{ "metadata": { "name": "", "signature": "sha256:24c423084544eb74b3903e47cf4df50df61e5b0825c0f2f723c85cbcdae27d10" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER 6: SEMICONDUCTOR DIODE" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.2, Page number 81" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration \n", "Vf =20; #Peak Input Voltage in V\n", "rf=10; #Forward Resistance in ohms\n", "RL=500.0; #Load Resistance in ohms\n", "V0=0.7; #Potential Barrier Voltage of the diodes in V\n", "\n", "#Calculation\n", "#(1)\n", "If_peak=(Vf-V0)/(rf+RL); #Peak current through the diode in A\n", "If_peak=If_peak*1000; #Peak current through the diode in mA\n", "#(2)\n", "V_out_peak =If_peak * RL/1000 ; #Peak output voltage in V\n", "\n", "#For an Ideal diode\n", "If_peak_ideal=Vf/RL; #Peak current through the ideal diode in A\n", "If_peak_ideal=If_peak_ideal*1000; #Peak current through the ideal diode in mA\n", "\n", "V_out_peak_ideal=If_peak_ideal * RL/1000; # Peak output voltage in case of the ideal diode in V\n", "\n", "#Result\n", "print '(i) Peak current through the diode = %.1f mA '%If_peak;\n", "print '(ii) Peak output voltage = %.1f V'%V_out_peak;\n", "print '(iii) Peak current through the ideal diode = %d mA '%If_peak_ideal;\n", "print '(iv) Peak output voltage in case of the ideal diode = %d V'%V_out_peak_ideal;\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(i) Peak current through the diode = 37.8 mA \n", "(ii) Peak output voltage = 18.9 V\n", "(iii) Peak current through the ideal diode = 40 mA \n", "(iv) Peak output voltage in case of the ideal diode = 20 V\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.3, Page number 82" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "V =10.0; #Battery voltage in V\n", "R1=50.0; #Resistor 1's resistance in ohms\n", "R2=5.0; #Resistor 2's resistance in ohms\n", "\n", "#Calculation\n", "#Using Thevenin's Theorem to find current in the diode\n", "E0=(R2/(R1+R2))*V; #Thevenin's Voltage in V\n", "R0=(R1*R2)/(R1+R2); #Thevenin's Resistance in ohms\n", "\n", "I0=E0/R0; #Current through the diode in A\n", "I0=I0*1000; #Current through the diode in mA\n", "\n", "#Result\n", "print 'Current through the diode = %d mA '%Io;\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Current through the diode = 200 mA \n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.4, Page number 82-83 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "V =10.0; #Battery voltage in V\n", "R0=48.0; #Resistance of the resistor in ohms\n", "Rd=1.0; #Forward resistance of the diodes in ohms\n", "Vd=0.7; #Potential barrier of the diodes in V\n", "#Calculation\n", "V_net=V-Vd-Vd; #Net voltage in the circuit in V\n", "R_net=R0+Rd+Rd #Net resistance of the circuit in ohms\n", "I_net=V_net/R_net; #Net current in the circuit in A\n", "I_net=I_net*1000; #Net current in mA\n", "\n", "#Result\n", "print 'Net current in the circuit = %d mA '%I_net;\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Net current in the circuit = 172 mA \n" ] } ], "prompt_number": 15 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.5, Page number 83" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "E1=24; #Voltage of first source in V\n", "E2=4; #Voltage of second source in V\n", "V0=0.7; #Potential barrier of diodes in V\n", "R=2000; #Resistance of the given resistor in ohms\n", "Rd=0; #Forward resistance of the diodes in ohms\n", "\n", "#Calculation\n", "I=(E1-E2-V0)/(R+Rd); #Current in the circuit in A\n", "I=I*1000; #Current in the circuit in mA \n", "\n", "#Result\n", "print 'Current in the circuit = %.2f mA '%I;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Current in the circuit = 9.65 mA \n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.6, Page number 83-84" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "V=20; #Voltage of source in V\n", "V0=0.3; #Potential barrier of Germanium diode in V\n", "V0_Si=0.7; #Potetial barrier of Silicon diode in V \n", "\n", "#Calculation\n", "#As only Ge diode is turned on due to less potential barrier,\n", "VA=V-V0; #Voltage VA acroos resistor of 3k