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diff --git a/Electronic_devices_and_circuits_by_I.J_Nagrath/chapter1.ipynb b/Electronic_devices_and_circuits_by_I.J_Nagrath/chapter1.ipynb new file mode 100755 index 00000000..f90f9ea5 --- /dev/null +++ b/Electronic_devices_and_circuits_by_I.J_Nagrath/chapter1.ipynb @@ -0,0 +1,1579 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:24516babfb49304af52e8c58a067e6fa8878e66215919fc53473493b16a99fdf"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 1: SEMICONDUCTORS,DIODE AND DIODE CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.1, Page number 5"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "A=6.022*10**23 #avagadro's number(/m^3)\n",
+ "d=2.7*10**6 #density of aluminium conductor(g/m^3)\n",
+ "a=26.98 # atomic weight aluminium conductor(g/g-atom)\n",
+ "D=10**4. #current density(A/m^2)\n",
+ "e=1.6*10**-19 #electronic charge(C)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "n=A*d/a #number of atoms(n/m^3)\n",
+ "\n",
+ "#Part b\n",
+ "u=D/(n*e) #drift velocity (m/s)\n",
+ "\n",
+ "#Results\n",
+ "print \"number of atoms per cubic meter is \",round(n/1e+28,3),\"*10^28 /m^3\"\n",
+ "print \"drift velocity is\",round(u/1e-6,2),\"*10^-6 m/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "number of atoms per cubic meter is 6.026 *10^28 /m^3\n",
+ "drift velocity is 1.04 *10^-6 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.2, Page number 6"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "n=10**23 #number of electrons(n/m^3)\n",
+ "e=1.6*10**-19 #electronic charge(C) \n",
+ "u=0.4 #mobility(m^2/Vs) \n",
+ "a=10**-7 #cross sectional area(m^2) \n",
+ "l=15*10**-2 #conductor length(m)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "G=n*e*u #conductivity(S/m)\n",
+ "\n",
+ "#Part b\n",
+ "R=l/(a*G) #resistance(ohm)\n",
+ "\n",
+ "#Results\n",
+ "print\"conductivity of the conductor is\",round((G/1E+3),1),\"*10**3 S/m\"\n",
+ "print\"resistance of the conductor is\",round(R,1),\"ohm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "conductivity of the conductor is 6.4 *10**3 S/m\n",
+ "resistance of the conductor is 234.4 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.3, Page number 9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "A=6.022*10**23 #avagadro's number\n",
+ "d=5.32*10**6 #density of Ge at 300k(g/m^3)\n",
+ "a=72.60 #atomic weight of Ge(g/g-atom)\n",
+ "e=1.6*10**-19 #electronic charge(C)\n",
+ "ni=2.4*10**19 #intrinsic concentration(electron-hole pairs/m^3)\n",
+ "un=0.39 #electron mobility(m^2/V.s)\n",
+ "up=0.19 #hole mobility(m^2/V.s)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "nA=A*d/a #number of atoms(nA/m^3)using avagadro's law\n",
+ "x=nA/ni #Germanium atoms/electron hole pair\n",
+ "\n",
+ "#Part b\n",
+ "g=(un+up)*e*ni #intrinsic conductivity(S/m)\n",
+ "r=1/g #intrinsic resistivity(ohm.m)\n",
+ "\n",
+ "#Results\n",
+ "print\"the relative concentration of Ge and electron hole pairs is\",round((x/1E9),2),\"*10^9 atoms/electron-hole pair\"\n",
+ "print\"the intrinsic resistivity of Ge is\",round(r,3),\"ohm.m\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the relative concentration of Ge and electron hole pairs is 1.84 *10^9 atoms/electron-hole pair\n",
+ "the intrinsic resistivity of Ge is 0.449 ohm.m\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.4,Page number 13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "ni=1.5*10**16 #intrinsic concentration(electron-hole pairs/m^3)\n",
+ "n=4.99*10**28 #number of Si atoms(atoms/m^3)\n",
+ "un=0.13 #electron mobility(m^2/V.s)\n",
+ "up=0.05 #hole mobility(m^2/V.s)\n",
+ "e=1.6*10**-19 #electronic charge(c)\n",
+ "\n",
+ "#Calculation\n",
+ "#Part a\n",
+ "g=e*ni*(un+up) #intrinsic conductivity(S/m)\n",
+ "r=1/g #interinsic resistivity(ohm.m)\n",
+ "Nd=n/10**8 #doped silicon(atoms/m^3)=nn,majority carriers\n",
+ "pn=ni**2/Nd #minority carrier density(holes/m^3)\n",
+ "\n",
+ "#Part b\n",
+ "k=e*un*Nd #conductivity(S/m)\n",
