #Series resistance in \u03a9\n", + "Iz=0.025; #Zener current rating in A\n", + "E0=30.0; #Output regulated voltage in V \n", + "\n", + "#Calculation\n", + "#Minimum input voltage will be required when Iz=0 A, and at this condition\n", + "IL=E0/RL; #Load current during Iz=0, in A\n", + "I=IL; #According to Kirchhoff's law, total current, in A\n", + "Ei_min=E0+(I*R); #Minimum input voltage in V\n", + "\n", + "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n", + "I=IL+Iz; #According to Kirchhoff's law, total current, in A\n", + "Ei_max=E0+(I*R); #maximum input voltage in V\n", + "\n", + "\n", + "#Result\n", + "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max); \n", + "\n", + "\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The required range of input voltage is from 33 V to 38 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.33, Page number 117-118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ei=16; #Unregulated input voltage in V\n", + "E0=12; #Output regulated voltage in V\n", + "IL_min=0; #Minimum load current in A\n", + "IL_max=0.2; #Maximum load current in A\n", + "Iz_min=0; #Minimum zener current in A\n", + "Iz_max=0.2; #Maximum zener current in A\n", + "\n", + "#Calculation\n", + "#As the regulated voltage required across the load is 12V\n", + "Vz=E0; #Voltage rating of zener diode in V\n", + "V_R=Ei-E0; #Constant Voltage that should remain across series resistance \n", + "#The minimum zener current will occur when the curent in the load in maximum\n", + "R=V_R/(Iz_min+IL_max); #Series resistance in \u03a9\n", + "\n", + "Max_power_rating=Vz*Iz_max; #Maximum power rating of zener diode in W\n", + "\n", + "#Result\n", + "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n", + "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n", + "The maximum power rating of the zener diode is = 2.4 W \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.34, Page number 118" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V=12; #Source voltage in V\n", + "R=1000; #Series resistance in \u03a9\n", + "RL=5000; #Load resistance in \u03a9\n", + "Vz=6; #Voltage rating of zener in V\n", + "\n", + "#Calculation\n", + "#Case i: zener is working properly\n", + "#The output voltage across the load will be equal to the zener voltage.\n", + "V0=Vz; # V\n", + "\n", + "#Result\n", + "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n", + "\n", + "#Case ii: zener is shorted\n", + "#As the zener is shorted, the potential difference across the load will be zero\n", + "V0=0; #V\n", + "\n", + "#Result\n", + "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n", + " \n", + "#Case iii: zener is open circuited\n", + "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n", + "V0=V*RL/(R+RL); #V\n", + "\n", + "#Result\n", + "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Case i: Output voltage when zener is working properly is 6 V\n", + "Case ii: Output voltage when zener is short circuited is 0 V\n", + "Case iii: Output voltage when zener is open circuited is 10 V\n" + ] + } + ], + "prompt_number": 4 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_6.ipynb new file mode 100644 index 00000000..537a179e --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter7_6.ipynb @@ -0,0 +1,212 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:e210474f5c4fc6668f4c7b5af2adf833a1c7f62577017a980ab8d11cd8ce2886" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "CHAPTER 7 : SPECIAL-PURPOSE DIODES" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1 : Page number 127-128\n" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_S=10.0; #Supply voltage in V\n", + "V_D=1.6; #Forward voltage drop of LED, in V\n", + "I_F=20.0; #Required limited current through LED, in mA\n", + "\n", + "#Calculations\n", + "R_S=(V_S-V_D)/(I_F/1000); #Series resistor required to limit the current through the LED, in \u2126\n", + "\n", + "#Result \n", + "print(\"The value of series resistor required to limit the current through the LED = %d \u2126.\"%R_S);\n", + " \n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of series resistor required to limit the current through the LED = 420 \u2126.\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2: Page number 128" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "V_S=15.0; #Supply voltage in V\n", + "V_D=2.0; #Forward voltage drop of LED, in V\n", + "R_S=2200.0; #Series resistor required to limit the current through the LED, in \u2126\n", + "\n", + "#Calculations\n", + "I_F=((V_S-V_D)/R_S)*1000; #Required limited current through LED, in mA\n", + "\n", + "#Result \n", + "print(\"The current through the LED in the circuit = %.2f mA\"%I_F);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through the LED in the circuit = 5.91 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3: Page number 132-133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "Ir=50.0; #Dark current as observed from the current Illumination curve, in mA \n", + "V_R=10.0; #Reverse voltage in V\n", + "\n", + "#Calculation\n", + "R_R=V_R/(Ir/pow(10,6)); #Dark Resistance in \u2126\n", + "\n", + "#Result\n", + "print(\"The dark resistance is=%d k\u2126\"%(R_R/1000));\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The dark resistance is=200 k\u2126\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4: Page number 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Variable declaration\n", + "E=2.5; #Illumination in mW/cm\u00b2\n", + "m=37.4; #sensitivity of the photodiode in \ud835\udf07A/mW/cm\u00b2\n", + "\n", + "#Calculations\n", + "I_R=m*E; #Reverse current in \ud835\udf07A\n", + "\n", + "#Result\n", + "print(\"The reverese current in the photodiode = %.1f \ud835\udf07A\"%I_R);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The reverese current in the photodiode = 93.5 \ud835\udf07A\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5: Page number 137" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import pi\n", + "from math import sqrt\t\n", + "#Variable declaration\n", + "L=1.0; #Inductance of the inductor in mH\n", + "C=100.0; #Capacitance of the varactor in pF\n", + "\n", + "#Result\n", + "f_r=1/(2*pi*sqrt(L*pow(10,-3)*C*pow(10,-12))); #Resonant frequency of the circuit in Hz\n", + "f_r=f_r/1000; #Resonant frequency of the circuit in kHz\n", + "\n", + "#Result\n", + "print(\"The resonant frequency of the circuit = %.1f kHz\"%f_r);\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resonant frequency of the circuit = 503.3 kHz\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "code", + "collapsed": false, + "input": [], + "language": "python", + "metadata": {}, + "outputs": [] + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_6.ipynb new file mode 100644 index 00000000..c13922ee --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter8_6.ipynb @@ -0,0 +1,1845 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1: Page number 147-148" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage amplification = 50. \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "Signal=500.0; #Signal voltage in V\n", + "Rin=20.0; #Input resistance in Ω \n", + "Rout=100.0; #Output resistance in Ω\n", + "R_C=1000.0; #Collector load in Ω\n", + "alpha_ac=1.0; #current amplification factor\n", + "\n", + "#Calculation\n", + "I_E=(Signal/1000)/Rin; \t#Input current in mA\n", + "I_C=I_E*alpha_ac; #Output current in mA\n", + "Vout=I_C*R_C; #Output voltage in V \n", + "Av=Vout/(Signal/1000); #Voltage amplification \n", + "\n", + "#Result\n", + "print(\"The voltage amplification = %d. \"%Av);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2: Page number 150" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The base current = 0.05 mA \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_E=1; #Emitter curent in mA\n", + "I_C=0.95; #Collector current in mA\n", + "\n", + "#Calculation\n", + "I_B=I_E-I_C; #Base current in mA\n", + "\n", + "#Result \n", + "print(\"The base current = %.2f mA \"%I_B);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 8.3: Page number 150\n" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The base current =0.1 mA\n" + ] + } + ], + "source": [ + "#variable declaration\n", + "alpha=0.9; #Current amplification factor\n", + "I_E=1; #Emitter current in mA\n", + "\n", + "#Calculation\n", + "I_C=alpha*I_E; #Collector current in mA\n", + "I_B=I_E-I_C; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"The base current =%.1f mA\"%I_B);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.4: Page number 150" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The current amplification factor = 0.95 .\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_C=0.95;\t\t\t#Collector current in mA\n", + "I_B=0.05;\t\t\t#Base current in mA\n", + "\n", + "#Calculation\n", + "I_E=I_B+I_C; #Emitter current in mA\n", + "alpha=I_C/I_E; #Current amplification factor \n", + "\n", + "#Result\n", + "print(\"The current amplification factor = %.2f .\"%alpha);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5: Page number 150" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The total collector current = 0.97 mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_E=1; #Emitter current in mA\n", + "I_CBO=50.0; #Collector current with emitter circuit open, in microAmp\n", + "alpha=0.92; #Current amplification factor\n", + "\n", + "#Calculation\n", + "I_C=alpha*I_E + (I_CBO/1000); #Total collector current in mA\n", + "\n", + "#Result\n", + "print(\"The total collector current = %.2f mA.