ohms\n", "\n", "#Result\n", "print 'Voltage VA = %.1f mA '%VA;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Voltage VA = 19.7 mA \n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.7, Page number 84" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "V=10; #Voltage of source in V\n", "V0=0.7; #Potetial barrier of Silicon diode in V \n", "# Resistance of all resistors in ohms\n", "R1=2000;\n", "R2=2000;\n", "R3=2000;\n", "\n", "#Calculation\n", "Id=(V-V0)/(R2+2*R3); #Current through the diodes in A\n", "VQ=2*Id*R3; #Voltage VQ across the grounded 2k ohm resistor in V\n", "Id=Id*1000; #Current through the diodes in mA\n", "\n", "#Result\n", "print 'Voltage VQ = %.1f V '%VQ;\n", "print 'Current through the diodes, Id = %.2f mA '%Id;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Voltage VQ = 6.2 V \n", "Current through the diodes, Id = 1.55 mA \n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.8, Page number 84" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "V=15; #Voltage of source in V\n", "V0=0.7; #Potetial barrier of Silicon diode in V \n", "R=500 # Resistance of all resistors in ohms\n", "\n", "#Calculation\n", "I1=(V-V0)/R; #total current in the circuit in A\n", "Id1=I1/2; #current in first diode in A\n", "Id1=Id1*1000; #current in first diode in mA\n", "Id2=Id1 #current in second diode in mA\n", "\n", "#Result\n", "print ('Current in first diode = %.1f mA'%Id1);\n", "print ('Current in second diode = %.1f mA'%Id2);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Current in first diode = 14.3 mA\n", "Current in second diode = 14.3 mA\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.9, Page number 85" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "E=20; #Voltage of source in V\n", "V0_d1=0.7; #Potetial barrier of first Silicon diode in V\n", "V0_d2=0.7; #Potetial barrier of second Silicon diode in V\n", "R1=5600; # Resistance of first resistor in ohms\n", "R2=3300; # Resistance of second resistor in ohms\n", "\n", "#Calculation\n", "I2=V0_d2/R2; #Current I2 through resistor R2 in A\n", "I2=round((I2*1000),3); #Current I2 through resistor R2 in mA\n", "I1=(E-V0_d1-V0_d2)/R1; #Current I1 through resistor R1 in A\n", "I1=round((I1*1000),2); #Current I1 through resistor R1 in mA\n", "I3=I1-I2; #Current I3 through diode D2 in mA\n", "\n", "#Result\n", "print 'Current I1= %.2f mA'%I1;\n", "print 'Current I1= %.3f mA'%I2;\n", "print 'Current I1= %.3f mA'%I3;\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Current I1= 3.32 mA\n", "Current I1= 0.212 mA\n", "Current I1= 3.108 mA\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.10, Page number 85-86" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable Declaration\n", "E=10.0; #Voltage of source in V\n", "V0=0.7; #Potetial barrier of Silicon diode in V\n", "R1=2000; # Resistance of first resistor in ohms\n", "R2=8000; # Resistance of second resistor in ohms\n", "R3=4000; #Resistance of third resistor in ohms\n", "R4=6000; #Resistance of fourth resistor in ohms\n", "\n", "#Calculation\n", "#Assuming the given diode to be reverse bised and calculating voltage across it's terminals\n", "V1=(E/(R1+R2))*R2; #voltage at the P side of the diode, i.e, voltage across R2 resistor,according to voltage divider rule, in V\n", "V2=(E/(R3+R4))*R4; #voltage at the N side of the diode, i.e, voltage across R4 resistor,according to voltage divider rule, in V\n", "\n", "#Result\n", "if((V1-V2)>=V0):\n", " print 'Our assumption was wrong and, the diode is forward biased';\n", "else:\n", " print 'The diode is reverse biased';\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Our assumption was wrong and, the diode is forward biased\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.11, Page number 86" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "V=2; #Supply voltage in V\n", "V0=0.7; #Potential barrier voltage of the diode in V \n", "R1=4000.0; #Resistance of first resistor in \u03a9\n", "R2=1000.0; ##Resistance of second resistor in \u03a9\n", "\n", "#Calculation\n", "#Assuming the diode to be in ON state\n", "I1=((V-V0)/R1)*1000; #Current