From 92cca121f959c6616e3da431c1e2d23c4fa5e886 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Tue, 7 Apr 2015 15:58:05 +0530 Subject: added books --- Electronic_Devices_and_Circuits/Chapter6_1.ipynb | 656 +++++++++++++++++++++++ 1 file changed, 656 insertions(+) create mode 100755 Electronic_Devices_and_Circuits/Chapter6_1.ipynb (limited to 'Electronic_Devices_and_Circuits/Chapter6_1.ipynb') diff --git a/Electronic_Devices_and_Circuits/Chapter6_1.ipynb b/Electronic_Devices_and_Circuits/Chapter6_1.ipynb new file mode 100755 index 00000000..85bada6c --- /dev/null +++ b/Electronic_Devices_and_Circuits/Chapter6_1.ipynb @@ -0,0 +1,656 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 06 : Ac analysis of BJT circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1, Page No 240" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vcc=12.0\n", + "R2=15.0*10**3\n", + "R1=33.0*10**3\n", + "rs=600\n", + "\n", + "#Calculations\n", + "print(\"with no signal source\")\n", + "Vb=(Vcc*R2)/(R1+R2)\n", + "print(\" base bais voltage when no signal source is present %3.2fV \" %Vb)\n", + "print(\" signal source directly connected\")\n", + "Vb=(Vcc*((rs*R2)/(rs+R2))/(R1+((rs*R2)/(rs+R2))))\n", + "\n", + "#Results\n", + "print(\"base bais voltage is %3.2fV \" %Vb)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "with no signal source\n", + " base bais voltage when no signal source is present 3.75V \n", + " signal source directly connected\n", + "base bais voltage is 0.21V \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2, Page No 244" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "\n", + "Rc=2.2*10**3\n", + "Re=2.7*10**3\n", + "R1=18.0*10**3\n", + "R2=8.2*10**3\n", + "Vbe=.7\n", + "\n", + "#Calculations\n", + "print(\"drawing dc load line\")\n", + "Rldc=Rc+Re\n", + "print(\" for Vce\")\n", + "Ic=0\n", + "Vcc=20\n", + "Vce=Vcc-Ic*(Rc+Re)\n", + "print(\"plot point A at\")\n", + "Ic=Vcc/(Rc+Re)\n", + "print(\"plot point B Ic=4.08mA and Vce=0\")\n", + "print(\" draw dc laod line through point A nad B\")\n", + "Vb=(Vcc*R2)/(R1+R2)\n", + "Ve=Vb-Vbe\n", + "Ic=Ve/Re\n", + "Ie=Ic\n", + "print(\"drawing the ac load line\")\n", + "Rlac=Rc#when there is no external Rl\n", + "Vce=Ic*Rc\n", + "\n", + "\n", + "#Results\n", + "print(\"The voltage is %.2f v \" %Vce)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "drawing dc load line\n", + " for Vce\n", + "plot point A at\n", + "plot point B Ic=4.08mA and Vce=0\n", + " draw dc laod line through point A nad B\n", + "drawing the ac load line\n", + "The voltage is 4.53 v \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3 Page No 251" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Vce=4.5\n", + "Ib=40.0*10**-6\n", + "\n", + "#Calculations\n", + "print(\"from current characteristic at Vce=4.5V and Ib=40uA\")\n", + "Ic=4.0*10**-3\n", + "Ib=30.0*10**-6\n", + "hFE=Ic/Ib\n", + "print(\" the value of hFE is %d \" %hFE)\n", + "print(\"from output characteristic at Vce=4.5 and Ib=40uA\")\n", + "Ic=0.2\n", + "Vce=6\n", + "hoe=(Ic/Vce)\n", + "R=1/hoe\n", + "\n", + "#Results\n", + "print(\"the value of hoe is %3.1fuS \" %(hoe*10**3))\n", + "print(\"the value of 1/hoe is %3.1fuS \" %(1/hoe))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "from current characteristic at Vce=4.5V and Ib=40uA\n", + " the value of hFE is 133 \n", + "from output characteristic at Vce=4.5 and Ib=40uA\n", + "the value of hoe is 33.3uS \n", + "the value of 