\"%I_C);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.6: Page number 150-151" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The base current = 0.05 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "alpha=0.95; #Current amplification factor\n", + "Rc=2.0; #Resistor connected to the collector, in kilo ohm\n", + "V_Rc=2.0; #Voltage drop across the resistor connected to the collector in V\n", + "\n", + "\n", + "#Calculation\n", + "I_C=V_Rc/Rc; #Collector current in mA\n", + "I_E=I_C/alpha; #Emitter current in mA\n", + "I_B=I_E-I_C; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"The base current = %.2f mA\"%I_B); \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.7: Page number 151" + ] + }, + { + "cell_type": "code", + "execution_count": 7, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector current =4.87 mA\n", + "The collector to base voltage = 12.16 V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_EE=8.0; #Supply voltage at the emitter in V\n", + "V_CC=18.0; #Supply voltage at the collector in V\n", + "V_BE=0.7; #Base to emitter voltage in V\n", + "R_E=1.5; #Emitter resistance in Ω\n", + "R_C=1.2; #Collector resistance in Ω\n", + "\n", + "#Calculations\n", + "I_E=(V_EE-V_BE)/R_E; #Emitter current in mA\n", + "I_C=I_E; #Collector current in mA (approximately equal to emitter current)\n", + "V_CB=V_CC-(I_C*R_C); #Collector to base voltage in V\n", + "\n", + "#Result\n", + "print(\"The collector current =%.2f mA\"%I_C);\n", + "print(\"The collector to base voltage = %.2f V\"%V_CB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.8:Page number 155" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Value of beta =9\n", + "(ii) Value of beta =49\n", + "(iii) Value of beta =99\n" + ] + } + ], + "source": [ + "#Function for calculating beta from alpha\n", + "def calc_beta(a): #a is the value of alpha\n", + "\treturn(a/(1-a));\n", + "\n", + "#Case (i)\n", + "alpha=0.9; #current amplification factor\n", + "beta=calc_beta(alpha);\t\t#Base current amplification factor \n", + "print(\"(i) Value of beta =%d\"%beta );\t\t\t\t\t\t\t\t\t\n", + "\n", + "#Case (ii)\n", + "alpha=0.98; #current amplification factor\n", + "beta=calc_beta(alpha); #Base current amplification factor\n", + "print(\"(ii) Value of beta =%.0f\"%beta );\n", + "\n", + "\n", + "#Case (iii)\n", + "alpha=0.99; #current amplification factor\n", + "beta=calc_beta(alpha); #Base current amplification factor \n", + "print(\"(iii) Value of beta =%.0f\"%beta );\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.9: Page number 155" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The emitter curent = 1.02 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "beta=50.0; #Base current amplification factor\n", + "I_B=20.0; #Base current in microAmp\n", + "\n", + "#Calculation\n", + "I_B=I_B/1000; #Base current in mA\n", + "I_C=beta*I_B; #Collector current in mA\n", + "I_E=I_B+I_C; #Emitter current in mA\n", + "\n", + "#Result\n", + "print(\"The emitter curent = %.2f mA\"%I_E);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.10: Page number 155" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "alpha=0.98.\n", + "Collector current determined using alpha =11.76 mA\n", + "Collector current determined using beta =11.76 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_B=240.0; #Base current in microAmp\n", + "I_E=12; #Emitter current in mA\n", + "beta=49.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "alpha=beta/(1+beta); #current amplification factor \n", + "I_C_alpha=alpha*I_E; #Collector current in mA calculated using alpha\n", + "I_C_beta=beta*(I_B/1000); #Collector current in mA calculated using beta\n", + "\n", + "#Results\n", + "print(\"alpha=%.2f.\"%alpha);\n", + "print(\"Collector current determined using alpha =%.2f mA\"%I_C_alpha);\n", + "print(\"Collector current determined using beta =%.2f mA\"%I_C_beta);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.11: Page number 156" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The base current =0.022 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "beta=45.0; #Base current amplification factor\n", + "R_C=1.0; #Resistance of the collector resistance in kΩ\n", + "V_R_C=1.0; #Voltage drop across the collector resistance in V\n", + "\n", + "#Calculation\n", + "I_C=V_R_C/R_C; #Collector current in mA\n", + "I_B=I_C/beta; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"The base current =%.3f mA\"%I_B);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.12: Page number 156" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector to emitter voltage = 7.5 V\n", + "Base current= 0.026 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=8.0; #Collector supply voltage in V\n", + "R_C=800.0; #Resistance of the collector resistance in Ω\n", + "V_R_C=0.5; #Voltage drop across collector resistance in V\n", + "alpha=0.96; #current amplification factor\n", + "\n", + "#Calculation\n", + "V_CE=V_CC-V_R_C; #Collector to emitter voltage in V\n", + "I_C=V_R_C/R_C; #Collector current in A\n", + "I_C=I_C*1000; #Collector current in mA\n", + "beta=alpha/(1-alpha); #Base current amplification factor\n", + "I_B=I_C/beta; #Base current in mA\n", + "\n", + "#Result\n", + "print(\"Collector to emitter voltage = %.1f V\"%V_CE);\n", + "print(\"Base current= %.3f mA\"%I_B);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.13: Page number 156-157" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Current amplification factor = 0.99 \n", + "The emitter curent =1010 μA \n", + "The base curent =10 μA \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=5; \t#Collector supply voltage in V\n", + "I_CBO=0.2; \t#Leakage current at collector base junction with emitter open, in μA\n", + "I_CEO=20.0; \t#Leakage current with base open, in μA\n", + "I_C=1.0; #Collector current in mA\n", + "I_C=I_C*1000; \t#Collector current in μA\n", + "\n", + "\n", + "#Calculation\n", + "alpha=1-(I_CBO/I_CEO);\t\t#current amplification factor\n", + "I_E=(I_C-I_CBO)/alpha; #Emitter current in μA\n", + "I_E=round(I_E,-1);\n", + "I_B=I_E-I_C; #Base current in μA\n", + "I_B=round(I_B,-1);\n", + "\n", + "#Result\n", + "print(\"Current amplification factor = %.2f \"%alpha);\n", + "print(\"The emitter curent =%d μA \"%I_E);\n", + "print(\"The base curent =%d μA \"%I_B);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.14: Page number 157" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vale of I_CBO= 2.4 μA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_CEO=300.0; #Leakage current in common emitter configuration, in μA\n", + "beta=120.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "alpha=beta/(1+beta); #Current amplification factor\n", + "alpha=round(alpha,3);\n", + "I_CBO=(1-alpha)*I_CEO; #Leakage current in common base configuration, in μA\n", + "\n", + "\n", + "#Result\n", + "print(\"Vale of I_CBO= %.1f μA\"%I_CBO);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.15: Page number 157" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of I_CBO=0.0048 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_B=20.0; #Base current in μA\n", + "I_C=2.0; #Collector current in mA\n", + "beta=80.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "I_CEO=I_C-(beta*I_B/1000); #Leakage current with base open, in mA \n", + "alpha=beta/(beta+1); #Current amplification factor\n", + "alpha=round(alpha,3);\n", + "I_CBO=(1-alpha)*I_CEO; #Leakage current with emitter open, in mA\n", + "\n", + "\n", + "#Result\n", + "print(\"Value of I_CBO=%.4f mA\"%I_CBO);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.17: Page number 158" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector to base voltage, V_CB= 2.85 V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "beta=150.0; \t#Base current amplification factor\n", + "R_B=10.0; \t#Base resistance in kilo ohm\n", + "R_C=100.0; \t#Collector resistance in kilo ohm\n", + "V_CC=10.0; #Collector supply voltage in V\n", + "V_BB=5.0; #Base supply voltage in V\n", + "V_BE=0.7; #Base to emitter voltage in V\n", + "\n", + "\n", + "#Calculation\n", + "I_B=(V_BB-V_BE)/R_B; #Base current in mA\n", + "I_C=beta*I_B; #Collector current in mA\n", + "V_CE=V_CC - (I_C/1000)*R_C; #Collector to emitter voltage in V\n", + "V_CB=V_CE-V_BE; #Collector to base voltage in V\n", + "\n", + "\n", + "#Result \n", + "print(\"Collector to base voltage, V_CB= %.2f V\"%V_CB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.18: Page number158-159" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector current determined using alpha rating =29.93 mA\n", + "Collector current determined using beta rating =29.92 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_B=68.0; #Base current in μA\n", + "I_E=30.0; #Emitter current in mA\n", + "beta=440.0;\t #Base current amplification factor\n", + "\n", + "#Calculation\n", + "alpha=beta/(beta + 1); #current amplification factor\n", + "I_C_alpha=alpha*I_E;\t\t#Collector current using alpha rating, in mA\n", + "I_C_beta=beta*(I_B/1000.0); #Collector current using beta rating, in mA\n", + "\n", + "#Result\n", + "print(\"Collector current determined using alpha rating =%.2f mA\"%I_C_alpha);\n", + "print(\"Collector current determined using beta rating =%.2f mA\"%I_C_beta);\n", + "\n", + "#Note: In the textbook, the collector current obtained from beta rating is approximated to 29.93 mA\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.19: Page number 159" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum allowable value of base current = 1.67 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "I_C_max=500.0; #Maximum collector current in mA\n", + "beta_max=300.0; #Maximum base current amplification factor\n", + "\n", + "#Calculation\n", + "I_B_max=I_C_max/beta_max; #Maximum base current in mA\n", + "\n", + "\n", + "#Result\n", + "print(\"The maximum allowable value of base current = %.2f mA\"%I_B_max);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.22 : Page number 167-168" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f9cdbfbafd0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=12.5; #Collector supply voltage, V\n", + "RC=2.5; #Collector resistor, kΩ\n", + "\n", + "#Calculation\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC-IC*RC; #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC-VCE)/RC; #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plt.plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,15])\n", + "limit.set_ylim([0,6])\n", + "plt.xlabel('VCE(V)');\n", + "plt.ylabel('IC(mA)');\n", + "plt.title('d.c load line');\n", + "plt.grid();\n", + "plt.show(p);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.23 : Page number 168" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f9cec03e9b0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + }, + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point: IC=1mA and VCE=6V.\n" + ] + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "RC=6.0; #Collector resistor, kΩ\n", + "IB=20.0; #Zero signal base current, μA\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "#Calculation\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC-IC*RC; #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC-VCE)/RC; #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plt.plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,15])\n", + "limit.set_ylim([0,5])\n", + "plt.xlabel('VCE(V)');\n", + "plt.ylabel('IC(mA)');\n", + "plt.title('d.c load line');\n", + "plt.grid();\n", + "plt.show(p);\n", + "\n", + "#Calculating Q-point\n", + "IC=beta*(IB/1000); #Collector current, mA\n", + "VCE=VCC-IC*RC; #Collector emitter voltage, V\n", + "\n", + "#Result\n", + "print(\"Operating point: IC=%dmA and VCE=%dV.\"%(IC,VCE));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.24 : Page number 168" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Operating point: VCE=6V and IC=1mA.\n", + "(ii) Operating point: VCE=5V and IC=1mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "RC=4.0; #Collector load, kΩ\n", + "IC_Q=1.0; #Quiescent current, mA\n", + "\n", + "#Calculation\n", + "#(i)\n", + "VCC=10; #Collector supply voltage, V\n", + "VCE=VCC-IC*RC; #Collector emitter voltage, V\n", + "\n", + "print(\"(i) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n", + "\n", + "#(ii)\n", + "RC=5.0; #Collector load, kΩ\n", + "VCE=VCC-IC*RC; #Collector emitter voltage, V\n", + "print(\"(ii) Operating point: VCE=%dV and IC=%dmA.\"%(VCE,IC) );\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Example 8.25 : Page number 168-169" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point: IC=39.6mA and VCE=6.93V.