through resistor R1, in mA\n", "I2=(V0/R2)*1000; #Current through resistor R2, in mA\n", "ID=I1-I2; #Diode current, in mA\n", "\n", "if(ID<0):\n", " #Since the diode current is negative, the diode must be OFF \n", " ID=0; #True value of diode current, mA\n", " \n", "#As the diode is in OFF state it can be replaced by an open ciruit equivalent \n", "VD=V*R2/(R1 +R2); #Voltage across the diode, in V\n", "\n", "#Result\n", "print 'ID =%d mA'%ID;\n", "print 'VD =%.1f V'%VD;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "ID =0 mA\n", "VD =0.4 V\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.12, Page number 89-90" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "AC_Input_Power=100.0; #Input AC Power in watts\n", "AC_Output_Power=40.0; #Output AC Power in watts\n", "Accepted_Power=50.0; #Power accepted by the half-wave rectifier in watt\n", "\n", "#Calculation\n", "R_eff=(AC_Output_Power/AC_Input_Power)*100; #Rectification efficiency of the half-wave rectifier\n", "Unused_power=AC_Input_Power-Accepted_Power; #Power not used by the half_wave rectifier due to open circuited condition of the diode in watt\n", "Power_dissipated=Accepted_Power-AC_Output_Power; #Power dissipated by the diode watt\n", "\n", "#Result\n", "print 'The rectification efficiency of the half-wave rectifier= %d%% '%R_eff;\n", "\n", "print 'Rest 60%% of the power is the unused power and power dissipated by the diode = %d watts and %d watts' %(Unused_power ,Power_dissipated);\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The rectification efficiency of the half-wave rectifier= 40% \n", "Rest 60% of the power is the unused power and power dissipated by the diode = 50 watts and 10 watts\n" ] } ], "prompt_number": 24 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.13, Page number 90" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import pi\n", "from math import sqrt\n", "#Variable declaration\n", "Vrms=230.0; #AC supply RMS voltage in V\n", "Turns_Ratio=10/1; #turn ratio of the transformer \n", "\n", "#Calculation\n", "Vpm=sqrt(2)*Vrms; #Maximum primary voltage in V\n", "Vsm=Vpm/Turns_Ratio; #Maximum secondary voltage in V\n", "#Case 1\n", "Vdc=Vsm/(round(pi,2)); #Output D.C voltage, which is the average voltage in V\n", "Vdc=round(Vdc,2);\n", "#Case 2\n", "PIV=Vsm; #Peak Inverse Voltage in V\n", "\n", "#Result\n", "print 'The output d.c voltage= %.2f V'%Vdc;\n", "print 'The peak inverse voltage= %.2f V'%PIV;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The output d.c voltage= 10.36 V\n", "The peak inverse voltage= 32.53 V\n" ] } ], "prompt_number": 27 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.14, Page number 90-91" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import pi\n", "#Variable declaration\n", "rf=20.0; #Internal resistance of the crystal diode in ohms\n", "Vm=50.0; #Maximum applied voltage in V\n", "RL=800.0; #Load Resistance in ohms\n", "\n", "#Calculation\n", "# 1\n", "Im=Vm/(rf+RL); #Maximum current in A\n", "Im=Im*1000; #Maximum current in \n", "Im=round(Im,0);\n", "Idc=Im/pi; #Average voltage in mA\n", "Idc=round(Idc,1);\n", "Irms=Im/2; #RMS value of the current in mA\n", "Irms=round(Irms,1)\n", "\n", "# 2\n", "AC_Input_Power=pow(Irms/1000,2)*(rf+RL); #Input a.c power in watt\n", "\n", "DC_Output_Power=pow(Idc/1000,2)*RL; #Output d.c power in watt\n", "\n", "# 3\n", "DC_Output_Voltage=(Idc/1000)*RL; #Output d.c voltage in V\n", "\n", "# 4\n", "Rectifier_efficiency=(DC_Output_Power/AC_Input_Power)*100; # Efficiency of rectification of the half-wave rectifier\n", "\n", "#Result\n", "print ' i:';\n", "print ' Im = %d mA'%Im;\n", "print ' Idc = %.1f mA'%Idc;\n", "print ' Irms = %.1f mA'%Irms;\n", "print ' ii: ';\n", "print ' a.c input power= %.3f watt'%AC_Input_Power;\n", "print ' d.c output power= %.3f watt'%DC_Output_Power;\n", "print ' iii: ';\n", "print ' d.c output voltage = %.2f volts'%DC_Output_Voltage;\n", "print ' iv: '\n", "print ' Efficiency of rectification = %.1f%%'%Rectifier_efficiency;\n", "\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " i:\n", " Im = 61 mA\n", " Idc = 19.4 