1/hoe is 30.0uS \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4, Page No 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "hfe=133.0\n", + "hoe=33.3*10**-6\n", + "hfc=1+hfe\n", + "\n", + "#Calculations\n", + "hob=hoe/(1+hfe)\n", + "A=hfe/(1+hfe)\n", + "\n", + "#Results\n", + "print(\"tye value of a is %3.1fuS \" %(A))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "tye value of a is 1.0uS \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5 Page No 253" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "\n", + "Ib=20.0*10**-6\n", + "Ic=1.0*10**-3\n", + "Ie=Ic\n", + "\n", + "#Calculations\n", + "re=(26*10**-3)/Ie\n", + "hfe=Ic/Ib\n", + "hie=(1+hfe)*re\n", + "r=hie\n", + "B=hfe\n", + "\n", + "#Results\n", + "print(\"the value of b is %3.1f \" %(B))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "the value of b is 50.0 \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6 Page No 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "hie=2.1*10**3\n", + "hfe=75.0\n", + "hoe=1*10**-6\n", + "R1=68.0*10**3\n", + "R2=56.0*10**3\n", + "Rc=3.9*10**3\n", + "Rl=82*10**3\n", + "\n", + "#Calculations\n", + "print(\" input impedance Zi=R1||R2||hie\")\n", + "Zi=((R1*R2*hie)/(R1+R2+hie))*10**-3\n", + "print(\" input impedance is %3.2fKohm \" %Zi)\n", + "print(\"output impedance is Zo=Rc||(1/hoe)\")\n", + "Zo=((Rc*(1/hoe))/(Rc+(1/hoe)))*10**-3\n", + "print(\" output impadance is %f3.2fKohm \" %Zo)\n", + "Av=-(hfe*((Rc*Rl)/(Rc+Rl)))/hie\n", + "\n", + "\n", + "#Results\n", + "print(\" voltage gain is %d \" %Av)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " input impedance Zi=R1||R2||hie\n", + " input impedance is 63416.34Kohm \n", + "output impedance is Zo=Rc||(1/hoe)\n", + " output impadance is 3.8848493.2fKohm \n", + " voltage gain is -132 \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7, Page No 259" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Ic=1.5*10**-3\n", + "Rc=4.7*10**3\n", + "Rl=56.0*10**3\n", + "\n", + "#Calculations\n", + "Ie=Ic\n", + "re=(26*10**-3)/Ie\n", + "Av=-(((Rc*Rl)/(Rc+Rl))/re)\n", + "\n", + "#Results\n", + "print(\" voltage gain is %d \" %Av)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " voltage gain is -250 \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8 Page No 262" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "hie=2.1*10**3\n", + "hfe=75.0\n", + "hoe=1.0*10**-6\n", + "Re=4.7*10**3\n", + "R1=68.0*10**3\n", + "R2=56.0*10**3\n", + "Rc=3.9*10**3\n", + "Rl=82.0*10**3\n", + "\n", + "#Calculations\n", + "Zb=hie+Re*(1+hfe)\n", + "print(\" input impedance is Zi=R1||R2||Zb\")\n", + "Zi=((R1*R2*Zb)/(R1+R2+Zb))\n", + "print(\" input circuit resistance is %3.3fKohm \" %Zi)\n", + "Zo=Rc\n", + "Av=-hfe*((Rc*Rl)/(Rc+Rl))/(hie+Re*(1+hfe))\n", + "\n", + "#Results\n", + "print(\"voltage gain is %3.3f \" %Av)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " input impedance is Zi=R1||R2||Zb\n", + " input circuit resistance is 2830983654.045Kohm \n", + "voltage gain is -0.777 \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9 Page No 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "hie=2.1*10**3\n", + "hfe=75.0\n", + "R1=10.0*10**3\n", + "R2=10.0*10**3\n", + "Re=4.7*10**3\n", + "Rl=12.0*10**3\n", + "rs=1.0*10**3\n", + "\n", + "#Calculations\n", + "print(\" Rl is not connected\")\n", + "hic=hie\n", + "hfc=1+hfe\n", + "Zb=hic+hfc*(Re)\n", + "Zi=(R1*R2*Zb)/(R1+R2+Zb)\n", + "Ze=(hic+(R1*R2*rs)/(R1+R2+rs))/hfc\n", + "Z0=(Ze*Re)/(Ze+Re)\n", + "print(\" when