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diff --git a/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb deleted file mode 100755 index cc8b5565..00000000 --- a/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb +++ /dev/null @@ -1,765 +0,0 @@ -{ - "metadata": { - "name": "" - }, - "nbformat": 3, - "nbformat_minor": 0, - "worksheets": [ - { - "cells": [ - { - "cell_type": "heading", - "level": 1, - "metadata": {}, - "source": [ - "Chapter - 2 : Diode Applications" - ] - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.1\n", - ": Page No 83" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from __future__ import division\n", - "from numpy import pi\n", - "from math import sqrt\n", - "# Given data\n", - "R_L = 1000 # in ohm\n", - "N2byN1= 4 \n", - "Vi= '10*sin(omega*t)'\n", - "# V2= N2byN1*V1\n", - "# V2= 40*sin(omega*t)\n", - "Vm= N2byN1*10 # in V\n", - "V_Lav= Vm/pi # in V\n", - "print \"The average load voltage = %0.2f volts\" %V_Lav\n", - "Im= Vm/R_L # in A\n", - "I_dc= Im/pi # in A\n", - "I_av = I_dc # in A\n", - "I_av= I_av*10**3 # in mA\n", - "print \"Average load current = %0.2f mA\" %I_av\n", - "V_Lrms = Vm/2 # in V\n", - "print \"RMS load voltage = %0.f V\" %V_Lrms\n", - "I_rms = V_Lrms/R_L # in A\n", - "I_rms= I_rms*10**3 # in mA\n", - "print \"RMS load current = %0.f mA\" %I_rms\n", - "Eta = I_av**2/I_rms**2*100 # in %\n", - "print \"Efficiency = %0.2f %%\" %Eta\n", - "V2rms= Vm/sqrt(2) # in V\n", - "TUF = ((I_av )**2)/(V2rms*I_rms)*100 # in %\n", - "print \"Transformer utilization factor = %0.2f %%\" %TUF\n", - "Gamma= sqrt(V_Lrms**2-I_av**2)/V_Lav*100 \n", - "print \"Ripple factor = %0.f %%\" %round(Gamma)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The average load voltage = 12.73 volts\n", - "Average load current = 12.73 mA\n", - "RMS load voltage = 20 V\n", - "RMS load current = 20 mA\n", - "Efficiency = 40.53 %\n", - "Transformer utilization factor = 28.66 %\n", - "Ripple factor = 121 %\n" - ] - } - ], - "prompt_number": 9 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.2\n", - ": Page No 96" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "R_L = 1 # in K ohm\n", - "R_L = R_L * 10**3 # in ohm\n", - "V_m = 15 # in V\n", - "V_i = '15*sin(314*t)' \n", - "I_m= V_m/R_L # in A\n", - "I_dc = I_m/pi # in A\n", - "I_dc = I_dc * 10**3 # in mA\n", - "print \"Average current through the diode = %0.2f mA\" %I_dc\n", - "I_drms = V_m/(2*R_L) \n", - "I_drms = I_drms * 10**3 # in mA\n", - "print \"RMS current = %0.1f mA\" %I_drms\n", - "I_m = V_m/R_L \n", - "I_m = I_m*10**3 # in mA\n", - "print \"Peak diode current = %0.f mA\" %I_m\n", - "PIV = 2*V_m # in V\n", - "print \"Peak inverse voltage = %0.f V\" %PIV" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Average current through the diode = 4.77 mA\n", - "RMS current = 7.5 mA\n", - "Peak diode current = 15 mA\n", - "Peak inverse voltage = 30 V\n" - ] - } - ], - "prompt_number": 10 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.3\n", - ": Page No 101" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "R1 = 2.2*10**3 # in ohm\n", - "R2 = 4.7*10**3 # in ohm\n", - "R_AB = (R1*R2)/(R1+R2) # in ohm\n", - "Vi = 20 # in V\n", - "V_o = (Vi * R_AB)/(R_AB+R1) # in V\n", - "PIV= Vi # in volts\n", - "print \"The output voltage = %0.1f V\" %V_o\n", - "print \"Peak inverse voltage = %0.f volts\" %PIV" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The output voltage = 8.1 V\n", - "Peak inverse voltage = 20 volts\n" - ] - } - ], - "prompt_number": 11 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - " Example 2.4.2 \n", - ": Page