mA\n", " Irms = 30.5 mA\n", " ii: \n", " a.c input power= 0.763 watt\n", " d.c output power= 0.301 watt\n", " iii: \n", " d.c output voltage = 15.52 volts\n", " iv: \n", " Efficiency of rectification = 39.5%\n" ] } ], "prompt_number": 16 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.15, Page number 91" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import pi\n", "#Variable declaration\n", "Vdc=50.0; #Output d.c voltage in V\n", "rf=25; #Diode resistance in ohm\n", "RL=800; #Load resistance in ohm\n", "\n", "\n", "#Calculation\n", "Vm=(pi*(rf+RL)*Vdc)/RL; #[ Vdc=Vm*RL/(pi*(rf+RL)) ]Maximum value of a.c voltage required to get a volatge of Vdc from the half-wave rectifier, in V\n", "Vm=round(Vm,0); \n", "#Result\n", "print 'The a.c voltage required should have maximum value of = %d V' %Vm;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The a.c voltage required should have maximum value of = 162 V\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.16, Page number 95" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import sqrt \n", "from math import pi\n", "#Variable declaration\n", "rf=20; #Internal resistance of the diodes in ohm\n", "Vrms=50; #RMS value of transformer's secondary voltage from centre tap to each end of secondary\n", "RL=980; #Load resistance in ohm\n", "\n", "#Calculation\n", "Vm=Vrms*sqrt(2); #Maximum a.c voltage in V\n", "Im=Vm/(rf+RL); #Maximum load current in A\n", "Im=Im*1000; #Maximum load current in mA\n", " \n", "# 1:\n", "Idc=2*Im/pi; #Mean load current\n", "\n", "# 2:\n", "Irms=Im/sqrt(2); #RMS value of load current in A\n", "\n", "#Result\n", "print 'i:';\n", "print' The mean load current= %d mA'%Idc;\n", "print 'ii:';\n", "print ' The r.m.s value of the load current = %d mA'%Irms; " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i:\n", " The mean load current= 45 mA\n", "ii:\n", " The r.m.s value of the load current = 50 mA\n" ] } ], "prompt_number": 20 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.17, Page number 95-96" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import sqrt \n", "#Variable declaration\n", "RL=100; #Load resistance in ohm \n", "rf=0; #Internal resistance of the diodes in ohm\n", "Turns_ratio=5/1; #Primary to secondary turns ratio of transformer \n", "P_Vrms=230; #R.M.S value of voltage in primary winding in V\n", "S_Vrms=P_Vrms/Turns_ratio; #R.M.S value of voltage in secondary winding in V\n", "S_Vm=S_Vrms*sqrt(2); #Maximum voltage across secondary winding in V\n", "Vm=S_Vm/2; #Maximum voltage across half seconfdary winding in V\n", "\n", "\n", "#Calculation\n", "# 1:\n", "Idc=2*Vm/(pi*RL); #Average current in A\n", "Vdc=Idc*RL; #d.c output voltage in V\n", "\n", "# 2:\n", "PIV=S_Vm; #Peak Invers Voltage(= Maximum secondary voltage) in V\n", "\n", "# 3:\n", "Pac=pow(Vm/(RL*sqrt(2)),2)*(rf+RL); #a.c input power in watt\n", "Pdc=(pow(Idc,2)*RL); #d.c output power in watt\n", "R_eff=(Pdc/Pac)*100; #Rectification efficiency\n", "R_eff=round(R_eff,1);\n", "\n", "#Result\n", "print 'i:';\n", "print ' The d.c output voltage= %.1f V'%Vdc;\n", "print 'ii:';\n", "print ' The peak inverse voltage= %d V'%PIV;\n", "print 'iii:';\n", "print ' Rectification efficiency= %.1f%%'%R_eff;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i:\n", " The d.c output voltage= 20.7 V\n", "ii:\n", " The peak inverse voltage= 65 V\n", "iii:\n", " Rectification efficiency= 81.1%\n" ] } ], "prompt_number": 27 }, { "cell_type": "markdown", "metadata": {}, "source": [ "NOTE: The value of rectification efficiency is calculated as 81.2% in the textbook using the formula 0.812/(1 + (rf/RL)), but by calculating using the correct values in the formula we get 81.1%." ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.18, Page number 96-97" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import sqrt \n", "#Variable declaration\n", "fin=50; #frequency of input ac source in Hz\n", "RL=200; #Load resistance in ohm\n", "Turns_ratio=4/1; #Transformers turns ratio, primary to secondary.