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{
"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": {}
}
]
}
|
