{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "CHAPTER 3 DIODE THEORY" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-2, Page 63" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "Pr=5 #Power rating(W)\n", "Vd=1.2 #diode voltage(V)\n", "Id=1.75 #diode current(A)\n", "\n", "PD=Vd*Id #Power dissipaion(W)\n", "\n", "print 'Power Dissipation =',PD,'W'\n", "print 'PD(',PD,'W) < ''Pr(',Pr,'W), So diode will not be destroyed.'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Power Dissipation = 2.1 W\n", "PD( 2.1 W) < Pr( 5 W), So diode will not be destroyed.\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-3, Page 65" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "Vs=10 #Source voltage(V)\n", "RL=1 #Load resistance(KOhm)\n", "VL=Vs #LOad voltage(V)\n", "\n", "IL=VL/RL #Load current(mA)\n", "\n", "print 'Vs will be appearing across RL'\n", "print 'Load voltage VL =',Vs,'V'\n", "print 'Load current IL =',IL,'mA'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Vs will be appearing across RL\n", "Load voltage VL = 10 V\n", "Load current IL = 10 mA\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-4, Page 65" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "print 'As per figure 3-6b, Thevenize the circuit to the left of the diode'\n", "R1=6 #Resistance(KOhm)\n", "R2=3 #Resistance(KOhm)\n", "RL=1 #Load Resistance(KOhm)\n", "Vs=36 #Supply voltage(V)\n", "\n", "Vth=R2*Vs/(R1+R2) #ThCevenin voltage(V)\n", "Rth=(R1*R2)/(R1+R2) #Thevenin resistance(KOhm)\n", "Rt=Rth+RL #total resistance(KOhm)\n", "IL=Vth/Rt #Load current(mA)\n", "VL=IL*RL #Load voltage(V)\n", "\n", "print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'\n", "print 'Visualize diode as closed switch,'\n", "print 'IL =',IL,'mA'\n", "print 'VL =',VL,'V'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "As per figure 3-6b, Thevenize the circuit to the left of the diode\n", "Vth = 12 V & Rth = 2 KOhm\n", "Visualize diode as closed switch,\n", "IL = 4 mA\n", "VL = 4 V\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-5, Page 67" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "print 'As per Second approximation in Fig.3-8,'\n", "Vd=0.7 #diode voltgage(V)\n", "Vs=10 #supply voltage(V)\n", "RL=1 #Load resistance(KOhm)\n", "\n", "VL=Vs-Vd #Load voltage(v)\n", "IL=VL/RL #Load current(mA)\n", "PD=Vd*IL #diode power(mW)\n", "\n", "print 'IL =',IL,'mA & VL =',VL,'V'\n", "print 'Diode power PD =',PD,'mW'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "As per Second approximation in Fig.3-8,\n", "IL = 9.3 mA & VL = 9.3 V\n", "Diode power PD = 6.51 mW\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-6, Page 67" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "print 'As per Second approximation in Fig.3-9a,'\n", "Vd=0.7 #diode voltgage(V)\n", "Vs=36 #supply voltage(V)\n", "R1=6 #Resistance(KOhm)\n", "R2=3 #Resistance(KOhm)\n", "RL=1 #Load resistance(KOhm)\n", "\n", "Vth=R2*Vs/(R1+R2)#Thevenin Voltage(V)\n", "Rth=(R1*R2)/(R1+R2)#Thevenin resistance(KOhm)\n", "IL=(Vth-Vd)/R2#Load current(mA)\n", "VL=IL*RL#Load voltage(V)\n", "PD=Vd*IL#diode power(mW)\n", "\n", "print 'Thevenize the circuit to the left of the diode'\n", "print 'Vth =',Vth,'V & Rth =',Rth,'KOhm'\n", "print 'VL =',round(VL,2),'V & IL =',round(IL,2),'mA'\n", "print 'Diode power PD =',round(PD,2),'mW'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "As per Second approximation in Fig.3-9a,\n", "Thevenize the circuit to the left of the diode\n", "Vth = 12 V & Rth = 2 KOhm\n", "VL = 3.77 V & IL = 3.77 mA\n", "Diode power PD = 2.64 mW\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-7, Page 68" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "print 'In Fig.3-11a,'\n", "Vd=0.7 #diode voltgage(V)\n", "Vs=10 #supply voltage(V)\n", "RL=1000L #Load resistance(Ohm)\n", "Rb=0.23 #bulk resistance\n", "\n", "print 'As per third approximation, we get fig.3-11b'\n", "if Rb<(RL/100):\n", " print'If Rb < 0.01RL than ignore Rb & use second approximation.'\n", " VL=Vs-Vd #Load voltage(V)\n", " IL=(VL/RL)*1000 #Load current(mA)\n", " PD=Vd*IL #diode power(mW)\n", "\n", " print 'IL =',IL,'mA & VL =',VL,'V'\n", " print 'Diode power PD =',PD,'mW'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "In Fig.3-11a,\n", "As per third approximation, we get fig.3-11b\n", "If Rb < 0.01RL than ignore Rb & use second approximation.\n", "IL = 9.3 mA & VL = 9.3 V\n", "Diode power PD = 6.51 mW\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 3-8, Page 69" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "print 'In Fig.3-11a, take RL = 10 Ohm'\n", "Vd=0.7 #diode voltgage(V)\n", "Vs=10 #supply voltage(V)\n", "RL=10 #Load resistance(Ohm)\n", "Rb=0.23 #bulk resistance\n", "RT=Rb+RL #Total reistance(Ohm)\n", "VT=Vs-Vd #total voltage(V)\n", "\n", "print 'RT =',RT,'Ohm & VT =',VT,'V'\n", "IL=VT/RT #Load current(mA)\n", "VL=IL*RL #Load voltage(V)\n", "VD=Vd+(IL*Rb) \n", "PD=VD*IL #diode power(W)\n", "\n", "print 'IL =',round(IL,2),'mA & VL =',round(VL,2),'V'\n", "print 'VD =',round(VD,2),'V'\n", "print 'Diode power PD =',round(PD,2),'W'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "In Fig.3-11a, take RL = 10 Ohm\n", "RT = 10.23 Ohm & VT = 9.3 V\n", "IL = 0.91 mA & VL = 9.09 V\n", "VD = 0.91 V\n", "Diode power PD = 0.83 W\n" ] } ], "prompt_number": 9 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }