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diff --git a/Electronic_Circuits_by_Dr._Sanjay_Sharma/Chapter3.ipynb b/Electronic_Circuits_by_Dr._Sanjay_Sharma/Chapter3.ipynb
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+{
+ "metadata": {
+  "name": "",
+  "signature": ""
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+  {
+   "cells": [
+    {
+     "cell_type": "heading",
+     "level": 1,
+     "metadata": {},
+     "source": [
+      "Chapter3 - Bipolar Junction Transistors(BJTs)"
+     ]
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.1 - page 182"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_E= -0.7 # in V\n",
+      "Bita=50 \n",
+      "RC= 5 # in k\u03a9\n",
+      "RE= 10 # in k\u03a9\n",
+      "RE= RE*10**3 # in \u03a9\n",
+      "RC= RC*10**3 # in \u03a9\n",
+      "V_CC= 10 # in V\n",
+      "V_BE= -10 # in volt\n",
+      "I_E= (V_E-V_BE)/RE # in A\n",
+      "print \"Emitter current = %0.2f mA\" %(I_E*10**3)\n",
+      "# I_E= I_B+I_C and I_C= Bita*I_B, so\n",
+      "I_B= I_E/(1+Bita) # in A\n",
+      "print \"Base current = %0.1f \u00b5A\" %(I_B*10**6)\n",
+      "I_C= I_E-I_B #in A\n",
+      "print \"Collector current = %0.2f mA\" %(I_C*10**3)\n",
+      "V_C= V_CC-I_C*RC # in V\n",
+      "print \"The value of V_C = %0.2f Volt\" %(V_C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Emitter current = 0.93 mA\n",
+        "Base current = 18.2 \u00b5A\n",
+        "Collector current = 0.91 mA\n",
+        "The value of V_C = 5.44 Volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 11
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.2 - page 183"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_E= 1.7 # in V\n",
+      "V_B= 1 # in V\n",
+      "RC= 5 # in k\u03a9\n",
+      "RE= 5 # in k\u03a9\n",
+      "RE= RE*10**3 # in \u03a9\n",
+      "RC= RC*10**3 # in \u03a9\n",
+      "RB= 100 #in k\u03a9\\\n",
+      "RB= RB*10**3 # in \u03a9\n",
+      "V_CC= 10 # in V\n",
+      "V_BE= -10 # in volt\n",
+      "I_E= (V_CC-V_E)/RE # in A\n",
+      "I_B= V_B/RB # in V\n",
+      "# Formula I_B=  (1-alpha)*I_E\n",
+      "alpha= 1-I_B/I_E \n",
+      "print \"Value of alpha = %0.3f \" %(alpha)\n",
+      "beta= alpha/(1-alpha) \n",
+      "print \"Value of beta =  %.01f \" %beta\n",
+      "V_C= (I_E-I_B)*RC-V_CC # in volt\n",
+      "print \"Collector voltage = %0.2f Volt\" %V_C\n",
+      "# Answer in the textbook is not accurate."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Value of alpha = 0.994 \n",
+        "Value of beta =  165.0 \n",
+        "Collector voltage = -1.75 Volt\n"
+       ]
+      }
+     ],
+     "prompt_number": 12
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.3 - page 187"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division# Given data \n",
+      "from numpy import log\n",
+      "# Given data\n",
+      "V_CC= 10 # in V\n",
+      "V_CE= 3.2 # in V\n",
+      "RC= 6.8 # in k\u03a9\n",
+      "RC= RC*10**3 # in \u03a9\n",
+      "I_S= 1*10**-15 # in A\n",
+      "V_T= 25*10**-3 # in V\n",
+      "I_C1= (V_CC-V_CE)/RC # in A\n",
+      "print \"Part(a) : \"\n",
+      "# Formula I_C= I_S*%e**(V_BE1/V_T)\n",
+      "V_BE1= V_T*log(I_C1/I_S) # in volt\n",
+      "print \"Collector current = %0.1f mA\" %(I_C1*10**3)\n",
+      "print \"Value of V_BE = %0.1f Volt\" %(V_BE1)\n",
+      "\n",
+      "print \"Part(b) : \"\n",
+      "v_in= 5*10**-3 # in V\n",
+      "Av= -(V_CC-V_CE)/V_T # in V/V\n",
+      "print \"Voltage gain = %0.1f V/V\" %(Av)\n",
+      "v_o= abs(Av )*v_in # in V\n",
+      "print \"Change in output voltage = %0.2f Volt\" %v_o\n",
+      "\n",
+      "print \"Part(c) : \"\n",
+      "#for V_CE= 0.3 V\n",
+      "V_CE= 0.3 # in V\n",