+ " #using Nd in place of nn as Nd=nn\n",
+ "rho=1/k #resistivity(ohm.m)\n",
+ "\n",
+ "#Results\n",
+ "print\"the minority carrier density of Si is\",round(pn/1e+11,2),\"*10^11 holes/m^3\"\n",
+ "print\"the resistivity of Si is\",round((rho/1E-2),2),\"*10**-2 ohm.m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the minority carrier density of Si is 4.51 *10^11 holes/m^3\n",
+ "the resistivity of Si is 9.63 *10**-2 ohm.m\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.5,Page number 17"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vo=0.7 #contact potential(V)\n",
+ "Vf=0.4 #forward biasing voltage(V) \n",
+ "\n",
+ "#Calculation\n",
+ "x=math.exp(-20*(Vo-Vf))/math.exp(-20*Vo) #increase in probability of majority carriers\n",
+ "\n",
+ "#Result\n",
+ "print\"increase in probability of majority carriers is\",round(x),\"times\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "increase in probability of majority carriers is 2981.0 times\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.6,Page number 18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "I=10 #Ge diode carries current(mA)\n",
+ "V=0.2 #forward bias voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "#Part a\n",
+ "Is=I/(math.expm1(40*V)-1) #reverse current(mA)\n",
+ "\n",
+ "#part b\n",
+ "I1=1*10**-3 #current(mA) \n",
+ "V1=(math.log((I1/Is)+1))/40 #voltage(V)\n",
+ "I2=100*10**-3 #current(mA) \n",
+ "V2=(math.log((I2/Is)+1))/40 #voltage(V) \n",
+ "\n",
+ "#Part c\n",
+ "Is1=4*Is #reverse saturation current doubles for every 10 degree celcius temp rise,so for 20 degree rise it will be 4 timese/ \n",
+ "x=37.44 #let x=e/kT\n",
+ "I3=Is1*(math.expm1(x*V)) #current when temp doubles(mA)\n",
+ "\n",
+ "#Results\n",
+ "print\"the reverse current is\",round(Is/1e-3,3),\"mA\" #incorrect units given in the textbook\n",
+ "print\"bias voltages are\",round(V1,3),\"V and\", round(V2,3),\"V resp\"\n",
+ "print\"Is at 20 degree is\",round(Is1/1e-3,2),\"uA and diode current at 0.2 V is\",round(I3,2),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the reverse current is 3.357 mA\n",
+ "bias voltages are 0.007 V and 0.086 V resp\n",
+ "Is at 20 degree is 13.43 uA and diode current at 0.2 V is 23.97 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.7,Page number 21"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "V=3. #Voltage(V)\n",
+ "Req=300. #total resistance as per circuit(ohm)\n",
+ "Rfa=20 #forward resistance(ohm) \n",
+ "Vt=0.7 #Thevinine's voltage(V)\n",
+ "Rfb=0 #forward resistance(ohm)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "I=V/Req #current(A)\n",
+ "\n",
+ "#Part b\n",
+ "Id=(V-Vt)/Req #diode current(mA)\n",
+ "\n",
+ "#Part c\n",
+ "Rf=20 #forward resistance(ohms) \n",
+ "Id1=(V-Vt)/(Req+Rfa) #diode current(mA)\n",
+ "\n",
+ "#Results\n",
+ "print\"current in this case is\",round(I,2),\"A\"\n",
+ "print\"diode current is\",round((Id/1E-3),2),\"mA\"\n",
+ "print\"diode current is\",round((Id1/1E-3),2),\"mA\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "current in this case is 0.01 A\n",
+ "diode current is 7.67 mA\n",
+ "diode current is 7.19 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.9,Page number 22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vx=1.4 #voltage at point X(V) \n",
+ "Vt=0.7 #diode voltage(V)\n",
+ "Vcc=5 #cathode voltage(V) \n",
+ "R=1 #circuit resistance(ohm) \n",
+ "Vs=Vx-Vt #supply voltage(V)\n",
+ "\n",
+ "#Calculations\n",
+ "I1=(Vcc-Vt-Vs)/R #current throgh D1(mA) for 0<Vs<0.7\n",
+ "I2=0 #current through D2 and D3\n",
+ "I1=I2=(Vcc-Vt-Vs)/R #for Vs>0.7 as D2 and D3 conducts\n",
+ "\n",
+ "#Results\n",
+ "print\"I1 for 0<Vs<0.7 is\",I1,\"mA\"\n",
+ "print\"I2 for 0<Vs<0.7 is\",I2,\"mA\"\n",
+ "print\"I1 and I2 for Vs>0.7 is\",I1,\"mA\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I1 for 0<Vs<0.7 is 3.6 mA\n",
+ "I2 for 0<Vs<0.7 is 0 mA\n",
+ "I1 and I2 for Vs>0.7 is 3.6 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.11,Page number 23"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vz=100 #zener voltage(V)\n",
+ "Rz=25 #diode resistance(ohm)\n",
+ "Il=0.05 #load current(A)\n",
+ "Iz=0.01 #zener diode current(A)\n",
+ "Rs=250 #supply resistance(ohm)\n",
+ "\n",
+ "#Calculations\n",
+ "Vl=Vz+(Iz*Rz) #load voltage(V)\n",
+ "Vs=Vl+(Il+Iz)*Rs #supply voltage(V)\n",
+ "VL=Vl*1.01 #increase in Vl(V)\n",
+ "IZ=(VL-Vz)/Rz #increase in zener current\n",
+ "VS=Vl+(Il+IZ)*Rs #increase in supply voltage(V)\n",
+ "Vss=(VS-Vs)/Vs #%increase in supply voltage(V)\n",
+ "P=Il*VL #power consumed(W) \n",
+ "\n",
+ "#Results\n",
+ "print\"load voltage is\",Vl,\"V\"\n",
+ "print\"supply voltage is\",Vs,\"V\"\n",
+ "print\"increase in supply voltage is\",VS,\"V\"\n",
+ "print\"power consumed is\",round(P,2),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " load voltage is 100.25 V\n",
+ "supply voltage is 115.25 V\n",
+ "increase in supply voltage is 125.275 V\n",
+ "power consumed is 5.06 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.12,Page number 25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vbb=5 #bias voltage(V)\n",
+ "Rl=1 #resistance(ohm)\n",
+ "Id=4.4 #from the figure(mA)\n",
+ "\n",
+ "#Part a\n",
+ "i=Vbb/Rl #load line intercepts the Id axis at i(mA)\n",
+ "Vl=Id*Rl #load voltage(V)\n",
+ "\n",
+ "#Part b\n",
+ "Vd=Vbb-Vl #diode voltage(V)\n",
+ "P=Vd*Id #power absorbed in diode(mW)\n",
+ "\n",
+ "#Part c \n",
+ "Ida=1.42 #diode current(mA)for 2V\n",
+ "Idb=7.35 #diode current(mA)for 8V\n",
+ "\n",
+ "#Part d\n",
+ "Idc=8.7 #diode current(mA)for Rl=0.5k ohm \n",
+ "Idd=2.2 #diode current(mA)for Rl=2k ohm\n",
+ "\n",
+ "#Results\n",
+ "print\"diode current is\",Id,\"mA and voltage across the load is\",Vl,\"V\"\n",
+ "print\"power absorbed in diode is\",P,\"mW\"\n",
+ "print\"diode current for Vbb=2V is\",Ida,\"mA\",\"and for Vbb=8V is\",Idb,\"mA\"\n",
+ "print\"diode current for Rl=0.5kohm is\",Idc,\"mA\",\"and for Rl=2kohm is\",Idd,\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "diode current is 4.4 mA and voltage across the load is 4.4 V\n",
+ "power absorbed in diode is 2.64 mW\n",
+ "diode current for Vbb=2V is 1.42 mA and for Vbb=8V is 7.35 mA\n",
+ "diode current for Rl=0.5kohm is 8.7 mA and for Rl=2kohm is 2.2 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.13,Page number 38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "T=300 #temperature(k)\n",
+ "Ig=100*10**-3 #current(mA)\n",
+ "Is=1*10**-9 #current(nA)\n",
+ "x=0.0259 #x=kT/e\n",
+ "\n",
+ "#Calculations\n",
+ "Voc=x*math.log(Ig/Is+1) #as Voc=kT/e*ln((Ig/Is)+1) where ln((Ig/Is)+1)=18.42 after solving \n",
+ "Isc=Ig\n",
+ "\n",
+ "#Result\n",
+ "print\"for a solar cell Voc is\",round(Voc,3),\"V and Isc is\",round(Isc/1E-3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for a solar cell Voc is 0.477 V and Isc is 100.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.14,Page number 38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Idc=0.1 #dc current(A)\n",
+ "Rf=0.5 #forward resistance(ohms)\n",
+ "Rl=20 #load resistance(ohm)\n",
+ "Rs=1 #secondary resistance of transformer(ohm)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Vdc=Idc*Rl #dc voltage(V)\n",
+ "Vm=(math.pi/2)*(Vdc+Idc*(Rs+Rf)) #mean voltage(V)\n",
+ "Vrms=Vm/math.sqrt(2) #rms value of voltage(V) \n",
+ "\n",
+ "#Part b\n",
+ "Pdc=Idc**2*Rl #dc power supplied to the load\n",
+ "\n",
+ "#Part c\n",
+ "PIV=2*Vm #PIV rating for each diode(V)\n",
+ "\n",
+ "#Part d\n",
+ "Im=(math.pi/2)*Idc #peak value of current(mA)\n",
+ "Irms=Im/math.sqrt(2) #rms calue of current(A)\n",
+ "Pac=Irms**2*(Rs+Rf+Rl) #ac power input(W)\n",
+ "\n",
+ "#Part e\n",
+ "eta=(Pdc/Pac)*100 #conversion efficiency\n",
+ "\n",
+ "#Part f\n",
+ "Vr=((Rs+Rf)/Rl)*100 #voltage regulation(V)\n",
+ "\n",
+ "#results\n",
+ "print\"rms value of voltage is\",round(Vrms,2),\"V\"\n",
+ "print\"dc power supplied to load is\",Pdc,\"W\"\n",
+ "print\"PIV rating for each diode\",round(PIV,2),\"V\"\n",
+ "print\"ac input power is\",round(Pac,3),\"W\"\n",
