From f270f72badd9c61d48f290c3396004802841b9df Mon Sep 17 00:00:00 2001 From: kinitrupti Date: Fri, 12 May 2017 18:53:46 +0530 Subject: Removed duplicates --- .../Chapter3.ipynb | 1201 ++++++++++++++++++++ 1 file changed, 1201 insertions(+) create mode 100755 Electronic_Devices_and_Circuits_by_David_A._Bell/Chapter3.ipynb (limited to 'Electronic_Devices_and_Circuits_by_David_A._Bell/Chapter3.ipynb') diff --git a/Electronic_Devices_and_Circuits_by_David_A._Bell/Chapter3.ipynb b/Electronic_Devices_and_Circuits_by_David_A._Bell/Chapter3.ipynb new file mode 100755 index 00000000..8f8a6a3b --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_David_A._Bell/Chapter3.ipynb @@ -0,0 +1,1201 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter 03 : Diode applications" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1, Page No 73" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vf=.7\n", + "Rl=500.0\n", + "Vi=22.0\n", + "Vpi=1.414*Vi\n", + "\n", + "#Calculations\n", + "Vpo=Vpi-Vf\n", + "print(\" peak vouput voltage is %3.2fV \" %Vpo)\n", + "Ip=Vpo/Rl\n", + "\n", + "#Results\n", + "print(\"peak load current is %3.4fA \" %Ip)\n", + "PIV=Vpi\n", + "print(\"diode paek reverse voltage %3.2fV \" %PIV)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " peak vouput voltage is 30.41V \n", + "peak load current is 0.0608A \n", + "diode paek reverse voltage 31.11V \n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.2, Page No 79" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Vi=30.0\n", + "Rl=300.0\n", + "Vf=0.7\n", + "\n", + "#Calculations\n", + "Vpi=1.414*Vi\n", + "Vpo=Vpi-2*Vf\n", + "print(\" peak output voltage %.3f V \" %Vpo)\n", + "Ip=Vpo/Rl\n", + "\n", + "#Results\n", + "print(\" current bridge is %.1f mA \" %(Ip*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " peak output voltage 41.020 V \n", + " current bridge is 136.7 mA \n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.3 Page No 83" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "C1=680.0*10**-6\n", + "Eo=28.0\n", + "Rl=200.0\n", + "f=60.0\n", + "\n", + "#Calculations\n", + "Il=Eo/Rl\n", + "T=1/f\n", + "t1=T\n", + "Vr=(Il*t1)/C1\n", + "\n", + "#Results\n", + "print(\"peak to peak ripple voltage is %.2f V \" %Vr)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "peak to peak ripple voltage is 3.43 V \n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.4, Page No 84" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Eo=20.0\n", + "Rl=500.0\n", + "f=60.0\n", + "\n", + "#Calculations\n", + "Vr=(10*Eo)/100#10% of Eo\n", + "Il=Eo/Rl\n", + "T=1/f\n", + "t1=T\n", + "C1=((Il*t1)/Vr)*10**6\n", + "\n", + "#Results\n", + "print(\"Reservior capacitance is %.2f uF \" %C1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Reservior capacitance is 333.33 uF \n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5 Page No 85" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Eo=20.0\n", + "f=60.0\n", + "Rl=500.0\n", + "Il=Eo/Rl\n", + "\n", + "#Calculations\n", + "Vr=(10.0*Eo)/100\n", + "print(\"10percent of Eo is %.2f V \" %Vr)\n", + "Eomin=Eo-0.5*Vr\n", + "Eomax=Eo+0.5*Vr\n", + "Q1=math.asin(Eomin/Eomax)\n", + "Q1=65\n", + "Q2=90-Q1\n", + "T=1/f\n", + "t2=(Q2*T)/360\n", + "print(\" charging time is %.2fs \" %t2)\n", + "t1=T-t2\n", + "print(\"discharging time is %.2fs \" %t1)\n", + "C1=((Il*t1)/Vr)*10**6\n", + "\n", + "#Results\n", + "print(\"reservior capacitance is %.2f uF \" %C1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "10percent of Eo is 2.00 V \n", + " charging time is 0.00s \n", + "discharging time is 0.02s \n", + "reservior capacitance is 310.19 uF \n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6 Page No 88" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Eo=21.0\n", + "Vf=0.7\n", + "\n", + "#Calculations\n", + "t1=1.16*10**-3\n", + "t2=15.54*10**-3\n", + "Vp=Eo+Vf\n", + "Vr=2*Vp\n", + "Il=40*10**-4\n", + "Ifrm=(Il*(t1+t2))/t2\n", + "Ifsm=30.0\n", + "Rs=Vp/Ifsm\n", + "\n", + "#Results\n", + "print(\" surge limiting resistance is %3.2fohm \" %Rs)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " surge