\n", "P_Vrms=230.0; #R.M.S value of voltage in primary winding in V\n", "S_Vrms=P_Vrms/Turns_ratio #R.M.S value of voltage in secondary winding in V\n", "Vm=S_Vrms*sqrt(2); #Maximum voltage across secondary winding in V\n", "\n", "#Calculation\n", "# 1:\n", "Idc=2*Vm/(pi*RL); # Average current in A\n", "Vdc=Idc*RL; #Output d.c voltage in V\n", "Vdc=round(Vdc,0);\n", "# 2:\n", "PIV= Vm; #Peak Inverse Voltage(= Maximum volutage across secondary winding) in V\n", "\n", "# 3:\n", "fout=2*fin; #Output frequency in Hz\n", "\n", "#Result\n", "print 'i:';\n", "print ' The d.c output voltage = %d V' %Vdc;\n", "print 'ii:';\n", "print ' The peak inverse voltage = %.1f V'%PIV;\n", "print 'iii:';\n", "print ' The output frequency = %d Hz'%fout;\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i:\n", " The d.c output voltage = 52 V\n", "ii:\n", " The peak inverse voltage = 81.3 V\n", "iii:\n", " The output frequency = 100 Hz\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.19, Page number 97" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import pi\n", "from math import sqrt\n", "\n", "#Variable declaration\n", "RL=100.0; #Load Resistance in ohm\n", "Turns_ratio=5/1; #Primary to secondary turns ratio of the transformer\n", "Vin=230.0; #R.M.S value of input voltage in V\n", "fin=50; #Input frequency in Hz\n", "\n", "#Calculation\n", "Vs_rms=Vin/Turns_ratio; #R.M.S value of the voltage in secondary winding, in v\n", "Vs_max=Vs_rms*sqrt(2); #Maximum voltage across secondary, in V\n", "\n", "# (i)\n", "#Case i: Centre-tap circuit\n", "Vm=Vs_max/2; #Maximum voltage across half secondary winding, in V \n", "Vdc=2*Vm*RL/(pi*RL); #DC output voltage, in V \n", "print 'The d.c output voltage for the centre-tap circuit = %.1f V'%Vdc;\n", "\n", "#Case ii:\n", "Vm=Vs_max; #Maximum voltage across secondary, in V\n", "Vdc=2*Vm*RL/(pi*RL); #DC output voltage, in V \n", "print 'The d.c output voltage for the bridge circuit = %.1f V'%Vdc; \n", "\n", "# ii:\n", "#Case i: Centre-tap circuit\n", "Turns_ratio=5/1;\n", "Vs_rms=Vin/Turns_ratio;\n", "Vs_max=Vs_rms*sqrt(2);\n", "Vm=Vs_max/2;\n", "PIV=2*Vm;\n", "print 'PIV in case of centre-tap circuit = %d V'%PIV;\n", "\n", "#Case ii: Bridge circuit\n", "Turns_ratio=10/1;\n", "Vs_rms=Vin/Turns_ratio;\n", "Vs_max=Vs_rms*sqrt(2);\n", "PIV=Vm;\n", "print 'PIV in case of bridge circuit = %.1f V'%PIV;\n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The d.c output voltage for the centre-tap circuit = 20.7 V\n", "The d.c output voltage for the bridge circuit = 41.4 V\n", "PIV in case of centre-tap circuit = 65 V\n", "PIV in case of bridge circuit = 32.5 V\n" ] } ], "prompt_number": 44 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.20, Page number 98" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import pi\n", "from math import sqrt\n", "#Variable declaration\n", "rf=1; #forward resistance of diodes of the rectifier in ohm\n", "RL=480; #Load resistance in ohm\n", "Vrms=240.0; #a.c supply voltage in V\n", "Vm=Vrms*sqrt(2); #Maximum a.c voltage in V \n", "\n", "#Calculation\n", "# 1:\n", "Rt=2*rf+RL; #Total circuit resistance at any instance in ohm\n", "Im=Vm/Rt; #Maximum load current in A\n", "Idc=2*Im/pi; #Mean load current in A\n", "\n", "# 2:\n", "Irms=Im/2; #R.M.S value of current in A\n", "P=pow(Irms,2)*rf; #Power dissipated in each diode in watt\n", "\n", "\n", "#Result\n", "print 'i:';\n", "print ' Mean load current = %.2f A'%Idc;\n", "print 'ii:';\n", "print ' Power dissipated in each diode= %.3f W'%P;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i:\n", " Mean load current = 0.45 A\n", "ii:\n", " Power dissipated in each diode= 0.124 W\n" ] } ], "prompt_number": 39 }, { "cell_type": "markdown", "metadata": {}, "source": [ "NOTE: The value of power dissipated is approximately 0.124 W , but in the textbook it is approximated as 0.123W." ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.21, Page number 98-99" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import sqrt,pi\n", "#Variable declaration\n", "RL=12000; #Load resistance in ohm\n", "V0=0.7; #Potential barrier voltage of diodes in V\n", "Vrms=12; #R.M.S value of input a.c voltage in V\n", "Vs_pk=Vrms*sqrt(2); #Peak secondary voltage in V\n", "\n", "#Calculation\n", "# 1:\n", "Vout_pk=Vs_pk-(2*V0); #Peak output voltage in V\n", "Vav=2*Vout_pk/pi; #Average output voltage in V\n", "Vav=round(Vav,2);\n", "\n", "# 2:\n", "Iav=Vav/RL; #Average output current in A\n", "Iav=Iav*pow(10,6); #Average output current in \u03bcA\n", "\n", "\n", "#Result\n", "print 'i:';\n", "print ' Average output voltage=%.2f V'%Vav;\n", "print 'ii:';\n", "print ' Average output current=%.1f \u03bcA'%Iav;\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i:\n", " Average output voltage=9.91 V\n", "ii:\n", " Average output current=825.8 \u03bcA\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.22, Page number 102" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Vdc_A=10; #Supply voltage of A in V\n", "Vdc_B=25; #Supply voltage of B in V\n", "Vac_rms_a=0.5; #Ripples in power supply A in V\n", "Vac_rms_b=0.001; #Ripples in power supply B in V\n", "\n", "#Calculation\n", "#For power supply A\n", "ripple_factor_A=Vac_rms_a/Vdc_A; #Ripple factor of power supply A\n", "\n", "#For power supply B\n", "ripple_factor_B=Vac_rms_b/Vdc_B; #Ripple factor of power supply B\n", "\n", "#Result\n", "if(ripple_factor_AVz):\n", " #Zener diode is in ON state\n", " # i:\n", " Output_voltage=Vz; #Voltage across load resistance, in V\n", " #ii:\n", " Voltage_R=Ei-Vz; #Voltage across the series resistance R, in V\n", " #iii:\n", " IL=Vz/RL; #Load current through RL in A\n", " IL=IL*1000; #Load current through RL in mA\n", " I=Voltage_R/R; #Current through the series resistance in A\n", " I=I*1000; #Current through the series resistance in mA\n", " Iz=I-IL; #Applying Kirchhoff's first law, Zener current in mA\n", " \n", " #Result\n", " print 'i) The output voltage across the load resistance RL = %d V'%Output_voltage;\n", " print 'ii) The voltage drop across the series resistance R = %d V'%Voltage_R;\n", " print 'iii) The current through the zener diode = %d mA'%Iz;\n", " " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i) The output voltage across the load resistance RL = 50 V\n", "ii) The voltage drop across the series resistance R = 70 V\n", "iii) The current through the zener diode = 9 mA\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.26, Page number 114-115" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Max_V=120.0; #Maximum input voltage in V\n", "Min_V=80.0; #Minimum input voltage in V\n", "R=5000.0; #Series resistance in ohm\n", "RL=10000.0; #Load resistance in ohm\n", "Vz=50.0; #Zener voltage in V\n", "\n", "\n", "#Calculation\n", "#Case i: Maximum zener current\n", "#Zener current will be maximum when the input voltage is maximum\n", "V_R_max=Max_V-Vz; #Voltage across series resistance R, in V\n", "I_max=V_R_max/R; #Current through series resistance R, in A\n", "I_max=I_max*1000; #Current through series resistance R, in mA\n", "IL_max=Vz/RL; #Load current in A\n", "IL_max=IL_max*1000; #Load current in mA\n", "Iz_max=I_max-IL_max; #Applying Kirchhoff's first law, Zener current in mA;\n", "\n", "#Case ii: Minimum zener current\n", "#The zener will conduct minimum current when the input voltage is minimum\n", "V_R_min=Min_V-Vz; #Voltage across series resistance R, in V\n", "I_min=V_R_min/R; #Current through series resistance R, in A\n", "I_min=I_min*1000; #Current through series resistance R, in mA\n", "IL_min=Vz/RL; #Load current in A\n", "IL_min=IL_min*1000; #Load current in mA\n", "Iz_min=I_min-IL_min; #Applying Kirchhoff's first law, Zener current in mA\n", "\n", "#Result\n", "print 'Case i: ';\n", "print 'Maximum zener current = %d mA'%Iz_max;\n", "print 'Case ii: ';\n", "print 'Minimum zener current = %d mA'%Iz_min;\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Case i: \n", "Maximum zener current = 9 mA\n", "Case ii: \n", "Minimum zener current = 1 mA\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.27, Page number 115" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Ei=12; #Input voltage in V\n", "Vz=7.2; #Zener voltage in V\n", "E0=Vz; #Voltage to be maintained across the load in V\n", "IL_max=0.1; #Maximum load current