{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter- 2 : Introduction To Operational Amplifiers" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.2 - Page No 70" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "Vin1= 5 # in \u00b5V\n", "Vin1= Vin1*10**-6 # in V\n", "Vin2= -7 #in \u00b5V\n", "Vin2= Vin2*10**-6 # in V\n", "Av= 2*10**5 # unit less\n", "Rin= 2 # in M\u03a9\n", "Vout= (Vin1-Vin2)*Av # in V\n", "print \"The output voltage = %0.1f volts\" %Vout" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The output voltage = 2.4 volts\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.4 - Page No 71" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from __future__ import division\n", "# Given data\n", "Rs= 2 # in k\u03a9\n", "RL= 5 # in k\u03a9\n", "A= 10**5 # unit less\n", "Rin= 100 #in k\u03a9\n", "Rout= 50 # in \u03a9\n", "Vout= 10 # in V\n", "# For Vout = 10 V, V1= V2 = Vout\n", "V1= Vout # in V\n", "V2= V1 # in V\n", "# From equation V1= Vs*Rin/(Rin+Rs)\n", "Vs= V1*(Rin+Rs)/Rin # in V\n", "Vout_by_Vs= Vout/Vs # value of Vout/Vs\n", "print \"The value of Vs =%0.1f volts\" %Vs\n", "print \"The value of Vout/Vs = %0.2f\" %Vout_by_Vs\n", "print \"The input resistance of the circuit = %0.f k\u03a9\" %Rin" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The value of Vs =10.2 volts\n", "The value of Vout/Vs = 0.98\n", "The input resistance of the circuit = 100 k\u03a9\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.6 - Page No 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "# Given data\n", "Ad= 100 # differential mode gain\n", "Acm= 0.01 # common mode gain\n", "CMRR= Ad/Acm \n", "CMRR_desh= 20*math.log(CMRR,10) # CMRR in dB\n", "print \"CMRR = %0.f dB\" %CMRR_desh" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "CMRR = 80 dB\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.7 - Page No 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "Ad= 10**5 # differential mode gain\n", "CMRR= 10**5 \n", "# Common-mode gain,\n", "Acm= Ad/CMRR \n", "print \"The common-mode gain = %0.f\" %Acm" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The common-mode gain = 1\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.8 - Page No 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "V1= 10 # in mV\n", "V2= 9 # in mV\n", "Ad= 60 # differential voltage gain in dB\n", "Ad= 10**(Ad/20) \n", "CMRR= 80 # in dB\n", "CMRR= 10**(CMRR/20) \n", "Vd= V1-V2 # difference signal in mV\n", "Vcm= (V1+V2)/2 # common-mode signal in mV\n", "# Output voltage,\n", "Vout= Ad*Vd*(1+1/CMRR*Vcm/Vd) # in mV\n", "AdVd= Ad*Vd # in mV\n", "# Error voltage\n", "Verror= Vout-AdVd # in mV\n", "Per_error= Verror/Vout*100 # percentage error\n", "print \"The error voltage = %0.2f mV\" %Verror\n", "print \"The percentage error in the output voltage = %0.3f \" %Per_error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The error voltage = 0.95 mV\n", "The percentage error in the output voltage = 0.095 \n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.9 - Page No 74" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "V1= 745 # in \u00b5V\n", "V2= 740 # in \u00b5V\n", "Ad= 5*10**5 # differential voltage gain\n", "CMRR= 80 # in dB\n", "CMRR= 10**(CMRR/20) \n", "Vd= V1-V2 # difference signal in \u00b5V\n", "Vcm= (V1+V2)/2 # common-mode signal in \u00b5V\n", "# Output voltage,\n", "Vout= Ad*Vd*(1+1/CMRR*Vcm/Vd) # in \u00b5V\n", "AdVd= Ad*Vd # in \u00b5V\n", "# Error voltage\n", "Verror= Vout-AdVd # in \u00b5V\n", "Vout= Vout*10**-6 # in V\n", "Verror= Verror*10**-6 # in V\n", "Per_error= Verror/Vout*100 # percentage error\n", "print \"The output voltage = %0.6f V\" %Vout\n", "print \"The percentage error in the output voltage= %0.4f\" %Per_error" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The output voltage = 2.537125 V\n", "The percentage error in the output voltage= 1.4633\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.10 - Page No 76" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "Vd= 25 #differential input voltage in \u00b5V\n", "Vd= Vd*10**-6 # in V\n", "A= 200000 # open loop gain\n", "# Output voltage,\n", "Vout= A*Vd # in V\n", "print \"The output voltage = \u00b1 %0.f \" %Vout" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The output voltage = \u00b1 5 \n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.11 - Page 83" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "dVout= 20 # change in output voltage in V\n", "dt= 4 # change in time in \u00b5s\n", "SR= dVout/dt # slew rate in V/\u00b5s\n", "print \"The slew rate = %0.f V/\u00b5s\" %SR" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The slew rate = 5 V/\u00b5s\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.12 - Page No 83" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "IB1= 10 # in \u00b5A\n", "IB2= 7.5 # in \u00b5A\n", "# Input bias current,\n", "I_in_bias= (IB1+IB2)/2 # in \u00b5A\n", "# Input offset current,\n", "I_in_offset= IB1-IB2 # in \u00b5A\n", "print \"The input bias current = %0.2f \u00b5A\" %I_in_bias\n", "print \"The input offset current = %0.1f \u00b5A\" %I_in_offset" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The input bias current = 8.75 \u00b5A\n", "The input offset current = 2.5 \u00b5A\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.13 - Page No 83" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import pi\n", "# Given data\n", "SR= 6 # slew rate in V/\u00b5s\n", "SR= 6*10**6 # in V/s\n", "\n", "# Part (i) For Vmax= 1V\n", "Vmax= 1 # in V\n", "fmax= SR/(2*pi*Vmax) # limiting frequency in Hz\n", "fmax= fmax*10**-6 # in MHz\n", "print \"Part (i) : The limiting frequency for maximum voltage of 1V = %0.3f MHz\" %fmax\n", "\n", "# Part (ii) For Vmax= 10V\n", "Vmax= 10 # in V\n", "fmax= SR/(2*pi*Vmax) # limiting frequency in Hz\n", "fmax= fmax*10**-3 # in kHz\n", "print \"Part (ii) : The limiting frequency for maximum voltage of 10V = %0.1f kHz\"%fmax" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Part (i) : The limiting frequency for maximum voltage of 1V = 0.955 MHz\n", "Part (ii) : The limiting frequency for maximum voltage of 10V = 95.5 kHz\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example : 2.14 - Page No 84" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "Vpp= 3 # output voltage in V\n", "del_t= 4 # in \u00b5s\n", "del_V= 90*Vpp/100-10*Vpp/100 # in V\n", "# Required slew rate,\n", "SR= del_V/del_t # in V/\u00b5s\n", "print \"The required slew rate = %0.1f V/\u00b5s\" %SR" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The required slew rate = 0.6 V/\u00b5s\n" ] } ], "prompt_number": 22 } ], "metadata": {} } ] }