limiting resistance is 0.72ohm \n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7, Page No 89 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vf=.7\n", + "Eo=21.0\n", + "\n", + "#Calculations\n", + "Il=40*10**-3\n", + "Vp=115.0\n", + "Vs=.707*(Vf+Eo)\n", + "print(\" Vrms voltage is %3.3fV \" %Vs)\n", + "Is=3.6*Il\n", + "print(\" rms current is %.2f mA \" %(Is*1000))\n", + "Ip=(Vs*Is)/Vp\n", + "\n", + "#Results\n", + "print(\"primary current is %.2f mA \" %(Ip*1000))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " Vrms voltage is 15.342V \n", + " rms current is 144.00 mA \n", + "primary current is 19.21 mA \n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8 Page No 92" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math \n", + "\n", + "#initialisation of variables\n", + "Vr=2.0\n", + "T=16.7*10**-3\n", + "t2=1.16*10**-3\n", + "\n", + "#Calculations\n", + "Il=40.0*10**-3#from example 3.5\n", + "t1=(T/2.0)-t2\n", + "C1=(Il*t1)/Vr\n", + "\n", + "#Results\n", + "print(\" resrvior capacitor is %.2f mF \" %(C1*10**6))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " resrvior capacitor is 143.80 mF \n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9 Page No 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vr=2.0\n", + "T=16.7*10**-3\n", + "Il=40.0*10**-3\n", + "\n", + "#Calculations\n", + "t1=T/2\n", + "C1=(Il*t1)/Vr\n", + "\n", + "#Results\n", + "print(\" reservior capacitance is %.1fF \" %(C1*10**6))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " reservior capacitance is 167.0F \n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10 Page No 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Eo=21.0\n", + "Vf=0.7\n", + "Il=40.0*10**-3\n", + "t1=7.19*10**-3\n", + "t2=1.16*10**-3\n", + "\n", + "#Calculations\n", + "Vp=Eo+(2*Vf)\n", + "Vr=Vp\n", + "If=Il/2\n", + "Ifrm=Il*(t1+t2)/t2\n", + "Ifsm=30\n", + "Rs=Vp/Ifsm\n", + "\n", + "#Results\n", + "print(\"surge limiting resistance is %.3fohm \" %Rs)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "surge limiting resistance is 0.747ohm \n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11, Page No 93" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "\n", + "#initialisation of variables\n", + "Eo=21.0\n", + "Vf=0.7\n", + "Il=40*10**-3\n", + "Vp=115.0\n", + "\n", + "#Calculations\n", + "Vs=0.707*(Eo+2*Vf)\n", + "Is=1.6*Il\n", + "Ip=(Vs*Is)/Vp\n", + "\n", + "#Results\n", + "print(\" supply current is %.1f mA \" %(Ip*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " supply current is 8.8 mA \n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12, Page No 97" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Eo=20.0\n", + "Il=40.0*10**-3\n", + "R1=22.0\n", + "Vr=2.0\n", + "C1=150*10**-6\n", + "C2=C1\n", + "fr=120\n", + "\n", + "#Calculations\n", + "Vo=Eo-Il*R1\n", + "vi=Vr/3.14\n", + "Xc2=1/(2*3.14*fr*C2)\n", + "vo=(vi*Xc2)/math.sqrt((R1**2) + (Xc2**2))\n", + "print(\" dc output voltage is %.3fV \" %vo)\n", + "Vpp=2*vo\n", + "\n", + "#Results\n", + "print(\" peak to peak voltage is %.1fV \" %(Vpp*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " dc output voltage is 0.238V \n", + " peak to peak voltage is 475.3V \n" + ] + } + ], + "prompt_number": 12 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13, Page No 98" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "C1=150*10**-6\n", + "C2=C1\n", + "vi=4.0\n", + "vo=1.0\n", + "f=120.0\n", + "\n", + "#Calculations\n", + "Xc2=8.84 #FROM EXAMPLE 3.12\n", + "Xl=Xc2*((vi/vo)+1)\n", + "L1=Xl/(2*3.14*f)\n", + "\n", + "#Results\n", + "print(\" suitable value of L1 is %.3fH \" %(L1*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " suitable value of L1 is 58.652H \n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14, Page No 101" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Edc=20.0\n", + "vo=0.24\n", + "Vo=20.0\n", + "Il=40*10**-3\n", + "fr=120.0\n", + "\n", + "#Calculations\n", + "Eomax=(3.14*Edc)/2\n", + "Epeak=(4*Eomax)/(3*3.14)\n", + "vi=Epeak\n", + "Rl=Vo/Il\n", + "Xlc=(2*Rl)/3\n", + "Lc=Xlc/(2*3.14*fr)\n", + "L=1.25*Lc\n", + "Xl=2*3.14*fr*L\n", + "Xc=Xl/((vi/vo)+1)\n", + "C1=1/(2*3.14*fr*Xc)\n", + "\n", + "#Results\n", + "print(\"The