in A\n", "IL_min=0.012; #Minimum load current in A\n", "Iz_min=0.01; #Minimum zener current in A\n", "\n", "#Calculation\n", "#When the load current is maximum at minimum value of RL, the zener current is minimum and, as the load current decreases due to increase in value of RL\n", "R=(Ei-E0)/(Iz_min+IL_max); #The value of series resistance R to maintain a voltage=E0 across load, in ohm\n", "\n", "#Result\n", "print 'The minimum value of series resistance R to maintain a constant value of 7.2 V is = %.1f \u03a9'%R;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The minimum value of series resistance R to maintain a constant value of 7.2 V is = 43.6 \u03a9\n" ] } ], "prompt_number": 16 }, { "cell_type": "markdown", "metadata": {}, "source": [ "NOTE: The actual value of R is 43.636363 (recurring) but, in the textbook the value of R is wrongly approximated 43.5 \u03a9" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.28, Page number 115" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Ei_min=22; #Minimum input voltage in V\n", "Ei_max=28; #Maximum input voltage in V\n", "Vz=18; #Zener voltage in V\n", "E0=Vz; #Constant voltage maintained across the load resistance in V\n", "Iz_min=0.2; #Minimum zener current in A\n", "Iz_max=2; #Maximum zener current in A\n", "RL=18; #Load resistance in \u03a9\n", "\n", "#Calculation\n", "IL=Vz/RL; #Constant value of load current in A\n", "#When the input voltage is minimum, the zener current will be minimum\n", "R=(Ei_min-E0)/(Iz_min+IL) #The value of series resistance so that the voltage E0 across RL remains constant\n", "\n", "print 'The value of series resistance R, to maintain constant voltage E0 across RL = %.2f \u03a9.'%R;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The value of series resistance R, to maintain constant voltage E0 across RL = 3.33 \u03a9.\n" ] } ], "prompt_number": 19 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.29, Page number 116 " ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Vz=10 #Zener voltage in V\n", "Ei_min=13; #Minimum input voltage in V\n", "Ei_max=16; #Maximum input voltage in V\n", "Iz_min=0.015; #Minimum zener current in A\n", "IL_min=0.01; #Minimum load current in A \n", "IL_max=0.085; #Maximum load curremt in A\n", "E0=Vz; #Constant voltage to be maintained in V \n", "\n", "#Calculation\n", "#The zener current will be minimum when the input voltage will be minimum and at that time the load current will be maximum\n", "R=(Ei_min-E0)/(Iz_min+IL_max); #The value of series resistance R to maintain a constant voltage across load\n", "\n", "\n", "#Result\n", "print 'The value of series resistance to maintain a constant voltage across the load resistance is = %d \u03a9'%R;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The value of series resistance to maintain a constant voltage across the load resistance is = 30 \u03a9\n" ] } ], "prompt_number": 23 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.30, Page number 116" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Iz=0.2; #Current rating of each zener in A\n", "Vz=15; #Voltage rating of each zener in V\n", "Ei=45; #Input voltage in V\n", "\n", "#Calculation\n", "# i: Regulated output voltage across the two zener diodes \n", "E0=2*Vz; # V\n", "\n", "# ii: Value of series resistance \n", "R=(Ei-E0)/Iz; # \u03a9\n", "\n", "#Result\n", "print 'i) The regulated output voltage = %d V'%E0;\n", "print 'ii) The value of the series resistance = %d \u03a9'%R;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "i) The regulated output voltage = 30 V\n", "ii) The value of the series resistance = 75 \u03a9\n" ] } ], "prompt_number": 28 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.31, Page number 116-117" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Vz=10; #Voltage rating of each zener in V\n", "Iz=1; #Current rating of each zener in A\n", "Ei=45; #Input unregulated voltage in V\n", "\n", "#Calculation\n", "#Regulated output voltage across the three zener diodes\n", "E0=3*Vz; # V\n", "\n", "#Value of series resistance to obtain a 30V regulated output