+      "I_C2= (V_CC-V_CE)/RC # in A\n",
+      "# I_C1= I_S*%e**(V_BE1/V_T)      (i)\n",
+      "# I_C2= I_S*%e**(V_BE2/V_T)      (ii)\n",
+      "# divide the equation (ii) by (i)\n",
+      "delta_V_BE= V_T*log(I_C2/I_C1) # in volt    ( where delta_V_BE = V_BE2-V_BE1 )\n",
+      "print \"The positive increament in V_BE = %0.1f mV\" %(delta_V_BE*10**3)\n",
+      "\n",
+      "print \"Part(d) : \"\n",
+      "v_o= 0.99*V_CC # in V\n",
+      "I_C3= (V_CC-v_o)/RC # in A\n",
+      "delta_V_BE= V_T*log(I_C3/I_C1) # in V\n",
+      "print \"The negative increament in V_BE = %0.1f mV\" %(delta_V_BE*10**3 )"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part(a) : \n",
+        "Collector current = 1.0 mA\n",
+        "Value of V_BE = 0.7 Volt\n",
+        "Part(b) : \n",
+        "Voltage gain = -272.0 V/V\n",
+        "Change in output voltage = 1.36 Volt\n",
+        "Part(c) : \n",
+        "The positive increament in V_BE = 8.9 mV\n",
+        "Part(d) : \n",
+        "The negative increament in V_BE = -105.5 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 13
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.4 - page 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_CC= 10 # in V\n",
+      "V_CE= 5 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "I_C= 5*10**-3 # in mA\n",
+      "bita= 100 \n",
+      "R_C= (V_CC-V_CE)/I_C # in \u03a9\n",
+      "I_B= I_C/bita # in A\n",
+      "R_B= (V_CC-V_BE)/I_B # in \u03a9\n",
+      "print \"The value of R_C = %0.1f k\u03a9\" %(R_C*10**-3)\n",
+      "print \"The value of I_B = %0.1f \u00b5A\" %(I_B*10**6)\n",
+      "print \"The value of R_B = %.01f k\u03a9\" %(R_B*10**-3)\n",
+      "\n",
+      "# Note: The value of base current in the book is wrong"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of R_C = 1.0 k\u03a9\n",
+        "The value of I_B = 50.0 \u00b5A\n",
+        "The value of R_B = 186.0 k\u03a9\n"
+       ]
+      }
+     ],
+     "prompt_number": 14
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.5 - page 197"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "%matplotlib inline\n",
+      "from matplotlib.pyplot import plot, text, xlabel, ylabel, show, axes, title, gca\n",
+      "from __future__ import division\n",
+      "from numpy import arange, nditer\n",
+      "# Given data \n",
+      "V_CC= 6 # 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",
+      "# when I_C=0\n",
+      "I_C=0 \n",
+      "V_CE= V_CC-I_C*R_C # in volt\n",
+      "V_CE= arange(0,7,0.1) # in Volt\n",
+      "# defining function to get the collector current\n",
+      "def current(V):\n",
+      "    it = nditer([V, None])\n",
+      "    for v_ce,i in it:\n",
+      "        i[...] = (V_CC-v_ce)/R_C*1000 \n",
+      "    return it.operands[1]\n",
+      "I_C=current(V_CE) # in mA\n",
+      "x=arange(-1,4,0.1)\n",
+      "y=arange(-0.5,1.02,0.1)\n",
+      "plot(V_CE,I_C,'r') \n",
+      "plot(4*(y/y),y,'--b')\n",
+      "plot(x,1*(x/x),'--b')\n",
+      "text(4,1.02,'Operating Point')\n",
+      "title(\"DC load line\")\n",
+      "xlabel(\"V_CE in volts\")\n",
+      "ylabel(\"I_C in mA\")\n",
+      "# Setting axes\n",
+      "axes = gca()\n",
+      "axes.set_xlim([0,6])\n",
+      "axes.set_ylim([0,3])\n",
+      "show()\n",
+      "# When V_CE= 0\n",
+      "I_C= V_CC/R_C #in A\n",
+      "# Operating point for silicon transistor \n",
+      "V_BE= 0.7 # in V\n",
+      "I_B= (V_CC-V_BE)/R_B #in A\n",
+      "I_CQ= bita*I_B # in A\n",
+      "V_CEQ= V_CC-I_CQ*R_C # in volt\n",
+      "print \"Operating point = (\",V_CEQ,\"V ,\",I_CQ*10**3,\"mA)\"\n",