No 102" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "from math import cos\n", - "# Given data\n", - "V_in = 10 # in V\n", - "R1 = 2000 # in ohm\n", - "R2 = 2000 # in ohm\n", - "V_o = V_in * (R1/(R1+R2) ) # in V\n", - "# Vdc= 5/(T/2)*integrate('sin(omega*t)','t',0,T/2) and omega*T= 2*pi, So\n", - "Vdc= -5/pi*(cos(pi)-cos(0)) # in V\n", - "print \"The value of Vdc = %0.3f volts\" %Vdc\n", - "PIV= V_in/2 # in V\n", - "print \"The PIV value = %0.f volts\" %PIV\n" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The value of Vdc = 3.183 volts\n", - "The PIV value = 5 volts\n" - ] - } - ], - "prompt_number": 23 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - " Example 2.4 (again 2.4)\n", - ": Page No 124" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_in = 10 # in V\n", - "R_L = 2000 # in ohm\n", - "R1 = 100 # in ohm\n", - "V_R= 0.7 # in V\n", - "V_o = V_in * ( (R_L)/(R1+R_L) ) # in V\n", - "print \"The peak magnitude of the positive output voltage = %0.2f V\" %V_o \n", - "Vo=-V_R # in V\n", - "print \"The peak magnitude of the negative output voltage = %0.1f V\" %Vo" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The peak magnitude of the positive output voltage = 9.52 V\n", - "The peak magnitude of the negative output voltage = -0.7 V\n" - ] - } - ], - "prompt_number": 27 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.7\n", - ": Page No 141" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V=240 # in V\n", - "R= 1 # in k\u03a9\n", - "R=R*10**3 # in \u03a9\n", - "Vsrms= V/4 # in V\n", - "Vm= sqrt(2)*Vsrms # in V\n", - "V_Ldc= -Vm/pi # in V\n", - "print \"The value of average load voltage = %0.f volts\" %V_Ldc" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The value of average load voltage = -27 volts\n" - ] - } - ], - "prompt_number": 28 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.8\n", - ": Page No 142" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V = 220 # in V\n", - "f=50 # in Hz\n", - "N2byN1=1/4 \n", - "R_L = 1 # in kohm\n", - "R_L= R_L*10**3 # in ohm\n", - "V_o = 220 # in V\n", - "V_s = N2byN1*V_o # in V\n", - "V_m = sqrt(2) * V_s # in V\n", - "V_Ldc = (2*V_m)/pi # in V\n", - "print \"Average load output voltage = %0.2f V\" %V_Ldc\n", - "P_dc = (V_Ldc)**2/R_L # in W\n", - "print \"DC power delivered to load = %0.2f watt\" %P_dc\n", - "PIV = V_m # in V\n", - "print \"Peak inverse Voltage = %0.2f V\" %PIV\n", - "f_o = 2*f # in Hz\n", - "print \"Output frequency = %0.f Hz\" %f_o" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Average load output voltage = 49.52 V\n", - "DC power delivered to load = 2.45 watt\n", - "Peak inverse Voltage = 77.78 V\n", - "Output frequency = 100 Hz\n" - ] - } - ], - "prompt_number": 32 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.10\n", - ": Page No 145" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "R_L = 20 # in ohm\n", - "I_Ldc = 100 # in mA\n", - "R2 = 1 # in ohm\n", - "R_F = 0.5 # in ohm\n", - "I_m = (pi * I_Ldc)/2 # in mA\n", - "V_m = I_m*10**-3*(R2+R_F+R_L) # in V\n", - "V_srms = V_m/sqrt(2) # in V\n", - "print \"RMS value of secondary signal voltage = %0.1f V\" %V_srms\n", - "P_Ldc = (I_Ldc*10**-3)**2*R_L # in Watt\n", - "print \"power delivered to load = %0.1f Watt\" %P_Ldc\n", - "PIV = 2*V_m # in V\n", - "print \"Peal inverse voltage = %0.2f V\" %PIV\n", - "P_ac = (V_m)**2/(2*(R2+R_F+R_L)) # in Watt\n", - "print \"Input power = %0.3f Watt\" %P_ac\n", - "Eta = P_Ldc/P_ac*100 # in %\n", - "print \"Conversion efficiency = %0.2f %%\" %Eta" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "RMS value of