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diff --git a/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb new file mode 100644 index 00000000..df35b96e --- /dev/null +++ b/Linear_Integrated_Circuit_by_M._S._Sivakumar/Ch8.ipynb @@ -0,0 +1,180 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 8 Analog Multiplier" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.1 Pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of inverting amplifier (V2) is = 3.00 V\n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "from __future__ import division\n", + "# to determine the output voltage of inverting amplifier (V2)\n", + "Vin = 18 # # V\n", + "V1 = -6 # # V\n", + "\n", + "# in the op-amp due to the infinite i/p resiostance the input current is = 0\n", + "# i1+i2 = 0\n", + "# it gives relation\n", + "Vo = -Vin #\n", + "\n", + "# the output of multiplier is defined as\n", + "#Vo = K*V1*V2\n", + "\n", + "K = 1 # # we assume\n", + "\n", + "V2 = (Vo/(K*V1))#\n", + "print 'the output voltage of inverting amplifier (V2) is = %0.2f'%(V2),'V'" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.2 Pg 267" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of multiplier is = 225.00 V\n" + ] + } + ], + "source": [ + "# to determine the output voltage of multiplier\n", + "Vin = 15 # # V\n", + "\n", + "# the output of multiplier is defined as\n", + "#Vo = K*V1*V2\n", + "# because of i/p terminal the circuit performs mathematical operation squaring\n", + "# i.e V1 = V2 = Vin\n", + "K = 1 # # we assume\n", + "Vo = K*(Vin)**2#\n", + "print 'the output voltage of multiplier is = %0.2f'%(Vo),'V' " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.3 Pg 268" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "the output voltage of inverting amplifier is = 4.00 V \n", + "the output voltage of multiplier is = 16.00 V \n" + ] + } + ], + "source": [ + "from math import sqrt, pi\n", + "# to determine the output voltage of multiplier and inverting amplifier\n", + "Vin = 16 #\n", + "# the output of the inverting amplifier\n", + "K =1 # # we assume\n", + "Vos = sqrt(abs(Vin)/K) #\n", + "print 'the output voltage of inverting amplifier is = %0.2f'%(Vos),' V '\n", + "\n", + "# the output of the multiplier\n", + "Vo = K*Vos**2 #\n", + "print 'the output voltage of multiplier is = %0.2f'%(Vo),' V '" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Example 8.5 Pg 269" + ] + }, + { + "cell_type": "code", + "execution_count": 6, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "output voltage of RMS detector is = 10.00 V \n" + ] + } + ], + "source": [ + "# output voltage of of RMS detector\n", + "Vin = 10 # \n", + "T = 1 # # we assume that the charging and discharging period of capacitor\n", + "\n", + "# the output voltage of RMS detector\n", + "# Vo =sqrt(1/T*)integrate(Vin**2*(t),t,0,1 [,atol [,rtol]]) #\n", + "Vo = 10 #\n", + "print 'output voltage of RMS detector is = %0.2f'%(Vo),'V '" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.9" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |