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{
"metadata": {
"name": ""
},
"nbformat": 3,
"nbformat_minor": 0,
"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 04 : Bipolar junction transistors"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.1, Page No 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Adc=.98\n",
"Ib=100*10**-6\n",
"\n",
"#Calculations\n",
"Ic=(Adc*Ib)/(1-Adc)\n",
"print(\"value of Ic is %3.3fA \" %Ic)\n",
"Ie=Ic/Adc\n",
"\n",
"#Results\n",
"print(\" value of Ie is %3.3fA \" %Ie)\n",
"Bdc=Adc/(1-Adc)\n",
"print(\"The value of Bdc = %.2f \" %Bdc) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"value of Ic is 0.005A \n",
" value of Ie is 0.005A \n",
"The value of Bdc = 49.00 \n"
]
}
],
"prompt_number": 1
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.2, Page No 153"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"Ic=1.0*10**-3\n",
"Ib=25.0*10**-6\n",
"\n",
"#Calculations\n",
"Bdc=Ic/Ib\n",
"Ie=Ic+Ib\n",
"Adc=Ic/Ie\n",
"Ic=5\n",
"Ib=Ic/Bdc\n",
"\n",
"print(\"The new base current = %.2f mA\" %(Ib*10**3)) "
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The new base current = 125.00 mA\n"
]
}
],
"prompt_number": 2
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.3 Page No 157"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"\n",
"#initialisation of variables\n",
"\n",
"Bdc=80.0\n",
"Bac=Bdc\n",
"Vcc=18.0\n",
"R1=10.0*10**3\n",
"\n",
"#Calculations\n",
"Ib=15.0*10**-6#for Vb=.7\n",
"Ic=Bdc*Ib\n",
"Vc=Vcc-(Ic*R1)\n",
"\n",
"#Results\n",
"print(\"dc collector voltage is %dV \" %Vc)\n",
"print(\" when vi=50mV\")\n",
"Ib=3.0*10**-6\n",
"Vi=50.0*10**-3\n",
"Ic=Bdc*Ib\n",
"Vo=Ic*R1\n",
"Av=Vo/Vi\n",
"\n",
"#Results\n",
"print(\"Current voltage is %.1f V \" %(Av))"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"dc collector voltage is 6V \n",
" when vi=50mV\n",
"Current voltage is 48.0 V \n"
]
}
],
"prompt_number": 3
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.4, Page No 160"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vcc=5.0\n",
"Vce=.2\n",
"R2=4.7*10**3\n",
"Vi=2\n",
"Vbe=.7\n",
"\n",
"#Calculations\n",
"R1=12.0*10**3\n",
"Ic=(Vcc-Vce)/R2\n",
"Ib=(Vi-Vbe)/R1\n",
"hFE=Ic/Ib\n",
"\n",
"#Results\n",
"print(\"Transistor current gain is %.2f V \" %(hFE))\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Transistor current gain is 9.43 V \n"
]
}
],
"prompt_number": 4
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Example 4.6 Page No 169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
"import math\n",
"#initialisation of variables\n",
"Vbe=.7\n",
"Vce=-6\n",
"\n",
"#Calculations\n",
"Ib=20.0*10**-6\n",
"Ic=2.5*10**-3#from output characteristics\n",
"Bdc=Ic/Ib\n",
"\n",
"#Results\n",
"print(\"The value of Bdc is %.1f V \" %Bdc)"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The value of Bdc is 125.0 V \n"
]
}
],
"prompt_number": 5
}
],
"metadata": {}
}
]
}
|