+ "print\"conversion efficiency\",round(eta,1),\"%\"\n",
+ "print\"voltage regulation\",Vr,\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "rms value of voltage is 2.39 V\n",
+ "dc power supplied to load is 0.2 W\n",
+ "PIV rating for each diode 6.75 V\n",
+ "ac input power is 0.265 W\n",
+ "conversion efficiency 75.4 %\n",
+ "voltage regulation 7.5 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.15,Page number 46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "from scipy import integrate\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vt=1 \n",
+ "Vl=12\n",
+ "Vm=63.63 #peak voltage(V) as Vm=sqr root of 2*45\n",
+ "Idc=8. #charging current(A)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "theta1=math.degrees(math.asin((Vt+Vl)/Vm))\n",
+ "theta2=180-theta1\n",
+ "Rl=((2*Vm*math.cos(theta1))-(2*(math.pi-2*theta1)*(Vt+Vl)))/(Idc*math.pi)\n",
+ "\n",
+ "#Part b\n",
+ "wt = lambda wt: (((((math.sqrt(2)*45*math.sin(wt))-(Vt+Vl))/Rl)*wt)**2)\n",
+ "integ,err = integrate.quad(wt, theta1 , theta2)\n",
+ "print integ\n",
+ "Irms = (integ/math.pi)**0.5\n",
+ "Pl=Irms**2*Rl #power loss in resistance(W)\n",
+ "\n",
+ "#Part c\n",
+ "P=Vl*Idc #power supplied to battery(W)\n",
+ "\n",
+ "#results\n",
+ "print\"Resistance to be added is\",round(Rl,2),\"Ohms\"\n",
+ "print\"\",Pl\n",
+ "print\"power supplied to battery is\",P,\"W\"\n",
+ "print\"\",Irms "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "5703935.44277\n",
+ "Resistance to be added is 24.75 Ohms\n",
+ " 44936628.7032\n",
+ "power supplied to battery is 96.0 W\n",
+ " 1347.44908683\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.16,Page number 47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Rf=5 #forward resistance(ohms)\n",
+ "Vo=20 #output voltage(V)\n",
+ "Rs=10 #secondary resistance of transformer(ohm)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Idc=0.1 #dc current(A) \n",
+ "Vm=Vo*(math.sqrt(2)) #mean voltage(V)\n",
+ "Vdc=(2*Vm/(math.pi))-Idc*(Rs+2*Rf) #dc voltage(V)\n",
+ "\n",
+ "#Part b\n",
+ "Idc1=0.2 #full load dc current(A)\n",
+ "Vdc2=((2*(math.sqrt(2))*Vo)/(math.pi))-Idc1*(Rs+2*Rf) #full load dc voltage(V)\n",
+ "Rl=Vdc2/Idc1 #load resistance(ohm)\n",
+ "x=((2*Rf+Rs)/Rl)*100 #% regulation \n",
+ "\n",
+ "#Part c\n",
+ "Idc=0.2 #dc current(A)\n",
+ "Im=(math.pi)*Idc/2 #peak current(mA)\n",
+ "Ilrms=Im/math.sqrt(2) #rms current(mA)\n",
+ "Vlrms=Ilrms*Rl #load rms voltage(V) \n",
+ "\n",
+ "#Part d\n",
+ "Vldc=14 #load dc voltage(V)\n",
+ "Vlacrms=math.sqrt(Vlrms**2-Vldc**2) #rms value of ac component(V)\n",
+ "\n",
+ "#Results\n",
+ "print\"dc voltage\",round(Vdc),\"V\"\n",
+ "print\"regulation is\",round(x,2),\"%\"\n",
+ "print\"rms value of output voltage at dc load current is\",round(Vlrms,2),\"V\"\n",
+ "print\"rms value of ac component of voltage\",round(Vlacrms,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "dc voltage 16.0 V\n",
+ "regulation is 28.56 %\n",
+ "rms value of output voltage at dc load current is 15.56 V\n",
+ "rms value of ac component of voltage 6.78 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.17,Page number 50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vh=60. #higher output voltage(V)\n",
+ "Vl=45. #lower output voltage(V) \n",
+ "fz=50. #frequency(Hz)\n",
+ "Vr=15. #peak to peak ripple voltage(V)\n",
+ "Rl=600. #resistance(ohms)\n",
+ " \n",
+ "#Calculations\n",
+ "Vldc=(Vh+Vl)/2 #avg load dc voltage(V) as voltage drops from 60 to 45\n",
+ "Idc=Vldc/Rl #dc current(A)\n",
+ "T=1/fz #discharging time(ms)\n",
+ "C=(Idc*T)/Vr #linear discharge rate(uF)\n",
+ "C1=C*2 #new capacitance(uF)\n",
+ "'''\n",
+ " CVr(p-p) 234Vr(p-p)*10^3\n",
+ "Idc = -------------- = --------------- ----(1)\n",
+ " T 20\n",
+ " \n",
+ " 60+[60-Vr(p-p) 120-Vr(p-p)\n",
+ "Idc = --------------- = ------------*1000 ----(2)\n",
+ " 2Rl 2*600\n",
+ " \n",
+ "Equating equations 1 & 2, we get, \n",
+ "'''\n",
+ "\n",
+ "Vr1 = (20*120*1000)/(1200*254)\n",
+ "Idc1=(Vh-(Vr1/2))/Rl #dc load current(mA)\n",
+ "\n",
+ "#Results\n",
+ "print\"value of capacitance is\",round(C/1E-6),\"uF\" \n",
+ "print\"Vr1 is\",Vr1,\"V\" \n",
+ "print\"dc load current Idc is\",round(Idc1/1E-3),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of capacitance is 117.0 uF\n",
+ "Vr1 is 7 V\n",
+ "dc load current Idc is 95.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.18,Page number 51"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vdc=30 #dc voltage(V)\n",
+ "V1=220 #source voltage(V)\n",
+ "f=50 #frequency(Hz)\n",
+ "Rl=1000 #load resistance(k ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "C=100/f*Rl #as Vdc/Vr=100\n",
+ "Vm=Vdc+(Vr/2) #peak voltage(V)\n",
+ "V2=Vm/(math.sqrt(2)) #secondary voltage(V)\n",
+ "r=V1/V2 #transformer turn ratio\n",
+ "\n",
+ "#Results\n",
+ "print\"capacitor filtor is\",C,\"uF\"\n",
+ "print\"transformer turn ratio is\",round(r,2),\"\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "capacitor filtor is 2000 uF\n",
+ "transformer turn ratio is 10.37 \n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.19,Page number 52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Idc=60*10**-3 #dc current(A)\n",
+ "Vm=60 #peak volage(V)\n",
+ "f=50 #frequency(Hz)\n",
+ "C=120*10**-6 #capacitance(F)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Vrms=Idc/(4*(math.sqrt(3))*f*C*Vm) #rms voltage(V)\n",
+ "Vr=2*(math.sqrt(3))*Vrms #ripple factor(V)\n",
+ "\n",
+ "#Part b\n",
+ "Vdc=Vm-(Vr/2) #by simplifying\n",
+ "\n",
+ "#Part c\n",
+ "r=(Vrms/Vdc)*100 #ripple factor\n",
+ "\n",
+ "#Results\n",
+ "print\"ripple factor is\",round(Vr,3),\"Vdc\"\n",
+ "print\"dc voltage is\",round(Vdc),\"V\"\n",
+ "print\"ripple factor\",round(r,4),\"%\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ripple factor is 0.083 Vdc\n",
+ "dc voltage is 60.0 V\n",
+ "ripple factor 0.0401 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.20,Page number 54"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "''' 200*1.141 4\n",
+ "v1(t)=-------------(1- - cos628t) \n",
+ " 3.14 3\n",
+ " 200*1.141 800*1.141 \n",
+ "v2(t)=----------- - ------------ cos(628t+<(V2/V1))\n",
+ " 3.14 3*3.14\n",
+ "\n",
+ "V2/V1|w=0 =0.8;V2/V1|w=628 =6.43*10^-4 <V2/V1|w=628 =180\n",
+ "v2(t)=72.02+0.0538 cos628t\n",
+ "'''\n",
+ "#Part b\n",
+ "vrms=0.0538\n",
+ "vdc=math.sqrt(2)*72.02\n",
+ "r=vrms/vdc\n",
+ "\n",
+ "#Results\n",
+ "print\"ripple factor is\",round((r/1E-4),2),\"*10**-4\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ripple factor is 5.28 *10**-4\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.24,Page number 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vz=2 #zener voltage(V)\n",
+ "r1=10 #resistance after reducing circuit by thevinin(ohms)\n",
+ "r2=20 #resistance after reducing circuit by thevinin(ohms)\n",
+ "V1=7.5 #voltage after circuit reduction(V)\n",
+ "V2=15 #voltage after circuit reduction(V)\n",
+ "Rz=100/3 #zener resistance(ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "Vab=V2-(((V2-V1)/(r1+r2))*r2) #thevinin voltage at ab(V)\n",
+ "Rth=(Vab*r2)/(Vab+r2) #thevinin resistance at ab(ohms)\n",
+ "Vd=Vab-Vz #diode voltage(V)\n",
+ "Id=Vd/(Rth+Rz) #diode current(A)\n",
+ "\n",
+ "#Results\n",
+ "print\"diode current is\",round(Id,2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "diode current is 0.2 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.25,Page number 61"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vd=0.7 #diode voltage(V)\n",
+ "Ro=18 #output resistance(k ohms)\n",
+ "R1=2 #diode1 resistance(k ohms)\n",
+ "R2=2 #diode2 resistance(k ohms) \n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "V1=10 #voltage to D1(V)\n",
+ "V2=0 #voltage to D2(V)\n",
+ "Io=(V1-Vd)/(R1+Ro) #output current(mA) \n",
+ "Vo=Io*Ro #output voltage(V)\n",
+ "\n",
+ "#Part b\n",
+ "V1=5 #voltage to D1(V)\n",
+ "V2=0 #voltage to D2(V)\n",
+ "Io=(V1-Vd)/(R1+Ro) #output current(mA) \n",
+ "Vo1=Io*Ro #output voltage(V)\n",
+ "\n",
+ "#Part c\n",
+ "V1=10 #voltage to D1(V)\n",
+ "V2=5 #voltage to D2(V)\n",
