From 79c59acc7af08ede23167b8455de4b716f77601f Mon Sep 17 00:00:00 2001 From: hardythe1 Date: Thu, 11 Jun 2015 17:31:11 +0530 Subject: add books --- .../chapter09.ipynb | 311 +++++++++++++++++++++ 1 file changed, 311 insertions(+) create mode 100755 Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/chapter09.ipynb (limited to 'Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/chapter09.ipynb') diff --git a/Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/chapter09.ipynb b/Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/chapter09.ipynb new file mode 100755 index 00000000..1d9f850d --- /dev/null +++ b/Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/chapter09.ipynb @@ -0,0 +1,311 @@ +{ + "metadata": { + "name": "", + "signature": "sha256:8ef26de80c6669dcd55ce081279263a5231dd84d81c7a5c7530a5f89da39e8f3" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "chapter09:Power Amplifiers" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E1 - Pg 327" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the turns ratio of the transformer\n", + "#given\n", + "import math\n", + "Rl=8.;#ohm\n", + "Rl_=5.*10.**3.;#ohm\n", + "TR=math.sqrt(Rl_/Rl); #Turns ratio\n", + "print '%s %.f %s' %(\"Turns Ratio =\",TR,\": 1\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Turns Ratio = 25 : 1\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E2 - Pg 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine the output impedance of the transistor\n", + "#given\n", + "TR=16./1.; #turn ratio\n", + "Rl=4.;#ohm #loudspeaker impedance\n", + "ro=(TR**2.)*Rl;\n", + "print '%s %.f %s' %(\"The output impedance of the transistor =\",ro,\"ohm\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output impedance of the transistor = 1024 ohm\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E3 - Pg 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "#Determine the efficiency of a single ended transformer\n", + "#given\n", + "Vceq=10.;#V #supply voltage\n", + "\n", + "#At Vp=10V\n", + "Vp=10.;#V\n", + "Vce_max1=Vceq+Vp;\n", + "Vce_min1=Vceq-Vp;\n", + "n1=50.*((Vce_max1-Vce_min1)/(Vce_max1+Vce_min1))**2.;\n", + "print '%s %.f %s' %(\"Efficiency (At Vp = 10V)=\",n1,\"percent\\n\");\n", + "\n", + "#At Vp=5V\n", + "Vp=5.;#V\n", + "Vce_max2=Vceq+Vp;\n", + "Vce_min2=Vceq-Vp;\n", + "n2=50.*((Vce_max2-Vce_min2)/(Vce_max2+Vce_min2))**2.;\n", + "print '%s %.1f %s' %(\"Efficiency (At Vp = 5V)=\",n2,\"percent\\n\");\n", + "\n", + "#At Vp=1V\n", + "Vp=1.;#V\n", + "Vce_max3=Vceq+Vp;\n", + "Vce_min3=Vceq-Vp;\n", + "n3=50.*((Vce_max3-Vce_min3)/(Vce_max3+Vce_min3))**2.;\n", + "print '%s %.1f %s' %(\"Efficiency (At Vp = 1V)=\",n3,\"percent\\n\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Efficiency (At Vp = 10V)= 50 percent\n", + "\n", + "Efficiency (At Vp = 5V)= 12.5 percent\n", + "\n", + "Efficiency (At Vp = 1V)= 0.5 percent\n", + "\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E4 - Pg 336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Determine input and output power and efficiency\n", + "#given\n", + "import math\n", + "Vcc=20.;#V#supply voltage\n", + "Rl=4.;#ohm\n", + "Vp=15.;#V\n", + "Ip=Vp/Rl;\n", + "Idc=Ip/math.pi;\n", + "Pi=Vcc*Idc;\n", + "Po=((Vp/2.)**2.)/Rl;\n", + "n=100.*Po/Pi;\n", + "print '%s %.1f %s' %(\"Input power =\",Pi,\"W\\n\");\n", + "print '%s %.2f %s' %(\"Output power =\",Po,\"W\\n\");\n", + "print '%s %.2f %s' %(\"Efficiency =\",n,\"percent\\n\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input power = 23.9 W\n", + "\n", + "Output power = 14.06 W\n", + "\n", + "Efficiency = 58.90 percent\n", + "\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E5 - Pg 337" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate the percentage increase in output power\n", + "#given\n", + "D=0.2;#harmonic distortion\n", + "P=(1.+D**2.);#Total power increase\n", + "\n", + "#percent increase= (Pi*(1+D**2)-Pi)*100/Pi;\n", + "#taking out and cancelling Pi\n", + "PI=(P-1.)*100.;\n", + "print '%s %.f %s' %(\"The percentage increase in output power=\",PI,\"percent\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The percentage increase in output power= 4 percent\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example E6 - Pg 338" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "#Calculate harmonic distortion and percentage increase in output voltage due to this\n", + "#given\n", + "import math\n", + "I1=60.;#A\n", + "I2=6.;#A\n", + "I3=1.2;#A\n", + "I4=0.6;#A\n", + "D2=I2/I1;\n", + "D3=I3/I1;\n", + "D4=I4/I1;\n", + "print '%s %.f %s %s %.f %s %s %.f %s' %(\"The Harmonic distortion of each component \\nD2=\",D2*100,\"percent\\n\",\"\\nD3=\",D3*100,\"percent\\n\",\"\\nD4=\",D4*100,\"percent\\n\");\n", + "D=math.sqrt((D2)**2.+(D3)**2.+(D4)**2.);\n", + "print '%s %.f %s' %(\"The Total Harmonic distortion =\",D*100,\"percent\\n\");\n", + "P=(1.+D**2.);#Total power increase\n", + "#percent increase= (Pi*(1+D**2)-Pi)*100/Pi;\n", + "#taking out and cancelling Pi\n", + "PI=(P-1.)*100.;\n", + "print '%s %.f %s' %(\"The percentage increase in output power =\",PI,\"percent\");\n", + "#Calculate harmonic distortion and percentage increase in output voltage due to this\n", + "#given\n", + "import math\n", + "I1=60.;#A\n", + "I2=6.;#A\n", + "I3=1.2;#A\n", + "I4=0.6;#A\n", + "D2=I2/I1;\n", + "D3=I3/I1;\n", + "D4=I4/I1;\n", + "print '%s %.f %s %s %.f %s %s %.f %s' %(\"The Harmonic distortion of each component \\nD2=\",D2*100,\"percent\\n\",\"\\nD3=\",D3*100,\"percent\\n\",\"\\nD4=\",D4*100,\"percent\\n\");\n", + "D=math.sqrt((D2)**2.+(D3)**2.+(D4)**2.);\n", + "print '%s %.f %s' %(\"The Total Harmonic distortion =\",D*100,\"percent\\n\");\n", + "P=(1.+D**2.);#Total power increase\n", + "#percent increase= (Pi*(1+D**2)-Pi)*100/Pi;\n", + "#taking out and cancelling Pi\n", + "PI=(P-1.)*100.;\n", + "print '%s %.f %s' %(\"The percentage increase in output power =\",PI,\"percent\");\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Harmonic distortion of each component \n", + "D2= 10 percent\n", + " \n", + "D3= 2 percent\n", + " \n", + "D4= 1 percent\n", + "\n", + "The Total Harmonic distortion = 10 percent\n" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n", + "The percentage increase in output power = 1 percent\n", + "The Harmonic distortion of each component \n", + "D2= 10 percent\n", + " \n", + "D3= 2 percent\n", + " \n", + "D4= 1 percent\n", + "\n", + "The Total Harmonic distortion = 10 percent\n", + "\n", + "The percentage increase in output power = 1 percent\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +} \ No newline at end of file -- cgit