Rl is connected\")\n", + "Zb=hic+hfc*((Re*Rl)/(Re+Rl))\n", + "Zi=(R1*R2*Zb)/(R1+R2+Zb)\n", + "hib=hie/(1+hfe)\n", + "Av=((Re*Rl)/(Re+Rl))/(hib+((Re*Rl)/(Re+Rl)))\n", + "\n", + "#Results\n", + "print(\"voltage gain is %3.3f \" %Av)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Rl is not connected\n", + " when Rl is connected\n", + "voltage gain is 0.992 \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10 Page No 273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "hie=2.1*10**3\n", + "hfe=75.0\n", + "Re=4.7*10**3\n", + "Rc=3.9*10**3\n", + "Rl=82.0*10**3\n", + "\n", + "#Calculations\n", + "hib=hie/(1+hfe)\n", + "hfb=hfe/(1+hfe)\n", + "Zi=(hib*Re)/(Re+hib)\n", + "print(\"input impedance is %3.2fohm \" %Zi)\n", + "Zo=Rc\n", + "print(\" output impedance is %3.2fohm \" %Zo)\n", + "Av=(hfb*((Rc*Rl)/(Rc+Rl)))/hib\n", + "\n", + "#Results\n", + "print(\" voltage gain is %3.2f \" %Av)\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "input impedance is 27.47ohm \n", + " output impedance is 3900.00ohm \n", + " voltage gain is 132.96 \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11, Page No 273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "hib=27.6\n", + "hfb=.987\n", + "R1=68.0*10**3\n", + "R2=56.0*10**3\n", + "Re=4.7*10**3\n", + "Rc=3.9*10**3\n", + "Rl=82.0*10**3\n", + "\n", + "#Calculations\n", + "Rb=(R1*R2)/(R1+R2)\n", + "Ze=hib+Rb*(1-hfb)\n", + "Zi=(Ze*Re)/(Ze+Re)\n", + "Av=(hfb*((Rc*Rl)/(Rc+Rl)))/(hib+Rb*(1-hfb))\n", + "\n", + "#Results\n", + "print(\"voltage gain is %3.3f \" %Av)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage gain is 8.609 \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12, Page No 277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Rc=5.6*10**3\n", + "Rl=33.0*10**3\n", + "rs=600.0\n", + "hfe=100\n", + "hie=1.5*10**3\n", + "vs=50.0*10**-3\n", + "\n", + "#Calculations\n", + "print(\" CE circuit operation with vs at transistor base and Re bypassed\")\n", + "Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie\n", + "Zb=hie\n", + "Rb=(R1*R2)/(R1+R2)\n", + "Zi=(Rb*Zb)/(Rb+Zb)\n", + "vi=(vs*Zi)/(rs+Zi)\n", + "vo=Av*vi\n", + "print(\"Cb circuit operation with vs at emitter and the base resistor bypassed\")\n", + "Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie\n", + "Ze=hie/(1+hfe)\n", + "Zi=(Ze*Re)/(Ze+Re)\n", + "vi=(vs*Zi)/(rs+Zi)\n", + "vo=Av*vi\n", + "\n", + "#Results\n", + "print(\"voltage vo is %3.2f \" %(vo*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " CE circuit operation with vs at transistor base and Re bypassed\n", + "Cb circuit operation with vs at emitter and the base resistor bypassed\n", + "voltage vo is 384.29 \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13, Page No 279" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Io=50.0*10**-9\n", + "Vbe=.7\n", + "Vbc=-10\n", + "Af=.995\n", + "Ar=.5\n", + "Vt=26.0*10**-3\n", + "n=2.0\n", + "Vd=-10.0\n", + "\n", + "#Calculations\n", + "x=Vd/(n*Vt)\n", + "Idc=(Io*((2.73**-x)-1))*10**9\n", + "Idc=Io*(-1)\n", + "y=Vbe/(n*Vt)\n", + "Ide=Io*((2.73**y)-1)\n", + "I1=Af*Ide\n", + "I2=Ar*Idc\n", + "Ic=I1-Idc\n", + "Ie=Ide-I2\n", + "Ib=Ie-Ic\n", + "\n", + "#Results\n", + "print(\"voltage gain is %3.3f \" %(Ib*10**6))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "voltage gain is 185.909 \n" + ] + } + ], + "prompt_number": 13 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit