{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 6: Bipolar junction Transistors (BJTs)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.1 page No.215" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Emitter current is 0.05 mA\n" ] } ], "source": [ "#Exa 6.1\n", "#find the Base current\n", "\n", "#given data\n", "Ic=9.95\t\t\t#in mA\n", "Ie=10 \t\t#in mA\n", "\n", "#Calculation\n", "Ib=Ie-Ic\t\t#in mA\n", "\n", "#result\n", "print\"Emitter current is \",Ib,\"mA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.2 page No. 216" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Emitter current is 1.0 mA\n", "Current amplification factor is 0.98\n", "Current gain factor is 49.0\n" ] } ], "source": [ "#Exa 6.2\n", "#Find (i)Emitter current (ii)Current amplification factor (iii)Current gain factor \n", "\n", "#given data\n", "IC=0.98\t\t\t#in mA\n", "IB=20.0\t\t\t#in uA\n", "IB=IB*10**-3\t\t#in mA\n", "\n", "#Calculation\n", "#part (i)\n", "IE=IB+IC\t\t#in mA\n", "\n", "#part (ii)\n", "alpha=IC/IE\t\t#unitless\n", "#part (iii)\n", "Beta=IC/IB\t\t#unitless\n", "\n", "#Result\n", "print\"Emitter current is\",IE,\"mA\"\n", "print\"Current amplification factor is \",alpha\n", "print\"Current gain factor is \",Beta" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.3 page No.216" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Emitter current is 2.7 mA\n", "Collector current is 2.65 mA\n" ] } ], "source": [ "#Exa 6.3\n", "#Emitter current and Collector current\n", "\n", "#given data\n", "alfaDC=0.98\t\t\t#unitless\n", "ICBO=4\t\t\t\t#in uA\n", "ICBO=ICBO*10**-3\t\t#in mA\n", "IB=50\t\t\t\t#in uA\n", "IB=IB*10**-3\t\t\t#in mA\n", "\n", "#calculation\n", "#Formula : IC=alfaDC*(IB+IC)+ICBO\n", "IC=alfaDC*IB/(1-alfaDC)+ICBO/(1-alfaDC)\t#in mA\n", "IE=IC+IB\t\t\t#in mA\n", "\n", "#Result\n", "print\"Emitter current is \",IE,\"mA\"\n", "print\"Collector current is \",IC,\"mA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.4 page No. 216" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Collector current in mA : 1.09 mA\n" ] } ], "source": [ "#Exa 6.4\n", "#Find the collector current\n", "\n", "#given data\n", "IB=10\t\t\t#in uA\n", "IB=IB*10**-3\t\t#in mA\n", "Beta=99\t\t\t#Unitless\n", "ICO=1\t\t\t#in uA\n", "ICO=ICO*10**-3\t\t#in mA\n", "\n", "#calculation\n", "#Formula : IC=alfa*(IB+IC)+ICO\n", "IC=Beta*IB+(1+Beta)*ICO\t#in mA\n", "\n", "#Result\n", "print\"Collector current in mA : \",IC,\"mA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.5 Page No.216" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(i) Current gain factor is 98.0\n", " Current amplification factor is 0.99\n", " Emitter Current is 5.05 mA\n", "(ii)New level of Ib is 101.0 micro A\n" ] } ], "source": [ "#Example 6.5\n", "#Find (i) alpha , beta and Ie \n", "#(ii)New level of Ib\n", "\n", "#Given\n", "Ic=5*10**-3 #mA collector current\n", "Ic_=10*10**-3 #mA collector current\n", "Ib=50*10**-6 #mA, Base current\n", "Icbo=1*10**-6 #micro A, Current to base open current\n", "\n", "#Calculation\n", "beta=(Ic-Icbo)/(Ib+Icbo)\n", "alpha=(beta/(1+beta))\n", "Ie=Ib+Ic\n", "\n", "Ib=(Ic_-(beta+1)*Icbo)/(beta)\n", "\n", "#Result\n", "print\"(i) Current gain factor is\",round(beta,0)\n", "print\" Current amplification factor is\",round(alpha,2)\n", "print\" Emitter Current is\",Ie*1000,\"mA\"\n", "print\"(ii)New level of Ib is\",round(Ib*10**6,0),\"micro A\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.6 page No. 222" ] }, { "cell_type": "code", "execution_count": 18, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Dynamic input resistance is 40 mohm\n" ] } ], "source": [ "#Exa 6.6\n", "#Find the dynamic input resistance\n", "\n", "#given data\n", "delVEB=200\t\t\t#in