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diff --git a/Electronics_Engineering_by_P._Raja/chapter_4.ipynb b/Electronics_Engineering_by_P._Raja/chapter_4.ipynb new file mode 100755 index 00000000..346bf695 --- /dev/null +++ b/Electronics_Engineering_by_P._Raja/chapter_4.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 0x7f9b900aba50>" + ] + } + ], + "prompt_number": 109 + }, + { + "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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\nSUVORK5CYII=\n", + "text": [ + "<matplotlib.figure.Figure at 0x7f9ba804b910>" + ] + } + ], + "prompt_number": 110 + }, + { + "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 0x7f9b92fd8850>" + ] + } + ], + "prompt_number": 91 + }, + { + "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": {} + } + ] +}
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