voltage\n", "R=(Ei-E0)/Iz; # \u03a9\n", "\n", "#Result\n", "print 'Value of series resistance to obtain a 30V regulated output voltage = %d \u03a9'%R;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Value of series resistance to obtain a 30V regulated output voltage = 15 \u03a9\n" ] } ], "prompt_number": 29 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.32, Page number 117" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#variable declaration\n", "RL=2000.0; #Load resistance in \u03a9\n", "R=200.0; #Series resistance in \u03a9\n", "Iz=0.025; #Zener current rating in A\n", "E0=30.0; #Output regulated voltage in V \n", "\n", "#Calculation\n", "#Minimum input voltage will be required when Iz=0 A, and at this condition\n", "IL=E0/RL; #Load current during Iz=0, in A\n", "I=IL; #According to Kirchhoff's law, total current, in A\n", "Ei_min=E0+(I*R); #Minimum input voltage in V\n", "\n", "#The maximum input voltage required will be when Iz=0.025 A, and at that condition \n", "I=IL+Iz; #According to Kirchhoff's law, total current, in A\n", "Ei_max=E0+(I*R); #maximum input voltage in V\n", "\n", "\n", "#Result\n", "print 'The required range of input voltage is from %d V to %d V'%(Ei_min,Ei_max); \n", "\n", "\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The required range of input voltage is from 33 V to 38 V\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.33, Page number 117-118" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "Ei=16; #Unregulated input voltage in V\n", "E0=12; #Output regulated voltage in V\n", "IL_min=0; #Minimum load current in A\n", "IL_max=0.2; #Maximum load current in A\n", "Iz_min=0; #Minimum zener current in A\n", "Iz_max=0.2; #Maximum zener current in A\n", "\n", "#Calculation\n", "#As the regulated voltage required across the load is 12V\n", "Vz=E0; #Voltage rating of zener diode in V\n", "V_R=Ei-E0; #Constant Voltage that should remain across series resistance \n", "#The minimum zener current will occur when the curent in the load in maximum\n", "R=V_R/(Iz_min+IL_max); #Series resistance in \u03a9\n", "\n", "Max_power_rating=Vz*Iz_max; #Maximum power rating of zener diode in W\n", "\n", "#Result\n", "print 'The regulator is designed using a Seris resistance of %d \u03a9 and a zener diode of zener voltage %d V'%(R,Vz);\n", "print 'The maximum power rating of the zener diode is = %.1f W '%Max_power_rating;" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The regulator is designed using a Seris resistance of 20 \u03a9 and a zener diode of zener voltage 12 V\n", "The maximum power rating of the zener diode is = 2.4 W \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 6.34, Page number 118" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Variable declaration\n", "V=12; #Source voltage in V\n", "R=1000; #Series resistance in \u03a9\n", "RL=5000; #Load resistance in \u03a9\n", "Vz=6; #Voltage rating of zener in V\n", "\n", "#Calculation\n", "#Case i: zener is working properly\n", "#The output voltage across the load will be equal to the zener voltage.\n", "V0=Vz; # V\n", "\n", "#Result\n", "print 'Case i: Output voltage when zener is working properly is %d V'%V0;\n", "\n", "#Case ii: zener is shorted\n", "#As the zener is shorted, the potential difference across the load will be zero\n", "V0=0; #V\n", "\n", "#Result\n", "print 'Case ii: Output voltage when zener is short circuited is %d V'%V0;\n", " \n", "#Case iii: zener is open circuited\n", "#If the zener is open circuited, the total voltage will drop across R and RL according to the voltage divider rule\n", "V0=V*RL/(R+RL); #V\n", "\n", "#Result\n", "print 'Case iii: Output voltage when zener is open circuited is %d V'%V0;\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Case i: Output voltage when zener is working properly is 6 V\n", "Case ii: Output voltage when zener is short circuited is 0 V\n", "Case iii: Output voltage when zener is open circuited is 10 V\n" ] } ], "prompt_number": 4 }, { "cell_type": "code", "collapsed": false, "input": [], "language": "python", "metadata": {}, "outputs": [] } ], "metadata": {} } ] }