From efb9ead5d9758d5d0bed7a22069320b14f972e40 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Fri, 25 Jul 2014 12:37:07 +0530 Subject: adding books --- .../chapter_18-checkpoint_2.ipynb | 535 +++++++++++++++++++++ 1 file changed, 535 insertions(+) create mode 100755 Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb (limited to 'Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb') diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb new file mode 100755 index 00000000..f3f727b4 --- /dev/null +++ b/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb @@ -0,0 +1,535 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:d16259f8430b68bdc5f61e2c9d378821ed7dc4c5b9f7144ffa85cfd5e017e63d" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Chapter 18: Operational amplifiers

" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 1, page no. 279

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the output voltage of the amplifier.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Vi2 = 2.45;# in Volts\n", + "Vi1 = 2.35;# in Volts\n", + "A0 = 120;# open-loop voltage gain\n", + "\n", + "#calculation:\n", + "Vo = A0*(Vi2 - Vi1)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n the output voltage is \",round(Vo,2),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " the output voltage is 12.0 V" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 2, page no. 281

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the common-mode gain of an op amp\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Vg = 150E3;# differential voltage gain \n", + "CMRR = 90;# in dB\n", + "\n", + "#calculation:\n", + "CMG = Vg/(10**(CMRR/20))\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n common-mode gain is \",round(CMG,2)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " common-mode gain is 4.74" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 3, page no. 282

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the common-mode gain and the CMRR.\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Vg = 120;# differential voltage gain \n", + "Vi = 3;# in Volts\n", + "Vo = 0.024;# in Volts\n", + "\n", + "#calculation:\n", + "CMG = Vo/Vi\n", + "CMRR = 20*(1/2.303)*math.log(Vg/CMG)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n common-mode gain is \",round(CMG,3),\" and CMRR is \",round(CMRR,2),\" dB\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " common-mode gain is 0.008 and CMRR is 83.51 dB" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 4, page no. 283

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the output voltage when the input voltage is: (a) +0.4 V (b) -1.2 V\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Rf = 2000;# in ohms\n", + "Ri = 1000;# in ohms\n", + "Vi1 = 0.4;# in Volts\n", + "Vi2 = -1.2;# in Volts\n", + "\n", + "#calculation:\n", + "Vo1 = -1*Rf*Vi1/Ri\n", + "Vo2 = -1*Rf*Vi2/Ri\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n output voltage when the input voltage is 0.4V is \",round(Vo1,2),\" V \"\n", + "print \" and when the input voltage is -1.2V is \",round(Vo2,2),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " output voltage when the input voltage is 0.4V is -0.8 V \n", + " and when the input voltage is -1.2V is 2.4 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 5, page no. 283

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate (a) the voltage gain, and\n", + "#(b) the output offset voltage due to the input bias current. \n", + "#(c) How can the effect of input bias current be minimised?\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Ii = 100E-9;# in Amperes\n", + "T = 20;# in \u00b0C\n", + "Rf = 1E6;# in ohms\n", + "Ri = 10000;# in ohms\n", + "\n", + "#calculation:\n", + "A = -1*Rf/Ri\n", + "Vos = Ii*Ri*Rf/(Ri+Rf)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)the voltage gain is \",round(A,2),\"\"\n", + "print \"\\n (b)output offset voltage is \",round(Vos*1000,2),\" mV\"\n", + "print \"\\n (c)The effect of input bias current can be minimised by ensuring \"\n", + "print \"that both inputs have the same driving resistance.\" \n", + "print \"This means that a resistance of value of 9.9 kohm (from part (b)) \"\n", + "print \"should be placed between the non-inverting (+) terminal and earth.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)the voltage gain is -100.0 \n", + "\n", + " (b)output offset voltage is 0.99 mV\n", + "\n", + " (c)The effect of input bias current can be minimised by ensuring \n", + "that both inputs have the same driving resistance.\n", + "This means that a resistance of value of 9.9 kohm (from part (b)) \n", + "should be placed between the non-inverting (+) terminal and earth.\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 6, page no. 284

