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diff --git a/Fundamental_of_Electronics_Devices/Ch4.ipynb b/Fundamental_of_Electronics_Devices/Ch4.ipynb new file mode 100644 index 00000000..e41f2209 --- /dev/null +++ b/Fundamental_of_Electronics_Devices/Ch4.ipynb @@ -0,0 +1,1079 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 4: Junction Properties"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.1 page No. 146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Majority carrier electron concentration is 5.97e+13 cm**-3\n",
+ "Minority carrier hole concentration is 9.7e+12 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.2 Page No.146"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Majority carrier electron concentration is 1e+16 cm**-3\n",
+ "Minority carrier hole concentration is 22500.0 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.3 Page No. 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Majority carrier hole concentration is 7e+15 cm**-3\n",
+ "Minority carrier electron concentration is 36571.0 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.4 Page No. 147"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum Temprature is 642.0 K\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.5 Page No. 151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Built in potential barrier is 0.7532 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.6 Page No.151"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Exa4.6\n",
+ "#What is Contact Potential.\n",
+ "\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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Contact potential is 0.5745 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.7 Page No. 154"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.8 Page No.160"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "New position of fermi level is 0.328 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.9 Page No. 161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Contact potential is 0.19 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.10 page No.161"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.10\n",
+ "#(i)Find the hole and electron concentration \n",
+ "#Is this Silicon P or N type\n",
+ "\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",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.11 Page No. 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current flowing through the circuit is 43.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.12 Page No. 168"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voltagee VA is 13.6 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.13 Page No.169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Voltage VA is 14.7 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.14 Page No.172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Temperature coefficient f zener diode is -0.053 percent\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.15 Page No. 174"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.16 Page No. 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum zener diode current is 9.0 mA\n",
+ "Minimum zener diode current is 1.0 mA\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.17 Page No. 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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. \""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.18 Page No. 175"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 52
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.19 Page No. 126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.20 Page No. 176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "zener diode is ON state\n",
+ "Hence the voltage dropp across the 5 Kohm resistor in Volts is 50 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 67
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.21 Page No. 176"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance Ri is 25.0 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 72
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.22 Page No. 177"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#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\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "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"
+ ]
+ }
+ ],
+ "prompt_number": 71
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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