\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f9cec03e978>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib import pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "VBB=10.0; #Base supply voltage, V\n", + "RC=330.0; #Collector resistor, Ω\n", + "RB=47.0; #Base resistoe, kΩ\n", + "beta=200.0; #Base current amplification factor\n", + "VBE=0.7; #Base -emitter voltage, V\n", + "\n", + "#Calculation\n", + "#VBB-IB*RB-VBE=0\n", + "IB=round(((VBB-VBE)/RB)*1000,0); #Base current, μA\n", + "IC=beta*IB/1000; #Collector current, mA\n", + "VCE=VCC-IC*(RC/1000); #Collector-emitter voltage, V\n", + "\n", + "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n", + "\n", + "#For d.c load line\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC-IC*RC; #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC-VCE)/(RC/1000.0); #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plt.plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,25])\n", + "limit.set_ylim([0,65])\n", + "plt.xlabel('VCE(V)');\n", + "plt.ylabel('IC(mA)');\n", + "plt.title('d.c load line');\n", + "plt.grid();\n", + "plt.show(p);\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.26 : Page number 169-170" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point: IC=1.8mA and VCE=9.74V.\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7f9cdbd4bdd8>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "from matplotlib import pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VEE=10.0; #Emitter supply voltage, V\n", + "RC=1.0; #Collector resistor, kΩ\n", + "RE=4.7; #Collector resistor, kΩ\n", + "RB=47.0; #Base resistoe, kΩ\n", + "beta=100.0; #Base current amplification factor\n", + "VBE=0.7; #Base -emitter voltage, V\n", + "\n", + "#Calculation\n", + "#-IB*RB-VBE-IE*RE+VEE=0\n", + "#AS, IC=beta*IB and IC~IE\n", + "IE=round((VEE-VBE)/(RE+(RB/beta)),1); #Emitter current, mA\n", + "IC=IE; #Collector current, mA\n", + "\n", + "#VCC-IC*RC-VCE-IE*RE+VEE=0\n", + "#IC~IE\n", + "VCE=VCC+VEE-IC*(RC+RE); #Collector-emitter voltage, V\n", + "\n", + "print(\"Operating point: IC=%.1fmA and VCE=%.2fV.\"%(IC,VCE));\n", + "\n", + "\n", + "#For d.c load line\n", + "#VCE=VCC-IC*RC\n", + "#For calculating VCE, IC=0\n", + "IC=0; #Collector current for maximum Collector-emitter voltage, mA\n", + "VCE_max=VCC+VEE-IC*(RC+RE); #Maximum collector-emitter voltage, V\n", + "\n", + "#For calculating VCE, IC=0\n", + "VCE=0; #Collector emitter voltage for maximum collector current, V\n", + "IC_max=(VCC+VEE-VCE)/(RC+RE); #Maximum collector current, mA\n", + "\n", + "\n", + "#Plotting of d.c load line\n", + "VCE_plot=[0,VCE_max]; #Plotting variable for VCE\n", + "IC_plot=[IC_max,0]; #Plotting variable for IC\n", + "p=plt.plot(VCE_plot,IC_plot);\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,25])\n", + "limit.set_ylim([0,5])\n", + "plt.xlabel('VCE(V)');\n", + "plt.ylabel('IC(mA)');\n", + "plt.title('d.c load line');\n", + "plt.grid();\n", + "plt.show(p);\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.27 : Page number 170-171" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Emitter voltage=-1.54V.\n", + "(i) Base voltage=10.7V.\n", + "(i) Collector voltage=8.2V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VEE=10.0; #Emitter supply voltage, V\n", + "IE=1.8; #Emitter current, mA\n", + "RE=4.7; #Emitter resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "VCC=10.0; #Collector supply voltage, V\n", + "IC=1.8; #Collector current, mA\n", + "RC=1.0; #Collector resistor, kΩ\n", + "\n", + "\n", + "#Calculation\n", + "#(i)\n", + "VE=-VEE+IE*RE; #Emitter voltage, V\n", + "\n", + "#(ii)\n", + "VB=VEE+VBE; #Base voltage, V\n", + "\n", + "#(iii)\n", + "VC=VCC-IC*RC; #Collector voltage, V\n", + "\n", + "\n", + "#Result\n", + "print(\"(i) Emitter voltage=%.2fV.\"%VE);\n", + "print(\"(i) Base voltage=%.1fV.\"%VB);\n", + "print(\"(i) Collector voltage=%.1fV.\"%VC);\n", + "\n", + "#Note: In the textbook, VB=VE+VBE has been written, which is worng. It should be VB=VEE+VBE. " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.28: Page number 173-174" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Input resistance =2 kΩ\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_BE_change=200.0; #Change in base-emitter voltage in mV\n", + "I_B_change=100.0; #Change in base current in μA\n", + "\n", + "#Calculations\n", + "Ri=V_BE_change/I_B_change; #Input resistance in kΩ\n", + "\n", + "#Result\n", + "print(\"Input resistance =%d kΩ\"%Ri);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.29; Page number 174" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The output resistance =8kΩ\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CE_final=10.0;\t\t\t#Final value of collector-emitter voltage in V\n", + "V_CE_initial=2.0; #Initial value of collector-emitter voltage in V\n", + "I_C_final=3.0; #Final value of collector current in mA\n", + "I_C_initial=2.0; #Initial value of collector current in mA\n", + "\n", + "#Calculations\n", + "V_CE_change=V_CE_final-V_CE_initial;\t\t#Change in collector to emitter voltage in V\n", + "I_C_change=I_C_final-I_C_initial; #Change in collector current in mA\n", + "R0=V_CE_change/I_C_change; #Output resistance in kΩ\n", + "\n", + "#Result\n", + "print(\"The output resistance =%dkΩ\"%R0);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.30: Page number 174" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The voltage gain of the amplifier =100 \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "R_C=2.0;\t\t#Collector load in kilo ohm\n", + "R_i=1.0;\t\t#Input resistance in kilo ohm\n", + "R_AC=R_C; #Effective collector load for single stage in kilo ohm(appoximately equal to collector load for single stage)\n", + "beta=50.0; #Current gain\n", + "\n", + "#Calculations\n", + "A_v=beta*(R_AC/R_i);\t\t#Voltage gain of the amplifier\n", + "\n", + "#Result \n", + "print(\"The voltage gain of the amplifier =%d \"%A_v);\t\t\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.31: Page number 175-176" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector current during saturation = 20 mA\n", + "Collector emitter voltage during cutoff = 20 V.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=20;\t\t#Collector supply voltage in V\n", + "R_C=1; #Collector resistance in kilo ohm\n", + "V_knee_Si=1;\t\t#Knee voltage of V_CE for Si in V \n", + "V_knee_Ge=0.5;\t\t#Knee voltage of V_CE for Ge in V\n", + "\n", + "#Calculations\n", + "I_C_sat_Si=(V_CC-V_knee_Si)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Si transistor)\n", + "I_C_sat_Ge=(V_CC-V_knee_Ge)/R_C;\t\t#Saturation (maximum) value of collector current in mA (for Ge transistor)\n", + "I_C_sat=(V_CC)/R_C;\t\t\t\t#Saturation (maximum) value of collector current in mA (neglecting knee voltage)\n", + "V_CE_cut_off=V_CC; #Collector to emitter voltage in cutoff when base current=0, in V\n", + "\n", + "#Result\n", + "print(\"Collector current during saturation = %d mA\"%I_C_sat);\n", + "print(\"Collector emitter voltage during cutoff = %d V.\"%V_CE_cut_off);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.32: Page number 176-177" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Vce(off)= 24V\n", + "Ic(sat) = 10.67 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=12.0;\t\t#Collector supply voltage in V\n", + "V_EE=12.0;\t\t#Emitter supply voltage in V\n", + "R_C=750.0;\t\t#Collector resistance in ohm\n", + "R_E=1.5;\t\t#Emitter resistance in kilo ohm\n", + "R_B=100.0;\t\t#Base resistance in ohm\n", + "beta=200;\t\t#base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Applying Kirchhoff's voltage law to the collector side of the circuit\n", + "#using the equation: Vcc -IcRc-Vce -IeRe+Vee=0\n", + "#we get Vce=Vcc+Vee-Ic(Rc+Re), [Ie=Ic, approximately]\n", + "#We get Vce(off), when Ic=0;\n", + "\n", + "I_C_Vce_off=0;\t\t\t\t\t#Collector current for Vce(off) in mA\n", + "V_CE_off=V_CC+V_EE -(I_C_Vce_off * (R_C +R_E));\t#Collector to emitter voltage in V, during transistor in off state\n", + "\n", + "#We get Ic(sat), when Vce=0\n", + "V_CE_Ic_sat=0;\t\t\t\t\t\t#Collector to emitter voltage for saturation current of collector in V\n", + "I_C_sat=(V_CC+V_EE-V_CE_Ic_sat)/(R_C+(R_E*1000));\t#Saturated collector current in A \n", + "I_C_sat=I_C_sat*1000;\t\t\t\t\t#Saturated collector current in mA\n", + "#Result\n", + "print(\"Vce(off)= %dV\"%V_CE_off);\n", + "print(\"Ic(sat) = %.2f mA\"%I_C_sat);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.33 : Page number 177" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_knee=0.2;\t\t\t\t#Knee voltage of collector-emitter voltage in V\n", + "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BB=3.0;\t\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V \t\n", + "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=1.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=50.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#applying Kirchhoff's voltage law along the collector side of the circuit,\n", + "#We get Vcc-Ic(sat)*Rc-V_knee=0\n", + "#From the above equation, we get:\n", + "I_C_sat=(V_CC-V_knee)/R_C;\t\t#Saturated collector current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law along base emitter side,\n", + "#We get VBB-IB*RB-VBE=0;\n", + "#From the above equation, we get:\n", + "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n", + "\n", + "\n", + "I_C=beta*I_B\t\t\t\t#Collector current in mA\n", + "\n", + "#Result\n", + "if(I_C>I_C_sat):\n", + "\tprint(\"The base current is large enough to produce Ic greater than Ic(sat), therefore the transistor is saturated.\");\n", + "else:\n", + "\tprint(\"The base current is not large enough to produce Ic greater than Ic(sat), therefore the transistor isn't saturated. \");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.34: Page number 177-178" + ] + }, + { + "cell_type": "code", + "execution_count": 37, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "V_CC=10.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BE=0.95;\t\t\t\t#Base-emitter voltage in V \t\n", + "I_B=100.0;\t\t\t\t#Base current in microAmp\n", + "R_C=970.0;\t\t\t\t#Collector resistor's resistance in ohm\n", + "beta=100.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculations\n", + "I_C=(I_B/1000)*beta;\t\t\t\t#Collector current in mA \n", + "\n", + "#Applying Kirchhoff's voltage law along collector side\n", + "#We get Vcc-IcRc-Vce=0\n", + "#From the above equation, we get:\n", + "\n", + "V_CE=V_CC-((I_C/1000)*R_C);\t\t\t\t#Collector-emitter voltage in V\n", + "\n", + "#From the equation, V_CE=V_CB+V_BE,\n", + "V_CB=V_CE-V_BE;\t\t\t\t\t\t#Collector-base voltage in V\n", + "\n", + "\n", + "#Result\n", + "if(V_CB<0 and V_BE >0):\n", + "\tprint(\"As both collector-base and emitter-base junction are forward biased, the transistor is operating in the saturation region. \");\n", + "else:\n", + "\tprint(\"No. The transistor isn't operating in the saturation region.