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Design an inverting amplifier\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Vg = 40;# in dB\n", + "bf = 5000;# in Hz\n", + "Ri = 10000;# in ohms\n", + "\n", + "#calculation:\n", + "A = 10**(Vg/20)\n", + "Rf = A*Ri\n", + "f = A*bf\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n the voltage gain is \",round(A,2),\", Rf = \",round(Rf/1000,2),\"kohm and frequency = \",round(f/1000,2),\" kHz\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " the voltage gain is 100.0 , Rf = 1000.0 kohm and frequency = 500.0 kHz" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 7, page no. 286

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine (a) the voltage gain (b) the output voltage\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "Vi = -0.4;# in Volts\n", + "R1 = 4700;# in ohms\n", + "R2 = 10000;# in ohms\n", + "\n", + "#calculation:\n", + "A = 1 + (R2/R1)\n", + "Vo = A*Vi\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n(a) the voltage gain is \",round(A,2),\"\"\n", + "print \"\\n(b) output voltageis \",round(Vo,2),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + "(a) the voltage gain is 3.13 \n", + "\n", + "(b) output voltageis -1.25 V" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 8, page no. 287

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#determine the output voltage, Vo\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V1 = 0.5;# in Volts\n", + "V2 = 0.8;# in Volts\n", + "V3 = 1.2;# in Volts\n", + "R1 = 10000;# in ohms\n", + "R2 = 20000;# in ohms\n", + "R3 = 30000;# in ohms\n", + "Rf = 50000;# in ohms\n", + "\n", + "#calculation:\n", + "Vo = -1*Rf*(V1/R1 + V2/R2 + V3/R3)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n output voltageis \",round(Vo,2),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " output voltageis -6.5 V" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 10, page no. 289

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "import math\n", + "from scipy import integrate\n", + "#initializing the variables:\n", + "Vs = -0.75;# in Volts\n", + "R = 200000;# in ohms\n", + "C = 2.5E-6;# in Farads\n", + "t = 0.1;# in secs\n", + "\n", + "#calculation:\n", + "f = lambda x,a : a*1\n", + "y, err = integrate.quad(f, 0, 0.1, args=(-0.75,))\n", + "Vo = (-1/(C*R))*y\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n output voltage is \",Vo,\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " output voltage is 0.15 V" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "

Example 11, page no. 290

" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the output voltage Vo\n", + "from __future__ import division\n", + "import math\n", + "#initializing the variables:\n", + "V1a = 0.005;# in Volts\n", + "V2a = 0;# in Volts\n", + "V1b = 0;# in Volts\n", + "V2b = 0.005;# in Volts\n", + "V1c = 0.05;# in Volts\n", + "V2c = 0.025;# in Volts\n", + "V1d = 0.025;# in Volts\n", + "V2d = 0.05;# in Volts\n", + "R1 = 10000;# in ohms\n", + "R2 = 10000;# in ohms\n", + "R3 = 100000;# in ohms\n", + "Rf = 100000;# in ohms\n", + "\n", + "#calculation:\n", + "Vo1 = -1*Rf*V1a/R1\n", + "Vo2 = (R3/(R2+R3))*(1 + (Rf/R1))*V2b\n", + "Vo3 = -1*Rf*(V1c-V2c)/R1\n", + "Vo4 = (R3/(R2+R3))*(1 + (Rf/R1))*(V2d-V1d)\n", + "\n", + "\n", + "#Results\n", + "print \"\\n\\n Result \\n\\n\"\n", + "print \"\\n (a)output voltage is \",round(Vo1,2),\" V\"\n", + "print \"\\n (b)output voltage is \",round(Vo2,2),\" V\"\n", + "print \"\\n (c)output voltage is \",round(Vo3,2),\" V\"\n", + "print \"\\n (d)output voltage is \",round(Vo4,2),\" V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "\n", + " Result \n", + "\n", + "\n", + "\n", + " (a)output voltage is -0.05 V\n", + "\n", + " (b)output voltage is 0.05 V\n", + "\n", + " (c)output voltage is -0.25 V\n", + "\n", + " (d)output voltage is 0.25 V" + ] + } + ], + "prompt_number": 10 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit