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author | hardythe1 | 2015-06-11 17:31:11 +0530 |
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committer | hardythe1 | 2015-06-11 17:31:11 +0530 |
commit | 79c59acc7af08ede23167b8455de4b716f77601f (patch) | |
tree | 2d6ff34b6f131d2671e4c6b798f210b3cb1d4ac7 /Electronic_Devices_and_Circuits_by_D._C._Kulshreshtha/chapter09.ipynb | |
parent | df60071cf1d1c18822d34f943ab8f412a8946b69 (diff) | |
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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": {}
+ }
+ ]
+}
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