Volts\n", "delIE=5\t\t\t\t#in mA\n", "\n", "#calculation\n", "rin=delVEB/delIE\t\t#in ohm\n", "\n", "#Result\n", "print\"Dynamic input resistance is \",rin,\"mohm\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.7 page No. 222" ] }, { "cell_type": "code", "execution_count": 32, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Current gain : 0.979\n", "Base current is 0.03 mA\n" ] } ], "source": [ "#Exa 6.7\n", "#Determine Current gain and base current\n", "\n", "\n", "#given data\n", "ICBO=12.5 \t\t\t#in uA\n", "ICBO=ICBO*10**-3 \t\t#in mA\n", "IE=2 \t\t\t\t#in mA\n", "IC=1.97 \t\t\t#in mA\n", "\n", "#calculation\n", "alfa=(IC-ICBO)/IE \t\t#unitless\n", "IB=IE-IC \t\t\t#in mA\n", "\n", "#result\n", "print\"Current gain : \",round(alfa,3)\n", "print\"Base current is \",IB,\"mA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.8 page No. 222" ] }, { "cell_type": "code", "execution_count": 37, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Base current ia 0.03 mA\n" ] } ], "source": [ "#Exa 6.8\n", "#given data\n", "RL=4.0 \t\t\t#in Kohm\n", "VL=3.0\t\t\t#in volt\n", "alfa=0.96 \t\t#unitless\n", "IC=VL/RL \t\t#in mA\n", "\n", "#calculation\n", "IE=IC/alfa \t\t#in mA\n", "IB=IE-IC \t\t#in mA\n", "\n", "#result\n", "print\"Base current ia\",round(IB,2),\"mA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.9 page No.227" ] }, { "cell_type": "code", "execution_count": 41, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Collector-emitter Voltage is 9.2 V\n", "Base current in uA : 41.67 microA\n" ] } ], "source": [ "#Exa 6.9\n", "#Determine Collector emitter voltage and base current\n", "\n", "#given data\n", "VCC=10\t\t\t #in volt\n", "RL=800\t\t\t #in ohm\n", "VL=0.8\t\t\t #in volt\n", "alfa=0.96\t\t #unitless\n", "\n", "#calculation\n", "#VR=IC*RL\n", "VCE=VCC-VL \t\t#in Volt\n", "IC=VL*1000/RL \t\t#in mA\n", "Beta=alfa/(1-alfa) \t#unitless\n", "IB=IC/Beta \t\t#in mA\n", "\n", "#Result\n", "print\"Collector-emitter Voltage is \",VCE,\"V\"\n", "print\"Base current in uA : \",round(IB*1000,2),\"microA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.10 page No. 227" ] }, { "cell_type": "code", "execution_count": 45, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Collector current is 11.28 mA\n" ] } ], "source": [ "#Exa 6.10\n", "#Determine Collector Current\n", "\n", "#given data\n", "alfao=0.98 \t\t#unitless\n", "ICO=10 \t\t\t#in uA\n", "ICO=ICO*10**-3 \t\t#in mA\n", "IB=0.22 \t\t#in mA\n", "\n", "#calculation\n", "IC=(alfao*IB+ICO)/(1-alfao) \t#in mA\n", "\n", "#result\n", "print\"Collector current is\",IC,\"mA\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.11 page No. 228" ] }, { "cell_type": "code", "execution_count": 47, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Dynamic input resistance is 250 ohm\n" ] } ], "source": [ "#Exa 6.11\n", "#determine Dynamic input resistance \n", "\n", "#given data\n", "delVEB=250 \t\t#in mVolts\n", "delIE=1 \t\t#in mA\n", "\n", "#calculation\n", "rin=delVEB/delIE \t#in ohm\n", "\n", "#result\n", "print\"Dynamic input resistance is\",rin,\"ohm\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.12 page No. 228" ] }, { "cell_type": "code", "execution_count": 49, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Dynamic output resistance is 6.25 kohm\n" ] } ], "source": [ "#Exa 6.12\n", "#Determine Dynamic output resistance\n", "\n", "#given data\n", "delVCE=10-5 \t\t#in Volts\n", "delIC=5.8-5\t \t#in mA\n", "\n", "#calculation\n", "rin=delVCE/delIC \t#in Kohm\n", "\n", "#result\n", "print\"Dynamic output resistance is \",rin,\"kohm\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.13 page No.232" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Operating point Q is ( 5.2 V, 0.6 mA)\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#Exa 6.13\n", "#Determine operating point\n", "%matplotlib inline\n", "import matplotlib.pyplot as plt\n", "#given data\n", "VCC=10 \t\t\t#in volt\n", "RC=8 \t\t\t#in Kohm\n", "Beta=40 \t\t#unitless\n", "IB=15 \t\t\t#in uA\n", "IB=IB*10**-3 \t\t#in mA\n", "\n", "#calculation\n", "# For VCE = 0 Volts\n", "IC=VCC/RC \t\t#in mA\n", "#For IC=0 VCE=VCC=10V :\n", "IC=Beta*IB \t\t#in mA\n", "VCE=VCC-IC*RC \t\t#in Volts\n", "\n", "#result\n", "print\"Operating point Q is (\",VCE,\"V,\",IC,\"mA)\"\n", "\n", "#Plot\n", "import matplotlib.pyplot as plt\n", "fig = plt.figure()\n", "ax = fig.add_subplot(111)\n", "\n", "Vce=[0,10]\n", "Ic=[1.25,0]\n", "plt.xlabel('Vce,V')\n", "plt.ylabel('Ic,mA')\n", "ax.plot([5.2], [0.6], 'o')\n", "ax.annotate('(5.2V,0.6 mA)', xy=(5.4,0.7))\n", "\n", "a=plt.plot(Vce,Ic)\n", "plt.show(a)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.14 page No. 232" ] }, { "cell_type": "code", "execution_count": 65, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Operating point at load resistance 5 kohm is ( 6.0 V, 1.2 mA)\n", "Operating point at load resistance 7.5 kohm is ( 3.0 V, 1.2 mA)\n" ] } ], "source": [ "#Exa 6.14\n", "#How will the Q point change when load resistance will be change\n", "\n", "#given data \n", "Vcc=12 \t\t#in Volt collector supply voltage\n", "Ic=1.2 #A, collector current\n", "Rl=5 #kohm load resistance\n", "\n", "#calculation\n", "Vce=Vcc-Ic*Rl #Collector emitter voltage\n", "Rl1=7.5\n", "Vce1=Vcc-Ic*Rl1\n", "\n", "#result\n", "print\"Operating point at load resistance 5 kohm is (\",Vce,\"V,\",Ic,\"mA)\"\n", "print\"Operating point at load resistance 7.5 kohm is (\",Vce1,\"V,\",Ic,\"mA)\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.15 Page No.233" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Collector to emitter voltage is (Vce) 20 V\n", "Collector current at saturation point is (Ic) 6.0 mA\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "#Example 6.15\n", "#Given\n", "Vcc=20 # V, collector voltage\n", "Rc=3.3*10**3\n", "\n", "#calculation\n", "#Appling kirchoff's Voltage Law\n", "Ic=0 #for cut off point\n", "Vce=Vcc\n", "Ic=Vcc/Rc\n", "print \"Collector to emitter voltage is (Vce)\",Vce,\"V\"\n", "print \"Collector current at saturation point is (Ic)\",round(Ic*1000,0),\"mA\"\n", "\n", "#Plot\n", "%matplotlib inline\n", "import matplotlib.pyplot as plt\n", "fig = plt.figure()\n", "ax = fig.add_subplot(111)\n", "\n", "Vce=[0,20]\n", "Ic=[6,0]\n", "plt.xlabel(\"Vce (V)\") \n", "plt.ylabel(\"Ic (mA)\") \n", "plt.xlim((0,25))\n", "plt.ylim((0,8))\n", "ax.plot([0], [6], 'o')\n", "ax.annotate('(0,6mA)', xy=(0,6))\n", "\n", "ax.plot([20], [0], 'o')\n", "ax.annotate('(20V,0)', xy=(20,0))\n", "a=plt.plot(Vce,Ic)\n", "plt.show(a)\n", "\n" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "### Example 6.16 page No. 233" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "collector voltage is -4.5 V\n", "Base voltage is -8.3 V\n" ] } ], "source": [ "#Exa 6.16\n", "#find collector voltage and base voltage\n", "\n", "#given data \n", "Beta=45 \t\t\t#Unitless\n", "VBE=0.7 \t\t\t#in Volt\n", "VCC=0 \t\t\t\t#in Volt\n", "RB=10**5 \t\t\t#in ohm\n", "RC=1.2*10**3 \t\t\t#in ohm\n", "VEE=-9 \t\t\t\t#in Volt\n", "\n", "#calculation\n", "#Applying Kirchoffs Voltage Law in input loop we have\n", "#IB*RB+VBE+VEE=0\n", "IB=-(VBE+VEE)/RB \t\t#in mA\n", "IC=Beta*IB \t\t\t#in mA\n", "VC=VCC-IC*RC \t\t\t#in Volts\n", "VB=VBE+VEE \t\t\t#in Volts\n", "\n", "#Result\n", "print\"collector voltage is \",round(VC,1),\"V\"\n", "print\"Base voltage is \",VB,\"V\"" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.6" } }, "nbformat": 4, "nbformat_minor": 0 }