value of c1 = %.2f mF \" %(C1*10**6))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of c1 = 180.11 mF \n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15, Page No 105" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Eo=20.0\n", + "E0=20-19.7 #load effect\n", + "\n", + "#Calculations\n", + "loadregulation =(E0*100)/Eo#percentage\n", + "sourceeffect=20.2-20\n", + "lineregulation =(sourceeffect*100)/Eo\n", + "\n", + "#Results\n", + "print(\"Line regulation = %.1f percent \" %lineregulation)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Line regulation = 1.0 percent \n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16, Page No 108" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vz=9.1\n", + "Izt=20*10**-3\n", + "Es=30.0\n", + "\n", + "#Calculations\n", + "R1=(Es-Vz)/Izt\n", + "Pr1=(Izt**2)*R1\n", + "Es=27\n", + "Iz=(Es-Vz)/R1\n", + "\n", + "#Results\n", + "print(\"The circuit current is %.2f mA \" %(Iz*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit current is 17.13 mA \n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17, Page No 110" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vz=6.2\n", + "Pd=400.0*10**-3\n", + "Es=16.0\n", + "\n", + "#Calculations\n", + "Izm=Pd/Vz\n", + "R1=(Es-Vz)/Izm\n", + "Pr1=(Izm**2)*R1\n", + "Izmin=5.0*10**-3\n", + "Izmax=Izm-Izmin\n", + "\n", + "#Results\n", + "print(\"maximum current is %3.2f mA \" %(Izmax*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "maximum current is 59.52 mA \n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18, Page No 112" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Zz=7.0\n", + "Es=16.0\n", + "Vo=6.2\n", + "Il=59.5*10**-3\n", + "\n", + "#Calculations\n", + "es=(10*Es)/100.0 #10% os Es\n", + "Rl=Vo/Il\n", + "print(\"es*Zz||Rl/R1+Zz||Rl\")\n", + "V0=es*((Zz*Rl)/(Zz+Rl))/(R1+((Zz*Rl)/(Zz+Rl)))\n", + "lineregulation=(V0*100)/Vo\n", + "print(\"line regulation voltage is %3.3fpercentage \" %lineregulation)\n", + "V0=Il*((Zz*R1)/(Zz+R1))\n", + "loadregulation=(V0*100)/Vo\n", + "print(\"loadregulation voltage is %3.3fpercentage \" %loadregulation)\n", + "Rr=((Zz*Rl)/(Zz+Rl))/(R1+(Zz*Rl)/(Zz+Rl))\n", + "\n", + "#Results\n", + "print(\"ripple rejection is %3.2f X 10^-2 \" %(Rr*10**2))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "es*Zz||Rl/R1+Zz||Rl\n", + "line regulation voltage is 1.068percentage \n", + "loadregulation voltage is 6.422percentage \n", + "ripple rejection is 4.14 X 10^-2 \n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.19, Page No 114" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "\n", + "E=9.0\n", + "Vf=.7\n", + "\n", + "#Calculations\n", + "If=1.0*10**-3\n", + "Vo=E-Vf\n", + "R1=Vo/If\n", + "Vr=E\n", + "\n", + "#Results\n", + "print(\"diode forward voltage is %3.2fohm \" %Vr)\n", + "print(\"diode forward current is %3.1fA \" %(If*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "diode forward voltage is 9.00ohm \n", + "diode forward current is 1.0A \n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.20, Page No 117" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "E=5.0\n", + "Vo=4.5\n", + "Il=2.0*10**-3\n", + "\n", + "#Calculations\n", + "R1=(E-Vo)/Il\n", + "print(\" suitable resistance is %dohm \" %R1)\n", + "Vr=E\n", + "print(\"when diode is forward baised\")\n", + "If=(E-Vf)/R1\n", + "\n", + "#Results\n", + "print(\" diode forward current is %3.2fA \" %(If*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " suitable resistance is 250ohm \n", + "when diode is forward baised\n", + " diode forward current is 17.20A \n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.21, Page No 119" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vo=2.7\n", + "Vf=.7\n", + "E=9.0\n", + "If=1*10**-3\n", + "\n", + "#Calculations\n", + "Il=If\n", + "Vb=Vo-Vf\n", + "R1=(E-Vo)/(Il+If)\n", + "\n", + "#Results\n", + "print(\"resistance is %.2f kOhm \" %(R1/10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "resistance