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authorhardythe12015-04-07 15:58:05 +0530
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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": {}
+ }
+ ]
+} \ No newline at end of file