+ "Vo=8.37 #as D1 only conducts,so,Vo is same as in part a\n",
+ "Vd1=V2-Vo #assume D1 conducts\n",
+ "Vo2=8.37 #D2 does not conduct as as Vd1 is negative\n",
+ "\n",
+ "#Part d\n",
+ "V1=V2=5 #voltage to D1 and D2(V) \n",
+ "Id1=(V1-Vd-Vo)/2 #diode1 current(mA) \n",
+ "Io=Vo/Ro #output current(mA) \n",
+ "Vo3=(Ro*(V1-Vd))/(Ro+1) #output voltage(V)\n",
+ "\n",
+ "print\"a)output voltage is\",Vo,\"V\"\n",
+ "print\"b)output voltage is\",Vo1,\"V\"\n",
+ "print\"c)output voltage is\",Vo2,\"V\"\n",
+ "print\"d)output voltage is\",round(Vo3,2),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "a)output voltage is 8.37 V\n",
+ "b)output voltage is 3.87 V\n",
+ "c)output voltage is 8.37 V\n",
+ "d)output voltage is 4.07 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.26,Page number 62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vs=10. #supply voltage(V) \n",
+ "Rs=1 #supply resistane(ohm)\n",
+ "Vl=10. #load voltage(V)\n",
+ "Vi=50. #nput voltage(V)\n",
+ "Iz=32 #zener diode current(mA)\n",
+ "Is=40 #supply current(mA) \n",
+ "\n",
+ "#Calculations\n",
+ "#Part a (Rl is min when Iz=0) \n",
+ "Is=(Vi-Vs)/Rs #source current(mA) \n",
+ "Rlmin=Vl/(Vi-Vs) #load resistance minimum(ohm) \n",
+ "\n",
+ "#Part b(Rl is maximum when Iz=32 mA) \n",
+ "Il=(Is-Iz)*10**-3 #load current(A) \n",
+ "Rlmax=Vl/Il #maximum load resistance(k ohms)\n",
+ "P=Vl*Iz #max diode wattage consumed(mW)\n",
+ "\n",
+ "#Results\n",
+ "print\"Range of Rl is\",round((Rlmin/1E-3),3),\"ohm\",\"to\", round((Rlmax/1E+3),2),\"k ohm\" \n",
+ "print \"Il = \",(Il/1E-3),\"*10**-3 A\"\n",
+ "print\"max power consumed is\",P,\"mW\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Range of Rl is 250.0 ohm to 1.25 k ohm\n",
+ "Il = 8.0 *10**-3 A\n",
+ "max power consumed is 320.0 mW\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.27,Page number 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Vz=20 #zener voltage(V)\n",
+ "Izmax=50 #maximum zener current(mA)\n",
+ "Rz=0 #zener resistance(ohms)\n",
+ "Rl=2. #load resistance(ohm)\n",
+ "Vl=20. #as Vz=Vl(V)\n",
+ "Rs=0.25 #source resistance(k ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "Il=Vl/Rl #load current(mA) \n",
+ "Vsmin=(Rs+Rl)*Il #as Iz is floating so Iz=0\n",
+ "\n",
+ "#Part b\n",
+ "Is=Izmax+Il #source current(mA) \n",
+ "Vsmax=Vz+(Is*Rs) #maximum source voltage(V)\n",
+ "\n",
+ "#Results\n",
+ "print\"Vsmin\",round(Vsmin,1),\"V\"\n",
+ "print\"Range of input voltage is\",round(Vsmin,1),\"to\", Vsmax,\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Il is 10.0 mA\n",
+ "Vsmin 22.5 V\n",
+ "Range of input voltage is 22.5 to 35.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.28,Page number 63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Variable declaration\n",
+ "Ilmax=100 #load maximum current(mA)\n",
+ "Ilmin=0 #load minimum current(mA)\n",
+ "Rz=0.05 #zener diode resistance(ohms)\n",
+ "Rs=10. #source resistance(k ohms)\n",
+ "Vl=16.015 #load voltage(V)\n",
+ "Vl1=16. #nominal load voltage(V) \n",
+ "Vs=20 #source voltage(V)\n",
+ "Vz=16 #zener diode voltage(V)\n",
+ "\n",
+ "#Calculations\n",
+ "#Case 1 (i)\n",
+ "Iz=(Vl-Vl1)/Rz #zener current(mA)\n",
+ "Is=Iz+Ilmax #supply current(A)\n",
+ "\n",
+ "#Case 1 (ii)\n",
+ "Is1=(Vs-Vz)/(Rs+Rz) #supply current(mA)\n",
+ "Vl2=Vl1+(Is1*Rz) #voltage(V) \n",
+ "Vr=((Vl2-Vl)/Vl1)*100 #voltage regulation\n",
+ "\n",
+ "#Case 2 (i)\n",
+ "Vs=18 #supply voltage(V)\n",
+ "Ilmax=0.1 #load current max(A)\n",
+ "Vl=16.005 #load voltage(V)\n",
+ "Iz=(Vl-Vl1)/Rz #zener current(mA)\n",
+ "Is2=Ilmax+Iz #supply current(A)\n",
+ "\n",
+ "#Case 2 (ii)\n",
+ "Ilmin=0\n",
+ "Iz1=(Vs-Vl1)/(Rs+Rz) #minimum diode current(mA) \n",
+ "Vl=Vl1+(Iz*Rz) #load voltage at Ilmin(V) \n",
+ "\n",
+ "#Part a\n",
+ "#Variable declaration\n",
+ "Is=0.4 #supply current(A)\n",
+ "Vs=20 #supply voltage(V) \n",
+ "Vl=16.015 #load voltage(V)\n",
+ "Iz=0.3 #zener current(mA)\n",
+ "\n",
+ "#Calculations\n",
+ "P=Is**2*Rs #power dissipated by Rs(W)\n",
+ "\n",
