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diff --git a/Electronic_Principles_/Chapter_4_New.ipynb b/Electronic_Principles_/Chapter_4_New.ipynb new file mode 100644 index 00000000..af0210e1 --- /dev/null +++ b/Electronic_Principles_/Chapter_4_New.ipynb @@ -0,0 +1,566 @@ +{
+ "metadata": {
+ "name": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "CHAPTER 4 DIODE CIRCUITS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-1, Page 92"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.1.py\n",
+ "#Calculate peak load voltage and the dc load voltage.\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=10 #RMS Value of sine wave(V)\n",
+ "f=60 #frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "Vp=Vrms/0.707 #peak source voltage(V)\n",
+ "Vpout=Vp #peak load voltage(V)\n",
+ "Vdc=Vp/math.pi #dc load voltage(V)\n",
+ "Vpouts=Vp-0.7 #peak load voltage in 2nd approx.\n",
+ "Vdc=Vpouts/math.pi #dc load voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'Vp=',round(Vp,2),'V'\n",
+ "print 'With an ideal diode, Vpout =',round(Vpout,2),'V'\n",
+ "print 'DC load voltage, Vdc =',round(Vdc,2),'V'\n",
+ "print 'With second approximation, Vpout =',round(Vpouts,2),'V'\n",
+ "print 'DC load voltage, Vdc =',round(Vdc,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Vp= 14.14 V\n",
+ "With an ideal diode, Vpout = 14.14 V\n",
+ "DC load voltage, Vdc = 4.28 V\n",
+ "With second approximation, Vpout = 13.44 V\n",
+ "DC load voltage, Vdc = 4.28 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-2, Page 94"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.2.py\n",
+ "#What are the peak load voltage and dc load voltage in Figure 4.5?\n",
+ "\n",
+ "import math\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vs=120 #supply voltage(V)\n",
+ "\n",
+ "#Calculation\n",
+ "V2=Vs/5 #Secondary voltage(V)\n",
+ "Vp=V2/0.707 #peak secondary voltage\n",
+ "Vpout=Vp #peak load voltage(V)\n",
+ "Vdc1=Vp/math.pi #dc load voltage(V)\n",
+ "Vpouts=Vp-0.7 #peak load voltage in 2nd approx.(V)\n",
+ "Vdc2=Vpouts/math.pi #dc load voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'As per fig.4-5, Transformer turns ratio is 5:1'\n",
+ "print 'V2=',round(V2,2),'V'\n",
+ "print 'Vp=',round(Vp,2),'V'\n",
+ "print 'With an ideal diode, Vpout =',round(Vpout,2),'V'\n",
+ "print 'DC load voltage, Vdc =',round(Vdc1,2),'V'\n",
+ "print 'With second approximation, Vpout =',round(Vpouts,2),'V'\n",
+ "print 'DC load voltage, Vdc =',round(Vdc2,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "As per fig.4-5, Transformer turns ratio is 5:1\n",
+ "V2= 24.0 V\n",
+ "Vp= 33.95 V\n",
+ "With an ideal diode, Vpout = 33.95 V\n",
+ "DC load voltage, Vdc = 10.81 V\n",
+ "With second approximation, Vpout = 33.25 V\n",
+ "DC load voltage, Vdc = 10.58 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-3, Page 97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.3.py\n",
+ "#Figure 4.7 shows full wave rectifier, Calculate the peak input and output voltages.\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=10 #turn ratio\n",
+ "\n",
+ "#Calculation\n",
+ "Vp1=Vrms/0.707 #peak primary voltage(V)\n",
+ "Vp2=Vp1/N12 #peak secondary voltage(V)\n",
+ "Vpin=0.5*Vp2 #input voltage(V)\n",
+ "Vpout=Vpin #Output voltage (V)\n",
+ "Vpouts=Vpin-0.7 #Output voltage in 2nd approx.(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'Peak primary voltage Vp1=',round(Vp1,2),'V'\n",
+ "print 'Peak secondary voltage Vp2=',round(Vp2,2),'V'\n",
+ "print 'Due to center-tap,' \n",
+ "print 'input voltage to each half-wave rectifier is only half the secondary voltage:'\n",
+ "print 'With an ideal diode, Vpout =',round(Vpout,2),'V'\n",
+ "print 'With second approximation, Vpout =',round(Vpouts,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak primary voltage Vp1= 169.73 V\n",
+ "Peak secondary voltage Vp2= 16.97 V\n",
+ "Due to center-tap,\n",
+ "input voltage to each half-wave rectifier is only half the secondary voltage:\n",
+ "With an ideal diode, Vpout = 8.49 V\n",
+ "With second approximation, Vpout = 7.79 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-4, Page 99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.4.py\n",
+ "#Figure 4.7 shows full wave rectifier. If one of the diodes were open then, calculate the peak input and output voltages.\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=10 #turn ratio\n",
+ "\n",
+ "#Calculation\n",
+ "Vp1=Vrms/0.707 #peak primary voltage(V)\n",