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.35: Page number 178" + ] + }, + { + "cell_type": "code", + "execution_count": 38, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Therefore, for putting transistor in saturation, VBB >= 1.95 V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=10.0;\t\t\t\t#Collector supplu voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n", + "R_B=50.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=2.0;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=200.0;\t\t\t\t#Base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Applying Kirchhoff's voltage law along the collector side,\n", + "#We get, Vcc-Ic(sat)*Rc-Vce=0;\n", + "#From the above equation, we get:\n", + "#I_C_sat=(V_CC-V_CE)/R_C, but as transistor goes into saturation, Vce=0;\n", + "\n", + "V_CE=0;\t\t\t\t\t\t#Collector-emiter voltage in V, for transistor in saturation \n", + "I_C_sat=(V_CC-V_CE)/R_C;\t\t\t#Saturated collector current in mA\n", + "\n", + "I_B=I_C_sat/beta;\t\t\t\t#Base current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law to the base circuit,\n", + "#We get, VBB - IB*RB - VBE=0\n", + "#From the above equation. we get:\n", + "V_BB=V_BE+ I_B*R_B;\t\t\t\t#Base supply voltage to put transistor in saturation, in V\n", + "\n", + "#Result\n", + "print(\"Therefore, for putting transistor in saturation, VBB >= %.2f V\"%V_BB);\n", + " \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.36: Page number 178-179" + ] + }, + { + "cell_type": "code", + "execution_count": 39, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\n", + "(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\n", + "(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=10.0;\t\t\t#Collector supply voltage in V\n", + "V_BB=2.7;\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n", + "beta=100.0;\t\t\t#Base current amplification factor\n", + "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n", + "\n", + "\n", + "#Calcultaion\t\n", + "V_B=V_BB;\t\t\t#Base voltage in V\n", + "V_E=V_B-V_BE;\t\t\t#Emitter voltage in V\n", + "I_E=V_E/R_E;\t\t\t#Emitter current in mA\n", + "I_C=I_E;\t\t\t#Collector current (approximately equal to emitter current) in mA\n", + "I_B=I_C/beta;\t\t\t#Base current in mA\n", + "\n", + "#Case (i):\n", + "R_C=2;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "#We get,Vcc-IcRc=Vc,\n", + "\n", + "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n", + "\n", + "if(V_C>V_E):\n", + "\tprint(\"(i)Our assumption was correct, the transistor is in active state for Rc=2 kilo ohm.\");\n", + "elif(V_C<V_E):\n", + "\tprint(\"(i)Our assumption was wrong, the transistor is in saturation for Rc=2 kilo ohm.\");\n", + "elif(V_C==V_E):\n", + "\tprint(\"(i)The transistor is at the edge of saturation for Rc=2 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n", + "\n", + "#Case (ii):\n", + "R_C=4;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "#We get,Vcc-IcRc=Vc,\n", + "\n", + "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n", + "if(V_C>V_E):\n", + "\tprint(\"(ii)Our assumption was correct, the transistor is in active state for Rc=4 kilo ohm.\");\n", + "elif(V_C==V_E):\n", + "\tprint(\"(ii)The transistor is at the edge of saturation for Rc=4 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n", + "elif(V_C<V_E):\n", + "\tprint(\"(ii)Our assumption was wrong, the transistor is in saturation for Rc=4 kilo ohm.\");\n", + "\n", + "\n", + "#Case (iii):\n", + "R_C=8;\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "#We get,Vcc-IcRc=Vc,\n", + "\n", + "V_C=V_CC-I_C*R_C;\t\t#Collector voltage in V\n", + "if(V_C>V_E):\n", + "\tprint(\"(iii)Our assumption was correct, the transistor is in active state for Rc=8 kilo ohm.\");\n", + "elif(V_C<V_E):\n", + "\tprint(\"(iii)Our assumption was wrong, the transistor is in saturation for Rc=8 kilo ohm.\");\n", + "elif(V_C==V_E):\n", + "\tprint(\"(iii)The transistor is at the edge of saturation for Rc=8 kilo ohm, therefore relation between transistor currents are same for both saturation and active state.\");\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.37 : Page number 179-180" + ] + }, + { + "cell_type": "code", + "execution_count": 40, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) Base voltage =0.5V is less than VBE=0.7V, therefore, transistor is cut-off.\n", + "0.8 0.8 7.0\n", + "(ii) VC=7V > VE=0.8V, therefore the transistor is active. Our assumption was correct.\n", + "(iii) VC=-8V < VE=2.3V, therefore the transistor is saturated. Our assumption was wrong.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=15.0;\t\t\t#Collector supply voltage in V\n", + "R_C=10.0;\t\t\t#Collector resistor's resistance in kilo ohm\n", + "V_BE=0.7;\t\t\t#Base-emitter voltage in V\n", + "beta=100.0;\t\t\t#Base current amplification factor\n", + "R_E=1.0;\t\t\t#Emitter resistor's resistance in kilo ohm\n", + "\n", + "\n", + "#Calculation\t\n", + "\n", + "#Case (i):\n", + "V_BB=0.5;\t\t\t#Base supply voltage in V\n", + "VB=V_BB; #Base voltage, V\n", + "print(\"(i) Base voltage =%.1fV is less than VBE=%.1fV, therefore, transistor is cut-off.\"%(VB,V_BE));\n", + "\n", + "\n", + "#Case (ii):\n", + "V_BB=1.5;\t\t\t#Base supply voltage in V\n", + "VB=V_BB; #Base voltage, V\n", + "VE=VB-V_BE; #Emitter voltage, V\n", + "IE=round(VE/R_E,1); #Emitter current, mA\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "IC=IE; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VC=V_CC-IC*R_C; #Collector voltage, V\n", + "print(VE,IE,VC);\n", + "print(\"(ii) VC=%dV > VE=%.1fV, therefore the transistor is active. Our assumption was correct.\"%(VC,VE));\n", + "\n", + "#Case (iii):\n", + "V_BB=3; \t\t\t#Base supply voltage in V\n", + "VB=V_BB; #Base voltage, V\n", + "VE=VB-V_BE; #Emitter voltage, V\n", + "IE=round(VE/R_E,1); #Emitter current, mA\n", + "#Assuming transistor to be in active state\n", + "#Applying Kirchhoff's voltage law along collector side,\n", + "IC=IE; #Collector current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VC=V_CC-IC*R_C; #Collector voltage, V\n", + "\n", + "print(\"(iii) VC=%dV < VE=%.1fV, therefore the transistor is saturated. Our assumption was wrong.\"%(VC,VE));" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.38: Page number 181" + ] + }, + { + "cell_type": "code", + "execution_count": 41, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum collector current that can be allowed without destruction of the transistor = 5 mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "P_D_max=100.0;\t\t\t#Maximum power dissipation of a transistor in mW\n", + "V_CE=20.0;\t\t\t#Collector emitter voltage in V\n", + "\n", + "#Calculation\n", + "#As power=curent*voltage\n", + "#P_D_max=I_C_max*V_CE\n", + "#From the above equation, we get:\n", + "\n", + "I_C_max=P_D_max/V_CE;\t\t#Maximum collector current that can be allowed without destruction of the transistor, in mA\n", + "\n", + "#Result\n", + "print(\"Maximum collector current that can be allowed without destruction of the transistor = %d mA.\"%I_C_max); \n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.39: Page number 181" + ] + }, + { + "cell_type": "code", + "execution_count": 42, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power dissipated = 4.3W\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BB=5.0;\t\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n", + "R_B=1.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=200.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculation\n", + "\n", + "#Applying Kirchhoff's voltage law along base circuit<\n", + "#We get, VBB- IB*RB - VBE=0.\n", + "#From the above equation, we get:\n", + "\n", + "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n", + "\n", + "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law along collector circuit:\n", + "\n", + "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n", + "\n", + "#As power=curent*voltage\n", + "#P_D=I_C*V_CE\n", + "#From the above equation, we get:\n", + "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n", + "P_D=P_D/1000;\t\t\t\t#Power dissipated in W\n", + "\n", + "#Result\n", + "print(\"Power dissipated = %.1fW\"%P_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.40: Page number 181-182" + ] + }, + { + "cell_type": "code", + "execution_count": 43, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power dissipated = 6mW\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=5.0;\t\t\t\t#Collector supply voltage in V\n", + "V_BB=1.0;\t\t\t\t#Base supply voltage in V\n", + "V_BE=0.7;\t\t\t\t#Base-emitter voltage in V\n", + "R_B=10.0;\t\t\t\t#Base resistor's resistance in kilo ohm\n", + "R_C=1.0;\t\t\t\t\t#Collector resistor's resistance in kilo ohm\n", + "beta=100.0;\t\t\t\t#base current amplification factor\n", + "\n", + "#Calculation\n", + "\n", + "#Applying Kirchhoff's voltage law along base circuit<\n", + "#We get, VBB- IB*RB - VBE=0.\n", + "#From the above equation, we get:\n", + "\n", + "I_B=(V_BB-V_BE)/R_B;\t\t\t#Base current in mA\n", + "\n", + "I_C=beta*I_B;\t\t\t\t#Collector current in mA\n", + "\n", + "#Applying Kirchhoff's voltage law along collector circuit:\n", + "\n", + "V_CE=V_CC-I_C*R_C;\t\t\t#Collector-emitter voltage in V\n", + "\n", + "#As power=curent*voltage\n", + "#P_D=I_C*V_CE\n", + "#From the above equation, we get:\n", + "P_D=V_CE*I_C;\t\t\t\t#Power dissipated in mW\n", + "\n", + "\n", + "#Result\n", + "print(\"Power dissipated = %.0fmW\"%P_D);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.41 : Page number 182" + ] + }, + { + "cell_type": "code", + "execution_count": 44, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IC=19.5mA is much less than IC_max=100mA. Therefore, will not change with VCC and current rating is not exceeded.\n", + "PD=293mW is less than PD_max=800mW. Therefore, power rating is not exceeded.\n", + "If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VBB=5.0; #Base supply voltage, V\n", + "RB=22.0; #Base resistor, kilo ohm\n", + "RC=1.0; #Collector resistor, kilo ohm\n", + "beta=100.0; #Base current amplification factor\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "PD_max=800.0; #Maximum power dissipation, mW\n", + "VCE_max=15.0; #Maximum collector-emitter voltage, V\n", + "IC_max=100.0; #Maximum collector current, mA\n", + "\n", + "#Calculation\n", + "IB=((VBB-VBE)/RB)*1000; #Base current, μA\n", + "IC=beta*IB/1000; #Collector current, mA\n", + "\n", + "print(\"IC=%.1fmA is much less than IC_max=%dmA. Therefore, will not change with VCC and current rating is not exceeded.\"%(IC,IC_max));\n", + "\n", + "#VCC=VCE+IC*RC\n", + "VCC_max=VCE_max+IC*RC; #Maximum value of Collector supply voltage, V\n", + "PD=VCE_max*IC; #Power dissipation, mW\n", + "\n", + "print(\"PD=%dmW is less than PD_max=%dmW. Therefore, power rating is not exceeded.\"%(PD,PD_max));\n", + "\n", + "print(\"If base current is removed, transistor will turn off. Hence, VCE_max will be exceeded because entire supply voltage VCC will be dropped across the transistor.\");" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "collapsed": false + }, + "outputs": [], + "source": [] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + }, + "widgets": { + "state": {}, + "version": "1.1.2" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_6.ipynb b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_6.ipynb new file mode 100644 index 00000000..6fa757b3 --- /dev/null +++ b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/chapter9_6.ipynb @@ -0,0 +1,1880 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# CHAPTER 9 : TRANSISTOR BIASING" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.1: Page number 195-196" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The maximum allowed collector current during application of signal for faithful amplification = 2 mA.\n", + "The minimum zero signal collector current required = 1 mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "V_CC=6.0; #Collector supply voltage\n", + "R_C=2.5; #Collector load in kΩ\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "#For faithful amplification Vce (collector-emitter voltage)> 1V for Si transistor.\n", + "V_CE_max=1; #Maximum allowed collector-emitter voltage for faithful amplification, in V.\n", + "V_Rc_max=V_CC-V_CE_max; #maximum voltage drop across collector load in V.\n", + "I_C_max=V_Rc_max/R_C; #Maximum allowed collector current in mA\n", + "\n", + "#(ii)\n", + "IC_min_zero_signal=I_C_max/2; #Minimum zero signal collector current in mA\n", + "\n", + "#Results\n", + "print(\"The maximum allowed collector current during application of signal for faithful amplification = %d mA.\"%I_C_max);\n", + "print(\"The minimum zero signal collector current required = %d mA.\"%IC_min_zero_signal);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.2: Page number 196" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Maximum base current =30 𝜇A.\n", + "Maximum input signal voltage =600 mV.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=13.0; #Collector supply voltage in V\n", + "V_knee=1.0; #Knee voltage in V\n", + "R_C=4.0; #Collector load in kΩ\n", + "rate_IC_VBE=5.0; #Rate of change of collector current IC with base-emitter voltage VBE in mA/V.\n", + "beta=100.0; #base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "V_Rc_max=VCC-V_knee; #Maximum allowed voltage across collector load in V\n", + "I_C_max=V_Rc_max/R_C; #Maximum allowed collector current in mA\n", + "I_B_max=I_C_max/beta; #Maximum base current in mA\n", + "I_B_max=I_B_max*1000; #Maximum base current in 𝜇A\n", + "\n", + "V_B_max=I_C_max/rate_IC_VBE; #Maximum base voltage signal in V\n", + "V_B_max=V_B_max*1000; #Maximum base voltage signal in mV\n", + "\n", + "#Results\n", + "print(\"Maximum base current =%d 𝜇A.\"%I_B_max);\n", + "print(\"Maximum input signal voltage =%d mV.\"%V_B_max);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.3: Page number 200-201" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector current = 1mA\n", + "Collector emitter voltage =7V.\n", + "The new operating point for base resistor RB=50 kΩ is, VCE=5V and IC=2mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=9.0; #Colector supply voltage in V\n", + "VBB=2.0; #Base supply voltage in V\n", + "R_B=100.0; #Base resistor's resistance in kΩ\n", + "R_C=2.0; #Collector load in kΩ\n", + "beta=50.0; #base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Case (i):\n", + "\n", + "#Applying Kirchhoff's law to the input circuit\n", + "#We get, IB*RB +VBE =VBB.\n", + "#Neglecting the small base-emitter voltage, we get:\n", + "I_B=VBB/R_B; #Base current in mA\n", + "I_C=beta*I_B; #Collector current in mA\n", + "\n", + "print(\"Collector current = %dmA\"%I_C);\n", + "\n", + "#Applying Kirchhoff's law to the output ciruit\n", + "#We get, IC*RC + VCE= VCC.\n", + "#From the above equation, we get:\n", + "V_CE=VCC-I_C*R_C; #Collector emitter voltage in V\n", + "\n", + "print(\"Collector emitter voltage =%dV.\"%V_CE);\n", + "\n", + "\n", + "#Case (ii):\n", + "\n", + "R_B=50.0;\n", + "I_B=VBB/R_B;\n", + "I_C=beta*I_B;\n", + "V_CE=VCC - I_C*R_C;\n", + "\n", + "print(\"The new operating point for base resistor RB=50 kΩ is, VCE=%dV and IC=%dmA.\"%(V_CE,I_C));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.4: Page number 201-202" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point: VCE= 4V and IC=1 mA\n", + "Stability factor= 101.\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fddb80f7b38>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "#variable declaration\n", + "beta=100.0; #base current amplification factor\n", + "VCC=6.0; #Collector suply voltagein V\n", + "VBE=0.7 #Base emitter voltage in V\n", + "R_B=530.0; #Base resistor's resistance in kΩ .\n", + "R_C=2.0; #Collector resistor's resistance in kΩ .\n", + "\n", + "#Calculation\n", + "#D.C load line equation : VCE=VCC-IC*RC;\n", + "#Calculating maximum VCE ,by IC=0;\n", + "I_C_Vce_max=0; #Collector current for maximum collector-emitter voltage, in mA\n", + "VCE_max=VCC;-I_C_Vce_max*R_C; #Maximum collector-emitter voltage in V\n", + "\n", + "\n", + "#Calculating maximum collector current IC,by VCE=0;\n", + "V_CE_IC_max=0; #Collector-emitter voltage for maximum collector current, in V \n", + "I_C_max=(VCC-V_CE_IC_max)/R_C; #Maximum collector current in mA\n", + "\n", + "\n", + "#Operating point:\n", + "#For input circuit, applying Kirchhoff's law, We get,\n", + "#VCC=IB*RB + VBE.\n", + "#From the above equation,\n", + "IB=(VCC-VBE)/R_B; #Base current in mA\n", + "IC=beta*IB; #Collector current\n", + "\n", + "#From the output circuit, applying Kirchhoff's law, we get:\n", + "VCE=VCC-IC*R_C; #Collector-emitter voltage in V\n", + "\n", + "\n", + "#Stability factor\n", + "SF=beta+1; \n", + "\n", + "#Result\n", + "print(\"Operating point: VCE= %dV and IC=%d mA\"%(VCE,IC));\n", + "print(\"Stability factor= %d.\"%SF);\n", + "\n", + "\n", + "#plot\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,10])\n", + "limit.set_ylim([0,5])\n", + "VCE=[i for i in range(0,(int)(VCC+1))]; #Plot variable for V_CE\n", + "IC=[((VCC-i)/(R_C)) for i in (VCE[:])]; #Plot variable for I_C\n", + "\n", + "p=plt.plot(VCE,IC);\n", + "plt.xlabel(\"VCE(V)\");\n", + "plt.ylabel(\"IC(mA)\");\n", + "plt.title(\"d.c load line\");\n", + "plt.show(p);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.5: Page number 202" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = 1170 kΩ\n", + "The new value of zero signal collector current =0.5mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "beta=100.0; #base current amplification factor\n", + "I_C_zero_signal=1.0; #zero signal collector current in mA\n", + "VBE=0.3; #Base-emitter voltage of Ge transistor in V\n", + "\n", + "#calculations\n", + "\n", + "#Case(i)\n", + "I_B_zero_signal=I_C_zero_signal/beta; #Zero signal base current in mA\n", + "\n", + "#applying the Kirchhoff's law along input circuit:\n", + "#We get, VCC=IB*RB +VBE\n", + "#From the above equation we get,\n", + "R_B=(VCC-VBE)/I_B_zero_signal; #Required base resistor's resistance in kΩ\n", + "\n", + "print(\"Value of base resistor for operating the given Ge transistor at zero signal IC=1mA is = %d kΩ\"%R_B);\n", + "\n", + "\n", + "\n", + "#Case(ii)\n", + "beta=50;\n", + "I_B=(VCC-VBE)/R_B; #Base current of another transistor with beta=50, in mA\n", + "I_C_zero_signal=beta*I_B; #Zero signal collector current for beta=50 , in mA\n", + "\n", + "print(\"The new value of zero signal collector current =%.1fmA\"%I_C_zero_signal);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.6:Page number 202-203" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Base current =0.0091 mA\n", + "Collector current =0.91 mA\n", + "Emitter current =0.919 mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage in V\n", + "VBE=0; #Base emitter voltage in V(considering itas zero due to it's small value)\n", + "R_B=1.0; #Base resistor's resistance in MΩ\n", + "R_C=2.0; #Collector resistor's resistance in kΩ \n", + "R_E=1.0; #Emitter resistor's resistance in kΩ\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#using Kirchhoff's law in the input circuit, we get:\n", + "#VCC=IB*RB +VBE +IE*RE\n", + "#Since, IE=(beta +1)*I_B\n", + "#From the above equation we get:\n", + "I_B=round((VCC-VBE)/((beta + 1)*R_E + R_B*1000),4); #Base current in mA\n", + "I_C=round(beta*I_B,2); #Collector current in mA\n", + "I_E=I_B+I_C; #Emitter current in mA\n", + "\n", + "#Result\n", + "print(\"Base current =%.4f mA\"%I_B);\n", + "print(\"Collector current =%.2f mA\"%I_C);\n", + "print(\"Emitter current =%.3f mA\"%I_E);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.7: Page number 203-204" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector load =3.5 kΩ .\n", + "Base resistor=720 kΩ .\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCE=8.0; #Collector-emitter voltage at operating point in V\n", + "IC=2.0; #Colector current at operating point in mA\n", + "VCC=15.0; #Collector supply voltagein V\n", + "beta=100.0; #base current amplification factor\n", + "VBE=0.6; #base emitter voltage in V\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC=VCE+IC*RC.\n", + "#So, from above equation we get:\n", + "RC=(VCC-VCE)/IC; #Collector resistor's resistance in kΩ .\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#Applying Kirchhoff's law along the input circuit,\n", + "#we get, VCC=IB*RB + VBE\n", + "#So, from the above equation:\n", + "RB=(VCC-VBE)/IB; #Base resistor's resistance in kΩ .\n", + "\n", + "\n", + "#Results\n", + "print(\"Collector load =%.1f kΩ .\"%RC);\n", + "print(\"Base resistor=%d kΩ .\"%RB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.8: Page number 204" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The percentage change in collector current =50%\n", + "The percentage change in collector-emitter voltage =-56.3%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "RB=100.0; #Base resistor's resistance in kΩ\n", + "RC=560.0; #Collector resistor's resistance in Ω\n", + "beta_25=100.0; #base current amplification factor at 25 degree celsius\n", + "beta_75=150.0; #base current amplification factor at 25 degree celsius\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "\n", + "#Applying Kirchhoff's law along input circuit, we get\n", + "#VCC=IB*RB+VBE\n", + "IB=(VCC-VBE)/RB; #Base current at 25 degree celsius, in mA\n", + "\n", + "\n", + "#For temperature 25 degree celsius\n", + "IC_25=beta_25*IB; #Collector current at 25 degree celsius, in mA\n", + "\n", + "\n", + "#Applying Kirchhoff's alw at the output circuit,\n", + "#we get: VCC=IC*RC + VCE\n", + "#From the above equation,\n", + "VCE_25=round(VCC-(IC_25/1000)*RC,2); #Collector emitter voltage at 25 degree celsius, in V\n", + "\n", + "\n", + "#For temperature 75 degree celsius\n", + "IC_75=round(beta_75*IB,0); #Collector current at 75 degree celsius, in mA\n", + "\n", + "#Applying Kirchhoff's alw at the output circuit,\n", + "#we get: VCC=IC*RC + VCE\n", + "#From the above equation,\n", + "VCE_75=round(VCC-(IC_75/1000)*RC,2); #Collector emitter voltage at 75 degree celsius, in V\n", + "\n", + "\n", + "change_IC=(IC_75-IC_25)*100.0/IC_25; #percentage change in collector current\n", + "change_VCE=(VCE_75-VCE_25)*100.0/VCE_25; #Percentage change in collector-emitter voltage \n", + "\n", + "#Results\n", + "print(\"The percentage change in collector current =%d%%\"%change_IC);\n", + "print(\"The percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.10: Page number 205" + ] + }, + { + "cell_type": "code", + "execution_count": 11, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector supply voltage = 20V\n", + "Collector load=2.5 kΩ .