+ "#Part b\n",
+ "Pd=Vl*Iz #power dissipated(W)\n",
+ "Po=(Vs**2)/Rs #output power(W)\n",
+ "\n",
+ "print\"maximum power dissipated by Rs is\",P,\"W\"\n",
+ "print\"maximum power dissipated by diode is\",round(Pd,3),\"W\"\n",
+ "print\"minimum diode current is\",round(Iz1,3),\"A\"\n",
+ "print\"voltage regulation is\",round(Vr,2),\"%\"\n",
+ "print\"output shorted will be\",Po,\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum power dissipated by Rs is 1.6 W\n",
+ "maximum power dissipated by diode is 4.804 W\n",
+ "minimum diode current is 0.199 A\n",
+ "voltage regulation is 0.03 %\n",
+ "output shorted will be 40.0 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.29,Page number 65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=20 #secondary voltage(V)\n",
+ "Rs=10 #Winding resistance(ohm)\n",
+ "Rf=5 #diode has forward resistance(ohms)\n",
+ "Idc=2*10**-3 #load current(mA) \n",
+ "\n",
+ "#Calculations\n",
+ "#Part a \n",
+ "Vdc=(Vrms*(math.sqrt(2)))/(math.pi) #no load Vdc\n",
+ "\n",
+ "#Part b \n",
+ "Vldc=Vdc-(Idc*(Rs+Rf)) #dc output voltage when load is 20mA\n",
+ "\n",
+ "#Part c\n",
+ "Rl=Vldc/Idc #load resistance(ohms)\n",
+ "r=((Rs+Rf)/Rl)*100 #percentage regulation(%)\n",
+ "\n",
+ "#Part d\n",
+ "Im=Idc*(math.pi) #peak current(mA)\n",
+ "Ilrms=Im/2 #rms load current(mA) \n",
+ "Vlrms=Ilrms*Rl #rms load voltage(V) \n",
+ "Vlrmsac=math.sqrt((Vlrms**2)-(Vldc**2)) #Ripple voltage rms(V)\n",
+ "f=50*2 #rippLe frequency(Hz)\n",
+ "\n",
+ "#Part e\n",
+ "eta=(((2*(math.pi))**2)/(1+((Rs+Rf)/Rl)))*100 #efficiency\n",
+ "\n",
+ "#Part f\n",
+ "PIV=Vm=Vrms*(math.sqrt(2)) #peak inverse voltage(V) \n",
+ "\n",
+ "#Results\n",
+ "print\"no load dc voltage is\",round(Vdc),\"V\" \n",
+ "print\"dc output voltage when the load is drawing 20 mA is\",round(Vldc,2),\"V\"\n",
+ "print\"percentage regulation at this load is\",round((r/1E-1),2),\"%\"\n",
+ "print\"ripple voltage rms is\",round(Vlrmsac,2),\"V and ripple frequency is\",f,\"Hz\"\n",
+ "print\"power conversion efficiency is\",round((eta/1E+2),1),\"%\"\n",
+ "print\"PIV is\",round(PIV),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "no load dc voltage is 9.0 V\n",
+ "dc output voltage when the load is drawing 20 mA is 8.97 V\n",
+ "percentage regulation at this load is 3.34 %\n",
+ "ripple voltage rms is 10.87 V and ripple frequency is 100 Hz\n",
+ "power conversion efficiency is 39.3 %\n",
+ "PIV is 28.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.30,Page number 66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vl=24 #battery voltage(V)\n",
+ "Vm=60*(math.sqrt(2)) #peak voltage(V)\n",
+ "Ip=2.5 #peak current(A)\n",
+ "c=20 #charge(Ah)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a\n",
+ "theta=math.asin(Vl/Vm) #angle at which conduction begins\n",
+ "Rs=(Vm-Vl)/Ip #source resistance(ohms) \n",
+ "\n",
+ "#Part b\n",
+ "Idc=(Vm/(math.pi)*Rs)*(math.cos(theta))-(((math.pi)-(2*theta))/2*math.pi)*(Vl/Rs) #load current(A)\n",
+ "T=c/Idc #time to deliver 20Ah(h)\n",
+ "\n",
+ "#Results\n",
+ "print\"resistance connected in series is\",round(Rs,1),\"ohm\"\n",
+ "print\"time required to deliver a charge of 20 Ah is\",round((T/1E-3),1),\"h\" \n",
+ "print\"Idc\",round((Idc/1E+3),2),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "resistance connected in series is 24.3 ohm\n",
+ "time required to deliver a charge of 20 Ah is 31.9 h\n",
+ "Idc 0.63 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.32,Page number 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "R=25. #external resistance(ohms)\n",
+ "Vm=200. #peak value of voltage(V) as vs=200 sinwt\n",
+ "Rf=50. #forward resistance(ohms)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a \n",
+ "Id=Vm/(2*Rf+R) #diode current(peak)\n",
+ "\n",
+ "#Part b \n",
+ "Idc=(2*Id)/math.pi #dc current(A)\n",
+ "\n",
+ "#Part c\n",
+ "PIV=Vm/2 #peak value of voltage across D1\n",
+ "PIVac=100/math.pi #average value of voltage across D1\n",
+ "\n",
+ "#Part d\n",
+ "Im=Id #peak value of current(A)\n",
