From 6846da7f30aadc7b3812538d1b1c9ef2e465a922 Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Fri, 25 Jul 2014 12:35:04 +0530 Subject: removing unwanted: --- .../chapter_18-checkpoint_2.ipynb | 535 --------------------- 1 file changed, 535 deletions(-) delete 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 deleted file mode 100755 index f3f727b4..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_18-checkpoint_2.ipynb +++ /dev/null @@ -1,535 +0,0 @@ -{ - "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