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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