+ "Irms=Im/(math.sqrt(2)) #rms value of current(A)\n",
+ "\n",
+ "#Results\n",
+ "print\"peak value of current is\",Id,\"A\"\n",
+ "print\"dc currect is\",round(Idc,2),\"A\"\n",
+ "print\"across D1 are peak voltage is\",PIV,\"V and average voltage is\",round(PIVac,1),\"V\"\n",
+ "print\"Irms is\",round(Irms,2),\"A\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "peak value of current is 1.6 A\n",
+ "dc currect is 1.02 A\n",
+ "across D1 are peak voltage is 100.0 V and average voltage is 31.8 V\n",
+ "Irms is 1.13 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.33,Page number 67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "f=50. #frequency(Hz)\n",
+ "dv=7. #difference between maximum and minimum(25-18)voltages across the load(V)\n",
+ "Ic=100. #load current(mA)\n",
+ "\n",
+ "#Calculations\n",
+ "dt=1/(2*f) #time of discharge(seconds)\n",
+ "C=Ic/(dv/dt) #capacitance(uF) \n",
+ "\n",
+ "#Results\n",
+ "print\"value of capacitor is\",round((C/1E-3),2),\"uF\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of capacitor is 142.86 uF\n"
+ ]
+ }
+ ],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.34,Page number 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vr=10. #peak to peak ripple voltage(V)\n",
+ "Vm=50. #peak output voltage(V) \n",
+ "C=300. #Capacitance(uF)\n",
+ "Rl=470. #load resistance(ohms)\n",
+ "f=50. #frequency(Hz)\n",
+ "\n",
+ "#Calculations\n",
+ "#Part a \n",
+ "Vdc=Vm-(Vr/2) #dc voltage(V) \n",
+ "C=Vdc/(f*Vr*Rl) #capacitance(mF)\n",
+ "\n",
+ "#Part b\n",
+ "C1=300*10**-6 #capacitance is increased(uF)\n",
+ "Vr=2*Vm/((2*f*C1*Rl)+1)\n",
+ "Vdc=Vm-Vr/2 #load voltage ripple(V)\n",
+ "Idc=Vdc/Rl #average load current(mA)\n",
+ "\n",
+ "#Results\n",
+ "print\"value of capacitor is\",round((C/1E-6),1),\"mF\" \n",
+ "print\"load voltage ripple is\",round(Vdc,2),\"V and average load current is\",round((Idc/1E-4),1),\"mA\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of capacitor is 191.5 mF\n",
+ "load voltage ripple is 46.69 V and average load current is 993.4 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.35,Page number 69"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "vo=7.5 #instantaneous voltage(V)\n",
+ "R1=15 #resistance(k ohms)\n",
+ "Von=0.5 #voltage of diode when on(V)\n",
+ "\n",
+ "#Calculations\n",
+ "Rth=(R1*vo)/(R1+vo) #equivalent resistance(V)\n",
+ "T=2*(math.pi)/10**4 #time period(ms)\n",
+ "t1=(math.asin(Von/2.5))/10**4 #timimgs when D1 conducts(ms)\n",
+ "t2=(T/2)-t1\n",
+ "\n",
+ "#Results\n",
+ "print\"time period is\",round((T/1E-3),3),\"ms\"\n",
+ "print\"t1 is\",t1,\"ms\"\n",
+ "print \"t2 is\",round((t2/1E-3),3),\"ms\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "time period is 0.628 ms\n",
+ "t1 is 2.0135792079e-05 ms\n",
+ "t2 is 0.294 ms\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1.36,Page number 70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "\n",
+ "#Variable declarations\n",
+ "v=12 #output voltage(V)\n",
+ "vm=20. #peak voltage(V)\n",
+ "v1=8 #output voltage(V) for negative half cycle\n",
+ "vm1=20. #peak voltage(V) for negative half cycle \n",
+ "\n",
+ "#Calculations\n",
+ "t1=(math.asin(v/vm))/10**4 #for positive half cycle when D1 conducts\n",
+ "t2=(0.1*math.pi)-t1/1e-3 \n",
+ "t3=(math.asin(v1/vm1))/10**4 #for negative half cycle when D2 conducts\n",
+ "t4=(0.1*(math.pi))+t3/1e-3 \n",
+ "t5=(0.2*(math.pi))-t3/1e-3\n",
+ "\n",
+ "#Results\n",
+ "print\"t1 is\",round(t1/1e-3,3),\"ms\"\n",
+ "print\"t2 is\",round(t2,2),\"ms\"\n",
+ "print\"t3 is\",round(t3/1e-3,3),\"ms\"\n",
+ "print\"t4 is\",round(t4,3),\"ms\"\n",
+ "print\"t5 is\",round(t5,3),\"ms\"\n",
+ "print\"vo is\",\"-5.33+6.66*sin(10**4*.15)\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "t1 is 0.064 ms\n",
+ "t2 is 0.25 ms\n",
+ "t3 is 0.041 ms\n",
+ "t4 is 0.355 ms\n",
+ "t5 is 0.587 ms\n",
+ "vo is -5.33+6.66*sin(10**4*.15)\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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