From 79c59acc7af08ede23167b8455de4b716f77601f Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Thu, 11 Jun 2015 17:31:11 +0530 Subject: add books --- Electronic_Devices_and_Circuits/Chapter6_2.ipynb | 656 ----------------------- 1 file changed, 656 deletions(-) delete mode 100755 Electronic_Devices_and_Circuits/Chapter6_2.ipynb (limited to 'Electronic_Devices_and_Circuits/Chapter6_2.ipynb') diff --git a/Electronic_Devices_and_Circuits/Chapter6_2.ipynb b/Electronic_Devices_and_Circuits/Chapter6_2.ipynb deleted file mode 100755 index 85bada6c..00000000 --- a/Electronic_Devices_and_Circuits/Chapter6_2.ipynb +++ /dev/null @@ -1,656 +0,0 @@ -{ - "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