\n", + "Base resistor's resistance=482.5 kΩ .\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCE_max=20.0; #Maximum collector-emitter voltage in V\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "IC_max=8.0; #Maximum collector current in mA\n", + "IB=40.0; #Base current in microampere\n", + "\n", + "#Calculations\n", + "\n", + "#During cut off state the collector-emitter voltage is maximum and equal to collector supply voltage\n", + "VCC=VCE_max; #Collector supply voltage in V\n", + "\n", + "#Maximum collector current IC_max=collector supply voltage(VCC)/collector load(RC)\n", + "#Collector load(RC)=VCC*IC_max\n", + "RC=VCC/IC_max; #Collector load in kΩ .\n", + "\n", + "#Applying Kirchhoff's law along input circuit,\n", + "#we get, VCC=IB*RB +VBE.\n", + "#From the above equation, we get:\n", + "RB=(VCC-VBE)/(IB/1000); #Base resistor's resistance in kΩ .\n", + "\n", + "#Results\n", + "print(\"Collector supply voltage = %dV\"%VCC);\n", + "print(\"Collector load=%.1f kΩ .\"%RC);\n", + "print(\"Base resistor's resistance=%.1f kΩ .\"%RB);\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.12: Page number 208" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The collector current = 1.73 mA\n", + "The emitter current = 1.73 mA\n", + "The voltage at collector terminal = 11.9 V\n", + "The collector-emitter voltage = 14.6 V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage in V\n", + "VEE=-20.0; #Emitter supply voltage in V\n", + "RB=100.0; #Base resistor's resistance in kΩ\n", + "RC=4.7; #Collector resistor's resistance in kΩ\n", + "RE=10.0; #Emitter resistor's resistance in kΩ\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "beta=85.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n", + "#we get,IB*RB +IE*RE +VBE -VEE=0.\n", + "#Since IB=IC/beta and IC~IE,\n", + "#(IE/beta)*RB + IE*RE + VBE + VEE =0.\n", + "IE=(-VEE-VBE)/(RE + RB/beta); #Emitter current in mA\n", + "IC=IE; #Collector current (approximately equal to emitter current) in mA\n", + "\n", + "#Applying Kirchhoff's law from VCC till collector terminal,\n", + "#we get, VCC - IC*RC =VC\n", + "VC=VCC-IC*RC; #voltage at collector terminal in V\n", + "\n", + "#Applying Kirchhoff's law from emitter terminal to VEE\n", + "#we get, VE -IE*RE =VEE\n", + "VE=VEE + IE*RE; #Voltage at emitter treminal in V\n", + "\n", + "VCE=VC-VE; #Collector-emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"The collector current = %.2f mA\"%IC);\n", + "print(\"The emitter current = %.2f mA\"%IE);\n", + "print(\"The voltage at collector terminal = %.1f V\"%VC);\n", + "print(\"The collector-emitter voltage = %.1f V\"%VCE);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.13: Page number 208-209" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Percentage change in collector current =1.7%\n", + "Percentage change in collector-emitter voltage =-3.5%\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage in V\n", + "VEE=-20.0; #Emitter supply voltage in V\n", + "RB=100.0; #Base resistor's resistance in kΩ\n", + "RC=4.7; #Collector resistor's resistance in kΩ\n", + "RE=10.0; #Emitter resistor's resistance in kΩ\n", + "beta1=85.0; #Base current amplification factor for case 1 \n", + "beta2=100.0; #Base current amplification factor for case 1\n", + "VBE_1=0.7; #Base emitter voltage for case 1 in V\n", + "VBE_2=0.6; #Base emitter voltage for case 2 in V\n", + "\n", + "\n", + "#Calculations\n", + "#For beta=85 and VBE=0.7,\n", + "#As calculated in the previous question,\n", + "IC_1=1.73; #Collector current in mA.\n", + "VCE_1=14.6; #Collector-emitter voltage in V.\n", + "\n", + "\n", + "#For case (ii)\n", + "#beta=100 and VBE=0.6\n", + "\n", + "#Applying Kirchhoff's voltage law along the base-emitter circuit (input circuit),\n", + "#we get,IB*RB +IE*RE +VBE -VEE=0.\n", + "#Since IB=IC/beta and IC~IE,\n", + "#(IE/beta)*RB + IE*RE + VBE +VEE =0.\n", + "IE_2=round((-VEE-VBE_2)/(RE + RB/beta2),2); #Emitter current in mA\n", + "IC_2=IE_2; #Collector current (approximately equal to emitter current) in mA\n", + "\n", + "#Applying Kirchhoff's law from VCC till collector terminal,\n", + "#we get, VCC - IC*RC =VC\n", + "VC=round(VCC-IC_2*RC,1); #voltage at collector terminal in V\n", + "\n", + "#Applying Kirchhoff's law from emitter terminal to VEE\n", + "#we get, VE -IE*RE =VEE\n", + "VE=round(VEE + IE_2*RE,1); #Voltage at emitter treminal in V\n", + "\n", + "VCE_2=VC-VE; #Collector-emitter voltage in V\n", + "\n", + "\n", + "change_IC= (IC_2-IC_1)*100/IC_1; #%age change in collector current\n", + "\n", + "change_VCE=(VCE_2-VCE_1)*100/VCE_2; #%age change in collector-emitter voltage\n", + "\n", + "\n", + "\n", + "#Results\n", + "print(\"Percentage change in collector current =%.1f%%\"%change_IC);\n", + "print(\"Percentage change in collector-emitter voltage =%.1f%%\"%change_VCE);" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.14: Page number 210" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The operating point : VCE=10.35V and IC=9.65mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage in V\n", + "VBE=0.7 #Base-emitter voltage in V\n", + "RB=100.0; #Base resistor's resistance in kΩ\n", + "RC=1.0; #Collector resistor's resistance in kΩ\n", + "beta=100.0; #base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along input circuit,\n", + "#we get, VCC -IC*RC -IB*RB -VBE=0.\n", + "#since IC= beta*IB,\n", + "#We get,\n", + "IB=(VCC-VBE)/(RB + beta*RC); #Base current in mA\n", + "IC=beta*IB; #Collector current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC=0.\n", + "#From the above equation,\n", + "VCE=VCC-IC*RC; #Collector emitter voltage in V\n", + "\n", + "\n", + "#Results\n", + "print(\"The operating point : VCE=%.2fV and IC=%.2fmA.\"%(VCE,IC));\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.15: Page number 210-211" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The resistance value of base resistor=770 kΩ and collector load= 4 kΩ.\n", + "The operating point : VCE=9.6V and IC=0.6mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "VBE=0.3; #Base emitter voltage in V\n", + "IC=1.0; #Collector current in mA\n", + "VCE=8.0; #Collector emitter voltage in V\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "\n", + "#Case(i)\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-IC*RC-VCE=0.\n", + "#from the above equation we get,\n", + "RC=(VCC-VCE)/IC; #Collector load in kilo ohm\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#Applying Kirchhoff's law along input circuit\n", + "#we get, VCC-VBE-(beta*IB*RC)-IB*RB=0.\n", + "#From the above equation we get,\n", + "RB=round((VCC-VBE-beta*IB*RC)/IB,0); #Base resistor's resistance in kΩ\n", + "\n", + "#Results\n", + "print(\"The resistance value of base resistor=%d kΩ and collector load= %d kΩ.\"%(RB,RC));\n", + "\n", + "#Case(ii)\n", + "\n", + "beta=50;\n", + "\n", + "#Applying Kirchhoff's law along input circuit,\n", + "#we get, VCC -IC*RC -IB*RB -VBE=0.\n", + "#since IC= beta*IB,\n", + "#We get,\n", + "IB=(VCC-VBE)/(RB + beta*RC); #Base current in mA\n", + "IC=beta*IB; #Collector current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC=0.\n", + "#From the above equation,\n", + "VCE=round(VCC-IC*RC,1); #Collector emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"The operating point : VCE=%.1fV and IC=%.1fmA.\"%(VCE,IC));\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.16 : Page number 211" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Value of base resistor's resistance=130 kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCE=2.0; #Collector-emitter voltage at operating point in V\n", + "VBE=0.7; #Base-emitter voltage in V \n", + "IC=1.0; #Collector current at operating point in mA\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#As, VCE=VCB +VBE\n", + "#we get,\n", + "VCB=VCE-VBE; #Collector-base voltage in V\n", + "RB=VCB/IB; #Base resistor's resistance in kΩ\n", + "\n", + "#Results\n", + "print(\"Value of base resistor's resistance=%d kΩ.\"%RB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.17 : Page number 211-212" + ] + }, + { + "cell_type": "code", + "execution_count": 17, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The operating point : VCE=5.7V and IC=1.26mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "VBE=0.7 #Base-emitter voltage in V\n", + "RB=400.0; #Base resistor's resistance in kΩ\n", + "RC=4.0; #Collector resistor's resistance in kΩ\n", + "RE=1.0; #Emitter resistor's resistance in kΩ\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along outut circuit,\n", + "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n", + "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n", + "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n", + "IC=(VCC-VBE)/(RB/beta + RC + RE); #Collector current current in mA.\n", + "IE=IC; #Emitter current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC -IE*RE=0. (IE~IC)\n", + "#From the above equation,\n", + "VCE=VCC-IC*(RC+RE); #Collector emitter voltage in V\n", + "\n", + "\n", + "#Results\n", + "print(\"The operating point : VCE=%.1fV and IC=%.2fmA.\"%(VCE,IC));\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.18 : Page number 212" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The d.c bias values are: VCE=1.55V and IC=0.845mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage in V\n", + "RB=100.0; #Base resistor's resistance in kΩ\n", + "RC=10.0; #Collector resistor's resistance in kΩ\n", + "RE=0; #Emitter resistor's resistance in kΩ\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "beta=100.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along outut circuit,\n", + "#we get, VCC -(IC+IB)*RC -IB*RB -VBE - IE*RE=0.\n", + "#since IC= beta*IB, IC+IB ~ IC and IE~IC,\n", + "#We get, VCC - IC*RC -(IC/beta)*RB -VBE - IE*RE\n", + "IC=(VCC-VBE)/(RC +RB/beta + RE); #Collector current in mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit,\n", + "#we get, VCC-VCE - IC*RC =0. (IE~IC)\n", + "#From the above equation,\n", + "VCE=VCC-IC*RC; #Collector-emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"The d.c bias values are: VCE=%.2fV and IC=%.3fmA\"%(VCE,IC));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.19: Page number 214-215" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector-emitter voltage at operating point=8.55V\n", + "Collector current at operating point = 2.15mA\n" + ] + }, + { + "data": { + "image/png": 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+ "text/plain": [ + "<matplotlib.figure.Figure at 0x7fddb98982b0>" + ] + }, + "metadata": {}, + "output_type": "display_data" + } + ], + "source": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage in V\n", + "R1=10.0; #Resistor R1's resistance in kΩ\n", + "R2=5.0; #Resistor R2's resistance in kΩ\n", + "RC=1.0; #Collector resistor's resistance in kΩ \n", + "RE=2.0; #Emitter resistor's resistance in kΩ\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law along output circuit\n", + "#VCE=VCC-IC*(RC+RE);\n", + "#IC=0, for VCE_max\n", + "VCE_max=VCC; #Maximum collector-emitter voltage in V\n", + "#VCE=0, for IC_max\n", + "IC_max=VCC/(RC+RE); #Maximum collector current in mA\n", + "\n", + "#Operating point\n", + "V2=(VCC*R2)/(R1+R2); #Voltage across R2 resistor V\n", + "IE=(V2-VBE)/RE; #Emitter current in mA\n", + "IC=IE; #Collector current(Approx. equal to emitter current) in mA\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage in V\n", + "\n", + "#Results\n", + "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n", + "print(\"Collector current at operating point = %.2fmA\"%IC);\n", + "\n", + "#plot\n", + "limit = plt.gca()\n", + "limit.set_xlim([0,20])\n", + "limit.set_ylim([0,6])\n", + "VCE=[i for i in range(0,(int)(VCC+1))]; #Plot variable for V_CE\n", + "IC=[((VCC-i)/(RC+RE)) for i in (VCE[:])]; #Plot variable for I_C\n", + "\n", + "p=plt.plot(VCE,IC);\n", + "plt.xlabel(\"VCE(V)\");\n", + "plt.ylabel(\"IC(mA)\");\n", + "plt.title(\"d.c load line\");\n", + "plt.show(p);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.20: Page number 215-216" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Collector-emitter voltage at operating point=8.55V\n", + "Collector current at operating point = 2.15mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage in V\n", + "R1=10.0; #Resistor R1's resistance in kΩ .\n", + "R2=5.0; #Resistor R2's resistance in kΩ .\n", + "RC=1.0; #Collector resistor's resistance in kΩ . \n", + "RE=2.0; #Emitter resistor's resistance in kΩ .\n", + "VBE=0.7; #Base-emitter voltage in V\n", + "\n", + "#Calculations\n", + "#Using Thevenin's Theorem for replacing circuit consisting of VCC,R1,R2\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage in V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's equivalent resistance in kΩ .\n", + "\n", + "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IC=beta*IB\n", + "#IC=(E0-VBE)/(R0/beta +RE);\n", + "IC=(E0-VBE)/RE; #(Since R0/beta << RE) collector current in mA\n", + "VCE=VCC-IC*(RC+RE); #Collector emitter voltage in V\n", + "\n", + "\n", + "#Results\n", + "print(\"Collector-emitter voltage at operating point=%.2fV\"%VCE);\n", + "print(\"Collector current at operating point = %.2fmA\"%IC);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.21: Page number 216-217" + ] + }, + { + "cell_type": "code", + "execution_count": 21, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i)Emitter current= 1.9mA\n", + "(ii)Emitter current= 1.7mA\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage in V\n", + "RE=1.0; #Emitter resistor, kΩ .\n", + "R1=50.0; #Resistor R1, kΩ .\n", + "R2=10.0; #Resistor R2, kΩ .\n", + "\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "VBE=0.1; #Base-emitter voltage in V\n", + "V2=(VCC*R2)/(R1+R2); #Voltage drop across resistor R2, V \n", + "IE=(V2-VBE)/RE; #Emitter current in mA\n", + "\n", + "print(\"(i)Emitter current= %.1fmA\"%IE);\n", + "\n", + "#(ii)\n", + "VBE=0.3; #Base-emitter voltage in V\n", + "V2=(VCC*R2)/(R1+R2); #Voltage drop across resistor R2, V\n", + "IE=(V2-VBE)/RE; #Emitter current in mA\n", + "\n", + "print(\"(ii)Emitter current= %.1fmA\"%IE);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.22: Page number 217" + ] + }, + { + "cell_type": "code", + "execution_count": 22, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Emitter current =2mA\n", + "Collector-emitter voltage=8V\n", + "Collector terminal's voltage=18V\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=20.0; #Collector supply voltage, V\n", + "R1=10.0; #Resistor R1, kΩ\n", + "R2=10.0; #Resistor R2, kΩ .\n", + "RC=1.0; #Collector resistor, kΩ .\n", + "RE=5.0; #Emitter resistor, kΩ .\n", + "\n", + "\n", + "#Calculations\n", + "V2=(VCC*R2)/(R1+R2); #Voltage drop across resistor R2, V\n", + "\n", + "#Applying kirchhoff's law from base terminal to emitter resistor\n", + "#V2=VBE+IE*RE\n", + "#VBE is neglected due to its small value\n", + "\n", + "IE=V2/RE; #Emitter current in mA\n", + "IC=IE; #Collector current (approx. equal to emitter current), mA\n", + "\n", + "#Applying Kirchhoff's law along output circuit\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage , V\n", + "VC=VCC-IC*RC; #Voltage at collector terminal,V\n", + "\n", + "\n", + "#Results\n", + "print(\"Emitter current =%dmA\"%IE);\n", + "print(\"Collector-emitter voltage=%dV\"%VCE);\n", + "print(\"Collector terminal's voltage=%dV\"%VC);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.23: Page number 219-220" + ] + }, + { + "cell_type": "code", + "execution_count": 23, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point : VCE= 3.72V and IC=1.2mA\n", + "Stability factor=18.4\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "VCC=12.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=50; #Base current amplification factor\n", + "R1=150; #Resistor R1, kΩ .\n", + "R2=100; #Resistor R2, kΩ .\n", + "RC=4.7; #Collector resistor, kΩ .\n", + "RE=2.2; #Emitter resistor, kΩ .\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage, V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance, kΩ .\n", + "\n", + "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IC=beta*IB\n", + "IB=round((E0-VBE)/(R0+beta*RE),3); #Base current in mA\n", + "IC=round(beta*IB,1); #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "VCE=VCC-IC*(RC+RE); #Collector-emitter voltage, V\n", + "\n", + "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE); #Stability factor\n", + "\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n", + "print(\"Stability factor=%.1f\"%S);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.24 : Page number 220" + ] + }, + { + "cell_type": "code", + "execution_count": 24, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point : VCE= 8.83V and IC=4.2mA\n", + "Stability factor=2.94\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage , V\n", + "beta=100.0; #Base current amplification factor\n", + "R1=6.0; #Resistor R1, kΩ .\n", + "R2=3.0; #Resistor R2, kΩ .\n", + "RC=470.0; #Collector resistor, Ω.\n", + "RE=1.0; #Emitter resistor, kΩ .\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage, V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance, kΩ .\n", + "\n", + "#Applying Kirchhoff' law along thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IC=beta*IB\n", + "IB=round((E0-VBE)/(R0+beta*RE),3); #Base current in mA\n", + "IC=round(beta*IB,1); #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "VCE=VCC-IC*(RC/1000+RE); #Collector-emitter voltage, V\n", + "\n", + "S=(beta+1)*(1+R0/RE)/(beta +1+R0/RE); #Stability factor\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.1fmA\"%(VCE,IC));\n", + "print(\"Stability factor=%.2f\"%S);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.25 : Page number 221-222" + ] + }, + { + "cell_type": "code", + "execution_count": 25, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RE=800 Ω., R1=17.75 kΩ . and R2=4.75 kΩ .\n" + ] + } + ], + "source": [ + "\n", + "#Varaible declaration\n", + "VCC=9; #Collector supply voltage, V\n", + "VCE=3; #Collector-emitter voltage, V\n", + "VBE=0.3; #Base-emitter voltage in V\n", + "RC=2.2; #Collector resistor , kΩ .\n", + "IC=2; #Collector current, mA\n", + "beta=50.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "IB=IC/beta; #Base current in mA\n", + "\n", + "#According to given relation, I1=10*IB\n", + "I1=IB*10; #Current through the resistor R1, mA\n", + "\n", + "#I1=VCC/(R1+R2), .'s LAW\n", + "R1_R2_sum=VCC/I1; #Sum of the resistor's R1 and R2, kΩ (OHM'S LAW).\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "#VCC=IC*RC+VCE+IE*RE\n", + "#IC~IE\n", + "RE=(VCC-IC*RC-VCE)/IC; #Emitter resistor, kΩ .\n", + "RE=round(RE*1000,0); #Emittter resistor, Ω .\n", + "\n", + "IE=IC; #Emittter current(approximately equal to collector current), mA\n", + "VE=IE*(RE/1000); #Voltage at emitter terminal (OHM's LAW), V\n", + "V2=VBE+VE; #Voltage drop across resistor R2, V\n", + "\n", + "R2=V2/I1; #Resistor R2,(OHM's LAW), kΩ .\n", + "R1=R1_R2_sum-R2; #Resistor R1, kΩ .\n", + "\n", + "\n", + "\n", + "#Results\n", + "print(\"RE=%d Ω., R1=%.2f kΩ . and R2=%.2f kΩ .\"%(RE,R1,R2));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.26 : Page number 222" + ] + }, + { + "cell_type": "code", + "execution_count": 26, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "R1=54.4 kΩ and RC=3 kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=16.0; #Collector supply voltage, V\n", + "R2=20.0; #Resistor R2, kΩ\n", + "RE=2.0; #Emitter resistor, kΩ\n", + "VCE=6.0; #Collector-emitter voltage, V\n", + "IC=2.0; #Collector current , mA\n", + "VBE=0.3; #Base-emitter voltage,V\n", + "alpha=0.985; #Current amplification factor\n", + "\n", + "#Calculations\n", + "beta=alpha/(1-alpha); #Base current amplificatioon factor\n", + "IE=IC; #Emitter current, mA\n", + "IB=IC/beta; #Base current, mA\n", + "VE=IE*RE; #Emitter voltage,(OHM's LAW) V\n", + "V2=VBE+VE; #Voltage drop across resistor R2,(Kirchhoff's law) V\n", + "V_R1=VCC-V2; #Voltage drop across resistor R1, V\n", + "I1=V2/R2; #Current through resistor R2 an R1,(OHM's LAW) mA\n", + "R1=V_R1/I1; #Resistor R1,(OHM's LAW) kΩ\n", + "\n", + "V_RC=(VCC-VCE-VE); #Voltage across collector resistor, V\n", + "RC=V_RC/IC; #Collector resistor,(OHM's LAW) kΩ\n", + "\n", + "\n", + "#Results\n", + "print(\"R1=%.1f kΩ and RC=%d kΩ.\"%(R1,RC));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.27 :Page number 222-223" + ] + }, + { + "cell_type": "code", + "execution_count": 27, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The exact value of emitter current in the circuit = 2.11mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=15.0; #Collector supply voltage, V\n", + "R1=10.0; #Resistor R1, kΩ \n", + "R2=5.0; #Resistor R2, kΩ \n", + "RC=1.0; #Collector resistor, kΩ \n", + "RE=2.0; #Emitter resistor, kΩ \n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Using Thevenin's theorem, calculating Thevenin's voltage and resistance\n", + "E0=(VCC*R2)/(R1+R2); #Thevenin's voltage, V\n", + "R0=(R1*R2)/(R1+R2); #Thevenin's resistance, kΩ \n", + "\n", + "#Applying Kirchhoff' law along Thevenin's equivalent circuit,\n", + "#E0=IB*R0+VBE+IE*RE;\n", + "#Since IE~IC and IB=IE/beta,\n", + "IE=(E0-VBE)/(R0/beta + RE); #Emitter current , mA\n", + "\n", + "\n", + "#Calculations\n", + "print(\"The exact value of emitter current in the circuit = %.2fmA.\"%IE);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 2.28: Page number 223-224" + ] + }, + { + "cell_type": "code", + "execution_count": 28, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The value of the resistor R1=28.89 kΩ.\n" + ] + } + ], + "source": [ + "\n", + "#Variable declaration\n", + "IE=2.0; #Emitter current, mA\n", + "IB=50.0; #Base current, mA\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VBE=0.2; #Base-emitter voltage, V\n", + "R2=10.0; #Resistor R2, kΩ\n", + "RE=1.0; #Emitter resistance, kΩ\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's law from the base to the emitter resistor,\n", + "V2=VBE+IE*RE; #Voltage at base terminal, V\n", + "I2=V2/R2; #Current through the resistor R2, mA\n", + "I1=I2+IB/1000; #Current through the resistor R2, mA\n", + "V1=VCC-V2; #Voltage drop across the resistor R2\n", + "R1=V1/I1; #Resistor R1, kΩ\n", + "\n", + "\n", + "#Results\n", + "print(\"The value of the resistor R1=%.2f kΩ.\"%R1);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.30 :Page number 225-226" + ] + }, + { + "cell_type": "code", + "execution_count": 29, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "VCE=3.94V, is approximately half of VCC=8V \n", + " therefore it is mid-point biased.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=8.0; #Collector supply voltage, V\n", + "RB=360.0; #Base resistor, kΩ\n", + "RC=2.0; #Collector resistor, kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=100.0; #base current amplification factor\n", + "\n", + "\n", + "#Calculations\n", + "IC_max=VCC/RC; #Maximum collector current, mA\n", + "VCE_max=VCC; #Maximum collector voltage, V\n", + "\n", + "#Operating point\n", + "#Applying Kirchhoff's law along the input circuit\n", + "IB=(VCC-VBE)/RB; #Base current, mA\n", + "IC=beta*IB; #Collector current, mA\n", + "\n", + "#Kirchhoff' law along the output circuit\n", + "VCE=VCC-IC*RC; #Collector-emitter voltage, V\n", + "\n", + "#Results\n", + "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.31: page number 226" + ] + }, + { + "cell_type": "code", + "execution_count": 30, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IC~IE=6.33mA.\n", + "VCE=4.94V, is approximately half of VCC=10V \n", + " therefore it is mid-point biased.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=50.0; #Base current amplification factor\n", + "R1=12.0; #Resistor R1, kΩ \n", + "R2=2.7; #Resistor R2, kΩ \n", + "RC=620.0; #Collector resistor, Ω \n", + "RE=180.0; #Emitter resistor, Ω\n", + "\n", + "\n", + "#Calculations\n", + "#Voltage divder rule across R1 and R2\n", + "V2=round((VCC*R2)/(R1+R2),2); #Voltage drop across resistor R2, V\n", + "IE=round(((V2-VBE)/RE)*1000,2); #Emitter current, mA\n", + "IC=IE; #Collector current(Approximately equal to emitter current), mA\n", + "print(\"IC~IE=%.2fmA.\"%IC);\n", + "\n", + "#Applying Kirchhoff's law along the output circuit\n", + "VCE=VCC-(IC/1000)*(RC+RE); #Collector-emitter voltage, V\n", + "\n", + "#Results\n", + "print(\"VCE=%.2fV, is approximately half of VCC=%dV \\n therefore it is mid-point biased.\"%(VCE,VCC));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.32 : Page number 227" + ] + }, + { + "cell_type": "code", + "execution_count": 31, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Base current= 49.75 𝜇A\n" + ] + } + ], + "source": [ + "from math import sqrt\n", + "\n", + "#Variable declaration\n", + "VCC=10.0; #Collector supply voltage, V\n", + "IC=10.0; #Collector current, mA \n", + "VBE=0.7; #Base-emitter voltage, V\n", + "R1=1.5; #Resistor R1, kΩ \n", + "R2=680.0; #Resistor R2, Ω \n", + "RC=260.0; #Collector resistor, Ω \n", + "RE=240.0; #Emitter resistor, Ω \n", + "beta_min=100; #Minimum value of base current amplification factor\n", + "beta_max=400; #Maximum value of base current amplification factor\n", + "\n", + "#Calculations\n", + "#Voltage divder rule across R1 and R2\n", + "V2=round((VCC*R2/1000)/(R1+R2/1000),2); #Voltage drop across resistor R2, V\n", + "IE=round((V2-VBE)/(RE/1000),0); #OHM' LAW, Emitter current, mA\n", + "IC=IE; #Collector current(approx. equal to emitter current),mA\n", + "beta_avg=sqrt(beta_min*beta_max); #Average value of base current amplification factor\n", + "IB=IE/(beta_avg +1); #Base current, mA\n", + "IB=IB*1000; #Base current, 𝜇A\n", + "\n", + "#Results\n", + "print(\"Base current= %.2f 𝜇A\"%IB);\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.33 : Page number 227-228" + ] + }, + { + "cell_type": "code", + "execution_count": 32, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point : VCE= 2.96V and IC=4.5mA.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VEE=12.0; #Emitter supply voltage, V\n", + "RC=1.5; #Collector resistor, kΩ\n", + "RB=120.0; #Base resistor kΩ\n", + "RE=510.0; #Emitter resistor, Ω \n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=60.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's voltage law,\n", + "#IB*RB - VBE - IE*RE +VEE=0\n", + "#Since IE~IC and IC=beta*IB,\n", + "IB=(VEE-VBE)/(RB + beta*RE/1000); #Base current , mA\n", + "IC=round(beta*IB,1); #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's voltage law along output circuit,\n", + "VCE=VEE-IC*(RC + RE/1000); #Collector-emitter voltage, V\n", + "\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.1fmA.\"%(VCE,IC));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.34 : Page number 228-229" + ] + }, + { + "cell_type": "code", + "execution_count": 33, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Operating point : VCE= 4.52V and IC=3.73mA.\n" + ] + } + ], + "source": [ + "from math import floor\n", + "\n", + "#Variable declaration\n", + "VEE=9.0; #Emitter supply voltage, V\n", + "RC=1.2; #Collector resistor, kΩ\n", + "RB=100.0; #Base resistor ,kΩ\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "beta=45.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#Applying Kirchhoff's voltage law,\n", + "#IB*RB + VBE=VEE\n", + "#Since IE~IC and IC=beta*IB,\n", + "IB=round((VEE-VBE)/RB,3); #Base current , mA\n", + "IC=floor(beta*IB*100)/100; #Collector current, mA\n", + "\n", + "#Applying Kirchhoff's voltage law along output circuit,\n", + "VCE=VEE-IC*RC; #Collector-emitter voltage, V\n", + "\n", + "#Results\n", + "print(\"Operating point : VCE= %.2fV and IC=%.2fmA.\"%(VCE,IC));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.35 : Page number 229" + ] + }, + { + "cell_type": "code", + "execution_count": 34, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "RE=1.6 kΩ , RC=8.4 kΩ, R1=143 kΩ and R2=24 kΩ.\n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "VCC=16.0; #Collector supply voltage, V\n", + "VBE=0.7; #Base-emitter voltage, V\n", + "IC=1.0; #Collector current, mA\n", + "VCE=6.0; #Collector-emitter voltage, V\n", + "beta=150.0; #Base current amplification factor\n", + "\n", + "#Calculations\n", + "#For a good design, VE=VCC/10;\n", + "VE=VCC/10; #Emitter terminal's voltage, V\n", + "#OHM's Law\n", + "#And, taking IE~IC\n", + "RE=VE/IC; #Emitter resistor, kΩ\n", + "\n", + "#Applying Kirchhoff's voltage law alog output circuit:\n", + "#VCC=IC*RC + VCE + VE\n", + "RC=(VCC-VCE-VE)/IC; #Collector resistor, kΩ\n", + "V2=VE+VBE; #Voltage drop across resistor R2,V\n", + "#From the relation I1=10*IB\n", + "R2=(beta*RE)/10; #Resistor R2, kilo ohm\n", + "\n", + "#From voltage divider rule across R1 and R2,\n", + "#V2=(VCC*R2)/(R1+R2)\n", + "R1=(VCC-V2)*R2/V2; #Resistor R1, kΩ \n", + "\n", + "#Results\n", + "print(\"RE=%.1f kΩ , RC=%.1f kΩ, R1=%.0f kΩ and R2=%d kΩ.\"%(RE,RC,R1,R2));\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.36 : Page number 230-231" + ] + }, + { + "cell_type": "code", + "execution_count": 35, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "(i) The percentage change in the zero signal collector current=82%. \n", + "(ii) The percentage change in the zero signal collector current=1.6%. \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ICBO=5.0; #Collector to base leakage current, microampere\n", + "beta=40.0; #Base current amplification factor\n", + "IC_zero_signal=2.0; #Zero signal collector current, mA\n", + "op_temp=25.0; #operating temperature, degree celsius\n", + "temp_risen=55.0; #Temperature risen, degree celsius\n", + "temp_ICBO_doubles=10.0; #Temperature after which ICBO doubles, degree celsius\n", + "\n", + "#Calculations\n", + "\n", + "#(i)\n", + "ICEO=(beta+1)*ICBO; #Collector to emitter leakage current, microampere\n", + "\n", + "#(ii)\n", + "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles; #Number of times ICBO doubles\n", + "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled; #Final value of collector to base leakage current, microampere\n", + "ICEO_final=ICBO_final*(beta + 1); #Final value of collector to emitter leakage current, microampere\n", + "\n", + "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal; #Zero signal collector current at 55 degree celius\n", + "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal; #Percentage change in zero signal collector current\n", + "\n", + "#Result\n", + "print(\"(i) The percentage change in the zero signal collector current=%.0f%%. \"%change)\n", + "\n", + "#(iii)\n", + "#For the silicon transistor\n", + "ICBO=0.1; #Collector to base leakage current, microampere\n", + "\n", + "ICEO=(beta+1)*ICBO; #Collector to emitter leakage current, microampere\n", + "\n", + "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles; #Number of times ICBO doubles\n", + "ICBO_final=ICBO*2**Number_of_times_ICBO_doubled; #Final value of collector to base leakage current, microampere\n", + "ICEO_final=ICBO_final*(beta + 1); #Final value of collector to emitter leakage current, microampere\n", + "\n", + "IC_zero_signal_55=(ICEO_final/1000) +IC_zero_signal; #Zero signal collector current at 55 degree celius\n", + "change=(IC_zero_signal_55-IC_zero_signal)*100/IC_zero_signal; #Percentage change in zero signal collector current\n", + "\n", + "\n", + "#Result\n", + "print(\"(ii) The percentage change in the zero signal collector current=%.1f%%. \"%change)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 9.37 : Page number 231" + ] + }, + { + "cell_type": "code", + "execution_count": 36, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Base current at 57 degree celsius=9.4 𝜇A \n" + ] + } + ], + "source": [ + "#Variable declaration\n", + "ICBO=0.02 #Collector to base leakage current, 𝜇A\n", + "alpha=0.99; #Current amplification factor\n", + "IE=1.0; #Emitter current, mA\n", + "op_temp=27.0; #operating temperature, degree celsius\n", + "temp_risen=57.0; #Temperature risen, degree celsius\n", + "temp_ICBO_doubles=6.0; #Temperature after which ICBO doubles, degree celsius\n", + "\n", + "#Calculations\n", + "Number_of_times_ICBO_doubled=(temp_risen - op_temp)/temp_ICBO_doubles; #Number of times ICBO doubles\n", + "ICBO_55=ICBO*2**Number_of_times_ICBO_doubled; #collector to base leakage current at 55 degree celsius, 𝜇A\n", + "IC=alpha*IE + ICBO_55/1000; #Collector current, mA\n", + "IB=IE-IC; #Base current, mA\n", + "IB=IB*1000; #Base current,𝜇A\n", + "\n", + "#Results\n", + "print(\"Base current at 57 degree celsius=%.1f 𝜇A \"%IB);\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3", + "language": "python", + "name": "python3" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 3 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython3", + "version": "3.5.1" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} diff --git a/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter10_ac_load_line_5.png b/Principles_of_Electronics_____by_V.K._Mehta_and_Rohit_Mehta/screenshots/chapter10_ac_load_line_5.png Binary files differnew file 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