+ "Vp2=Vp1/N12 #peak secondary voltage(V)\n",
+ "Vpin=0.5*Vp2 #input voltage(V)\n",
+ "Vpout=Vpin #Output voltage(V)\n",
+ "Vpouts=Vpin-0.7 #Output voltage in 2nd approx.(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'Peak primary voltage Vp1=',round(Vp1,2),'V'\n",
+ "print 'Peak secondary voltage Vp2=',round(Vp2,2),'V'\n",
+ "print 'Due to one of the diodes were open, load voltage will be the half wave signal'\n",
+ "print 'But still peak of half wave signal is same as prior case'\n",
+ "print 'With an ideal diode, Vpout =',round(Vpout,2),'V'\n",
+ "print 'With second approximation, Vpout =',round(Vpouts,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak primary voltage Vp1= 169.73 V\n",
+ "Peak secondary voltage Vp2= 16.97 V\n",
+ "Due to one of the diodes were open, load voltage will be the half wave signal\n",
+ "But still peak of half wave signal is same as prior case\n",
+ "With an ideal diode, Vpout = 8.49 V\n",
+ "With second approximation, Vpout = 7.79 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-5, Page 102"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.5.py\n",
+ "#Calculate peak & output voltages in figure 4.9\n",
+ "\n",
+ "#variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=10 #turn ratio\n",
+ "\n",
+ "#Calculation\n",
+ "Vp1=Vrms/0.707 #peak primary voltage(V)\n",
+ "Vp2=Vp1/N12 #peak secondary voltage(V)\n",
+ "Vpout=Vp2 #Output voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'Peak primary voltage Vp1=',round(Vp1,2),'V'\n",
+ "print 'Peak secondary voltage Vp2=',round(Vp2,2),'V'\n",
+ "print 'secondary voltage is input of rectifier'\n",
+ "print 'With an ideal diode, Vpout =',round(Vpout,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak primary voltage Vp1= 169.73 V\n",
+ "Peak secondary voltage Vp2= 16.97 V\n",
+ "secondary voltage is input of rectifier\n",
+ "With an ideal diode, Vpout = 16.97 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-6, Page 108"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.6.py\n",
+ "#What is the dc load voltage and ripple in figure 4.14?\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=5 #turn ratio\n",
+ "RL=5 #Load resistance(KOhm)\n",
+ "C=100 #Capacitance(uF)\n",
+ "f=60 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V2=Vrms/N12 #RMS secondary voltage(V)\n",
+ "Vp=V2/0.707 #peak secondary voltage(V)\n",
+ "VL=Vp #dc load voltage(V)\n",
+ "IL=VL/RL #Load current(mA)\n",
+ "VR=(IL/(f*C))*(10**3) #ripple voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'RMS secondary voltage V2=',V2,'V'\n",
+ "print 'Peak secondary voltage Vp=',round(Vp,2),'V'\n",
+ "print 'with ideal diode and small ripple, dc load voltage, VL =',round(VL,2),'V'\n",
+ "print 'To calculate ripple get, dc load current,'\n",
+ "print 'DC Load current IL=',round(IL,2),'mA'\n",
+ "print 'As per ripple formula,'\n",
+ "print 'Ripple voltage VR=',round(VR,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS secondary voltage V2= 24 V\n",
+ "Peak secondary voltage Vp= 33.95 V\n",
+ "with ideal diode and small ripple, dc load voltage, VL = 33.95 V\n",
+ "To calculate ripple get, dc load current,\n",
+ "DC Load current IL= 6.79 mA\n",
+ "As per ripple formula,\n",
+ "Ripple voltage VR= 1.13 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-7, Page 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.7.py\n",
+ "#What is the dc load voltage and ripple in figure 4.15?\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=5 #turn ratio\n",
+ "RL=5 #Load resistance(KOhm)\n",
+ "C=100 #Capacitance(uF)\n",
+ "f=60 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V2=Vrms/N12 #RMS secondary voltage(V)\n",
+ "Vp=V2/0.707 #peak secondary voltage(V)\n",
+ "VL=Vp/2 #dc load voltage(V)\n",
+ "IL=VL/RL #Load current(mA)\n",
+ "VR=(IL/(2*f*C))*(10**3) #ripple voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'RMS secondary voltage V2=',V2,'V'\n",
+ "print 'Peak secondary voltage Vp=',round(Vp,2),'V'\n",
+ "print 'Half this voltage is input to each half-wave section, with ideal diode and small ripple, dc load voltage, VL =',round(VL,2),'V'\n",
+ "print 'But, due to 0.7V across conducting diode actual dc voltage is, VL =',round((VL-0.7),2),'V'\n",
+ "print 'To calculate ripple get, dc load current,'\n",
+ "print 'DC Load current IL=',round(IL,2),'mA'\n",
+ "print 'As per ripple formula,'\n",
+ "print 'Ripple voltage VR=',round(VR,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS secondary voltage V2= 24 V\n",
