{ "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": {} } ] }