{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "CHAPTER 2 SEMICONDUCTORS" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2-5, Page 49" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "Tj=100 #junction temperature 1(C)\n", "Vb=0.7 #Barrier potential(V)\n", "Tamb=25 #Ambient temperature(C)\n", "T1=0 #junction temperature 2(C)\n", "\n", "Vd=-0.002*(Tj-Tamb) #Change in barrier potential(V)\n", "Vbn=Vb+Vd #Barrier potential(V)\n", "Vd1=-0.002*(T1-Tamb) #Change in barrier potential(V)\n", "Vb1n=Vb+Vd1 #Barrier potential(V)\n", "\n", "print 'Change in barrier potential = Vd =',Vd,'V'\n", "print 'Brrier potential = Vbn =',Vbn,'V'\n", "print 'For junction temperature 0 C'\n", "print 'Change in barrier potential = Vd1 =',Vd1,'V'\n", "print 'New barrier potential = Vb1n =',Vb1n,'V'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Change in barrier potential = Vd = -0.15 V\n", "Brrier potential = Vbn = 0.55 V\n", "For junction temperature 0 C\n", "Change in barrier potential = Vd1 = 0.05 V\n", "New barrier potential = Vb1n = 0.75 V\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2-6, Page 51" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "\n", "T2=100 #Temperature(C)\n", "T1=25 #Temperature(C)\n", "Isat1=5 #Current(nA)\n", "\n", "Td=T2-T1 #Temperature change(C)\n", "Is1=(2**7)*Isat1 #Current(nA)\n", "Is2=(1.07**5)*Is1 #Current(nA)\n", "\n", "print 'Change in temperature = Td = T2-T1 =',Td,'C'\n", "print 'For first 70 C change seven doublings are there.'\n", "print 'Is1 = (2^7)*Isat1 =',Is1,'nA'\n", "print 'For additional 5 C, 7% per C'\n", "print 'Is2 = ',round(Is2,2),'nA'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Change in temperature = Td = T2-T1 = 75 C\n", "For first 70 C change seven doublings are there.\n", "Is1 = (2^7)*Isat1 = 640 nA\n", "For additional 5 C, 7% per C\n", "Is2 = 897.63 nA\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 2-7, Page 52" ] }, { "cell_type": "code", "collapsed": false, "input": [ "\n", "Isl=2*(10**-9) #surface leakage current(nA)\n", "Vr=25 #Reverse voltage(V)\n", "Vr1=35 #Reverse voltage(V)\n", "\n", "Rsl=Vr/Isl #surface leakage reistance(Ohm)\n", "Isl1=(Vr1/Rsl)*(10**9) #surface leakage current(nA)\n", "\n", "print 'surface-leakage reistance Rsl = Vr/Isl =',Rsl*10**-6,'MOhm'\n", "print 'for Vr1 = 35 V,'\n", "print 'surface-leakage reistance Isl1 = Vr1/Rsl =',Isl1,'nA'" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "surface-leakage reistance Rsl = Vr/Isl = 12500.0 MOhm\n", "for Vr1 = 35 V,\n", "surface-leakage reistance Isl1 = Vr1/Rsl = 2.8 nA\n" ] } ], "prompt_number": 17 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }