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| -rw-r--r-- | Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb | 183 | ||||
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diff --git a/Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb new file mode 100644 index 00000000..18461132 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_1_2.ipynb @@ -0,0 +1,1165 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 1 : Introduction To Electronics Diode Fundamentals" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.1\n", + ": Page No 27 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "T1 = 25 # in degree C\n", + "T2 = 100 # in degree C\n", + "del_T = T2-T1 # in degree C\n", + "V= 0.7 # barrier potential t 25\u00b0C in V\n", + "del_V = -(2)*del_T # in mV\n", + "del_V= del_V*10**-3 # in V\n", + "V_B = V- abs(del_V) # in V\n", + "print \"(i) When the junction temperature is 100 \u00b0C, the barrier potential of a silicon diode = %0.2f V\" %V_B\n", + "T2 = 0 # in degree C\n", + "del_T = T2-T1 # in degree C\n", + "del_V = -(2)*del_T # in mV\n", + "del_V= del_V*10**-3 #in V\n", + "V_B = V+del_V # in V\n", + "print \"(ii) When the junction temperature is 0 \u00b0C, the barrier potential of a silicon diode = %0.2f V\" %V_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "(i) When the junction temperature is 100 \u00b0C, the barrier potential of a silicon diode = 0.55 V\n", + "(ii) When the junction temperature is 0 \u00b0C, the barrier potential of a silicon diode = 0.75 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.2\n", + ": Page No 30" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "T1 = 25 # in degree C\n", + "T2 = 100 # in degree C\n", + "del_T = T2-T1 # in degree C\n", + "I_S = (2)**7 *5 # in nA\n", + "I_S = (1.07)**5*I_S # in nA\n", + "print \"The saturation current at 100 degree C = %0.f nA\" %round(I_S)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The saturation current at 100 degree C = 898 nA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.3\n", + ": Page No 33" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_L = 10 # in V\n", + "R_L = 1*10**3 # in \u03a9\n", + "I_L = V_L/R_L # in A\n", + "I_L = I_L*10**3 # mA\n", + "print \"The load voltage = %0.f volts\" %V_L\n", + "print \"The load current = %0.f mA\" %I_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load voltage = 10 volts\n", + "The load current = 10 mA\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.4\n", + ": Page No 37" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "v1 = 10 # in V\n", + "v2 = 0.7 # in V\n", + "V_L = v1-v2 # in V\n", + "print \"The load voltage = %0.1f V\" %V_L\n", + "R_L = 1*10**3 # in \u03a9\n", + "I_L = V_L/R_L # in A\n", + "print \"The load current = %0.1f mA\" %(I_L*10**3)\n", + "P_D = v2*I_L # in watt\n", + "print \"The diode Power = %0.2f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load voltage = 9.3 V\n", + "The load current = 9.3 mA\n", + "The diode Power = 6.51 mW\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.5\n", + ": Page No 39" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L1 = 1*10**3 # in ohm\n", + "R_L2 = 0.23 # in ohm\n", + "R_T = R_L1+R_L2 # in ohm\n", + "v1 = 10 # in V\n", + "v2 = 0.7 # in V\n", + "V_T = v1-v2 # in V\n", + "I_L = V_T/R_T # in A\n", + "print \"The load current = %0.2f mA\" %(I_L*10**3)\n", + "V_L = I_L*R_L1 # in V\n", + "print \"The load voltage = %0.1f V\" %V_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load current = 9.30 mA\n", + "The load voltage = 9.3 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.6\n", + ": Page No 45" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_o = 0.7 # in V\n", + "print \"The value of V_o = %0.1f V\" %V_o\n", + "E = 10 # in V\n", + "V_D = V_o # in V\n", + "R = 330 # in ohm\n", + "I1 = (E - V_D)/R # in A\n", + "I1 = I1*10**3 # in mA\n", + "print \"The value of I1 = %0.2f mA\" %I1\n", + "I_D1 = I1/2 # in mA\n", + "print \"The value of I_D1 = %0.2f mA\" %I_D1\n", + "I_D2 = I_D1 # in mA\n", + "print \"The value of I_D2 = %0.2f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_o = 0.7 V\n", + "The value of I1 = 28.18 mA\n", + "The value of I_D1 = 14.09 mA\n", + "The value of I_D2 = 14.09 mA\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.7\n", + ": Page No 46" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_i = 12 # in V\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.3 # in V\n", + "R = 5.6*10**3 # in ohm\n", + "V_o = V_i - V_D1 - V_D2 # in V\n", + "print \"The value of Vo voltage = %0.f V\" %V_o\n", + "I_D = V_o/R # in A\n", + "I_D = I_D*10**3 # in mA\n", + "print \"The value of I_D = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Vo voltage = 11 V\n", + "The value of I_D = 1.96 mA\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.8\n", + ": Page No 47" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 24 # in V\n", + "V2 = 6 # in V\n", + "V_D1 = 0.7 # in V\n", + "R = 3*10**3 # in ohm\n", + "I = (V1 - V2 - V_D1)/R # in A\n", + "I = I * 10**3 # in mA\n", + "print \"The current = %0.2f mA\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current = 5.77 mA\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.9\n", + ": Page No 48" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r= 20 # in \u03a9\n", + "R_B= 15 # in \u03a9\n", + "V_K1= 0.2 # in V\n", + "V_K2= 0.6 # in V\n", + "V= 100 # in V\n", + "R1= 10*10**3 # in \u03a9\n", + "# Vo= V_K1+r*I1 = V_K2+R_B*I2\n", + "# Resulting current I= I1+I2 or\n", + "# (V-Vo)/(R1) = (Vo-V_K1)/r + (Vo-V_K2)/R_B\n", + "Vo= (r*R_B*V+R1*R_B*V_K1+R1*r*V_K2)/(R1*R_B+R1*r+r*R_B) # in V\n", + "I1= (Vo-V_K1)/r # in A\n", + "I2= (V_K2-Vo)/R_B # in A\n", + "print \"The value of I1 = %0.2f mA\" %(I1*10**3)\n", + "print \"The value of I2 = %0.2f mA\" %(I2*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I1 = 15.69 mA\n", + "The value of I2 = 5.74 mA\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.10\n", + ": Page No 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 10 # in mA\n", + "I_D = I_D * 10**-3 # in A\n", + "V_D = 0.5 # in V\n", + "r_F1 = V_D/I_D # in ohm\n", + "print \"The value of r_F1 = %0.f ohm\" %r_F1\n", + "I_D = 20 # in mA\n", + "I_D = I_D * 10**-3 # in A\n", + "V_D = 0.8 # in V\n", + "r_F2 = V_D/I_D # in ohm\n", + "print \"The value of r_F2 = %0.f ohm\" %r_F2\n", + "I_D = -1 # in \u00b5A\n", + "I_D = I_D * 10**-6 # in A\n", + "V_D = -10 # in V \n", + "r_R = V_D/I_D # in ohm\n", + "print \"The value of r_R = %0.f Mohm\" %(r_R*10**-6)\n", + "\n", + "# Note: There is calculation error to evaluate the value of r_F1. So the asnwer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of r_F1 = 50 ohm\n", + "The value of r_F2 = 40 ohm\n", + "The value of r_R = 10 Mohm\n" + ] + } + ], + "prompt_number": 15 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.11\n", + ": Page No 49" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R= 5.6*10**3 # in \u03a9\n", + "I_D = 0 # in A\n", + "V_D = 0 # in V\n", + "E= 12 # in V\n", + "Vo= I_D*R # in V\n", + "print \"The value of I_D = %0.f A\" %I_D\n", + "print \"The value of Vo = %0.f V\" %Vo\n", + "V_D1 = 0 # in V\n", + "V_D2 = E -V_D1 - Vo # in V\n", + "print \"The value of V_D2 = %0.f V\" %V_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_D = 0 A\n", + "The value of Vo = 0 V\n", + "The value of V_D2 = 12 V\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.12\n", + ": Page No 50" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "E = 20 # in V\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.7 # in V\n", + "V2 = E - V_D1 - V_D2 # in V\n", + "R1 = 3.3*10**3 # in ohm\n", + "R2 = 5.6*10**3 # in ohm\n", + "I2 = V2/R2 # in A\n", + "I2 = I2*10**3 # in mA\n", + "print \"The current through resistor R2 = %0.2f mA\" %I2\n", + "I1 = V_D2/R1 \n", + "I1 = I1 * 10**3 # in mA\n", + "print \"The current through resistor R1 = %0.2f mA\" %I1\n", + "I_D2 = I2-I1 # in mA\n", + "print \"The current through diode D2 = %0.2f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through resistor R2 = 3.32 mA\n", + "The current through resistor R1 = 0.21 mA\n", + "The current through diode D2 = 3.11 mA\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.13\n", + ": Page No 51" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 12 # in V\n", + "V2 = 0.3 # in V\n", + "V_o = V1-V2 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 11.7 V\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.14\n", + ": Page No 52" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "print \"Part (a) Analysis using approximate diode model\"\n", + "V_D = 0.7 # in V\n", + "print \"The value of V_D = %0.1f V\" %V_D\n", + "E = 30 # in V\n", + "V_R = E-V_D # in V\n", + "print \"The value of V_R = %0.1f V\" %V_R\n", + "R = 2.2 * 10**3 # in ohm\n", + "I_D = V_R/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The value of I_D = %0.2f mA\" %I_D\n", + "print \"Part (b) Analysis using ideal diode model\"\n", + "V_D = 0 # in V\n", + "print \"The value of V_D = %0.f V\" %V_D\n", + "V_R = E # in V\n", + "print \"The value of V_R = %0.f V\" %V_R\n", + "I_D = V_R/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The value of I_D = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) Analysis using approximate diode model\n", + "The value of V_D = 0.7 V\n", + "The value of V_R = 29.3 V\n", + "The value of I_D = 13.32 mA\n", + "Part (b) Analysis using ideal diode model\n", + "The value of V_D = 0 V\n", + "The value of V_R = 30 V\n", + "The value of I_D = 13.64 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.15\n", + ": Page No 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 20 # in V\n", + "V2 = 0.7 # in V\n", + "V = V1-V2 # in V\n", + "R = 20 # in ohm\n", + "I = V/R # in A\n", + "print \"The current through resistance = %0.3f A\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current through resistance = 0.965 A\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.16\n", + ": Page No 54" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1= 2 # in k\u03a9\n", + "R2= 2 # in k\u03a9\n", + "V=19 # in V\n", + "V_o = (V*R1)/(R1+R2) # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 9.5 V\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.17\n", + ": Page No 55" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 0.7 # in V\n", + "V2 = 5 # in V\n", + "V_o = V1-V2 # in V\n", + "R = 2.2*10**3 # in ohm\n", + "I_D = -V_o/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The output voltage = %0.1f volts\" %V_o\n", + "print \"The current through diode = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -4.3 volts\n", + "The current through diode = 1.95 mA\n" + ] + } + ], + "prompt_number": 24 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.18\n", + ": Page No 56" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_gamma = 0.7 # in V\n", + "R1 = 5*10**3 # in ohm\n", + "R2 = 10*10**3 # in ohm\n", + "V=5 # in V\n", + "print \"Part (a)\"\n", + "I_R2 = (V-V_gamma-(-V))/(R1+R2) # in A\n", + "I_D2 = I_R2 # in A\n", + "print \"The value of I_D1 and I_D2 = %0.2f mA\" %(I_D2*10**3)\n", + "V_o = V - (I_D2 * R1) # in V\n", + "print \"The value of Vo = %0.1f V\" %V_o\n", + "V_A = V_o - V_gamma # in V\n", + "print \"The value of V_A = %0.1f V\" %V_A\n", + "print \"Part (b)\"\n", + "V_I = 4 # in V\n", + "V_A= V_I-V_gamma # in V\n", + "Vo= V_A+V_gamma # in V\n", + "I_R1= (V-Vo)/R1 # in A\n", + "I_D2= I_R1 # in A\n", + "print \"The value of I_D2 = %0.1f mA\" %(I_D2*10**3)\n", + "I_R2= (V_A-(-V))/R2 # in A\n", + "I_D1= I_R2-I_R1 # in A\n", + "print \"The value of I_D1 = %0.2f mA\" %(I_D1*10**3)\n", + "print \"The value of Vo = %0.f volts\" %Vo" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The value of I_D1 and I_D2 = 0.62 mA\n", + "The value of Vo = 1.9 V\n", + "The value of V_A = 1.2 V\n", + "Part (b)\n", + "The value of I_D2 = 0.2 mA\n", + "The value of I_D1 = 0.63 mA\n", + "The value of Vo = 4 volts\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.19\n", + ": Page No 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 6 # in V\n", + "V_D = 0.7 # in V\n", + "R = 10 # in K ohm\n", + "R = R*10**3 # in ohm\n", + "I_T = (V_S-V_D)/R # in A\n", + "print \"The total current = %0.f \u00b5A\" %(I_T*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total current = 530 \u00b5A\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.20\n", + ": Page No 58" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 5 # in V\n", + "V_D = 0.7 # in V\n", + "R1 = 1.2 * 10**3 # in ohm\n", + "R2 = 2.2 * 10**3 # in ohm\n", + "I_T = (V_S-V_D)/(R1+R2) \n", + "I_T = I_T * 10**3 # in mA\n", + "print \"The total circuit current = %0.2f mA\" %I_T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total circuit current = 1.26 mA\n" + ] + } + ], + "prompt_number": 32 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.21\n", + ": Page No 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 4 # in V\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.7 # in V\n", + "R = 5.1*10**3 # in ohm\n", + "I_T = (V_S-V_D1-V_D2)/R # in A\n", + "print \"The total current = %0.f \u00b5A\" %round(I_T*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The total current = 510 \u00b5A\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.22\n", + ": Page No 59" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 10 # in V\n", + "R1 = 1.5*10**3 # in ohm\n", + "R2 = 1.8*10**3 # in ohm\n", + "I_T = V_S/(R1+R2) # in A\n", + "print \"Using the ideal diode, the total current = %0.2f mA\" %(I_T*10**3)\n", + "V_D1 = 0.7 # in V\n", + "V_D2 = 0.7 # in V\n", + "I_T = (V_S-V_D1-V_D2)/(R1+R2) # in A\n", + "print \"Using the pracitcal diode, the total current = %0.2f mA\" %(I_T*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Using the ideal diode, the total current = 3.03 mA\n", + "Using the pracitcal diode, the total current = 2.61 mA\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.23\n", + ": Page No 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 5 # in V\n", + "V2 = 3 # in V\n", + "R = 500 # in ohm\n", + "I_D2 = (V_S-V2)/R # in A\n", + "I_D2 = I_D2 * 10**3 # in mA\n", + "print \"The diode current = %0.f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode current = 4 mA\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.24\n", + ": Page No 60" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 2 # in V\n", + "R = 100 # in ohm\n", + "I_D = V_S/R \n", + "I_D = I_D * 10**3 # in mA\n", + "print \"Part (a)\"\n", + "print \"The diode current = %0.f mA\" %I_D\n", + "V_K = 0.7 # in V\n", + "I_D1 = (V_S-V_K)/R \n", + "I_D1 = I_D1*10**3 # in mA\n", + "print \"Part (b)\"\n", + "print \"The diode current = %0.f mA\" %I_D1\n", + "R_f = 30 # in ohm\n", + "I_D2 = (V_S - V_K)/(R+R_f) \n", + "I_D2 = I_D2 * 10**3 # in mA\n", + "print \"Part (c)\"\n", + "print \"The diode current = %0.f mA\" %I_D2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The diode current = 20 mA\n", + "Part (b)\n", + "The diode current = 13 mA\n", + "Part (c)\n", + "The diode current = 10 mA\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.25\n", + ": Page No 61" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1= 1 # in k\u03a9\n", + "R2= 0.47 # in k\u03a9\n", + "V_o1 = 0.7 # in V\n", + "print \"The value of Vo1 = %0.1f V\" %V_o1\n", + "V_o2 = 0.3 # in V\n", + "print \"The value of Vo2 = %0.1f V\" %V_o2\n", + "I1 = (20-V_o1)/R1 # in mA\n", + "I2 = (V_o2-V_o1)/R2 # in mA\n", + "I = I1 + I2 # in mA\n", + "print \"The current = %0.2f mA\" %I" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of Vo1 = 0.7 V\n", + "The value of Vo2 = 0.3 V\n", + "The current = 18.45 mA\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.26\n", + ": Page No 62" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 10 # in V\n", + "V2 = 0.7 # in V\n", + "R1 = 1*10**3 # in ohm\n", + "R2 = 2*10**3 # in ohm\n", + "I = (V1-V2)/(R1+R2) # in A\n", + "V_o = I * R2 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o\n", + "I_D = I/2 # in A\n", + "print \"The diode current = %0.2f mA\" %(I_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 6.2 V\n", + "The diode current = 1.55 mA\n" + ] + } + ], + "prompt_number": 42 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.27\n", + ": Page No 63" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 20 # in V\n", + "V2 = 0.7 # in V\n", + "R = 4.7*10**3 # in ohm\n", + "I = (V1-V2)/R # in A\n", + "I_D = I/2 # in A\n", + "print \"The diode current = %0.2f mA\" %(I_D*10**3)\n", + "V_o = I_D*R # in V\n", + "print \"The output