is 3.15 kOhm \n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.22, Page No 120" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vo=5.0\n", + "Vf=0.7\n", + "Iz=5.0\n", + "Il=1.0\n", + "E=20.0\n", + "\n", + "#Calculations\n", + "Vz=Vo-Vf\n", + "R1=(E-Vo)/(Il+Iz)\n", + "\n", + "#Results\n", + "print(\"zener diode resistance si %.2f ohm \" %R1)\n", + "#Answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "zener diode resistance si 2.50 ohm \n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.23, Page No 122" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "E=10.0\n", + "R1=56.0*10**3\n", + "f=1000.0\n", + "C1=1.0*10**-6\n", + "\n", + "#Calculations\n", + "Vo=2*E\n", + "Ic=Vo/R1\n", + "t=1/(2*f)\n", + "Vc=(Ic*t)/C1\n", + "\n", + "#Results\n", + "print(\" tilt output voltage is %3.2fV \" %(Vc*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " tilt output voltage is 178.57V \n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.24, Page No 124" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "f=500.0\n", + "Rs=600.0\n", + "E=8.0\n", + "\n", + "#Calculations\n", + "t=1.0/(2*f)\n", + "PW=t\n", + "C1=PW/Rs\n", + "Vo=2.0*E\n", + "Vc=(1*Vo)/100#1% of the Vo\n", + "Ic=(Vc*C1)/t\n", + "R1=(2*E)/(Ic*1000)\n", + "\n", + "#Results\n", + "print(\"suitable value of R1 is %.2f ohm \" %R1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "suitable value of R1 is 60.00 ohm \n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.25, Page No 125" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "Vf=0.7\n", + "E=6.0\n", + "Vb1=3.0\n", + "\n", + "#Calculations\n", + "Vc=Vb1-Vf-(-E)\n", + "Vo=Vb1-Vf\n", + "print(\"when input is -E\")\n", + "Vo=E+Vc\n", + "Vo=Vb1+Vf\n", + "print(\"Capicitor voltage is %.2f ohm \" %Vc)\n", + "print(\"when input is +E\")\n", + "Vo=E+(Vc)\n", + "\n", + "#Results\n", + "print(\"Capicitor voltage is %.2f ohm \" %Vo)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "when input is -E\n", + "Capicitor voltage is 8.30 ohm \n", + "when input is +E\n", + "Capicitor voltage is 14.30 ohm \n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.26, Page No 130" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "E=12.0\n", + "Vf=0.7\n", + "Rl=47*10**3\n", + "f=5000.0\n", + "\n", + "#Calculations\n", + "Vo=2*(E-Vf)\n", + "Il=Vo/Rl\n", + "print(\" capacitor discharge time\")\n", + "t=1.0/(2*f)\n", + "print(\" for 1% ripple allow .5% due to discharge of C2 %.5%due to discharge of C1\")\n", + "Vc=(.5*Vo)/100\n", + "C2=((Il*t)/Vc)*10**6\n", + "print(\" value of capacitor C2 is %3.2fuF \" %C2)\n", + "C1=2*C2\n", + "\n", + "#Results\n", + "print(\"value of capacitor C1 is %3.2fuF \" %C1)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + " capacitor discharge time\n", + " for 1% ripple allow .5% due to discharge of C2 %.5%due to discharge of C1\n", + " value of capacitor C2 is 0.43uF \n", + "value of capacitor C1 is 0.85uF \n" + ] + } + ], + "prompt_number": 26 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.27, Page No 133" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "import math\n", + "#initialisation of variables\n", + "\n", + "Vcc=5.0\n", + "Vf=.7\n", + "R1=3.3*10**3\n", + "\n", + "#Calculations\n", + "print(\"A)\")\n", + "Ir1=(Vcc-Vf)/R1\n", + "print(\"diode forward current when all input are low is %3.4fA \" %Ir1)\n", + "print(\"for each diode\")\n", + "If=Ir1/3\n", + "print(\"B)\")\n", + "If2=Ir1/2\n", + "If3=If2\n", + "print(\" forward current when input A is high is %3.5fA \" %If3)\n", + "print(\"C)\")\n", + "If3=Ir1\n", + "\n", + "#Results\n", + "print(\" forward current when input A and B are high and C is low %3.2fA \" %(If3*10**3))" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "A)\n", + "diode forward current when all input are low is 0.0013A \n", + "for each diode\n", + "B)\n", + " forward current when input A is high is 0.00065A \n", + "C)\n", + " forward current when input A and B are high and C is low 1.30A \n" + ] + } + ], + "prompt_number": 27 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit