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-{
- "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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ETU1NSE9Px6xZszBlyhRcunQJly9fxqVLl7Bt27a+qpGoWxw7IDKdwTGCAQMG4Mknn8Sr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\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",
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isdvt2O32Ij1GsVx4lZ6eTq9evdi2bdvVBejCKxFxk7NnzZm6M2aYbZ1/+xtU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\n",
- "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