+ "Peak secondary voltage Vp= 33.95 V\n",
+ "Half this voltage is input to each half-wave section, with ideal diode and small ripple, dc load voltage, VL = 16.97 V\n",
+ "But, due to 0.7V across conducting diode actual dc voltage is, VL = 16.27 V\n",
+ "To calculate ripple get, dc load current,\n",
+ "DC Load current IL= 3.39 mA\n",
+ "As per ripple formula,\n",
+ "Ripple voltage VR= 0.28 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-8, Page 110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.8.py\n",
+ "#What is the dc load voltage and ripple in figure 4.16?\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=5 #turn ratio\n",
+ "RL=5 #Load resistance(KOhm)\n",
+ "C=100 #Capacitance(uF)\n",
+ "f=60 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V2=Vrms/N12 #RMS secondary voltage(V)\n",
+ "Vp=V2/0.707 #peak secondary voltage(V)\n",
+ "VL=Vp #dc load voltage(V)\n",
+ "IL=VL/RL #Load current(mA)\n",
+ "VR=(IL/(2*f*C))*(10**3) #ripple voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'RMS secondary voltage V2=',V2,'V'\n",
+ "print 'Peak secondary voltage Vp=',round(Vp,2),'V'\n",
+ "print 'with ideal diode and small ripple, dc load voltage, VL =',round(VL,2),'V'\n",
+ "print 'But, due to 1.4V across two conducting diodes actual dc voltage is, VL =',round((VL-1.4),2),'V'\n",
+ "print 'To calculate ripple get, dc load current,'\n",
+ "print 'DC Load current IL=',round(IL,2),'mA'\n",
+ "print 'As per ripple formula,'\n",
+ "print 'Ripple voltage VR=',round(VR,2),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS secondary voltage V2= 24 V\n",
+ "Peak secondary voltage Vp= 33.95 V\n",
+ "with ideal diode and small ripple, dc load voltage, VL = 33.95 V\n",
+ "But, due to 1.4V across two conducting diodes actual dc voltage is, VL = 32.55 V\n",
+ "To calculate ripple get, dc load current,\n",
+ "DC Load current IL= 6.79 mA\n",
+ "As per ripple formula,\n",
+ "Ripple voltage VR= 0.57 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-9, Page 111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.9.py\n",
+ "#Calculate the dc load voltage and ripple in figure 4.17\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=15 #turn ratio\n",
+ "RL=0.5 #Load resistance(KOhm)\n",
+ "C=4700 #Capacitance(uF)\n",
+ "f=60 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V2=Vrms/N12 #RMS secondary voltage(V)\n",
+ "Vp=V2/0.707 #peak secondary voltage(V)\n",
+ "VL=Vp-1.4 #dc load voltage(V)\n",
+ "IL=VL/RL #Load current(mA)\n",
+ "VR=(IL/(2*f*C))*(10**3) #ripple voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'RMS secondary voltage V2=',V2,'V'\n",
+ "print 'Peak secondary voltage Vp=',round(Vp,2),'V'\n",
+ "print 'with ideal diode and small ripple & due to 1.4V across two conducting diodes actual dc voltage is, VL =',round(VL,2),'V'\n",
+ "print 'To calculate ripple get, dc load current,'\n",
+ "print 'DC Load current IL=',round(IL,2),'mA'\n",
+ "print 'As per ripple formula,'\n",
+ "print 'Ripple voltage VR=',round((VR*1000),2),'mV'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS secondary voltage V2= 8 V\n",
+ "Peak secondary voltage Vp= 11.32 V\n",
+ "with ideal diode and small ripple & due to 1.4V across two conducting diodes actual dc voltage is, VL = 9.92 V\n",
+ "To calculate ripple get, dc load current,\n",
+ "DC Load current IL= 19.83 mA\n",
+ "As per ripple formula,\n",
+ "Ripple voltage VR= 35.16 mV\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4-10, Page 114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#Example 4.10.py\n",
+ "#What is the peak inverse voltage for turns ratio is 8:1?\n",
+ "#Breakdown voltage of 50V.\n",
+ "\n",
+ "#Variable declaration\n",
+ "Vrms=120 #RMS value of supply(V)\n",
+ "N12=8 #turn ratio\n",
+ "f=60 #Frequency(Hz)\n",
+ "\n",
+ "#Calculation\n",
+ "V2=Vrms/N12 #RMS secondary voltage(V)\n",
+ "Vp=V2/0.707 #peak secondary voltage(V)\n",
+ "PIV = Vp #Peak Inverse Voltage(V)\n",
+ "\n",
+ "#Result\n",
+ "print 'RMS secondary voltage V2=',V2,'V'\n",
+ "print 'Peak inverse voltage PIV =',round(PIV,2),'V'\n",
+ "print 'PIV << breakdown voltage(50V), So, it is safe to use IN4001'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "RMS secondary voltage V2= 15 V\n",
+ "Peak inverse voltage PIV = 21.22 V\n",
+ "PIV << breakdown voltage(50V), So, it is safe to use IN4001\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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