secondary signal voltage = 2.4 V\n", - "power delivered to load = 0.2 Watt\n", - "Peal inverse voltage = 6.75 V\n", - "Input power = 0.265 Watt\n", - "Conversion efficiency = 75.40 %\n" - ] - } - ], - "prompt_number": 34 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.16\n", - ": Page No 151" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_dc = 12 # in V\n", - "R_L = 500 # in ohm\n", - "R_F = 25 # in ohm\n", - "I_dc = V_dc/R_L # in A\n", - "V_m = I_dc * pi * (R_L+R_F) # in V\n", - "V_rms = V_m/sqrt(2) # in V\n", - "V = V_rms # in V\n", - "print \"The voltage = %0.f V\" %round(V)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The voltage = 28 V\n" - ] - } - ], - "prompt_number": 35 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.17\n", - ": Page No 152" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_dc = 100 # in V\n", - "V_m = (V_dc*pi)/2 # in V\n", - "PIV = 2*V_m # in V\n", - "print \"Peak inverse voltage for center tapped FWR = %0.2f V\" %PIV\n", - "PIV1 = V_m # in V\n", - "print \"Peak inverse voltage for bridge type FWR = %0.2f V\" %PIV1" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Peak inverse voltage for center tapped FWR = 314.16 V\n", - "Peak inverse voltage for bridge type FWR = 157.08 V\n" - ] - } - ], - "prompt_number": 36 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.19\n", - ": Page No 153" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_Gamma = 0.7 # in V\n", - "R_f = 0 # in ohm\n", - "V_rms = 120 # in V\n", - "V_max = sqrt(2)*V_rms # in V\n", - "R_L = 1 # in K ohm\n", - "R_L = R_L * 10**3 # in ohm\n", - "I_max = (V_max - (2*V_Gamma))/R_L # in A\n", - "I_dc = (2*I_max)/pi # in mA\n", - "V_dc = I_dc * R_L # in V\n", - "print \"The dc voltage available at the load = %0.2f V\" %V_dc\n", - "PIV = V_max # in V\n", - "print \"Peak inverse voltage = %0.1f V\" %PIV\n", - "print \"Maximum current through diode = %0.1f mA\" %(I_max*10**3)\n", - "P_max = V_Gamma * I_max # in W\n", - "print \"Diode power rating = %0.2f mW\" %(P_max*10**3)" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The dc voltage available at the load = 107.15 V\n", - "Peak inverse voltage = 169.7 V\n", - "Maximum current through diode = 168.3 mA\n", - "Diode power rating = 117.81 mW\n" - ] - } - ], - "prompt_number": 38 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.20\n", - ": Page No 154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V1 = 10 # in V\n", - "V2 = 0.7 # in V\n", - "V3 = V2 # in V\n", - "V = V1-V2-V3 # in V\n", - "R1 = 1 # in ohm\n", - "R2 = 48 # in ohm\n", - "R3 = 1 # in ohm\n", - "R = R1+R2+R3 # in ohm\n", - "I = V/R # in A\n", - "I = I * 10**3 # in mA\n", - "print \"Current = %0.f mA\" %I" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Current = 172 mA\n" - ] - } - ], - "prompt_number": 39 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.21\n", - ": Page No 154" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_m = 50 # in V\n", - "r_f = 20 # in ohm\n", - "R_L = 800 # in ohm\n", - "I_m = V_m/(R_L+r_f) # in A\n", - "I_m = I_m * 10**3 # in mA\n", - "print \"The value of Im = %0.f mA\" %round(I_m)\n", - "I_dc = I_m/pi # in mA\n", - "print \"The value of I_dc = %0.1f mA\" %I_dc\n", - "I_rms = I_m/2 # in mA\n", - "print \"The value of Irms = %0.1f mA\" %I_rms\n", - "P_ac = (I_rms * 10**-3)**2 * (r_f + R_L) # in Watt\n", - "print \"AC power input = %0.3f Watt\" %P_ac\n", - "V_dc = I_dc * 10**-3*R_L # in V\n", - "print \"DC output voltage = %0.2f V\" %V_dc\n", - "P_dc = (I_dc * 10**-3)**2 * (r_f + R_L) # in Watt\n", - "Eta = (P_dc/P_ac)*100 # in %\n", - "print \"The efficiency of rectification = %0.1f %%\" %Eta\n", - "\n", - "# Note: There is calculation error to evaluate the ac power input (i.e. P_ac), so the value of Eta is also wrong" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The value of Im = 61 mA\n", - "The value of I_dc = 19.4 mA\n", - "The value of Irms = 30.5 mA\n", - "AC power input = 0.762 Watt\n", - "DC output voltage = 15.53 V\n", - "The efficiency of rectification = 40.5 %\n" - ] - } - ], - "prompt_number": 41 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.22\n", - ": Page No 155" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "R_L = 1 # in K ohm\n", - "R_L = R_L * 10**3 # in ohm\n", - "r_d = 10 # in ohm\n", - "V_m = 220 # in V\n", - "I_m = V_m/(r_d+R_L) # in A\n", - "print \"Peak value of current = %0.2f A\" %I_m\n", - "I_dc = (2*I_m)/pi # in A\n", - "print \"DC value of current = %0.2f A\" %I_dc\n", - "Irms= I_m/sqrt(2) # in A\n", - "r_f = sqrt((Irms/I_dc)**2-1)*100 # in %\n", - "print \"Ripple factor = %0.1f %%\" %r_f\n", - "Eta = (I_dc)**2 * R_L/((Irms)**2*(R_L+r_d))*100 # in %\n", - "print \"Rectification efficiency = %0.1f %%\" %Eta" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Peak value of current = 0.22 A\n", - "DC value of current = 0.14 A\n", - "Ripple factor = 48.3 %\n", - "Rectification efficiency = 80.3 %\n" - ] - } - ], - "prompt_number": 44 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.23\n", - ": Page No 156" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_s = 12 # in V\n", - "R_L = 5.1 # in k ohm\n", - "R_L = R_L * 10**3 # in ohm\n", - "R_s = 1 # in K ohm\n", - "R_s = R_s * 10**3 # in ohm\n", - "V_L = (V_s*R_L)/(R_s+R_L) # in V\n", - "print \"Peak load voltage = %0.f V\" %V_L" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "Peak load voltage = 10 V\n" - ] - } - ], - "prompt_number": 45 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.24\n", - ": Page No 157" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_s = 10 # in V\n", - "R_L = 100 # in ohm\n", - "I_L = V_s/R_L # in A\n", - "print \"The load current during posotive half cycle = %0.1f A\" %I_L\n", - "I_D2 = 0 # in A\n", - "R2 = R_L # in ohm\n", - "I_L1 = -(V_s)/(R2+R_L) # in A\n", - "print \"The load current during negative half cycle = %0.2f A\" %I_L1" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The load current during posotive half cycle = 0.1 A\n", - "The load current during negative half cycle = -0.05 A\n" - ] - } - ], - "prompt_number": 46 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.25\n", - ": Page No 158" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "V_m = 50 # in V\n", - "V_dc = (2*V_m)/pi # in V\n", - "print \"The dc voltage = %0.2f V\" %V_dc" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The dc voltage = 31.83 V\n" - ] - } - ], - "prompt_number": 47 - }, - { - "cell_type": "heading", - "level": 2, - "metadata": {}, - "source": [ - "Example 2.26\n", - ": Page No 159" - ] - }, - { - "cell_type": "code", - "collapsed": false, - "input": [ - "# Given data\n", - "R1 = 1.1 # in K ohm\n", - "R2 = 2.2 # in K ohm\n", - "Vi= 170 # in V\n", - "V_o = (Vi*R1)/(R1+R2) # in V\n", - "print \"The output voltage = %0.2f V\" %V_o\n", - "V_dc = (2*V_o)/pi # in V\n", - "print \"The dc voltage = %0.2f V\" %V_dc" - ], - "language": "python", - "metadata": {}, - "outputs": [ - { - "output_type": "stream", - "stream": "stdout", - "text": [ - "The output voltage = 56.67 V\n", - "The dc voltage = 36.08 V\n" - ] - } - ], - "prompt_number": 48 - } - ], - "metadata": {} - } - ] -}
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