voltage = %0.2f V\" %V_o\n", + "#Note : The answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode current = 2.05 mA\n", + "The output voltage = 9.65 V\n" + ] + } + ], + "prompt_number": 48 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.28\n", + ": Page No 64" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 15 # in V\n", + "V2 = 0.7 # in V\n", + "V3 = 5 # in V\n", + "R = 2.2 # in K ohm\n", + "I_D = (V1-V2+V3)/R # in mA\n", + "print \"The diode current = %0.2f mA\" %I_D\n", + "V_o = (R * I_D) - V3 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The diode current = 8.77 mA\n", + "The output voltage = 14.3 V\n" + ] + } + ], + "prompt_number": 50 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 1.29\n", + ": Page No 65" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 16 # in V\n", + "V2 = 0.7 # in V\n", + "V3 = V2 # in V\n", + "V4 = 12 # in V\n", + "R = 4.7 # in K ohm\n", + "I = (V1-V2-V3-V4)/R # in mA\n", + "print \"The current = %0.3f mA\" %I\n", + "V_o = (I * 10**-3 * R * 10**3) + V4 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current = 0.553 mA\n", + "The output voltage = 14.6 V\n" + ] + } + ], + "prompt_number": 51 + } + ], + "metadata": {} + } + ] +} diff --git a/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb new file mode 100644 index 00000000..cc8b5565 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_2_2.ipynb @@ -0,0 +1,765 @@ +{ + "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": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb new file mode 100644 index 00000000..600220b3 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_3_2.ipynb @@ -0,0 +1,710 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 3 : Special-Purpose Diode" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.1\n", + ": Page No 179" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V1 = 18 # in V\n", + "V2 = 10 # in V\n", + "R = 270 # in ohm\n", + "I_S = (V1-V2)/R # in A\n", + "V_L = 10 # in V\n", + "R_L = 1 # in K ohm\n", + "R_L = R_L*1000 # in ohm\n", + "I_L = V_L/R_L # in A\n", + "I_Z = I_S-I_L # in A\n", + "print \"The zener current = %0.1f mA\" %(I_Z*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The zener current = 19.6 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.5\n", + ": Page No 186" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_Z = 2*10**-3 # in A\n", + "R_Z = 8.5 # in V\n", + "del_VL = I_Z*R_Z # in V\n", + "V1 = 10 # in V\n", + "print \"Change in load voltage = %0.2f V\" %del_VL\n", + "V_L = V1 + del_VL # in V\n", + "print \"The load voltage = %0.2f V\" %V_L\n", + "\n", + "# Note: There is calculation error to evaluate the value of del_VL. So the answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Change in load voltage = 0.02 V\n", + "The load voltage = 10.02 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.6\n", + ": Page No 194" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 1.2 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "V_i = 16 # in V\n", + "R_i = 1 # in K ohm\n", + "R_i = R_i * 10**3 # in ohm\n", + "V = (R_L * V_i)/(R_L + R_i) # in V\n", + "V_L = V # in V\n", + "print \"The load voltage = %0.2f V\" %V_L\n", + "V_R = V_i - V_L # in V\n", + "print \"The voltage = %0.2f V\" %V_R\n", + "I_Z = 0 # A\n", + "print \"The zener diode current = %0.f A\" %I_Z\n", + "V_Z = 10 # in V\n", + "P_Z = V_Z*I_Z # in W\n", + "print \"Power dissipation = %0.f watt\" %P_Z" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The load voltage = 8.73 V\n", + "The voltage = 7.27 V\n", + "The zener diode current = 0 A\n", + "Power dissipation = 0 watt\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.7\n", + ": Page No 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_Z1 = 20 # in mA\n", + "I_Z1= I_Z1*10**-3 # in A\n", + "I_Z2 = 30 # in mA\n", + "I_Z2= I_Z2*10**-3 # in A\n", + "V_Z1 = 5.6 # in V\n", + "V_Z2 = 5.75 # in V\n", + "del_IZ = I_Z2-I_Z1 # in A\n", + "del_VZ = V_Z2-V_Z1 # in V\n", + "r_Z = del_VZ/del_IZ # in ohm\n", + "print \"Resistance of zener diode = %0.f ohm\" %r_Z" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Resistance of zener diode = 15 ohm\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.8\n", + ": Page No 195" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R = 1 # in K ohm\n", + "R = R * 10**3 # in ohm\n", + "V_Z = 10 # in V\n", + "V_i = 50 # in V\n", + "I_ZM = 32 # in mA\n", + "I_ZM= I_ZM*10**-3 # in A\n", + "R_Lmin = (R*V_Z)/(V_i-V_Z) # in ohm\n", + "print \"The minimum value of R_L = %0.f ohm\" %R_Lmin\n", + "V_R = V_i-V_Z # in V\n", + "I_R = V_R/R # in A\n", + "I_Lmin = I_R-I_ZM # in A\n", + "print \"The minimum value of I_L = %0.f mA\" %(I_Lmin*10**3)\n", + "R_Lmax = V_Z/I_Lmin # in ohm\n", + "print \"The maximum value of R_L = %0.2f kohm\" %(R_Lmax*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of R_L = 250 ohm\n", + "The minimum value of I_L = 8 mA\n", + "The maximum value of R_L = 1.25 kohm\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.9\n", + ": Page No 196" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_Z = 20 # in V\n", + "R_L = 1.2 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "R = 220 # in ohm\n", + "I_ZM = 60 # in mA\n", + "I_ZM= I_ZM*10**-3 # in A\n", + "Vi_min = (R_L + R)/R_L*V_Z # in V\n", + "print \"The minimum value of Vi = %0.2f V\" %Vi_min\n", + "V_L= V_Z # in V\n", + "I_L= V_L/R_L # in A\n", + "Vi_max= (I_ZM+I_L)*R+V_Z # in V\n", + "print \"The maximum value of Vi = %0.2f V\" %Vi_max" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum value of Vi = 23.67 V\n", + "The maximum value of Vi = 36.87 V\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.10\n", + ": Page No 197" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 18 # in V\n", + "V2 = 270 # in V\n", + "R = 1 # in K ohm\n", + "R = R*1000 # in ohm\n", + "V = (V1*R)/(V2+R) # in V\n", + "print \"The open circuit voltage = %0.1f volts\" %V\n", + "if V>=10 :\n", + " print \"The zener diode is operating in the breakdown region.\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The open circuit voltage = 14.2 volts\n", + "The zener diode is operating in the breakdown region.\n" + ] + } + ], + "prompt_number": 11 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.11\n", + ": Page No 198" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 300 # in ohm\n", + "R = 200 # in ohm\n", + "V_i = 20 # in V\n", + "V = (R_L/(R_L+R))*V_i # in V\n", + "print \"The value of V_L = %0.f Volts\" %V\n", + "V_L = 10 # in V\n", + "V_Z= V_L # in V\n", + "I_L = V_L/R_L # A\n", + "print \"The value of I_L = %0.2f mA\" %(I_L*10**3)\n", + "I_R = (V_i-V_L)/R # in A\n", + "print \"The value of I_R = %0.f mA\" %(I_R*10**3)\n", + "I_Z = I_R-I_L # in A\n", + "print \"The value of I_Z = %0.2f mA\" %(I_Z*10**3)\n", + "# Formula V_Z= R_L*V_i/(R_L+R)\n", + "R_L= R*V_Z/(V_i-V_Z) # in ohm\n", + "print \"The value of R_L = %0.f ohm\" %R_L" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_L = 12 Volts\n", + "The value of I_L = 33.33 mA\n", + "The value of I_R = 50 mA\n", + "The value of I_Z = 16.67 mA\n", + "The value of R_L = 200 ohm\n" + ] + } + ], + "prompt_number": 13 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.12\n", + ": Page No 199" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_Z = 5 # in V\n", + "I_Zmin = 2 # in mA\n", + "I_Zmin= I_Zmin*10**-3 # in A\n", + "I_Zmax = 20 # in mA\n", + "I_Zmax=I_Zmax*10**-3 # in A\n", + "R_L = 1 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "I_L = V_Z/R_L # in A\n", + "I = I_L + I_Zmin # in A\n", + "Vin_min = V_Z + (I*R_L) # in V\n", + "print \"The minimum input voltage = %0.f V\" %Vin_min\n", + "I = I_L + I_Zmax # in A\n", + "Vin_max = V_Z + I* R_L # in V\n", + "print \"The maximum input voltage = %0.f V\" %Vin_max" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum input voltage = 12 V\n", + "The maximum input voltage = 30 V\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.13\n", + ": Page No 200" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_in1 = 18 # in V\n", + "V_in2 = 22 # in V\n", + "V_o = 6 # in V\n", + "I_L = 50 # in mA\n", + "I_L= I_L*10**-3 # in A\n", + "I_Zmin = 5 # in mA\n", + "I_Zmin= I_Zmin*10**-3 # in A\n", + "P_Z = 0.5 # in Watt\n", + "V_Z= 6 # in V\n", + "I_Zmax = P_Z/V_Z # in A\n", + "print \"Zener diode current = %0.2f mA\" %(I_Zmax*10**3)\n", + "R_S1 = (V_in2 - V_Z)/(I_L+I_Zmax) # in ohm\n", + "print \"The minimum value of Rs = %0.f ohm\" %R_S1\n", + "R_S2 = (V_in1-V_Z)/(I_L+I_Zmin) # in ohm\n", + "print \"The maximum value of Rs = %0.1f ohm\" %R_S2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Zener diode current = 83.33 mA\n", + "The minimum value of Rs = 120 ohm\n", + "The maximum value of Rs = 218.2 ohm\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.14\n", + ": Page No 201" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_S = 91 # in ohm\n", + "V_Z = 8 # in V\n", + "P_Z = 400 # in mW\n", + "P_Z= P_Z*10**-3 # in W\n", + "R_L = 0.22 # in K ohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "I_L = V_Z/R_L # in A\n", + "I_Z = P_Z/V_Z # in A\n", + "print \"The value of I_Zmax = %0.f mA\" %(I_Z*10**3)\n", + "Vin_min = (V_Z*(R_S+R_L))/R_L # in V\n", + "print \"The minimum input voltage = %0.2f V\" %Vin_min\n", + "I_R = I_L + I_Z # in A\n", + "Vin_max = V_Z + (I_R*R_S) # in V\n", + "print \"The maximum input voltage =%0.2f V\" %Vin_max" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_Zmax = 50 mA\n", + "The minimum input voltage = 11.31 V\n", + "The maximum input voltage =15.86 V\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.15\n", + ": Page No 202" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_L = 12 # in V\n", + "I_Lmin = 0 # in mA\n", + "I_Lmin =I_Lmin *10**-3 # in A\n", + "I_Lmax = 200 # in mA\n", + "I_Lmax =I_Lmax *10**-3 # in A\n", + "I_Zmin = 5 # in mA\n", + "I_Zmin= I_Zmin*10**-3 # in A\n", + "I_Zmax = 200 # in mA\n", + "I_Zmax= I_Zmax*10**-3 # in A\n", + "V_i = 16 # in V\n", + "V_Z = V_L # in V\n", + "print \"The value of V_Z = %0.f V\" %V_Z\n", + "R_Lmin = V_L/I_Lmax # in ohm\n", + "print \"The minimum value of R_L = %0.f ohm\" %R_Lmin\n", + "# R_L2 = V_L/I_Lmin # in ohm\n", + "print \"The maximum value of R_L = infinite\" \n", + "I_Z = I_Zmin+I_Zmax # in A\n", + "print \"The zener current = %0.f mA\" %(I_Z*10**3)\n", + "P_Zmax = V_Z*I_Z # in Watt\n", + "print \"The maximum value of Pz = %0.2f Watt\" %P_Zmax\n", + "R_S = (V_i-V_L)/(I_Zmin+I_Lmax) # in ohm\n", + "print \"The value of R_S = %0.2f ohm\" %R_S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_Z = 12 V\n", + "The minimum value of R_L = 60 ohm\n", + "The maximum value of R_L = infinite\n", + "The zener current = 205 mA\n", + "The maximum value of Pz = 2.46 Watt\n", + "The value of R_S = 19.51 ohm\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.16\n", + ": Page No 203" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_in = 20 # in V\n", + "R_S = 220 # in ohm\n", + "V_Z = 10 # in V\n", + "P_Z = 400 # in mW\n", + "# Part (I)\n", + "R_L = 200 # in ohm\n", + "print \"Part (I) For R_L= 200 \u03a9\"\n", + "V_L = V_Z # in V\n", + "print \"Load voltage = %0.f V\" %V_L\n", + "I_L = V_L/R_L # in A\n", + "print \"Load current = %0.2f A\" %I_L\n", + "I_R = (V_in-V_Z)/R_S # in A\n", + "print \"The current through resistor = %0.3f A\" %I_R\n", + "I_Z = I_R-I_L # in A\n", + "print \"The value of I_Z = %0.2e A\" %I_Z\n", + "# Part (II)\n", + "R_L = 50 # in ohm\n", + "print \"Part (II) For R_L= 50 \u03a9\"\n", + "V_L = V_Z #\n", + "print \"Load voltage = %0.f V\" %V_L\n", + "I_L = V_L/R_L # in A\n", + "print \"Load current = %0.1f A\" %I_L\n", + "I_R = (V_in-V_Z)/R_S # in A\n", + "print \"The current through resistor = %0.3f A\" %I_R\n", + "I_Z = I_R-I_L # in A\n", + "print \"Zener current = %0.3f A\" %I_Z\n", + "print \"For both values of R_L, the current I_R is less than I_L and I_Z is negative.\"\n", + "print \"It shows that given circuit can not work successfully as a voltage regulator\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (I) For R_L= 200 \u03a9\n", + "Load voltage = 10 V\n", + "Load current = 0.05 A\n", + "The current through resistor = 0.045 A\n", + "The value of I_Z = -4.55e-03 A\n", + "Part (II) For R_L= 50 \u03a9\n", + "Load voltage = 10 V\n", + "Load current = 0.2 A\n", + "The current through resistor = 0.045 A\n", + "Zener current = -0.155 A\n", + "For both values of R_L, the current I_R is less than I_L and I_Z is negative.\n", + "It shows that given circuit can not work successfully as a voltage regulator\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.17\n", + ": Page No 204" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_Zmin = 1 # in mA\n", + "I_Zmin=I_Zmin*10**-3 # in A\n", + "I_Zmax = 5 # in mA\n", + "I_Zmax=I_Zmax*10**-3 # in A\n", + "I_Lmin = 0 # in mA\n", + "I_Lmin=I_Lmin*10**-3 # in A\n", + "I_Lmax = 4 # in mA\n", + "I_Lmax=I_Lmax*10**-3 # in A\n", + "R = 5 # in kohm\n", + "R = R * 10**3 # in ohm\n", + "V_Z = 50 # in V\n", + "print \"Part (A)\"\n", + "V_max = (I_Zmax+ I_Lmin)*R+V_Z # in V\n", + "print \"The maximum Voltage = %0.f V\" %V_max\n", + "V_min = (I_Zmin+I_Lmax)*R + V_Z # in V\n", + "print \"The minimum Voltage = %0.f V\" %V_min\n", + "print \"Part (B)\"\n", + "V_L = 50 # in V\n", + "V_in = 75 # in V\n", + "R_L = 15 # in kohm\n", + "R_L= R_L*10**3 # in ohm\n", + "I_L = V_L/R_L # in A\n", + "V_max = (I_Zmax+I_L)*R+V_Z # in V\n", + "print \"The maximum Voltage = %0.f V\" %round(V_max)\n", + "V_min = (I_Zmin+I_L)*R+V_Z # in V\n", + "print \"The minimum Voltage = %0.f V\" %round(V_min)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (A)\n", + "The maximum Voltage = 75 V\n", + "The minimum Voltage = 75 V\n", + "Part (B)\n", + "The maximum Voltage = 92 V\n", + "The minimum Voltage = 72 V\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 3.18\n", + ": Page No 205" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 7.5 # in V\n", + "V_Z = 5 # in V\n", + "R_S = 4.75 # in ohm\n", + "I_Zmin= 0.05 # in A\n", + "I_Zmax=1.0 # in A\n", + "I_S = (V_S-V_Z)/R_S # in A\n", + "I_Lmax= I_S-I_Zmin # in A\n", + "print \"The maximum value of load current = %0.3f A\" %I_Lmax\n", + "# when\n", + "V_S= 10 # in V\n", + "I_S = (V_S-V_Z)/R_S # in A\n", + "I_Lmin= I_S-I_Zmax # in A\n", + "print \"The minimum value of load current = %0.2f A\" %I_Lmin\n", + "print \"Thus, the range of load current for regulation =\",round(I_Lmin,2),\"<= I_L <=\",round(I_Lmax,3),\"A\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum value of load current = 0.476 A\n", + "The minimum value of load current = 0.05 A\n", + "Thus, the range of load current for regulation = 0.05 <= I_L <= 0.476 A\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb new file mode 100644 index 00000000..ad82474d --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_4_2.ipynb @@ -0,0 +1,1826 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 4 : Bipolar Junction Transistors" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.1\n", + ": Page No 223" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "I_C= 0.9 # in mA\n", + "I_E=1 # in mA\n", + "alpha = I_C/I_E \n", + "print \"Current gain = %0.1f\" %alpha\n", + "# Formula I_E= I_B+I_C\n", + "I_B= I_E-I_C # in mA\n", + "print \"The base current = %0.1f mA\" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Current gain = 0.9\n", + "The base current = 0.1 mA\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.2\n", + ": Page No 224" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "alpha= 0.97 \n", + "I_E=1 # in mA\n", + "# Formula alpha = I_C/I_E \n", + "I_C= alpha*I_E # in mA\n", + "# Formula I_E= I_B+I_C\n", + "I_B= I_E-I_C # in mA\n", + "print \"The base current = %0.2f mA\" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 0.03 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.3\n", + ": Page No 226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "# Part (i)\n", + "a= 0.90 \n", + "B=a/(1-a) \n", + "print \"At alpha= 0.90, the value of Bita = %0.f\" %B\n", + "# Part (ii)\n", + "a= 0.99 \n", + "B=a/(1-a) \n", + "print \"At alpha= 0.99, the value of Bita = %0.f\" %B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At alpha= 0.90, the value of Bita = 9\n", + "At alpha= 0.99, the value of Bita = 99\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.4\n", + ": Page No 226" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 50 \n", + "I_E= 10 # in mA\n", + "I_B= 200*10**-3 # in mA\n", + "alfa= bita/(1+bita)\n", + "print \"The value of alfa = %0.2f\" %alfa\n", + "I_C= alfa*I_E # in mA\n", + "print \"The value of I_C = %0.1f mA using the value of alpha\" %I_C\n", + "I_C= bita*I_B # in mA\n", + "print \"The value of I_C = %0.f mA using the value of bita\" %I_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of alfa = 0.98\n", + "The value of I_C = 9.8 mA using the value of alpha\n", + "The value of I_C = 10 mA using the value of bita\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.5\n", + ": Page No 233" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB= 10 # in V\n", + "V_CC= 10 # in V\n", + "V_BE= 0.7 # in V\n", + "R_B= 1 # in M\u03a9\n", + "R_B= R_B*10**6 # in \u03a9\n", + "R_C= 2 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "bita= 300 \n", + "I_B= (V_BB-V_BE)/R_B # in A\n", + "I_C= bita*I_B # in A\n", + "V_CE= V_CC-I_C*R_C # in V\n", + "P_D= V_CE*I_C # in W\n", + "print \"The value of I_B = %0.1f \u00b5A\" %(I_B*10**6)\n", + "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n", + "print \"The value of V_CE = %0.2f volts\" %V_CE\n", + "print \"The value of P_D = %0.1f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 9.3 \u00b5A\n", + "The value of I_C = 2.79 mA\n", + "The value of V_CE = 4.42 volts\n", + "The value of P_D = 12.3 mW\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.6\n", + ": Page No 241" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BE= 0 # in V\n", + "V_BB= 15 # in V\n", + "R_B= 470 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "V_CC= 15 # in V\n", + "R_C= 3.6 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "I_B= (V_BB-V_BE)/R_B # in A\n", + "I_C= bita*I_B # in A\n", + "V_CE= V_CC-I_C*R_C # in V\n", + "I_E= I_C+I_B # in A\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "print \"The value of V_CE = %0.2f volts\" %V_CE\n", + "print \"The emitter current = %0.2f mA\" %(I_E*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 31.9 \u00b5A\n", + "The collector current = 3.19 mA\n", + "The value of V_CE = 3.51 volts\n", + "The emitter current = 3.22 mA\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.7\n", + ": Page No 242 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita= 100 \n", + "V_BE= 0.7 # in V\n", + "V_BB= 15 # in V\n", + "R_B= 470 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "V_CC= 15 # in V\n", + "R_C= 3.6 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "I_B= (V_BB-V_BE)/R_B # in A\n", + "I_C= bita*I_B # in A\n", + "V_CE= V_CC-I_C*R_C # in V\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "print \"The value of V_CE = %0.2f volts\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 30.4 \u00b5A\n", + "The collector current = 3.04 mA\n", + "The value of V_CE = 4.05 volts\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.8\n", + ": Page No 255" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_CC= 15 # in V\n", + "V_BE= 0.7 # in V\n", + "R_C= 1 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "R_E= 2 # in k\u03a9\n", + "R_E= R_E*10**3 # in \u03a9\n", + "R1= 10 # in k\u03a9\n", + "R1= R1*10**3 # in \u03a9\n", + "R2= 5 # in k\u03a9\n", + "R2= R2*10**3 # in \u03a9\n", + "V_CE= np.arange(0,V_CC,0.1)\n", + "I_C= (V_CC-V_CE)/(R_C+R_E)*10**3 # in mA\n", + "plt.plot(V_CE,I_C) \n", + "plt.plot([0,8.55],[2.15,2.15], '--',)\n", + "plt.plot([8.55,8.55],[0,2.15], '--')\n", + "plt.xlabel('V_CE in volts')\n", + "plt.ylabel('I_C in mA')\n", + "plt.title('DC load line') \n", + "V_B= V_CC*R2/(R1+R2) # in V\n", + "I_E= (V_B-V_BE)/R_E # in A\n", + "I_C= I_E # in A\n", + "I_CQ= I_C # in A\n", + "V_CE= V_CC-I_C*(R_C+R_E) # in V\n", + "print \"Q-point is : \",round(V_CE,2),\" V\",round(I_CQ*10**3,2),\" mA\"\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q-point is : 8.55 V 2.15 mA\n", + "DC load line shown in figure\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7ff2041da990>" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.9\n", + ": Page No 258" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB= 1.8 # in V\n", + "V_BE= 0.7 # in V\n", + "R1= 10 # in k\u03a9\n", + "R2= 2.2 # in k\u03a9\n", + "R_E= 1 # in k\u03a9\n", + "bita= 200 \n", + "R= R1*R2/(R1+R2) # in k\u03a9\n", + "R=R*10**3 # in \u03a9\n", + "R_E= R_E*10**3 # in \u03a9\n", + "I_E= (V_BB-V_BE)/(R_E+R/bita) # in mA\n", + "print \"The emitter current = %0.2f mA\" %(I_E*10**3)\n", + "print \"This is extremely close to 1.1 mA, the value we get with the simplified analysis.\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter current = 1.09 mA\n", + "This is extremely close to 1.1 mA, the value we get with the simplified analysis.\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.10\n", + ": Page No 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC= 10 # in V\n", + "V_BE= 0.7 # in V\n", + "V_CE= 5 # in V\n", + "bita= 100 \n", + "I_C= 5 # in mA\n", + "# Applying KVL to collector circuit, V_CC-V_CE-I_C*R_C =0\n", + "R_C= (V_CC-V_CE)/I_C # in k\u03a9\n", + "print \"The value of R_C = %0.f k\u03a9\" %R_C\n", + "I_B= I_C/bita # in mA\n", + "print \"The value of I_B = %0.f \u00b5A\" %(I_B*10**3)\n", + "# Applying KVL to base circuit, V_CC-I_B*R_B-V_BE= 0\n", + "R_B= (V_CC-V_BE)/I_B # in k\u03a9\n", + "print \"The value of R_B = %0.f k\u03a9\" %R_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R_C = 1 k\u03a9\n", + "The value of I_B = 50 \u00b5A\n", + "The value of R_B = 186 k\u03a9\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.11\n", + ": Page No 261" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_CC= 6 # in V\n", + "V_BE= 0.7 # in V\n", + "bita= 100 \n", + "R_C= 2 # in k\u03a9\n", + "R_C= R_C*10**3 # in \u03a9\n", + "R_B= 530 # in k\u03a9\n", + "R_B= R_B*10**3 # in \u03a9\n", + "R1= 10 # in k\u03a9\n", + "R1= R1*10**3 # in \u03a9\n", + "R2= 5 # in k\u03a9\n", + "R2= R2*10**3 # in \u03a9\n", + "V_CE= np.arange(0,V_CC,0.1) # in V\n", + "I_C= (V_CC-V_CE)/(R_C)*10**3 # in mA\n", + "plt.plot(V_CE,I_C) \n", + "plt.xlabel('V_CE in volts') \n", + "plt.ylabel('I_C in mA')\n", + "plt.plot([0,4],[1,1], '--',)\n", + "plt.plot([4,4],[0,1], '--')\n", + "plt.title('DC load line') \n", + "I_B= (V_CC-V_BE)/R_B # in A\n", + "I_CQ= I_B*bita # in A\n", + "V_CE= V_CC-I_CQ*R_C # in V\n", + "print \"Q-point is : (\",round(V_CE,),\"V\",round(I_CQ*10**3),\"mA )\"\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q-point is : ( 4.0 V 1.0 mA )\n", + "DC load line shown in figure\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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xYMEC+Pn5wd/fH4WFhd03VpuxlpYWtY+Pj/ry5cvq5uZmdXBwsPr8+fNSlyWY\n48ePq4uLi9WBgYFSlyKKq1evqktKStRqtVp969Yt9ahRo6zq70+tVqvv3LmjVqvV6vv376tDQ0PV\n+fn5ElckrG3btqmfffZZdUxMjNSlCM7Ly0t9/fp1qcsQzdKlS9UffPCBWq1u/++zsbGx27ZmnQiM\nWZBmycLCwuDi4iJ1GaIZNmwYQkJCAACOjo7w8/NDbW2txFUJy8HBAQDQ3NyM1tZWyOVyiSsSTnV1\nNXJycrBq1Sqr3fDRWj/XzZs3kZ+fjxUrVgBoH691cnLqtr1ZdwTGLEgjy1BRUYGSkhKEhoZKXYqg\n2traEBISAjc3N0RERMDf31/qkgTz4osvIiUlBXZ2Zv010WsymQxPPPEEJkyYgD179khdjqAuX76M\nRx55BMuXL8e4ceOwevVq3L17t9v2Zv03bO3rBmzF7du3sWDBArzzzjtwdHSUuhxB2dnZ4dSpU6iu\nrsbx48ctfrsCjezsbAwdOhQKhcJq/9X87bffoqSkBLm5udi1axfy8/OlLkkwLS0tKC4uxrp161Bc\nXIyBAwfi9ddf77a9WXcExixII/N2//59PPPMM/j1r3+NefPmSV2OaJycnBAdHY3vv/9e6lIEUVBQ\ngKysLHh7eyMuLg5HjhzB0qVLpS5LUI8++igA4JFHHsHTTz+NIlN2bTMzHh4e8PDwwMSJEwEACxYs\nQHFxcbftzboj0F2Q1tzcjMzMTMydO1fqsshIarUaK1euhL+/PxITE6UuR3D19fVobGwEAPz000/4\n6quvoFAoJK5KGFu2bEFVVRUuX76MjIwMTJ8+Hfv27ZO6LMHcvXsXt27dAgDcuXMHX375pVXN3hs2\nbBg8PT1RVlYGADh8+DACAgK6bS/qgjJTdbcgzVrExcXh2LFjuH79Ojw9PfGnP/0Jy5cvl7oswXz7\n7bf4+OOPtVP0AGDr1q2YPXu2xJUJ4+rVq4iPj0dbWxva2tqwZMkSzJgxQ+qyRGFtt2nr6urw9NNP\nA2i/jbJ48WLMnDlT4qqE9e6772Lx4sVobm6Gj4+PwcW6XFBGRGTjzPrWEBERiY8dARGRjWNHQERk\n49gREBHZOHYEREQ2jh0BEZGNY0dARGTj2BGQxZs+fTq+/PLLDo+9/fbbWLduXbevKSsrQ1RUFEaN\nGoXx48dj0aJFuHbtGlQqFZycnKBQKLQ/R44c0Xt9dHQ0mpqaBP8sGsuWLcM//vEP7Wf56aefRHsv\nIrNeWUzWNRiZAAADd0lEQVRkjLi4OGRkZHRYGZqZmYmUlJQu29+7dw9z5szBjh07EB0dDQA4duwY\nfvzxR8hkMkybNg2ff/65wfc8dOiQcB+gCzKZTLua95133sGSJUswYMAAUd+TbBcTAVm8Z555BocO\nHUJLSwuA9i2va2trMXXq1C7b//3vf8fjjz+u7QQAIDw8HAEBAUbvtOnl5YWGhgZUVFTAz88Pzz33\nHAIDAzFr1izcu3evQ9ubN2/Cy8tL+/udO3cwYsQItLa24tSpU5g8eTKCg4Mxf/587d5FQPteTe++\n+y5qa2sRERGBGTNmoK2tDcuWLUNQUBDGjh2Lt99+29j/mYi6xY6ALJ5cLsekSZOQk5MDAMjIyMCi\nRYu6bV9aWorx48d3+3x+fn6HW0OXL1/Wa6O79055eTnWr1+Pc+fOwdnZWXtLR8PJyQkhISHaLaqz\ns7Mxe/Zs9OvXD0uXLkVKSgpOnz6NoKAgJCcnd3iP559/HsOHD4dKpcLXX3+NkpIS1NbW4uzZszhz\n5oxV7U1F0mFHQFZBc3sIaL8tFBcXZ7C9oX/5h4WFoaSkRPvj7e1t8Fre3t4YO3YsAGD8+PGoqKjQ\na7No0SJkZmYC+F9HdfPmTdy8eRNhYWEAgPj4eBw/ftzge/n4+ODSpUvYsGEDvvjiCwwePNhgeyJj\nsCMgqzB37lztv5jv3r1rcDvogIAAnDx5UrD3fuihh7R/7tevn/YWla6YmBjk5eXhxo0bKC4uxvTp\n0/XaGHNbytnZGWfOnIFSqcT777+PVatWmVY8EdgRkJVwdHREREQEli9fjmeffdZg22effRYFBQXa\nW0kAcPz4cZSWlopa38SJE7FhwwbExMRAJpPByckJLi4u+OabbwAAf/3rX6FUKvVeO2jQIO0MpevX\nr6OlpQXz58/Hn//8Z4OHjRAZi7OGyGrExcVh/vz5OHDggMF2Dz/8MLKzs5GYmIjExET0798fwcHB\nePvtt1FfX68dI9D44x//iPnz53e4hu4YQee9+rvbu3/RokVYuHBhh+MsP/roI6xZswZ3797tds/4\n5557DrNnz4a7uzt27NiB5cuXo62tDQAMHj9IZCyeR0BEZON4a4iIyMbx1hBZrbNnz+oduP7www/j\nxIkTElVEZJ54a4iIyMbx1hARkY1jR0BEZOPYERAR2Th2BERENo4dARGRjft/F92OqLa0R54AAAAA\nSUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7ff1eee33190>" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.12\n", + ": Page No 265" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC= 12 # in V\n", + "V_BE= 0.7 # in V\n", + "bita= 100 \n", + "R_C= 10 # in k\u03a9\n", + "R_C=R_C*10**3 # in \u03a9\n", + "R_B= 100 # in \u03a9\n", + "R_B=R_B*10**3 # in \u03a9\n", + "I_BQ= (V_CC-V_BE)/((1+bita)*R_C+R_B) # in A\n", + "I_CQ= bita*I_BQ # in A\n", + "V_CEQ= V_CC-(I_CQ+I_BQ)*R_C # in volts\n", + "print \"Q-Point value for the circuit =\",round(V_CEQ,3),\"V and\",round(I_CQ*10**3,3),\"mA\"\n", + "# For dc load line when \n", + "I_C=0 \n", + "V_CE= V_CC-(I_C+I_BQ)*R_C # in V\n", + "print \"At I_C=0, the value of V_CE = %0.2f volts\" %V_CE\n", + "# When\n", + "V_CE= 0 \n", + "I_C= (V_CC-I_BQ*R_C)/R_C # in A\n", + "print \"At V_CE=0, the value of I_C = %0.1f mA\" %(I_C*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Q-Point value for the circuit = 1.718 V and 1.018 mA\n", + "At I_C=0, the value of V_CE = 11.90 volts\n", + "At V_CE=0, the value of I_C = 1.2 mA\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.13\n", + ": Page No 266" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE= 0.7 # in V\n", + "V_CC= 15 # in V\n", + "V_CE= 5 # in V\n", + "I_C= 5 # in mA\n", + "I_C=I_C*10**-3 # in A\n", + "bita= 100 \n", + "I_B= I_C/bita # in A\n", + "# Applying KVL to collector circuit, V_CC= (I_C+I_B)*R_C+V_CE\n", + "R_C= (V_CC-V_CE)/(I_C+I_B) # in \u03a9\n", + "# Applying KVL to base circuit, V_CC= (I_C+I_B)*R_C+I_B*R_B+V_BE\n", + "R_B= (V_CC-V_BE-R_C*(I_C+I_B))/I_B # in \u03a9\n", + "print \"The value of R_C = %0.2f k\u03a9\" %(R_C*10**-3)\n", + "print \"The value of R_B = %0.f k\u03a9\" %(R_B*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R_C = 1.98 k\u03a9\n", + "The value of R_B = 86 k\u03a9\n" + ] + } + ], + "prompt_number": 25 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.14\n", + ": Page No 267" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_B= 20*10**-6 # in A\n", + "V_CE= 7.3 # in V\n", + "V_BE= 0.6 # in V\n", + "V_E= 2.1 # in V\n", + "R_E= 0.68*10**3 # in \u03a9\n", + "R_C= 2.7*10**3 # in \u03a9\n", + "I_E= V_E/R_E # in A\n", + "I_C= I_E # in A (approx)\n", + "bita= round(I_C/I_B) \n", + "V_CC= V_CE+I_C*R_C+I_E*R_E # in V\n", + "# From V_CC= I_B*R_B+V_BE+V_E\n", + "R_B= (V_CC-(V_BE+V_E))/I_B # in \u03a9\n", + "print \"The value of bita = %0.f\" %bita\n", + "print \"The value of V_CC = %0.1f volts\" %V_CC\n", + "print \"The value of R_B = %0.f k\u03a9\" %(R_B*10**-3)\n", + "\n", + "# Note: In the book, there is an error to calculate the value of R_B, hence the value of R_B in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of bita = 154\n", + "The value of V_CC = 17.7 volts\n", + "The value of R_B = 752 k\u03a9\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.15\n", + ": Page No 268" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 18 # in V\n", + "bita = 90 \n", + "R_C = 2.2 * 10**3 # in ohm\n", + "R_E = 1.8*10**3 # in ohm\n", + "R_B = 510*10**3 # in ohm\n", + "I_B = V_CC/( (bita*(R_C+R_E))+R_B ) # in A\n", + "I_C = bita*I_B # in A\n", + "print \"The value of I_C = %0.1f mA\" %(I_C*10**3)\n", + "V_CE = I_B*R_B # in V\n", + "print \"The value of V_CE = %0.1f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_C = 1.9 mA\n", + "The value of V_CE = 10.6 V\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.16\n", + ": Page No 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "bita = 50 \n", + "V_CC = 12 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 240 # in kohm\n", + "R_B = R_B*10**3 # in ohm\n", + "I_C = 2.35 * 10**-3 # in A\n", + "R_C = 2.2 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "I_BQ = (V_CC - V_BE)/R_B # in A\n", + "print \"The value of I_BQ = %0.2f \u00b5A\" %(I_BQ*10**6)\n", + "I_CQ = bita*I_BQ # in A\n", + "print \"The value of I_CQ = %0.2f mA\" %(I_CQ*10**3)\n", + "V_CEQ = V_CC - (I_C*R_C) # in V\n", + "print \"The value of V_CEQ = %0.2f V\" %V_CEQ\n", + "V_B = V_BE # in V\n", + "print \"The value of V_B = %0.1f V\" %V_B\n", + "V_BC = V_B -V_CEQ # in V\n", + "print \"The voltage = %0.2f V\" %V_BC\n", + "\n", + "# Note: In the book, there is a calculation error to evaluating the value of V_CEQ. So the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_BQ = 47.08 \u00b5A\n", + "The value of I_CQ = 2.35 mA\n", + "The value of V_CEQ = 6.83 V\n", + "The value of V_B = 0.7 V\n", + "The voltage = -6.13 V\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.17\n", + ": Page No 269" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 18 # in V\n", + "V_BE = 0.7 # in V\n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "R_B = 210 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "bita = 75 \n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "R_E = 510 # in ohm\n", + "I_B = (V_CC-V_BE)/( R_C+R_B+bita*(R_C+R_E) ) # A\n", + "print \"The value of I_B = %0.f \u00b5A\" %round(I_B*10**6)\n", + "I_C = bita*I_B # in A\n", + "print \"The value of I_C = %0.1f mA\" %(I_C*10**3)\n", + "V_C = V_CC - (I_C*R_C) # in V\n", + "print \"The voltage = %0.2f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 35 \u00b5A\n", + "The value of I_C = 2.6 mA\n", + "The voltage = 9.42 V\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.18\n", + ": Page No 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.7 # in V\n", + "I_B = 40 * 10**-6 # in A\n", + "V_CC = 20 # in V (From the load line)\n", + "print \"The voltage = %0.f V\" %V_CC\n", + "I_C = 8 # in mA\n", + "R_C = V_CC/I_C # in kohm\n", + "print \"The resistance = %0.1f kohm\" %R_C\n", + "R_B = (V_CC - V_BE)/I_B # in ohm\n", + "print \"The resistance = %0.1f kohm\" %(R_B*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage = 20 V\n", + "The resistance = 2.5 kohm\n", + "The resistance = 482.5 kohm\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.19\n", + ": Page No 271" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 47 # in kohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 10 # in kohm\n", + "R2= R2*10**3 # in ohm\n", + "R_E = 1.1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R_C = 2.4 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_CC = -18 # in V\n", + "V_B = (R2*V_CC)/(R1+R2) # in V\n", + "V_BE = -0.7 # in V\n", + "V_E = V_B - V_BE # in V\n", + "I_E = abs(V_E)/R_E # in A\n", + "V_CE = V_CC + (I_E)*(R_C+R_E) # in V\n", + "print \"The value of V_B = %0.2f volts\" %V_B\n", + "print \"The value of I_E = %0.2f mA\" %(I_E*10**3)\n", + "print \"The value of V_CE = %0.2f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_B = -3.16 volts\n", + "The value of I_E = 2.23 mA\n", + "The value of V_CE = -10.18 V\n" + ] + } + ], + "prompt_number": 37 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.20\n", + ": Page No 273" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.8 # in V\n", + "V_CE = 0.2 # in V\n", + "V1 = 5 # in V\n", + "R_B = 50 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "R_C = 3 # in K ohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "bita = 100 \n", + "R_E = 2 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_B = (V1-V_BE)/(R_B+(1+bita)*R_E) # in A\n", + "print \"The value of I_B = %0.2f \u00b5A\" %(I_B*10**6)\n", + "V_CC = 10 # in V\n", + "I_Csat = (V_CC - V_CE - (I_B*R_E))/(R_C+R_E) #in A\n", + "print \"The value of I_C(sat) = %0.3f mA\" %(I_Csat*10**3)\n", + "I_Bmin = I_Csat /bita # in A\n", + "print \"The minimum value of I_B = %0.3f \u00b5A\" %(I_Bmin*10**6)\n", + "\n", + "# Note: There is calculation error to evaluate the value of I_Csat in the book, so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 16.67 \u00b5A\n", + "The value of I_C(sat) = 1.953 mA\n", + "The minimum value of I_B = 19.533 \u00b5A\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.21\n", + ": Page No 274" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 5 # in kohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 5 # in kohm\n", + "R2= R2*10**3 # in ohm\n", + "R_B = R1*R2/(R1+R2) # in ohm\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "V_EE = 3 # in V\n", + "V_Th = (R2*V_EE)/(R1+R2) # in V\n", + "V_BE = 0.7 # in V\n", + "bita = 44 \n", + "I_B = (V_EE - V_BE - V_Th)/( ((1+bita)*R_E)+R_B) # in A\n", + "I_BQ = I_B # in A\n", + "print \"The value of I_BQ = %0.2f \u00b5A\" %(I_BQ*10**6)\n", + "I_C = bita*I_BQ # in A\n", + "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n", + "I_E = (1+bita)*I_B # in A\n", + "print \"The value of I_E = %0.3f mA\" %(I_E*10**3)\n", + "V_EC = (I_E*R_E)-V_EE # in V\n", + "print \"The value of V_EC = %0.3f V\" %V_EC\n", + "print \"Q-point = (\",round(V_EC,3),\"V\",round(I_C*10**3,2),\"mA )\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_BQ = 16.84 \u00b5A\n", + "The value of I_C = 0.74 mA\n", + "The value of I_E = 0.758 mA\n", + "The value of V_EC = -2.242 V\n", + "Q-point = ( -2.242 V 0.74 mA )\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.22\n", + ": Page No 275" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.7 # in V\n", + "V_BB = 5 # in V\n", + "R_B = 100 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "R_E = 2 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "bita = 100 \n", + "I_B = (V_BB-V_BE)/( R_B+((1+bita)*R_E) ) # in A\n", + "print \"The value of I_B = %0.3f mA\" %(I_B*10**3)\n", + "V_B = V_BB-(I_B*10**-3*R_B) # in V\n", + "I_C = bita*I_B # in A\n", + "print \"The value of I_C = %0.1f mA\" %(I_C*10**3)\n", + "V_CC = 10 # in V\n", + "V_C = V_CC-(I_C*R_E) # in V\n", + "print \"The voltage = %0.1f V\" %V_C\n", + "print \"Transistor is in active region is valid\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_B = 0.014 mA\n", + "The value of I_C = 1.4 mA\n", + "The voltage = 7.2 V\n", + "Transistor is in active region is valid\n" + ] + } + ], + "prompt_number": 49 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.23\n", + ": Page No 276 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 20 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 430 # in kohm\n", + "R_B = 430 * 10**3 # in ohm\n", + "bita = 50 \n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R_C = 2 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "I_B = (V_CC - V_BE)/(R_B +(1+bita)*R_E) # in A\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "I_C = bita*I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CE = V_CC - I_C*(R_C+R_E) # in V\n", + "print \"The value of V_CE = %0.2f V\" %V_CE\n", + "V_C = V_CC - (I_C*R_C) # in V\n", + "print \"The value of V_C = %0.2f V\" %V_C\n", + "V_E = V_C - V_CE # in V\n", + "print \"The value of V_E = %0.2f V\" %V_E\n", + "V_B = V_BE+V_E # in V\n", + "print \"The value of V_B = %0.2f V\" %V_B\n", + "V_BC = V_B-V_C # in V\n", + "print \"The value of V_BC = %0.2f V\" %V_BC" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 40.1 \u00b5A\n", + "The collector current = 2.01 mA\n", + "The value of V_CE = 13.98 V\n", + "The value of V_C = 15.99 V\n", + "The value of V_E = 2.01 V\n", + "The value of V_B = 2.71 V\n", + "The value of V_BC = -13.28 V\n" + ] + } + ], + "prompt_number": 51 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.24\n", + ": Page No 277" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 20 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 680 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "R_C = 4.7 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "bita = 120 \n", + "I_B = (V_CC - V_BE)/(R_B+bita*R_C) # in A\n", + "I_CQ = bita*I_B # in A\n", + "print \"The value of I_CQ = %0.2f mA\" %(I_CQ*10**3)\n", + "V_CEQ = V_CC - (I_CQ*R_C) # in V\n", + "print \"The value of V_CEQ = %0.2f V\" %V_CEQ\n", + "V_B = V_BE # in V\n", + "V_C = 11.26 # in V\n", + "V_E = 0 # in V\n", + "print \"The value of V_E = %0.f V\" %V_E\n", + "V_BC = V_B - V_C # in V\n", + "print \"The value of V_BC = %0.2f V\" %V_BC" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_CQ = 1.86 mA\n", + "The value of V_CEQ = 11.25 V\n", + "The value of V_E = 0 V\n", + "The value of V_BC = -10.56 V\n" + ] + } + ], + "prompt_number": 52 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.25\n", + ": Page No 278" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 16 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 470 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 120 \n", + "R_C = 3.6 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_E = 0.51 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_B = (V_CC - V_BE)/(R_B+bita*(R_C+R_E)) # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)\n", + "I_C = bita*I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_C = V_CC - I_C*R_C # in V\n", + "print \"The collector voltage = %0.2f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 15.88 \u00b5A\n", + "The collector current = 1.91 mA\n", + "The collector voltage = 9.14 V\n" + ] + } + ], + "prompt_number": 53 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.26\n", + ": Page No 279" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 10 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 250 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 90 \n", + "R_C = 4.7 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_E = 1.2 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_BQ = (V_CC - V_BE)/(R_B + bita*(R_C+R_E)) # in A\n", + "print \"The base current at Q-point = %0.2f \u00b5A\" %(I_BQ*10**6)\n", + "I_CQ = bita*I_BQ # in A\n", + "print \"The collector current at Q-point = %0.2f mA\" %(I_CQ*10**3)\n", + "V_CEQ = V_CC - (I_CQ*(R_C+R_E)) # in V\n", + "print \"Collector emitter voltage at Q point = %0.3f V\" %V_CEQ" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current at Q-point = 11.91 \u00b5A\n", + "The collector current at Q-point = 1.07 mA\n", + "Collector emitter voltage at Q point = 3.677 V\n" + ] + } + ], + "prompt_number": 55 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.27\n", + ": Page No 281" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 12 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 150 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 180 \n", + "R_C = 4.7 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_E = 3.3 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "I_B = (V_CC-V_BE)/(R_B + bita*(R_C+R_E)) # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 7.11 \u00b5A\n" + ] + } + ], + "prompt_number": 58 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.28\n", + ": Page No 282" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_B = 4 # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1.2 # in kohm\n", + "R_E= R_E*10**3 # in ohm\n", + "V_E = V_B-V_BE # in V\n", + "R_C = 2.2 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_B= 330 # in kohm\n", + "R_B= R_B*10**3 # in ohm\n", + "bita = 180 \n", + "I_B = 7.11 * 10**-6 # in A\n", + "V_CC = 18 # in V\n", + "print \"Part (a)\"\n", + "print \"The value of V_E = %0.1f V\" %V_E\n", + "I_C = V_E/R_E # in A\n", + "print \"Part (b)\"\n", + "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n", + "V_C =V_CC - (I_C*R_C) # in V\n", + "print \"Part (c)\"\n", + "print \"The value of V_C = %0.2f V\" %V_C\n", + "V_CE = V_C-V_E # in V\n", + "print \"Part (d)\"\n", + "print \"The value of V_CE = %0.2f V\" %V_CE\n", + "I_B = (V_CC - (I_C*R_C) - V_BE - V_E)/R_B # in A\n", + "print \"Part (e)\"\n", + "print \"Base current = %0.2f \u00b5A\" %(I_B*10**6)\n", + "bita = I_C/I_B \n", + "print \"Part (f)\"\n", + "print \"Current gain = %0.f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The value of V_E = 3.3 V\n", + "Part (b)\n", + "The value of I_C = 2.75 mA\n", + "Part (c)\n", + "The value of V_C = 11.95 V\n", + "Part (d)\n", + "The value of V_CE = 8.65 V\n", + "Part (e)\n", + "Base current = 24.09 \u00b5A\n", + "Part (f)\n", + "Current gain = 114\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.29\n", + ": Page No 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_E = 10 # in mA\n", + "I_C = 9.95 # in mA\n", + "I_B = I_E-I_C # in mA\n", + "print \"The base current = %0.2f mA\" %I_B" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 0.05 mA\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.30\n", + ": Page No 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_C = 10 # in mA\n", + "I_B = 0.1 # in mA\n", + "bita = I_C/I_B \n", + "print \"The current gain = %0.f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The current gain = 100\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.31\n", + ": Page No 284" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0.7 # in V\n", + "V_BB = 10 # in V\n", + "R_B = 470 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 19.79 \u00b5A\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.32\n", + ": Page No 285" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 10 # in V\n", + "V_BE = 0 # in V\n", + "R_B = 470 # in kohm\n", + "R_B = R_B * 10**3 # in ohm \n", + "I_B = (V_BB - V_BE)/R_B # in A\n", + "bita = 200 \n", + "I_C = bita*I_B # in A\n", + "V_CC = 10 # in V\n", + "R_C = 820 # in ohm\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"Part (a) : For ideal approximation\"\n", + "print \"The collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in W\n", + "print \"Power dissipation = %0.2f mW\" %(P_D*10**3)\n", + "print \"Part (b) : For second approximation\"\n", + "V_BE = 0.7 # in V\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "I_C = bita*I_B # in A\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"The collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in W\n", + "print \"Power dissipation = %0.2f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) : For ideal approximation\n", + "The collector emitter voltage = 6.51 V\n", + "Power dissipation = 27.70 mW\n", + "Part (b) : For second approximation\n", + "The collector emitter voltage = 6.75 V\n", + "Power dissipation = 26.73 mW\n" + ] + } + ], + "prompt_number": 69 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.33\n", + ": Page No 286" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BE = 0 # in V\n", + "V_BB = 12 # in V\n", + "R_B = 680 # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "beta_dc = 175 \n", + "I_C = beta_dc*I_B # in A\n", + "V_CC = 12 # in V\n", + "R_C = 1.5 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"Part (a) For ideal approximation\"\n", + "print \"The collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in mW\n", + "print \"Transistor power = %0.2f mW\" %(P_D*10**3)\n", + "print \"Part (b) For second approximation\"\n", + "V_BE1 = 0.7 # in V\n", + "I_B = (V_BB-V_BE1)/R_B # in A\n", + "I_C = beta_dc * I_B # in A\n", + "V_CE = V_CC - (I_C*R_C) # in V\n", + "print \"Collector emitter voltage = %0.2f V\" %V_CE\n", + "P_D = V_CE * I_C # in W\n", + "print \"Power dissipation = %0.2f mW\" %(P_D*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) For ideal approximation\n", + "The collector emitter voltage = 7.37 V\n", + "Transistor power = 22.75 mW\n", + "Part (b) For second approximation\n", + "Collector emitter voltage = 7.64 V\n", + "Power dissipation = 22.21 mW\n" + ] + } + ], + "prompt_number": 71 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.34\n", + ": Page No 288" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "V_CC = 20 # in V\n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "I_C = V_CC/R_C # in A\n", + "print \"Collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CE = V_CC # in V\n", + "print \"Collector emitter voltage = %0.f V\" %V_CE\n", + "V_CE=np.arange(0,20,0.1) # in V\n", + "I_C= (V_CC-V_CE)/(R_C*10**-3) # in mA\n", + "plt.plot(V_CE,I_C) \n", + "plt.xlabel('V_CE in volts')\n", + "plt.ylabel('I_C in mA')\n", + "plt.title('DC load line')\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Collector current = 6.06 mA\n", + "Collector emitter voltage = 20 V\n", + "DC load line shown in figure" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7ff2041850d0>" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.35\n", + ": Page No 289" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 10 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 1 # in kohm\n", + "R_B = 1 * 10**6 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.1f \u00b5A\" %(I_B*10**6)\n", + "beta_dc = 200 \n", + "I_C = beta_dc * I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CC = 20 # in V\n", + "R_C = 3.3 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_CE = V_CC - I_C*R_C # in V\n", + "print \"The collector voltage = %0.3f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 9.3 \u00b5A\n", + "The collector current = 1.86 mA\n", + "The collector voltage = 13.862 V\n" + ] + } + ], + "prompt_number": 73 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.36\n", + ": Page No 290" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 5 # in V\n", + "V_BE = 0.7 # in V\n", + "R_B = 680 # in kohm\n", + "R_B = 680*10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.2f \u00b5A\" %(I_B*10**6)\n", + "beta_dc= 150 \n", + "I_C = beta_dc * I_B # in A\n", + "print \"The collector current = %0.2f mA\" %(I_C*10**3)\n", + "V_CC = 5 # in V\n", + "R_C = 470 # in ohm\n", + "V_CE = V_CC-(I_C*R_C) # in V\n", + "print \"Voltage between collector and ground = %0.2f V\" %V_CE" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 6.32 \u00b5A\n", + "The collector current = 0.95 mA\n", + "Voltage between collector and ground = 4.55 V\n" + ] + } + ], + "prompt_number": 75 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.37\n", + ": Page No 291" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 2.5 # in V\n", + "V_BE = 0.7 # in V\n", + "V_E = V_BB-V_BE # in V\n", + "print \"The emitter voltage = %0.1f V\" %V_E\n", + "R_E = 1.8 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "I_E = V_E/R_E # in A\n", + "I_C= I_E # in A\n", + "V_CC = 20 # in V\n", + "R_C = 10 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_C = V_CC-(I_C*R_C) # in V\n", + "print \"The collector voltage = %0.f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter voltage = 1.8 V\n", + "The collector voltage = 10 V\n" + ] + } + ], + "prompt_number": 77 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.38\n", + ": Page No 292" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CC = 25 # in V\n", + "R2 = 2.2 # in kohm\n", + "R1 = 10 # in kohm\n", + "V_BB = (V_CC * R2)/(R1+R2) # in V\n", + "V_BE = 0.7 # in V\n", + "V_E = V_BB - V_BE # in V\n", + "print \"The emitter voltage = %0.1f V\" %V_E\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "I_E = V_E/R_E # in A\n", + "I_C= I_E # in A\n", + "V_CC = 25 # in V\n", + "R_C = 3.6 # in kohm\n", + "R_C = R_C * 10**3 # in ohm\n", + "V_C = V_CC - (I_C*R_C) # in V\n", + "print \"Collector voltage = %0.2f V\" %V_C" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The emitter voltage = 3.8 V\n", + "Collector voltage = 11.29 V\n" + ] + } + ], + "prompt_number": 78 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 4.39\n", + ": Page No 293 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_BB = 4.50 # in V\n", + "V_E = 3.8 # in V\n", + "V_C = 11.32 # in V\n", + "I_C = 3.8 # in mA\n", + "I_C=I_C*10**-3 # in A\n", + "V_BE = 0.7 # in V\n", + "R1 = 10 # in kohm\n", + "R2 = 2.2 # in kohm\n", + "R_B = (R1*R2)/(R1+R2) # in kohm\n", + "R_B = R_B * 10**3 # in ohm\n", + "I_B = (V_BB-V_BE)/R_B # in A\n", + "print \"The base current = %0.2f mA\" %(I_B*10**3)\n", + "V_CE = V_C-V_E # in V\n", + "print \"Collector emitter voltage = %0.2f V\" %V_CE\n", + "print \"Thus the Q-point is :\",round(V_CE,2),\"V\",round(I_B*10**3,2),\"mA\"\n", + "\n", + "# Note: There is calculation error to evaluate the value of I_B. So the answer in the book is wrong." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base current = 2.11 mA\n", + "Collector emitter voltage = 7.52 V\n", + "Thus the Q-point is : 7.52 V 2.11 mA\n" + ] + } + ], + "prompt_number": 80 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb new file mode 100644 index 00000000..54ba0032 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_5_2.ipynb @@ -0,0 +1,517 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 5 : Transistor Circuits" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.1\n", + ": Page No 309 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "from numpy import pi\n", + "# Given data\n", + "R1 = 600 # in ohm\n", + "R2 = 1000 # in ohm\n", + "R_TH = (R1*R2)/(R1+R2) # in ohm\n", + "X_C = 37.5 # in ohm\n", + "f = 1 # in kHz\n", + "f= f*10**3 # in Hz\n", + "C = 1/(2*pi * f*X_C) # in F\n", + "print \"Value of C = %0.1f \u00b5F\" %(C*10**6)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Value of C = 4.2 \u00b5F\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.2\n", + ": Page No 323" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_C= 3.6*10**3 # in ohm\n", + "R_L= 10*10**3 # in ohm\n", + "r_c = (R_C*R_L)/(R_C+R_L) # in ohm\n", + "V_CC = 10 # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R1 = 10 # in kohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 2.2 # in kohm\n", + "R2= R2*10**3 # in ohm\n", + "V_B = (V_CC*R2)/(R1+R2) # in V\n", + "I_E = (V_B-V_BE)/R_E # in A\n", + "V = 25*10**-3 # in V # only value is given in the book \n", + "r_e = V/I_E # in ohm\n", + "A_V = round(r_c/r_e) \n", + "print \"The voltage gain = %0.f\" %A_V\n", + "V_in = 2 #in mV\n", + "V_out = A_V*V_in # in mV\n", + "print \"The output voltage = %0.f mV\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = 117\n", + "The output voltage = 234 mV\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.3\n", + ": Page No 324" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "A_V = 117 \n", + "r_e = 22.7 # in ohm\n", + "bita = 300 \n", + "Zin_base = bita*r_e # in ohm\n", + "R1 = 2.2*10**3 # in ohm\n", + "R2 = 10*10**3 # in ohm\n", + "Zin_stage = (Zin_base*R1*R2)/(Zin_base*R1+R1*R2+R2*Zin_base) # in ohm \n", + "R = 600 # in ohm\n", + "V = 2 # in mV\n", + "V_in = (Zin_stage/(R+Zin_stage))*V # in mV\n", + "V_out = A_V * V_in # in mV\n", + "print \"The output voltage = %0.f mV\" %round(V_out)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 165 mV\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.4\n", + ": Page No 328" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R1 = 4.3 # in K ohm\n", + "R1= R1*10**3 # in ohm\n", + "R2 = 10 # in K ohm\n", + "R2= R2*10**3 # in ohm\n", + "r_e = (R1*R2)/(R1+R2) # in ohm\n", + "bita = 200 \n", + "V=25 # in mV\n", + "I= 1 # in mA\n", + "r_e_desh= V/I # in ohm\n", + "Zin_base = bita*(r_e + r_e_desh) # in ohm\n", + "print \"The input impedence of the base = %0.f k\u03a9\" %(Zin_base*10**-3)\n", + "R3 = 10*10**3 # in ohm\n", + "Zin_stage = (R2*R3*Zin_base)/(R2*Zin_base+R3*Zin_base+R2*R3) # in ohm\n", + "print \"The input impedance of the stage = %0.2f k\u03a9\" %(Zin_stage*10**-3)\n", + "print \"Because the input impedence of base is much larger than the input impedence of the stage,\"\n", + "print \"usually approximate the input impedence of the stage as the parallel of the biasing resistor only %0.f k\u03a9\" %(Zin_stage*10**-3)\n", + "Zin_stage= R2*R3/(R2+R3) # in ohm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input impedence of the base = 606 k\u03a9\n", + "The input impedance of the stage = 4.96 k\u03a9\n", + "Because the input impedence of base is much larger than the input impedence of the stage,\n", + "usually approximate the input impedence of the stage as the parallel of the biasing resistor only 5 k\u03a9\n" + ] + } + ], + "prompt_number": 16 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.5\n", + ": Page No 332" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_CE = 0.2 # in V\n", + "V_BE= 0.7 # in V\n", + "R = 1 # in kohm\n", + "R = R * 10**3 # in ohm\n", + "V = 10 # in V\n", + "I_C = (V-V_CE)/R # in A\n", + "beta_min = 50 \n", + "I_B = I_C/beta_min # in A\n", + "I_B1 = V*I_B # in A\n", + "V1 = 5 # in V\n", + "R_B = (V1-V_BE)/I_B1 # in ohm\n", + "print \"The base resistance = %0.1f k\u03a9\" %(R_B*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The base resistance = 2.2 k\u03a9\n" + ] + } + ], + "prompt_number": 17 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.6\n", + ": Page No 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R = 10 # in K ohm\n", + "R = R * 10**3 # in ohm\n", + "X_C = 0.1 * R \n", + "C = 47 # in \u00b5F\n", + "C = C * 10**-6 # in F\n", + "f = 1/(2*pi * X_C *C) # in Hz\n", + "print \"Lowest frequency = %0.2f Hz\" %f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Lowest frequency = 3.39 Hz\n" + ] + } + ], + "prompt_number": 18 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.7\n", + ": Page No 333" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "C = 220 # in \u00b5F\n", + "C = C * 10**-6 # in F\n", + "R1 = 10 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "R2 = 2.2 # in kohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R_TH = (R1*R2)/(R1+R2) # in ohm\n", + "X_C = 0.1*R_TH # in ohm\n", + "f = 1/(2*pi*C*X_C) # in Hz\n", + "print \"The lowest frequency = %0.2f Hz\" %f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The lowest frequency = 4.01 Hz\n" + ] + } + ], + "prompt_number": 19 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.8\n", + ": Page No 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "i_c = 15 # in mA\n", + "i_c = i_c * 10**-3 # in A\n", + "i_b = 100 # in \u00b5A\n", + "i_b = i_b * 10**-6 # in A\n", + "bita = i_c/i_b \n", + "print \"The value of ac bita = %0.f\" %bita" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of ac bita = 150\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.9\n", + ": Page No 334" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_C = 3.6 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "R_L = 10 # in kohm\n", + "R_L=R_L*10**3 # in ohm\n", + "R_TH = (R_C*R_L)/(R_C+R_L) # in ohm\n", + "V_CC = 10 # in V\n", + "R2 = 2.2 # in kohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R1 = 10 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "V_BE = 0.7 # in V\n", + "V_B = (V_CC*R2)/(R1+R2) # in V\n", + "R_E = 1 # in kohm \n", + "R_E = R_E *10**3 # in ohm\n", + "I_E = (V_B-V_BE)/R_E # in A\n", + "V1 = 25 # in mV\n", + "V1 = V1*10**-3 # in V\n", + "r_e = V1/(I_E) # in ohm\n", + "A_v = (R_TH)/r_e \n", + "V_in = 2 # in mV\n", + "V_in = V_in * 10**-3 # in V\n", + "V_out = A_v*V_in # in V\n", + "print \"The output voltage = %0.2f V\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 0.23 V\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.10\n", + ": Page No 336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_L = 10 # in kohm\n", + "R_L= R_L*10**3 # in ohm\n", + "R_C = 3.6 # in kohm\n", + "R_C= R_C*10**3 # in ohm\n", + "r_e_desh = 22.73 # in ohm \n", + "R_L_desh = R_L/2 # in ohm\n", + "A_v = ( (R_C*R_L_desh)/(R_C+R_L_desh))/r_e_desh \n", + "print \"The voltage gain = %0.2f\" %A_v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = 92.08\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.11\n", + ": Page No 336" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_E = 1 # in kohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "R_L = 3.3 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_e = (R_E*R_L)/(R_E+R_L) # in ohm\n", + "V_CC = 15 # in V\n", + "R2 = 2.2 # in K ohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R1 = R2 # in ohm\n", + "V_B = (V_CC*R2)/(R1+R2) # in V\n", + "V_BE = 0.7 # in V\n", + "R_E = 1 # in K ohm\n", + "R_E = R_E * 10**3 # in ohm\n", + "I_E = (V_B-V_BE)/R_E # in A\n", + "V1 = 25*10**-3 # in V\n", + "r_e1 = V1/I_E \n", + "bita = 200 \n", + "Zin_base = bita*(r_e+r_e1) # in ohm\n", + "print \"The input impedence of the base = %0.2f k\u03a9\" %(Zin_base*10**-3)\n", + "Zin_stage = (R1*R2*Zin_base)/(R1*R2+R2*Zin_base+R1*Zin_base) # in ohm\n", + "print \"The input impedance of the stage = %0.2f k\u03a9\" %(Zin_stage*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input impedence of the base = 154.22 k\u03a9\n", + "The input impedance of the stage = 1.09 k\u03a9\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 5.12\n", + ": Page No 337 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r_e = 767.44 \n", + "r_e1 = 3.68 \n", + "V_in = 1 # in V\n", + "A_v = round(r_e/(r_e+r_e1)) \n", + "print \"The voltage gain = %0.f\" %A_v\n", + "V_o = A_v*V_in # in V\n", + "print \"The load voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = 1\n", + "The load voltage = 1 V\n" + ] + } + ], + "prompt_number": 24 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb new file mode 100644 index 00000000..27931e37 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_6_2.ipynb @@ -0,0 +1,1319 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 6 : Field Effect Devices (JFET)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.1\n", + ": Page No 351" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V_D = 10 # in V\n", + "R = 10*10**3 # in ohm\n", + "I_D = V_D/R # in A\n", + "V_P = 4 # in V\n", + "I_DSS = 10 # in mA\n", + "I_DSS = I_DSS * 10**-3 # in A\n", + "R_DS = V_P/I_DSS # in ohm\n", + "V_D = (R_DS/(R+R_DS))*V_D # in V\n", + "print \"The drain voltage = %0.3f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain voltage = 0.385 V\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.2\n", + ": Page No 353" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_P = 4 # in V\n", + "I_DSS = 10 # in mA\n", + "I_DSS =I_DSS *10**-3 # in A\n", + "R_DS = V_P/I_DSS # in ohm\n", + "V_DD = 30 # in V\n", + "I_D = 2.5 # in mA\n", + "R_D = 2 # in kohm\n", + "V_D = V_DD - (I_D*R_D) # in V\n", + "print \"The drain voltage = %0.f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain voltage = 25 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.3\n", + ": Page No 355" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "R2 = 1 # in M ohm\n", + "R2 = R2*10**6 # in ohm\n", + "R1 = 2 # in M ohm\n", + "R1 = R1*10**6 # in ohm\n", + "V_DD = 30 # in V\n", + "R_D= 1*10**3 # in ohm\n", + "V_G = (R2/(R1+R2))*V_DD # in V\n", + "R_S= 2*10**3 # in ohm\n", + "I_D= V_G/R_S # in A\n", + "V_D= V_DD-I_D*R_D # in V\n", + "V_DS= V_D-V_G # in V\n", + "R_D= R_D+R_S # in ohm\n", + "I_Dsat=V_DD/R_D*10**3 # in mA\n", + "print \"The value of I_D = %0.f mA\" %(I_D*10**3)\n", + "print \"The value of V_DS = %0.f volts\" %V_DS\n", + "print \"Thus the Q-point = (\",int(V_DS),\"V,\",int(I_D*10**3),\"mA)\" \n", + "V_D= np.arange(0,V_DD,0.1) # in V\n", + "I_D= (V_DD-V_D)/R_D*10**3 # in mV\n", + "plt.plot(V_D,I_D) \n", + "plt.plot([0,15],[5,5], '--')\n", + "plt.plot([15,15],[0,5], '--')\n", + "plt.ylabel(\"I_D in mA\")\n", + "plt.xlabel(\"V_DS in volts\")\n", + "plt.title(\"DC load line\")\n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_D = 0 mA\n", + "The value of V_DS = 30 volts\n", + "Thus the Q-point = ( 30 V, 0 mA)\n", + "DC load line shown in figure" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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d5wgEAkkAQcvh0MTbDVIBgpJfmsCUKVO0e/duZWRkcAE5wOGQDINZAYKSX1YH9ejRQzfc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\nbIwmAAA2RhMAABv7f2bQfoJhPpDQAAAAAElFTkSuQmCC\n", + "text": [ + "<matplotlib.figure.Figure at 0x7ffba61e4e10>" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.4\n", + ": Page No 358" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_DD = 15 # in V\n", + "R = 3 # in kohm\n", + "I_D = V_DD/R # in mA\n", + "R_D = 1 # in kohm\n", + "V_D = V_DD - (I_D*R_D) # in V\n", + "print \"The drain voltage = %0.f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain voltage = 10 V\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.5\n", + ": Page No 362" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_D = 3.6 # in K ohm\n", + "R_L = 10 # in K ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in K ohm\n", + "g_m = 5000 # in \u00b5S\n", + "g_m= g_m*10**-6 # in S\n", + "A_v = g_m *r_d \n", + "V_out = A_v # in V\n", + "print \"The output volatge = %0.1f mV\" %(V_out*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output volatge = 13.2 mV\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.6\n", + ": Page No 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = -2 # in V\n", + "V_P = -5 # in V\n", + "V_DS = V_GS-V_P # in V\n", + "I_DSS = 8 # in mA\n", + "I_DS = I_DSS*( 1-(V_GS/V_P) )**2 # in mA\n", + "print \"The drain current = %0.2f mA\" %I_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 2.88 mA\n" + ] + } + ], + "prompt_number": 23 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.7\n", + ": Page No 370" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 8.4 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "V_P = -3 # in V\n", + "V_GS = -1.5 # in V\n", + "I_D = I_DSS*( 1-(V_GS/V_P) )**2 # in A\n", + "print \"The drain current = %0.1f mA\" %(I_D*10**3)\n", + "V_GS1 = 0 # in V\n", + "g_mo = -( (2*I_DSS)/V_P ) # in A/V\n", + "g_m = g_mo*(1-(V_GS/V_P)) # in A/V\n", + "print \"Transconductacne = %0.1f mA/V\" %(g_m*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 2.1 mA\n", + "Transconductacne = 2.8 mA/V\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.8\n", + ": Page No 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_P = -4 # in V\n", + "V_GS = -2 # in V\n", + "I_DSS = 10 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in A\n", + "print \"The drain current = %0.1f mA\" %(I_D*10**3)\n", + "V_DS = V_P # in V\n", + "print \"The minimum value of V_DS = %0.f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 10.0 mA\n", + "The minimum value of V_DS = -4 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.9\n", + ": Page No 371" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from math import sqrt\n", + "# Given data\n", + "I_DSS = -40 # in mA\n", + "V_P = 5 # in V\n", + "I_D = -15 # in mA\n", + "V_GS = V_P*(1-sqrt(I_D/I_DSS)) # in V\n", + "print \"The gate source voltage = %0.3f V\" %V_GS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate source voltage = 1.938 V\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.10\n", + ": Page No 372" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 4 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "V_P = -4 # in V\n", + "V_GG = -2 # in V\n", + "V_GS = V_GG # in V\n", + "print \"The value of V_GS = %0.f V\" %V_GS\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in A\n", + "print \"The value of I_D = %0.f mA\" %(I_D*10**3)\n", + "V_DD = 10 # in V\n", + "R_D = 5 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The value of V_DS = %0.f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_GS = -2 V\n", + "The value of I_D = 1 mA\n", + "The value of V_DS = 5 V\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.11\n", + ": Page No 373" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols, solve, N#Given data\n", + "I_D= symbols('I_D')\n", + "# Given data\n", + "V_DD= 20 # in V\n", + "R1= 2.1*10**6 # in \u03a9\n", + "R2= 270*10**3 # in \u03a9\n", + "R_D= 4.7 # in k\u03a9\n", + "R_S= 1.5 # in k\u03a9\n", + "I_DSS= 8 # in mA\n", + "V_P= -4 # in V\n", + "V_G= V_DD*R2/(R1+R2) # in V\n", + "# V_GS= V_G-R_S*I_D (as Vs= I_D*R_S) and \n", + "# I_D= I_DSS*(1-V_GS/V_P)**2 # in A\n", + "# I_D= I_DSS*(1-(V_G-R_S*I_D)/V_P)**2 # in mA or\n", + "# I_D= I_D**2*I_DSS*R_S**2/V_P**2 + I_D*(2*R_S*I_DSS/V_P-2*V_G*R_S*I_DSS/V_P**2-1) + I_DSS*(1+V_G**2/V_P**2-2*V_G/V_P)\n", + "expr= I_D**2*I_DSS*R_S**2/V_P**2 + I_D*(2*R_S*I_DSS/V_P-2*V_G*R_S*I_DSS/V_P**2-1) + I_DSS*(1+V_G**2/V_P**2-2*V_G/V_P)\n", + "I_D , x1= solve(expr, I_D)\n", + "I_DQ= I_D # in mA\n", + "print \"The value of I_DQ = %0.2f mA\" %I_DQ\n", + "V_GSQ= V_G-R_S*I_D # in V\n", + "print \"The value of V_GSQ = %0.3f V\" %V_GSQ\n", + "V_DSQ= V_DD-I_DQ*(R_D+R_S) # in V\n", + "print \"The value of V_DSQ = %0.2f V\" %V_DSQ\n", + "V_S= I_D*R_S # in V\n", + "V_D= V_S+V_DSQ #in V\n", + "V_DS= V_D-V_G # in V\n", + "print \"The value of V_S = %0.3f V\" %V_S\n", + "print \"The value of V_D = %0.3f V\" %V_D\n", + "print \"The value of V_DS = %0.3f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_DQ = 2.65 mA\n", + "The value of V_GSQ = -1.698 V\n", + "The value of V_DSQ = 3.57 V\n", + "The value of V_S = 3.976 V\n", + "The value of V_D = 7.542 V\n", + "The value of V_DS = 5.263 V\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.12\n", + ": Page No 374" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "I_D= symbols('I_D')\n", + "# Given data\n", + "V_DD= 20 # in V\n", + "I_DSS= 9 # in mA\n", + "V_BB= -10 # in V\n", + "R_S= 1.5 # in k\u03a9\n", + "R_D= 1.8 # in k\u03a9\n", + "V_P= -3 # in V\n", + "V_G=0 \n", + "# V_S= I_D*R_S+V_BB \n", + "# V_GS= V_G-V_S or\n", + "# V_GS= V_G-(I_D*R_S+V_BB)\n", + "# I_D= I_DSS*(1-V_GS/V_P)**2 or\n", + "# I_D**2*R_S**2 + I_D*[2*R_S*V_BB+2*V_P*R_S-V_P**2/I_DSS]+[V_P**2+V_BB**2+2*V_BB*V_P]\n", + "expr = I_D**2*R_S**2 + I_D*(2*R_S*V_BB+2*V_P*R_S-V_P**2/I_DSS)+(V_P**2+V_BB**2+2*V_BB*V_P)\n", + "I_D , x1= solve(expr, I_D)\n", + "I_DQ= I_D # in mA\n", + "print \"The value of I_DQ = %0.2f mA\" %I_DQ\n", + "V_GS= V_G-(I_D*R_S+V_BB) # in V\n", + "V_GSQ= V_GS # in V\n", + "print \"The value of V_GSQ = %0.3f volts\" %V_GSQ\n", + "V_DS= V_DD-I_D*(R_D+R_S)-V_BB # in V\n", + "print \"The value of V_DS = %0.3f volts\" %V_DS\n", + "V_S= I_D*R_S+V_BB # in V\n", + "print \"The value of V_S = %0.3f volts\" %V_S\n", + "V_D= V_S+V_DS # in V\n", + "print \"The value of V_D = %0.3f volts\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_DQ = 6.91 mA\n", + "The value of V_GSQ = -0.371 volts\n", + "The value of V_DS = 7.185 volts\n", + "The value of V_S = 0.371 volts\n", + "The value of V_D = 7.555 volts\n" + ] + } + ], + "prompt_number": 45 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.13\n", + ": Page No 376" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_S = 1.7 # in V\n", + "R_S = 0.51 # in kohm\n", + "R_S= R_S*10**3 # in ohm\n", + "V_DD = 18 # in V\n", + "R_D = 2*10**3 # in ohm\n", + "V_GS = -1.7 # in V\n", + "V_P = - 4.5 # in V\n", + "I_DQ = V_S/R_S #in A\n", + "print \"The value of I_DQ = %0.2f mA\" %(I_DQ*10**3)\n", + "V_GSQ = -V_S # in V\n", + "print \"The value of V_GSQ = %0.1f V\" %V_GSQ\n", + "I_DSS = I_DQ/( (1-(V_GS/V_P))**2 ) # in A\n", + "print \"The value of I_DSS = %0.1f mA\" %(I_DSS*10**3)\n", + "V_D = V_DD - (I_DQ*R_D) # in V\n", + "print \"The value of V_D = %0.2f V\" %V_D\n", + "V_DS = V_D-V_S # in V\n", + "print \"The value of V_DS = %0.2f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_DQ = 3.33 mA\n", + "The value of V_GSQ = -1.7 V\n", + "The value of I_DSS = 8.6 mA\n", + "The value of V_D = 11.33 V\n", + "The value of V_DS = 9.63 V\n" + ] + } + ], + "prompt_number": 46 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.14\n", + ": Page No 377" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "I_DSS = 12 # in mA\n", + "V_GS = 0 # in V\n", + "I_D = 0 # in mA\n", + "V_P = -6 # in V\n", + "V_GS= np.arange(0,V_P,-0.1) # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # mA\n", + "plt.subplot(1,2,1)\n", + "plt.plot(V_GS,I_D) \n", + "plt.xlabel('V_GS in volts')\n", + "plt.ylabel('I_D in mA')\n", + "plt.title('n-channel device')\n", + "V_P = 6 # in V\n", + "V_GS= np.arange(0,V_P,0.1) # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # mA\n", + "plt.subplot(1,2,2)\n", + "plt.plot(V_GS,I_D) \n", + "plt.xlabel(\"V_GS in volts\")\n", + "plt.ylabel(\"I_D in mA\")\n", + "plt.title(\"p-channel device\")\n", + "print \"\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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SJaBtW75ffWWncZrbUA8AnDwJhIcDaWnmv/U0kQ8N9RBVio/n++4rbe6+3Pz8+LYUO3aI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\n/IkRN7BkXwAAAABJRU5ErkJggg==\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f2d0df80250>" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.15\n", + ": Page No 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 30 # in mA\n", + "V_GS = -5 # in V\n", + "V_GS_off = -8 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_GS_off))**2 # in mA\n", + "print \"The drain current = %0.3f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 4.219 mA\n" + ] + } + ], + "prompt_number": 62 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.16\n", + ": Page No 378" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 1.5 # in mA\n", + "I_DSS = 5 # in mA\n", + "V_P = -2 # in V\n", + "V_GS = V_P*(1-sqrt(I_D/I_DSS)) # in V\n", + "V_G = 0 # in V\n", + "V_S = V_G-V_GS # in V\n", + "R_S = V_S/I_D # in kohm\n", + "print \"The source resistance = %0.f ohm\" %(R_S*10**3)\n", + "V_DD = 20 # in V\n", + "V_DS= 10 # in V\n", + "R_D = (V_DD-(V_DS+(I_D*R_S)))/(I_D) # in kohm\n", + "print \"The diode resistance = %0.f K ohm\" %R_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The source resistance = 603 ohm\n", + "The diode resistance = 6 K ohm\n" + ] + } + ], + "prompt_number": 63 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.17\n", + ": Page No 379" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 0.8 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "I_DSS = 1.645 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "V_P = -2 # in V\n", + "V_GS = V_P * (1-sqrt(I_D/I_DSS)) # in V\n", + "print \"The gate source voltage = %0.2f V\" %V_GS\n", + "g_mo = -((2*I_DSS)/V_P) # in A/V\n", + "g_m = g_mo*(1-(V_GS/V_P)) # in A/V\n", + "print \"The transconductance = %0.2f mA/V\" %(g_m*10**3)\n", + "R_S = -(V_GS/I_D) # in ohm\n", + "print \"The source resistance = %0.f ohm\" %R_S\n", + "AdB= 20 # in dB\n", + "A= 10**(AdB/20) \n", + "R_D= A/g_m # in ohm\n", + "print \"The value of R_D = %0.2f k\u03a9\" %(R_D*10**-3)\n", + "\n", + "# Note: There is calculation error to find the value of R_S in the book . So the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate source voltage = -0.61 V\n", + "The transconductance = 1.15 mA/V\n", + "The source resistance = 757 ohm\n", + "The value of R_D = 8.72 k\u03a9\n" + ] + } + ], + "prompt_number": 64 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.18\n", + ": Page No 381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GG = 2 # in V\n", + "V_GS = -V_GG # in V\n", + "print \"The value of V_GS = %0.f V\" %V_GS\n", + "I_DSS = 10 # in mA\n", + "V_P = -8 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in mA\n", + "I_DQ= I_D # in mA\n", + "print \"The value of I_DQ = %0.3f mA\" %I_DQ\n", + "R_D = 2 # in K ohm\n", + "V_DD = 16 # in V\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The value of V_DS = %0.2f V\" %V_DS\n", + "V_D = V_DS # in V\n", + "print \"The value of V_D = %0.2f V\" %V_D\n", + "V_G = V_GS # in V\n", + "print \"The value of V_G = %0.f V\" %V_G\n", + "V_S = 0 # in V\n", + "print \"The value of V_S = %0.f V\" %V_S" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_GS = -2 V\n", + "The value of I_DQ = 5.625 mA\n", + "The value of V_DS = 4.75 V\n", + "The value of V_D = 4.75 V\n", + "The value of V_G = -2 V\n", + "The value of V_S = 0 V\n" + ] + } + ], + "prompt_number": 65 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.19\n", + ": Page No 381" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 10 # in V\n", + "I_G = 0.001 # in \u00b5A\n", + "R_GS = V_GS/I_G # in M\u03a9\n", + "print \"The gate source resistance = %0.f M\u03a9\" %R_GS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The gate source resistance = 10000 M\u03a9\n" + ] + } + ], + "prompt_number": 66 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.20\n", + ": Page No 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "del_VDS = 1.5 # in V\n", + "del_ID = 120 * 10**-6 # in A\n", + "r_d = del_VDS/del_ID # in ohm\n", + "r_d = r_d * 10**-3 # in kohm\n", + "print \"The drain resistance of the JFET = %0.1f K ohm\" %r_d" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain resistance of the JFET = 12.5 K ohm\n" + ] + } + ], + "prompt_number": 67 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.21\n", + ": Page No 382" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 8.4 # in mA\n", + "V_P = -3 # in V\n", + "V_GS = -1.5 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # in mA\n", + "g_mo = -( (2*I_DSS)/V_P ) # in mA/V\n", + "g_m = g_mo*(1-(V_GS/V_P)) # in mA/V\n", + "print \"The value of g_m = %0.1f mA/V\" %g_m" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of g_m = 2.8 mA/V\n" + ] + } + ], + "prompt_number": 68 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.22\n", + ": Page No 383" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from sympy import symbols, solve, N#Given data\n", + "V_GS= symbols('V_GS')\n", + "# Given data\n", + "V_DD= 20 # in V\n", + "I_DSS= 9 # in mA\n", + "V_P= -3 # in V\n", + "R1= 0.3*10**3 # in k\u03a9\n", + "R2= 1.7*10**3 #in k\u03a9\n", + "R_D= 3.2 # in k\u03a9\n", + "R=1 # in k\u03a9\n", + "V_G= V_DD*R1/(R1+R2) # in V\n", + "#I_D= I_DSS*[1-V_GS/V_P]**2 (i)\n", + "# V_G= V_GS+I_D*R or I_D= (V_G-V_GS)/R (ii)\n", + "# From (i) and (ii)\n", + "#V_GS*1/V_P**2+V_GS*[1/(R*I_DSS)-2/V_P]+[1-V_G/(R*I_DSS)]=0\n", + "expr= V_GS**2*(R*I_DSS/V_P**2)+V_GS*(1-2*R*I_DSS/V_P)+(R*I_DSS-V_G)\n", + "x1 , V_GS= solve(expr, V_GS)\n", + "I_D= I_DSS*(1-V_GS/V_P)**2 # in mA\n", + "print \"The value of I_D = %0.f mA\" %I_D\n", + "V_S= I_D*R #in V\n", + "V_D= V_DD-I_D*R_D # in V\n", + "V_DS= V_D-V_S # in V\n", + "gm= -2*I_DSS/V_P*(1-V_GS/V_P) # in mA/V\n", + "print \"The value of V_DS = %0.1f volts\" %V_DS\n", + "print \"The transconductance = %0.f mA/V\" %gm" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of I_D = 4 mA\n", + "The value of V_DS = 3.2 volts\n", + "The transconductance = 4 mA/V\n" + ] + } + ], + "prompt_number": 87 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.23\n", + ": Page No 385" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "r_d = 25 # in k\u03a9\n", + "R1 = r_d # in k\u03a9\n", + "R2 = r_d # in k\u03a9\n", + "g_m = 2 #mA/V\n", + "g_m= g_m*10**-3 # in A/V\n", + "R_L = (r_d*R1*R2)/(r_d*R1+R1*R2+R2*r_d) # in k\u03a9\n", + "R_L= R_L*10**3 # in \u03a9\n", + "A_v = -g_m*R_L \n", + "print \"The voltage gain = %0.2f\" %A_v" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The voltage gain = -16.67\n" + ] + } + ], + "prompt_number": 88 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.24\n", + ": Page No 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 15 # in V\n", + "I_G = 1 # in nA\n", + "I_G =I_G * 10**-9 # in A\n", + "R_in = V_GS/I_G # in \u03a9\n", + "print \"Input resistance = %0.f G\u03a9\" %(R_in*10**-9)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Input resistance = 15 G\u03a9\n" + ] + } + ], + "prompt_number": 90 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.25\n", + ": Page No 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 20 # in mA\n", + "V_P = 4 # in V\n", + "I_D = I_DSS # in mA\n", + "print \"The maximum drain current = %0.f mA\" %I_D\n", + "V_GS = -V_P # in V\n", + "print \"The gate source cut off voltage = %0.f volts\" %V_GS\n", + "R_DS = V_P/I_DSS # in k\u03a9\n", + "print \"The value of ohmic resistance = %0.f \u03a9\" %(R_DS*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The maximum drain current = 20 mA\n", + "The gate source cut off voltage = -4 volts\n", + "The value of ohmic resistance = 200 \u03a9\n" + ] + } + ], + "prompt_number": 92 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.26\n", + ": Page No 386" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS= 16*10**-3 # in A\n", + "V_GSoff= -6 #in V\n", + "V_GS= V_GSoff/2 # in V\n", + "I_D= I_DSS*(1-V_GS/V_GSoff)**2 # in A\n", + "print \"The drain current = %0.f mA\" %(I_D*10**3)\n", + "V_GS= abs(V_GSoff)/2 # in V\n", + "print \"The gate voltage = %0.f volts\" %V_GS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 4 mA\n", + "The gate voltage = 3 volts\n" + ] + } + ], + "prompt_number": 93 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.27\n", + ": Page No 387" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_DD = 15 # in V\n", + "R_D = 10 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "I_D = V_DD/R_D # in A\n", + "print \"The drain current = %0.1f mA\" %(I_D*10**3)\n", + "V_D = V_DD - I_D*R_D # in V\n", + "print \"The drain voltage = %0.f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 1.5 mA\n", + "The drain voltage = 0 V\n" + ] + } + ], + "prompt_number": 94 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.28\n", + ": Page No 388" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "R2 = 1 # in M ohm\n", + "R2 = R2 * 10**6 # in ohm\n", + "R1 = 1.5 # in M ohm\n", + "R1 = R1 * 10**6 # in ohm\n", + "V_DD = 25 # in V\n", + "V_G = (R2*V_DD)/(R1+R2) # in V\n", + "R_S = 22 # in kohm\n", + "R_S = R_S * 10**3 # in ohm\n", + "I_D = V_G/R_S # in A\n", + "print \"The drain current = %0.2f mA\" %(I_D*10**3)\n", + "R_D = 10 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_D = V_DD - (I_D*R_D) #in V\n", + "V_S = 10 # in V\n", + "V_DS = V_D - V_S # in V\n", + "print \"The Drain source voltage = %0.1f V\" %V_DS\n", + "print \"Thus the Q-point is : (\",round(V_DS,1),\"V,\",round(I_D*10**3,2),\"mA)\"\n", + "I_Dsat = V_DD/R_D # in A\n", + "V_DS = V_DD # in V\n", + "V_D= np.arange(0,25,0.1) # in V\n", + "I_D= (V_DD-V_D)/R_D*10**3 # in mA\n", + "plt.plot(V_D,I_D) \n", + "plt.xlabel(\"V_DS in volts\") \n", + "plt.ylabel(\"I_D in mA\") \n", + "plt.title(\"DC load line\") \n", + "print \"DC load line shown in figure\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 0.45 mA\n", + "The Drain source voltage = 10.5 V\n", + "Thus the Q-point is : ( 10.5 V, 0.45 mA)\n", + "DC load line shown in figure" + ] + }, + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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+ "text": [ + "<matplotlib.figure.Figure at 0x7f2d246a6ad0>" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.29\n", + ": Page No 389" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_SS = 25 # in V\n", + "V_GS = 0 # in V\n", + "R_S = 18 # in kohm\n", + "R_S = R_S * 10**3 # in ohm\n", + "I_D = (V_SS-V_GS)/R_S # in A\n", + "print \"The drain current = %0.2f mA\" %(I_D*10**3)\n", + "V_DD = 25 # in V\n", + "R_D = 7.5 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_D = V_DD - (I_D*R_D) # in V\n", + "print \"The drain voltage = %0.2f V\" %V_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 1.39 mA\n", + "The drain voltage = 14.58 V\n" + ] + } + ], + "prompt_number": 102 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 6.30\n", + ": Page No 390 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_D = 1 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "V_in = 2 # in mV\n", + "V_in = V_in * 10**-3 # in V\n", + "R_L = 10 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in ohm\n", + "g_m = 3000 #in \u00b5S\n", + "g_m = g_m * 10**-6 # in S\n", + "A_v = g_m*r_d \n", + "V_out = A_v*V_in # in V\n", + "V_out = V_out * 10**3 # in mV\n", + "print \"The output Voltage = %0.2f mV\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output Voltage = 5.45 mV\n" + ] + } + ], + "prompt_number": 103 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb new file mode 100644 index 00000000..299a6c38 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_7_2.ipynb @@ -0,0 +1,519 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 7 : Metal Oxide Semiconductor Field Effect Transistors (MOSFET)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.1\n", + ": Page No 403" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "V_GS = 0 # in V\n", + "I_D = 4 # in mA\n", + "R = 2 # in kohm\n", + "V_DD = 15 # in V\n", + "V_DS = V_DD - (I_D*R) # in V\n", + "g_m = 2000 # in \u00b5S\n", + "g_m= g_m*10**-6 # in S\n", + "g_mo = g_m # in S\n", + "R_D = 2 # in kohm\n", + "R_D = R_D * 10**3 # in ohm\n", + "R_L = 10 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in ohm\n", + "A_v = g_m*r_d \n", + "V_in = 20 # in mV\n", + "V_out = A_v * V_in # in mV\n", + "print \"The output voltage = %0.1f mV\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 66.7 mV\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.2\n", + ": Page No 411" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 20 # in V\n", + "V2 = 2 # in V\n", + "V = V1-V2 # in V\n", + "R = 1 # in kohm\n", + "R = R * 10**3 # in ohm\n", + "I_D = V/R # in A\n", + "I_D = I_D * 10**3 # in mA\n", + "print \"The drain current = %0.f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 18 mA\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.3\n", + ": Page No 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "\n", + "\n", + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "I_DSS = 10 # in mA\n", + "V_GS = 0 # in V\n", + "I_D = 0 # in mA\n", + "V_P = -4 # in V\n", + "V_GS= np.arange(0,V_P,-0.1) # in V\n", + "I_D = I_DSS*(1-(V_GS/V_P))**2 # mA\n", + "plt.plot(V_GS,I_D) \n", + "plt.xlabel(\"V_gs in volts\") \n", + "plt.ylabel(\"I_D in mA\") \n", + "plt.title(\"Transfer characteristics for an n-channel depletion type MOSFET\")\n", + "print \"Transfer characteristics for an n-channel depletion type MOSFET Shown in figure\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Transfer characteristics for an n-channel depletion type MOSFET Shown in figure\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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WIio0NmwA1qwBzp4FbG3VjqbgYh8LERUKBw8CI0YAv/4KeHioHU3e0ULdyRYL\nERV4/v7KVfW7dxfspKIV7LYiogLtr7+U+6msWwc0b652NIUDEwsRFVghIcq1Kv/7H9Cjh9rRFB5M\nLERUIEVEKEnlnXeUe9aT+bDznogKnNhY4I03gJYtgSVLCs5V9bmhhbqTiYWICpTERKB7d6BiRaVf\npTAlFUAbdScTCxEVGHo9MHiwkly2bweKFsLzXrVQdxbCYieigkgEePdd4MkT5ZqVwphUtIJFT0QF\nwuzZwG+/KRdAliypdjSFGxMLEeV7K1YA27YpY4DZ2KgdDTGxEFG+9t13yv1UTp8GnJzUjoYAJhYi\nysd27gSmTQN++QVwdVU7GkrFxEJE+dLBg8C4ccChQxz/S2uYWIgo3zlxAvD1VQaV9PZWOxrKiEO6\nEFG+cuEC0K8fsHUr0KyZ2tFQVlRLLJGRkejbty88PDzg6emJ8+fPqxUKEeUTly8rg0muXw+0bat2\nNGSMaofCJkyYgC5dumDHjh1ITk7G8+fP1QqFiPKBW7eAzp2BL74AunZVOxrKjipDukRFRcHb2xt/\n//230Xm0MCwBEWlDYCDQpg3w8cdK3woZp4W6U5VDYYGBgXB0dMSIESNQv359jB49GnFxcWqEQkQa\nFxKijFQ8fTqTSn6hSmJJTk7GxYsXMW7cOFy8eBGlS5fGokWL1AiFiDTsyRMlqYwdq5xaTPnDS/Wx\nxMfHY9++fejXr99LbdTFxQUuLi5o1KgRAKBv375ZJpa5c+canvv4+MDHx+eltkdE+U9YGNCuHTBw\nIDB1qtrRaJefnx/8/PzUDiOdXPex6PV6HDp0CJs3b8bRo0fRsmVL7Ny586U33Lp1a6xduxY1atTA\n3LlzER8fj8WLF/8bmAaOExKROsLDgddfV84A+9//Ct89VV6FFurObBOLiODEiRPYvHkzDhw4gCZN\nmuDUqVMIDAyEpaXlK2348uXLGDVqFBITE+Hu7o7169fD1tb238A0UDhEZH4REUpLpVMnYMECJpUX\npYW6M9vE4uLiAk9PT7z99tvo3r07SpcujSpVqiAwMDDvA9NA4RCReUVGKn0qPj6F75bCpqKFujPb\nzvu+ffsiICAAW7duxd69e3mtCRHlmagooGPHwnmf+oImxz6WlJQU+Pn5YfPmzTh48CAiIyOxbt06\ndO3aFVZWVnkXmAayLhGZR3S0klQaNAA+/5xJ5VVooe58oQskExMTcfjwYWzevBmHDx9GeHh43gWm\ngcIhorxS7uRZAAAZrElEQVQXG6v0p9SpA6xaxaTyqrRQd770lffx8fEoVaqUqeMx0ELhEFHeev4c\n6NIFqFEDWLMGsOCwuK9MC3Vnrr7GvXv3wtvbG/b29rC2toa1tTWcnZ3zOjYiKsBiY5WkUrUqk0pB\nk6sWi7u7O37++WfUrl0bFmb69rWQdYkob8TEKANKvvYa8PXXTCqmpIW6M1dfp4uLC2rVqmW2pEJE\nBVfq2V+1azOpFFS5arGcP38es2fPRtu2bVG8eHFlQZ0OkydPzrvANJB1ici0IiOVpNKoEc/+yita\nqDtzNVbYrFmzYG1tjYSEBCQmJuZ1TERUAEVEAB06KNepLF/OpFKQ5arFUrt2bVy7ds0c8RhoIesS\nkWmEhytX1Ldrx4sf85oW6s5cHd3s0qULDh8+nNexEFEBFBamDCjZsSOTSmGRqxaLlZUV4uLiULx4\ncRQrVkxZUKdDdHR03gWmgaxLRK8mNFRppfTsqdz9kUkl72mh7lTl1sS5oYXCIaKXl3rnxwEDgDlz\nmFTMRQt1J0/0IyKTCwwEWrcG3noLmDuXSaWwYWIhIpO6dUtJKlOmANOmqR0NqeGlbk1MRJSVP/9U\nrqhftAjw9VU7GlJLrhOLXq9HaGgokpOTDe+5urrmSVBElP+cPw+8+Sbw5ZdA375qR0NqylVi+fzz\nz/HRRx/ByckJRYoUMbx/9erVPAuMiPKPX39VOuk3bFAGlqTCLdeDUPr7+8PBwcEcMQHQxpkNRJSz\n/fuBESOAbduUWwqTurRQd+aq897V1RU2NjZ5HQsR5TPbtwNvvw3s3cukQv/K1aGwKlWqoG3btuja\ntavZBqEkIm1btw6YNQs4cgTw8lI7GtKSXCUWV1dXuLq6IjExEYmJiRAR6HhiOlGhJAIsXqzcnMvP\nT7n7I1FavPKeiHItJQWYOlVppRw+DFSooHZElJEW6s5sWywTJkzAZ599hu7du2eaptPpsGfPnjwL\njIi0JSkJGDUKCAgATp4E7O3Vjoi0KtvEMnz4cADAlClTMk3joTCiwiMuDujfXzkMdvQoYGmpdkSk\nZTwURkTZevYM6N4dqFIF+PZb4P8GOCeN0kLdybHCiMiohw+BNm2Axo2BjRuZVCh3mFiIKEsBAcpt\nhAcNApYtAyxYW1AucVchokwuXABatQJmzFAe7FKlF5FjYtmwYQPq168PS0tLWFpaomHDhti4caM5\nYiMiFezZo/SpfPMNMHq02tFQfpTtWWEbN27EZ599hk8//RTe3t4QEVy6dAlTp06FTqcznDVGRAXD\n6tXA//6njP/VqJHa0VB+le1ZYU2aNMGWLVtQpUqVdO8HBQVhwIABuHDhQt4FpoEzG4gKCxFg5kxg\nxw7g4EHA3V3tiOhlaaHuzLbFEhMTkympAICbmxtiYmLyLCgiMp/ERGDkSKWz/swZwNFR7Ygov8s2\nsZQsWfKlphFR/hAVBfTpA1hZAceP88JHMo1sD4WVKlUK1apVy3La3bt3ERcXl3eBaaA5R1SQhYQo\nN+Vq2RJYuRJIcw8/yse0UHdm22K5efOmueIgIjO6fBno0QMYNw54/32eTkymZZIhXZo1a4Zz5869\n8HJ6vR4NGzaEi4sL9u7dmz4wDWRdooIo9Y6PX3yhjP9FBYsW6k6TXCCZkJDwUst99tln8PT05ICW\nRGYgAnz2mXJtyp49TCqUd1S78v7Bgwc4cOAARo0apXp2JSrokpOB8eOVix7PngWaNlU7IirIcnUH\nybwwadIkLFmyBNHR0WqFQFQoREUBAwYoz8+cAWxt1Y2HCj5VWiz79u2Dk5OT4Wp+IsobQUFAixbK\nBY/79jGpkHmYpMXy3XffvdD8Z8+exZ49e3DgwAEkJCQgOjoaw4cPz7SeuXPnGp77+PjAx8fHBNES\nFQ7nzwO9ewPTpwPvvcczvwoqPz8/+Pn5qR1GOtmeFWZlZWW0Y12n05nkMNaJEyewdOlSnhVGZEKb\nNwP//S+wfj3QrZva0ZA5aaHuzLbFEhsba5YgeFYYkWno9cqYX9u2KVfS162rdkRUGPHWxEQFRFQU\nMHiwcn/67duBsmXVjojUoIW6kzf6IioA7txRTiGuUgU4coRJhdTFxEKUzx05otztcdIk5Wp63pee\n1KbadSxE9GpEgOXLgSVLlPuotGqldkRECiYWonwoIQF45x3gyhXltOLKldWOiOhfPBRGlM/cvw+0\naQPExwOnTzOpkPYwsRDlI7/8AjRuDPTtC2zdCpQurXZERJnxUBhRPiCi9KUsXw78+CPw+utqR0Rk\nHBMLkcZFRyv3T3nwAPD3BypVUjsiouzxUBiRht24oRz6cnQETp5kUqH8gYmFSKO2b1c66adNA1av\nBkqUUDsiotzhoTAijUlKAmbMAHbuBA4fBurXVzsiohfDxEKkIffvAwMHAjY2wO+/Aw4OakdE9OJ4\nKIxII/bvBxo1Anr0UJ4zqVB+xRYLkcqSkoAPP1TuobJjB9CypdoREb0aJhYiFaUe+rK1BS5e5KjE\nVDDwUBiRStIe+tq3j0mFCg62WIjMLO2hr507gRYt1I6IyLSYWIjM6O5dYMgQoEwZHvqigouHwojM\nQAT47jvlLo+DBvHQFxVsbLEQ5bGoKOA//wEuXwaOHQO8vNSOiChvscVClIfOnAHq1QPs7IDffmNS\nocKBLRaiPJCcDMybp4zx9fXXyplfRIUFEwuRiQUFKR30lpZKB32FCmpHRGRePBRGZCIiwPr1yrUp\nPXsqA0gyqVBhxBYLkQk8fgyMGQPcuwccPw7Urat2RETqYYuF6BVt36500Nepo9zhkUmFCju2WIhe\nUkQEMH488McfwK5dyjUqRMQWC9FLOXhQaZmULQtcusSkQpQWWyxELyA6Gpg6FTh0CNi4EWjXTu2I\niLSHLRaiXDp4UOlHSU4GrlxhUiEyhi0WohyEhwOTJgGnTgHr1gFvvKF2RETaxhYLkREiwLZtQO3a\nymjEV68yqRDlBlssRFl49AgYNw64fRv46SegWTO1IyLKP9hiIUpDBPj2W2WwyNq1lTO+mFSIXgxb\nLET/5/ZtpZUSFQUcPcqRiIleFlssVOjFxwOzZyu3CO7RAzh/nkmF6FWoklju37+Ptm3bolatWqhd\nuzZWrlypRhhEOHxYOYX45k3lRlwTJgBF2Y4neiU6ERFzb/Tx48d4/Pgx6tWrh9jYWDRo0AC7du2C\nh4fHv4HpdFAhNCokHj5UTiH+7Tfgiy+ALl3UjojINLRQd6rSYilXrhzq1asHALCysoKHhwcePnyo\nRihUyOj1wOefK4e6qlcHrl1jUiEyNdUb/UFBQbh06RKaNGmidihUwJ09C7z3HmBtDZw8CaRpIBOR\nCamaWGJjY9G3b1989tlnsLKyyjR97ty5huc+Pj7w8fExX3BUYISEANOmAX5+wOLFwODBgE6ndlRE\npuHn5wc/Pz+1w0hHlT4WAEhKSkK3bt3QuXNnTJw4MdN0LRwnpPwtIQFYvhxYtgx45x1gxgwgi98v\nRAWKFupOVVosIoKRI0fC09Mzy6RC9CpEgL17gcmTlYscL1wA3N3Vjoqo8FClxXL69Gm0bt0adevW\nhe7/jkksXLgQnTp1+jcwDWRdyn9u3gQmTlRuEfzZZ0CHDmpHRGReWqg7VTsUlhMtFA7lH2FhwMcf\nA5s3AzNnAu++CxQrpnZUROanhbqTV95TvhYfDyxc+O8ZXjduKC0WJhUi9ah+ujHRy9DrgR9+AGbN\nAho3Bs6dU65LISL1MbFQvnPkCPD++0Dp0sDWrRx9mEhrmFgo37h8WUkogYHAokVAr168HoVIi9jH\nQpp36xYwYADQqRPQvTtw/TrQuzeTCpFWMbGQZv39N+DrC7RuDdSvDwQEAOPHs2OeSOuYWEhzHjwA\nxo5VOuWrVgX++ksZkqV0abUjI6LcYGIhzQgNVYay9/IC7OyUOzrOmQPY2qodGRG9CCYWUt3Dh8rw\nKx4eQEqK0oeyaBHg4KB2ZET0MphYSDVBQco95mvXVsb3unpVGYalXDm1IyOiV8HEQmZ35w4wYgTQ\noIFyyOvWLWUU4ooV1Y6MiEyB17GQ2Vy5AixYAPzyi3LDrYAAwN5e7aiIyNTYYqE8JQL8+ivQrRvQ\nsaPSSrl7VxmKhUmFqGBii4XyRFISsH07sHSpMlDk5MnK61Kl1I6MiPIah80nk4qKAr75Bli5Urm5\n1v/7f0DnzoAF28ZEZqGFupMtFjKJ4GAlmaxfrySSXbuUq+WJqPDh70h6aSkpykjDPXsqSUSnA/78\nE/jxRyYVosKMLRZ6Yc+eARs2AF99BVhaKndr/PFHDrlCRAomFsq1S5eAL78Edu4EunRRDns1b85R\nhokoPSYWylZsLLBtG7B2rTI45DvvKBc0OjurHRkRaRXPCqNMRICzZ4FvvwV++kkZtv7tt4GuXYGi\n/ClCpGlaqDtZTZDBo0fAd98pCUWnA0aOBG7e5NhdRPRimFgKufh4YP9+YONG4PRpoE8fpe+kWTP2\nnRDRy2FiKYSSk4Hjx4FNm4A9e5RhVoYMATZvBqys1I6OiPI79rEUEiLAuXNKMtm+HXBzAwYPBvr3\nB8qXVzs6IjIVLdSdbLEUYCkpwO+/Kx3wW7cq43QNHgycOQNUq6Z2dERUUDGxFDBJScDJk8DPPyvD\nqlhZAb16Kc/r1mW/CRHlPSaWAiA+Xhla5eefgX37gCpVlGRy9Khyu18iInNiH0s+FRgIHDwIHDoE\nnDihdMD36qWM21WpktrREZFatFB3MrHkE/HxSgI5dEhJKFFRQKdOyqN9e8DBQe0IiUgLtFB3MrFo\nlF6v3MrXz085zHXmDFCvnpJIOncGvLx4jxMiykwLdScTi0akTSR+fsCpU8p4XD4+QLt2wBtvAHZ2\nKgdJRJqnhbqTiUUlCQnAxYvA+fPKIa6TJ5WhU3x8lEebNhxKhYhenBbqTiYWMxAB7t5VksiFC8rf\n69eVM7aaNFGSCBMJEZmCFupOJhYTS0lRksjly8rj4kUlmZQqBTRtqiSSpk2VOyxaWqodLREVNFqo\nO1VLLIcOHcLEiROh1+sxatQoTJs2LX1gGiicnERHA9eu/ZtELl9WXjs4KJ3rXl5Kh3uTJkDFimpH\nS0SFgRbqTlUSi16vR82aNXHs2DFUrFgRjRo1wubNm+GR5mo+LRQOoFzJ/vffwJ07wO3byt/U59HR\nQKVKfmjZ0seQSOrW1WYnu5+fH3x8fNQOI0f5Ic78ECPAOE0tv8SphbpTlRNW/f39Ua1aNbi5uaFY\nsWIYOHAgdu/erUYoiI9XEsWxY8pw8R99pNyHpH17oHp1wNpaucHV6tVASAjg7Q3Mng389ptyd8WB\nA/2wdi3w3nvKDbG0mFQA5Z8iP8gPceaHGAHGaWr5JU4tUGVIl5CQEFRKc3m4i4sLLly48Mrr1euV\nyj4mRnmEhwNhYcCTJ8rfjM8fPVJaHS4ugKvrv49mzYABA5TnVasCxYu/cmhERIWGKolF94ojId6/\nr9w/JG0SiYlRWh+lSyutDGtrpa/D0RFwclL+urkBjRsrzx0dlbOwnJ15oSERkUmJCs6dOycdO3Y0\nvF6wYIEsWrQo3Tzu7u4CgA8++OCDjxd4uLu7m7tKz0SVzvvk5GTUrFkTx48fR4UKFdC4ceNMnfdE\nRJQ/qXIorGjRovjiiy/QsWNH6PV6jBw5kkmFiKiA0OwFkkRElD9prtt62bJlsLCwQERERJbTDx06\nhNdeew3Vq1fH4sWLzRwdMGvWLHh5eaFevXpo164d7t+/n+V8bm5uqFu3Lry9vdG4cWNNxqh2WU6d\nOhUeHh7w8vJC7969ERUVleV8apYlkPs41S7P7du3o1atWihSpAguXrxodD61yzO3capdnhEREWjf\nvj1q1KiBDh06IDIyMsv51CjP3JTNf//7X1SvXh1eXl64dOmSWeIyULeLJ7179+5Jx44dxc3NTcLD\nwzNNT05OFnd3dwkMDJTExETx8vKSGzdumDXG6Ohow/OVK1fKyJEjs5zP2Gcwh9zEqIWyPHLkiOj1\nehERmTZtmkybNi3L+dQsS5HcxamF8rx586bcvn1bfHx85I8//jA6n9rlmZs4tVCeU6dOlcWLF4uI\nyKJFizSzf+ambPbv3y+dO3cWEZHz589LkyZNzBafiIimWiyTJ0/GJ598YnS6Fi6stLa2NjyPjY1F\n2bJljc4rKh1lzE2MWijL9u3bw+L/zvVu0qQJHjx4YHRetcoSyF2cWijP1157DTVq1MjVvGqWZ27i\n1EJ57tmzB76+vgAAX19f7Nq1y+i85izP3JRN2tibNGmCyMhIhIaGmi1GzSSW3bt3w8XFBXXr1jU6\nT1YXVoaEhJgjvHRmzpwJV1dXbNy4EdOnT89yHp1OhzfeeAMNGzbEN998Y+YIc45RK2WZ6ttvv0WX\nLl2ynKZ2WaZlLE6tlWd2tFSexmihPENDQ+Hs7AwAcHZ2Nloxm7s8c1M2Wc2T3Q83UzPrWWHt27fH\n48ePM70/f/58LFy4EEeOHDG8l9UvgFe9sDK3jMW5YMECdO/eHfPnz8f8+fOxaNEiTJo0CevXr880\n75kzZ1C+fHmEhYWhffv2eO2119CqVSvNxKiVsgSU77948eIYPHhwluvI67I0RZxaKs+caKU8s6N2\nec6fPz9TPMZiMkd5ZowlNzLWoeYqU8DMieXo0aNZvn/t2jUEBgbCy8sLAPDgwQM0aNAA/v7+cHJy\nMsxXsWLFdB3R9+/fh4uLi9nizGjw4MFGf2WXL18eAODo6IhevXrB39/fpDvbq8aolbLcsGEDDhw4\ngOPHjxudJ6/L0hRxaqU8c0ML5ZkTLZSns7MzHj9+jHLlyuHRo0fp6qK0zFGeaeWmbDLO8+DBA1Q0\n5xDrZu3RySVjnWFJSUlStWpVCQwMlH/++UeVDr07d+4Ynq9cuVKGDh2aaZ7nz58bOtBjY2OlefPm\ncvjwYU3FqIWyPHjwoHh6ekpYWJjRedQuS5HcxamF8kzl4+Mjv//+e5bTtFCeqbKLUwvlOXXqVMOI\nIAsXLsyy816N8sxN2aTtvD937pzZO+81mViqVKliSCwhISHSpUsXw7QDBw5IjRo1xN3dXRYsWGD2\n2Pr06SO1a9cWLy8v6d27t4SGhmaK8+7du+Ll5SVeXl5Sq1Yts8eZmxhF1C/LatWqiaurq9SrV0/q\n1asn//nPfzLFqXZZ5jZOEfXL86effhIXFxcpWbKkODs7S6dOnTLFqYXyzE2cIuqXZ3h4uLRr106q\nV68u7du3l2fPnmWKU63yzKpsVq9eLatXrzbM8+6774q7u7vUrVs327ME8wIvkCQiIpPSzFlhRERU\nMDCxEBGRSTGxEBGRSTGxEBGRSTGxEBGRSTGxEBGRSTGxEBGRSTGxUL70+uuvpxtbDgBWrFiBcePG\nmXxbe/fuzdP7gQQFBaFOnToAgMuXL+PgwYN5ti0ic2BioXxp0KBB2LJlS7r3tm7danQgy1fRvXt3\nTJs2zeTrzcqlS5dw4MABs2yLKK8wsVC+1KdPH+zfvx/JyckAlF/9Dx8+RMuWLbOcX0Qwbtw4eHh4\noEOHDujatSt27twJAJg+fTpq1aoFLy8vTJ06NdOyGzZswHvvvQcAeOuttzBhwgS0aNEC7u7uhnWk\nNWPGDKxatcrweu7cuVi2bBkA5W6UderUQd26dbFt27Z0yyUlJWH27NnYunUrvL29sW3bNpw4cQLe\n3t7w9vZG/fr1ERsb+xKlRWReZh3dmMhUypQpg8aNG+PAgQPo0aMHtmzZggEDBhidf+fOnQgODsbN\nmzcRGhoKDw8PjBw5EuHh4di1axdu3boFAIiOjs60bMbhxh8/fowzZ87g5s2b6NGjB/r06ZNu+oAB\nAzBx4kTDYbnt27fjyJEj2LlzJy5fvowrV64gLCwMjRo1Qps2bQzLFStWDB9//DH++OMPrFy5EgDQ\no0cPrFq1Cs2aNUNcXBxKlCjxcgVGZEZssVC+lfZw2NatWzFo0CCj8545cwb9+/cHoAyH3rZtWwCA\nnZ0dSpYsiZEjR+Lnn39GqVKlst2mTqdDz549AQAeHh5Z3vypXr16ePLkCR49eoTLly/D3t4eFStW\nxOnTpzF48GDodDo4OTmhTZs28Pf3T7esKAPDGl63aNECkyZNwueff45nz56hSJEiuSgZInUxsVC+\n1aNHDxw/fhyXLl1CXFwcvL29s50/q/FWixQpAn9/f/Tt2xf79u1Dp06dctxu8eLFs10nAPTr1w87\nduzAtm3bMHDgQABKUso4f043X5o2bRrWrVuH+Ph4tGjRArdv384xPiK1MbFQvmVlZYW2bdtixIgR\nOXbat2jRAjt37oSIIDQ0FH5+fgCA58+fIzIyEp07d8ann36Ky5cvZ1r2ZQYAHzBgADZv3owdO3ag\nX79+AIBWrVph69atSElJQVhYGE6ePInGjRunW87GxgYxMTGG13fv3kWtWrXw/vvvo1GjRkwslC+w\nj4XytUGDBqF3796ZOsIz6tOnD44fPw5PT09UqlQJ9evXh62tLWJiYvDmm28iISEBIoLly5dnWjbj\nbWmNPU/L09MTsbGxcHFxMdw3vVevXjh37hy8vLyg0+mwZMkSODk5ISgoyLCetm3bYtGiRfD29saM\nGTNw+vRp/Prrr7CwsEDt2rXRuXPnFy4jInPj/Vio0Hj+/DlKly6N8PBwNGnSBGfPnjV6u1kienls\nsVCh0a1bN0RGRiIxMRGzZ89mUiHKI2yxUIFy9epVDB8+PN17JUuWxLlz51SKiKjwYWIhIiKT4llh\nRERkUkwsRERkUkwsRERkUkwsRERkUkwsRERkUv8fqco5ssBOZXcAAAAASUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f9785cf0a50>" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.4\n", + ": Page No 414" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 0 # in V\n", + "I_DSS = 10 # in mA\n", + "I_D = I_DSS # in mA\n", + "R_D = 1.5 # in kohm\n", + "V_DD = 20 # in V\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The value of V_DS = %0.f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of V_DS = 5 V\n" + ] + } + ], + "prompt_number": 20 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.5\n", + ": Page No 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 5 # in mA\n", + "V_GS1 = 8 # in V\n", + "V_GS2 = 4 # in V\n", + "V_GS = 6 # in V\n", + "K = I_D/(V_GS1-V_GS2)**2 # in mA/V**2\n", + "I_D = K*(V_GS-V_GS2)**2 # in mA\n", + "print \"The drain current = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The drain current = 1.25 mA\n" + ] + } + ], + "prompt_number": 21 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.6\n", + ": Page No 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_T = 1 # in V\n", + "I_D = 4 # in mA\n", + "V_GS = 5 # in V\n", + "V_GSth = 1 # in V\n", + "K = I_D/(V_GS-V_GSth)**2 # in mA/V**2\n", + "print \"The value of K = %0.2f mA/V**2\" %K" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of K = 0.25 mA/V**2\n" + ] + } + ], + "prompt_number": 22 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.7\n", + ": Page No 415" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_GS = 3 # in V\n", + "V_GSth=2 # inV\n", + "# Part (a)\n", + "print \"Part (a) : For V_DS= 0.5 V\"\n", + "V_DS= 0.5 # in V\n", + "if V_DS<(V_GS-V_GSth) :\n", + " print \"Transistor is in ohmic region\"\n", + "else :\n", + " print \"Transistor is in saturation region\"\n", + "\n", + "# Part (b)\n", + "print \"Part (b) : For V_DS= 1 V\"\n", + "V_DS= 1 # in V\n", + "if V_DS<(V_GS-V_GSth) :\n", + " print \"Transistor is in ohmic region\"\n", + "else :\n", + " print \"Transistor is in saturation region\"\n", + "\n", + "# Part (c)\n", + "print \"Part (c) : For V_DS= 5 V\"\n", + "V_DS= 5 # in V\n", + "if V_DS<(V_GS-V_GSth) :\n", + " print \"Transistor is in ohmic region\"\n", + "else :\n", + " print \"Transistor is in saturation region\"\n" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) : For V_DS= 0.5 V\n", + "Transistor is in ohmic region\n", + "Part (b) : For V_DS= 1 V\n", + "Transistor is in saturation region\n", + "Part (c) : For V_DS= 5 V\n", + "Transistor is in saturation region\n" + ] + } + ], + "prompt_number": 27 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.8\n", + ": Page No 416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 4 # in mA\n", + "V_GSoff = -2 # in V\n", + "V_GS = -0.5 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_GSoff))**2 # in mA\n", + "print \"At V_GS=-0.5 V, the drain current = %0.2f mA\" %I_D\n", + "V_GS = -1 #in V\n", + "I_D = I_DSS*(1-(V_GS/V_GSoff))**2 # in mA\n", + "print \"At V_GS=-1.0 V, the drain current = %0.f mA\" %I_D\n", + "V_GS = -1.5 # in V\n", + "I_D = I_DSS*(1-(V_GS/V_GSoff))**2 # in mA\n", + "print \"At V_GS=-1.5 V, the drain current = %0.2f mA\" %I_D" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "At V_GS=-0.5 V, the drain current = 2.25 mA\n", + "At V_GS=-1.0 V, the drain current = 1 mA\n", + "At V_GS=-1.5 V, the drain current = 0.25 mA\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.9\n", + ": Page No 416" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_DSS = 12 # in mA\n", + "I_DSS= I_DSS*10**-3 # in A\n", + "I_D = I_DSS # in A\n", + "V_DD = 12 # in V\n", + "R_D = 470 # in ohm\n", + "V_DS = V_DD - (I_D*R_D) # in V\n", + "print \"The circuit drain current = %0.f mA\" %(I_D*10**3)\n", + "print \"The drain source voltage = %0.2f V\" %V_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The circuit drain current = 12 mA\n", + "The drain source voltage = 6.36 V\n" + ] + } + ], + "prompt_number": 29 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.10\n", + ": Page No 417" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 12 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "I_DSS = I_D # in A\n", + "V_DS = 6.36 # in V\n", + "g_mo = 4000 # in \u00b5S\n", + "g_mo=g_mo*10**-6 # in S\n", + "g_m = g_mo # in S\n", + "R_D = 470 # in ohm\n", + "R_L = 2 # in kohm\n", + "R_L = R_L * 10**3 # in ohm\n", + "r_d = (R_D*R_L)/(R_D+R_L) # in ohm\n", + "print \"The value of r_d = %0.2f \u03a9\" %r_d\n", + "A_v = g_m*r_d \n", + "print \"The value of A_v = %0.2f\" %A_v\n", + "V_in = 100 # in mV\n", + "V_in = V_in *10**-3 # in V\n", + "V_out = A_v*V_in # in V\n", + "print \"The value of Vout = %0.2f V\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of r_d = 380.57 \u03a9\n", + "The value of A_v = 1.52\n", + "The value of Vout = 0.15 V\n" + ] + } + ], + "prompt_number": 30 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.11\n", + ": Page No 417 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_DS = 0.1 # in V\n", + "I_D = 10 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "R_DS = V_DS/I_D # in ohm\n", + "print \"Part (a) The value of R_DS(on) = %0.f ohm\" %R_DS\n", + "V_DS = 0.75 # in V\n", + "I_D = 100 # in mA\n", + "I_D= I_D*10**-3 # in A\n", + "R_DS = V_DS/I_D # in ohm \n", + "print \"Part (b) The value of R_DS(on) = %0.1f ohm\" %R_DS" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a) The value of R_DS(on) = 10 ohm\n", + "Part (b) The value of R_DS(on) = 7.5 ohm\n" + ] + } + ], + "prompt_number": 31 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 7.12\n", + ": Page No 418 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "I_D = 500 # in mA\n", + "V_GS = 3 # in V\n", + "R_DS = 2 # in ohm\n", + "V_DD = 20 # in V\n", + "R1 = 1 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "V_out = (R_DS/(R1+R_DS))*V_DD # in V\n", + "print \"The output voltage = %0.2f V\" %V_out" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 0.04 V\n" + ] + } + ], + "prompt_number": 32 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb new file mode 100644 index 00000000..869fcc8c --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_8_2.ipynb @@ -0,0 +1,860 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 8 : Operational Amplifiers (OPAMPs)" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.1\n", + ": Page No 432 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "A_V = -100 \n", + "R1 = 2.2 # in kohm\n", + "R1 = R1*10**3 # in ohm\n", + "R_f =-( A_V*R1) # in ohm\n", + "R_f = R_f * 10**-3 # in kohm\n", + "print \"The resistance value = %0.f k\u03a9\" %R_f" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The resistance value = 220 k\u03a9\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.2\n", + ": Page No 432" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 200 # in kohm \n", + "R1 = 2 # in kohm\n", + "A_V = - (R_f/R1) \n", + "V_in = 2.5 # in mV\n", + "V_in= V_in*10**-3 # in V\n", + "V_o = (A_V * V_in) # in V\n", + "print \"The output voltage = %0.2f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -0.25 V\n" + ] + } + ], + "prompt_number": 2 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.3\n", + ": Page No 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 2 # in V\n", + "R_f = 500 # in kohm\n", + "R_f = R_f*10**3 # in ohm\n", + "R1 = 100 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "V_o = (1+(R_f/R1))*V1 # in V\n", + "print \"The output voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 12 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.4\n", + ": Page No 461" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 1 # in Mohm\n", + "R_f = R_f * 10**6 # in ohm\n", + "print \"Part (a)\"\n", + "V1 = 1 # in V\n", + "V2 = 2 # in V\n", + "V3 = 3 # in V\n", + "R1 = 500 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "R2 = 1 # in Mohm\n", + "R2 = R2 * 10**6 # in ohm\n", + "R3 = 1 # in Mohm\n", + "R3 = R3 * 10**6 # in ohm\n", + "V_o = -(R_f) * ( (V1/R1)+(V2/R2)+(V3/R3) ) # in V\n", + "print \"The output voltage = %0.f V\" %V_o\n", + "\n", + "print \"Part (b)\"\n", + "V1 = -2 # in V\n", + "V2 = 3 # in V\n", + "V3 = 1 # in V\n", + "R1 = 200 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "R2 = 500 # in kohm\n", + "R2 = R2 * 10**3 # in ohm\n", + "R3 = 1 # in Mohm\n", + "R3 = R3 * 10**6 # in ohm\n", + "V_o = -(R_f) * ( (V1/R1)+(V2/R2)+(V3/R3) ) # in V\n", + "print \"The output voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (a)\n", + "The output voltage = -7 V\n", + "Part (b)\n", + "The output voltage = 3 V\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.6\n", + ": Page No 462" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 0 # in V\n", + "R1 = 2 # in kohm\n", + "R1 = R1 * 10**3 # in ohm\n", + "A_vmin = (1+(R_f/R1)) \n", + "print \"The minimum closed loop voltage gain = %0.f\" %A_vmin\n", + "R_f1 = 100 # in kohm\n", + "R_f1 = R_f1 * 10**3 # in ohm\n", + "A_vmax = (1+(R_f1/R1)) \n", + "print \"The maximum closed loop voltage gain = %0.f\" %A_vmax" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum closed loop voltage gain = 1\n", + "The maximum closed loop voltage gain = 51\n" + ] + } + ], + "prompt_number": 6 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.7\n", + ": Page No 463" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V1 = 745 # in \u00b5V\n", + "V1 = V1 * 10**-6 # in V\n", + "V2 = 740 # in \u00b5V\n", + "V2 = V2 * 10**-6 # in V\n", + "Av = 5*10**5 \n", + "CMRR = 80 # in dB\n", + "# Formula CMRR in dB= 20*log(Av/Ac)\n", + "Ac= Av/(10**(CMRR/20)) \n", + "print \"The common mode gain = %0.2f\" %Ac\n", + "V_o = Av*(V1-V2)+Ac*((V1+V2)/2) # in V\n", + "print \"The output voltage = %0.2f V\" %V_o\n", + "\n", + "# Note: In the book the calculation of finding the value of common mode gain (i.e. Ac) is wrong, \n", + "# so the answer in the book is wrong" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The common mode gain = 50.00\n", + "The output voltage = 2.54 V\n" + ] + } + ], + "prompt_number": 7 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.8\n", + ": Page No 464" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 1 # in Mohm\n", + "R_f = R_f * 10**6 # in ohm\n", + "Ri= 1*10**6 # in ohm\n", + "R1 = Ri # in ohm\n", + "A_VF = -(R_f/R1) \n", + "print \"The Voltage gain = %0.f\" %A_VF" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The Voltage gain = -1\n" + ] + } + ], + "prompt_number": 8 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.10\n", + ": Page No 465" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_F = 3 # in kohm\n", + "R1 = 1 # in kohm\n", + "V1 = 2 # in V\n", + "V2 = 3 # in V\n", + "V_o1 = (1+(R_F/R1))*V1 # in V\n", + "V_o2 = (1+(R_F/R1))*V2 # in V\n", + "V_o = V_o1+V_o2 # in V\n", + "print \"The output voltage = %0.f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 20 V\n" + ] + } + ], + "prompt_number": 9 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.11\n", + ": Page No 466" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_i = 10 # in k\u03a9 \n", + "R_im = 20 # in k\u03a9\n", + "R_f = 500 # in k\u03a9\n", + "A_vmin = -(R_f/R_i) \n", + "print \"Closed loop voltage gain corresponding to minimum resistance = %0.f\" %A_vmin\n", + "A_vmax = -(R_f/R_im) \n", + "print \"Closed loop voltage gain corresponding to maximum resistance = %0.f\" %A_vmax" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Closed loop voltage gain corresponding to minimum resistance = -50\n", + "Closed loop voltage gain corresponding to maximum resistance = -25\n" + ] + } + ], + "prompt_number": 10 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.12\n", + ": Page No 467" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 200 # in kohm\n", + "R1 = 20 # in kohm\n", + "A_v = -(R_f/R1) \n", + "V_i = 0.1 # in V\n", + "V_im = 0.5 # in V\n", + "V_omin = -10*V_i # in V\n", + "print \"The minimum output voltage = %0.f V\" %V_omin\n", + "V_omax = -10*V_im # in V\n", + "print \"The maximum output voltage = %0.f V\" %V_omax\n", + "print \"Output voltage ranges : from\",int(V_omin,),\"to\",int(V_omax)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum output voltage = -1 V\n", + "The maximum output voltage = -5 V\n", + "Output voltage ranges : from -1 to -5\n" + ] + } + ], + "prompt_number": 14 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.13\n", + ": Page No 467" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "%matplotlib inline\n", + "import matplotlib.pyplot as plt\n", + "import numpy as np\n", + "# Given data\n", + "R = 133 # in kohm\n", + "R = R *10**3 # in ohm\n", + "C = 0.1 # in \u00b5F\n", + "C = 0.1 * 10**-6 # in F\n", + "Vi= 15 # in V\n", + "plt.subplot(2,1,1)\n", + "plt.plot([0,10],[1.5,1.5]) \n", + "plt.ylabel(\"Vi in volts\")\n", + "plt.xlabel(\"t\")\n", + "plt.title(\"Input voltage\") \n", + "t=np.arange(0,1,0.1)\n", + "Vo= -1/(R*C)*t \n", + "plt.subplot(2,1,2)\n", + "plt.plot(t,Vo)\n", + "plt.xlabel(\"t\") \n", + "plt.ylabel(\"Vo in volts\") \n", + "plt.title(\"Output voltage\")\n", + "print \"\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "\n" + ] + }, + { + "metadata": {}, + "output_type": "display_data", + "png": 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dXbt20bhxY/bu3cv06dP1Pp5RS4MUV9++fenbt6+xwzC6Tp065fslWt506NCB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\nIYTQiyQMIYQQevn/1U0AkpdVVxgAAAAASUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7fdbd4185210>" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.14\n", + ": Page No 468" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "Rf = 250 # in kohm\n", + "Vo= '-5*Va+3*Vb' # given expression\n", + "# But output voltage of difference amplifier is \n", + "# Vo= -Rf/R1*Va+(R2/(R1+R2))*(1+Rf/R1)*Vb (i)\n", + "# By comparing (i) with given expression\n", + "R1 = Rf/5 # in kohm\n", + "print \"The value of R1 = %0.f k\u03a9\" %R1\n", + "# (R2/(R1+R2))*(1+Rf/R1)= 3\n", + "R2= 3*R1**2/(R1+Rf-3*R1) # in k\u03a9\n", + "print \"The value of R2 = %0.f k\u03a9\" %R2\n", + "\n", + "# Note: There is calculation error to find the value of R2 in the book." + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The value of R1 = 50 k\u03a9\n", + "The value of R2 = 50 k\u03a9\n" + ] + } + ], + "prompt_number": 28 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.15\n", + ": Page No 469" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_i1 = 150 # in \u00b5V\n", + "V_i2 = 140 # in \u00b5V\n", + "V_d = V_i1-V_i2 # in \u00b5V\n", + "V_C = (1/2)*(V_i1+V_i2) # in \u00b5V\n", + "print \"Part (i)\"\n", + "CMRR = 100 \n", + "A_d = 4000 \n", + "V_o = (A_d * V_d)*(1+(1/CMRR)*(V_C/V_d)) # in \u00b5V\n", + "V_o = V_o * 10**-3 # in mV\n", + "print \"The output voltage = %0.1f mV\" %V_o\n", + "print \"Part(ii)\"\n", + "CMRR = 10**5 \n", + "V_o = (A_d * V_d)*(1+(1/CMRR)*(V_C/V_d)) # in \u00b5V\n", + "V_o = V_o * 10**-3 # in mV\n", + "print \"The output voltage = %0.3f mV\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Part (i)\n", + "The output voltage = 45.8 mV\n", + "Part(ii)\n", + "The output voltage = 40.006 mV\n" + ] + } + ], + "prompt_number": 33 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.16\n", + ": Page No 470" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 470 # in k\u03a9\n", + "R1 = 4.3 # in k\u03a9\n", + "R2 = 33 # in k\u03a9\n", + "R3 = R2 # in k\u03a9\n", + "A1 = (1+R_f/R1) \n", + "A2 = -(R_f/R2) \n", + "A3 = -(R_f/R3) \n", + "A = A1*A2*A3 \n", + "V_i = 80 # in \u00b5V\n", + "V_i= 80*10**-6 # in V\n", + "V_o = A*V_i \n", + "print \"The output voltage = %0.2f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = 1.79 V\n" + ] + } + ], + "prompt_number": 34 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.18\n", + ": Page No 472" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R4 = 300 # in k\u03a9\n", + "R2 = 150 # in k\u03a9\n", + "R3 = 10 # in k\u03a9\n", + "R1 = 10 # in k\u03a9\n", + "V1 = 1 # in V\n", + "V2 = 2 # in V\n", + "V_o = ( (1+(R4/R2))*((R3/(R1+R3))*V1)-((R4/R2)*V2) ) # in V\n", + "print \"The output voltage = %0.1f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -2.5 V\n" + ] + } + ], + "prompt_number": 35 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.19\n", + ": Page No 472" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "V_o = 2 # in V\n", + "R_i = 20 # in k\u03a9\n", + "R_f = 1 # in M\u03a9\n", + "V_i = -((V_o*R_i)/R_f) # in mV\n", + "print \"The input volatge = %0.f mV\" %V_i" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The input volatge = -40 mV\n" + ] + } + ], + "prompt_number": 36 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.20\n", + ": Page No 473" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 200 # in k\u03a9\n", + "R_i = 30 # in k\u03a9\n", + "V_i = 0.1 # in V\n", + "V_im = 0.5 # in V\n", + "Vo_min = -((R_f/R_i)*V_i) # in V\n", + "print \"The minimum output voltage = %0.2f V\" %Vo_min\n", + "Vo_max = -((R_f/R_i)*V_im) # in V\n", + "print \"The minimum output voltage = %0.2f V\" %Vo_max\n", + "print \"The output voltage range is : \",round(Vo_min,2),\"V to\",round(Vo_max,2),\"V\"" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The minimum output voltage = -0.67 V\n", + "The minimum output voltage = -3.33 V\n", + "The output voltage range is : -0.67 V to -3.33 V\n" + ] + } + ], + "prompt_number": 38 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.21\n", + ": Page No 473" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 360 # in kohm\n", + "R_i = 12 # in kohm\n", + "V1 = - 0.3 # in V\n", + "V_o = (1+(R_f/R_i))*V1 # in V\n", + "print \"The output voltage = %0.1f V\" %V_o\n", + "V_o1 = 2.4 # in V\n", + "V_i = V_o1/(1+(R_f/R_i)) # in V\n", + "print \"The input voltage = %0.2f mV\" %(V_i*10**3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -9.3 V\n", + "The input voltage = 77.42 mV\n" + ] + } + ], + "prompt_number": 39 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.22\n", + ": Page No 474" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = -68 # in kohm\n", + "R1 = 33 # in kohm\n", + "R2 = 22 # in kohm\n", + "R3 = 12 # in kohm\n", + "V1 = 0.2 # in V\n", + "V2 = - 0.5 # in V\n", + "V3 = 0.8 # in V\n", + "V_o = ((R_f/R1)*V1) + ((R_f/R2)*V2) + ((R_f/R3)*V3) # in V\n", + "print \"The output voltage = %0.3f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The output voltage = -3.400 V\n" + ] + } + ], + "prompt_number": 40 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.23\n", + ": Page No 475" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 1.8 # in kohm\n", + "R_f = R_f * 10**3 # in ohm\n", + "R1 = 180 # in ohm\n", + "A_v = (R_f/R1) \n", + "print \"Closed loop gain = %0.f\" %A_v\n", + "F = 1 # in MHz\n", + "F = F * 10**6 # in Hz\n", + "f2 = F/A_v # in Hz\n", + "print \"Closed loop bandwidth = %0.f Hz\" %f2\n", + "V_in = 25 # in mV\n", + "V_in = V_in * 10**-3 # in V\n", + "V_o = A_v*V_in # in V\n", + "print \"The output voltage = %0.2f V\" %V_o" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Closed loop gain = 10\n", + "Closed loop bandwidth = 100000 Hz\n", + "The output voltage = 0.25 V\n" + ] + } + ], + "prompt_number": 41 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 8.24\n", + ": Page No 475 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "R_f = 3 # in K ohm\n", + "R_f = R_f * 10**3 # in ohm\n", + "R1 = 150 # in ohm\n", + "A_v = (R_f/R1) + 1 \n", + "print \"Close loop gain for inverting amplifier = %0.f\" %A_v\n", + "f = 1 # in MHz\n", + "f = f * 10**6 # in Hz\n", + "f2 = f/A_v # in Hz\n", + "f2 = f2 * 10**-3 # in KHz\n", + "print \"The closed loop bandwidth = %0.2f KHz\" %f2" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Close loop gain for inverting amplifier = 21\n", + "The closed loop bandwidth = 47.62 KHz\n" + ] + } + ], + "prompt_number": 42 + } + ], + "metadata": {} + } + ] +}
\ No newline at end of file diff --git a/Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb b/Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb new file mode 100644 index 00000000..bbc73967 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_9_2.ipynb @@ -0,0 +1,183 @@ +{ + "metadata": { + "name": "" + }, + "nbformat": 3, + "nbformat_minor": 0, + "worksheets": [ + { + "cells": [ + { + "cell_type": "heading", + "level": 1, + "metadata": {}, + "source": [ + "Chapter - 9 : Electronic Instrumentation And Measurements" + ] + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.1\n", + ": Page No 512 " + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "from __future__ import division\n", + "# Given data\n", + "scale= 5 # in mV/cm\n", + "gh= 5.2 #amplitude of the graph in cm\n", + "PtoPamplitude= gh*scale # in mV\n", + "print \"Peak-to-peak amplitude = %0.f mV\" %PtoPamplitude" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak-to-peak amplitude = 26 mV\n" + ] + } + ], + "prompt_number": 1 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.2\n", + ": Page No 512" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 100 # in mV/cm\n", + "gh= 5.2 #amplitude of the graph in cm\n", + "PtoPamplitude= gh*scale # in mV\n", + "print \"Peak-to-peak amplitude = %0.2f V\" %(PtoPamplitude*10**-3)" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "Peak-to-peak amplitude = 0.52 V\n" + ] + } + ], + "prompt_number": 3 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.3\n", + ": Page No 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 20 # in \u00b5S/cm\n", + "gh= 3.2 #amplitude of the graph in cm\n", + "T= gh*scale # in mV\n", + "print \"The period of the waveform = %0.f \u00b5S\" %T" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The period of the waveform = 64 \u00b5S\n" + ] + } + ], + "prompt_number": 4 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.4\n", + ": Page No 513" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 50 # in \u00b5S/cm\n", + "gh= 2 #amplitude of the graph in cm\n", + "T_PD= gh*scale # in mV\n", + "print \"The pulse delay for the waveform = %0.f \u00b5s\" %T_PD" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pulse delay for the waveform = 100 \u00b5s\n" + ] + } + ], + "prompt_number": 5 + }, + { + "cell_type": "heading", + "level": 2, + "metadata": {}, + "source": [ + "Example 9.5\n", + ": Page No 514" + ] + }, + { + "cell_type": "code", + "collapsed": false, + "input": [ + "# Given data\n", + "scale= 2 # in \u00b5S/cm\n", + "gh= 4.6 #amplitude of the graph in cm\n", + "T_PQ= gh*scale # in mV\n", + "print \"The pulse width of the waveform = %0.1f \u00b5s\" %T_PQ" + ], + "language": "python", + "metadata": {}, + "outputs": [ + { + "output_type": "stream", + "stream": "stdout", + "text": [ + "The pulse width of the waveform = 9.2 \u00b5s\n" + ] + } + ], + "prompt_number": 6 + } + ], + "metadata": {} + } + ] +}
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