+      "print \"DC load line shown in figure\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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dTYmaTG0juH//PhITEzFkyBD0798fV69eRW5uLq5evYpVq1a1VI1EDePcAVGzqZ0jaNu2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\nbARERCaOjYCIyMT9P7uiwsTxuJHPAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7ff7b76ad790>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Operating point = ( 4.0 V , 1.0 mA)\n",
+        "DC load line shown in figure\n"
+       ]
+      }
+     ],
+     "prompt_number": 7
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.6 - page 203"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# 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 k\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 volt\n",
+      "# For dc load line\n",
+      "# When\n",
+      "I_C=0 \n",
+      "V_CE= V_CC-(I_C+I_BQ)*R_C # in volt\n",
+      "# When\n",
+      "V_CE= 0 \n",
+      "I_C= (V_CC-I_BQ*R_C)/R_C #in A\n",
+      "print \"Q- point values for circuit is\",round(V_CEQ,2),\"V and\",round(I_CQ*10**3),\"mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Q- point values for circuit is 1.72 V and 1.0 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 16
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.7 - page 204"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_CC= 15 # in V\n",
+      "V_BE= 0.7 # 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",
+      "print \"Base current = %0.f \u00b5A\" %(I_B*10**6)\n",
+      "#Apply 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",
+      "print \"The value of R_C = %0.2f k\u03a9\" %(R_C*10**-3)\n",
+      "#Apply KVL to base or input circuit,  V_CC= (I_C+I_B)*R_C+V_CE + I_B*R_B\n",
+      "R_B= (V_CC-V_BE-(I_C+I_B)*R_C)/I_B # in ohm\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": [
+        "Base current = 50 \u00b5A\n",
+        "The value of R_C = 1.98 k\u03a9\n",
+        "The value of R_B = 86 k\u03a9 \n"
+       ]
+      }
+     ],
+     "prompt_number": 17
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.8 - page 205"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_BE= 0.7 # in V\n",
+      "V_CE= 3 # in V\n",
+      "I_C= 1 # in mA\n",
+      "I_C=I_C*10**-3 # in A\n",
+      "bita= 100 \n",
+      "I_B= I_C/bita # in A\n",
+      "# V_CE= V_BE+V_CB and V_CB= I_B*R_B\n",
+      "R_B= (V_CE-V_BE)/I_B # in \u03a9\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_B = 230 k\u03a9\n"
+       ]
+      }
+     ],
+     "prompt_number": 18
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.9 - page 208"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "%matplotlib inline\n",
+      "from numpy import nditer, arange\n",
+      "from matplotlib.pyplot import plot, text, xlabel, ylabel, show, axes, title, gca\n",
+      "# Given data \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",
+      "RC= 1;# in k\u03a9\n",
+      "RC=RC*10**3;# in \u03a9\n",
+      "RE= 2;# in k\u03a9\n",
+      "RE=RE*10**3;# in \u03a9\n",
+      "V_CC= 15;# in V\n",
+      "V_BE= 0.7;# in V\n",
+      "# When\n",
+      "I_C=0;\n",
+      "V_CE= V_CC-I_C*(RC+RE);# in V\n",
+      "# When V_CE= 0\n",
+      "I_C= V_CC/(RC+RE);# in A\n",
+      "V_B= V_CC*R2/(R1+R2);# in V\n",
+      "I_E= (V_B-V_BE)/RE;# in A\n",
+      "I_C= I_E;# in A (approx)\n",
+      "I_CQ= I_C;# in A\n",
+      "V_CE= V_CC-I_C*(RC+RE);# in V\n",
+      "V_CEQ= V_CE;# in V\n",
+      "#############\n",
+      "V_CE= arange(0,16,0.1);# in Volt\n",
+      "def current(v):\n",
+      "    it = nditer([v, None])\n",
+      "    for x,y in it:\n",
+      "        y[...]= (V_CC-x)/(RC+RE)*1000\n",
+      "    return it.operands[1]\n",
+      "I_C = current(V_CE)\n",
+      "\n",
+      "#I_C= (V_CC-V_CE)/(RC+RE)*1000;# in mA\n",
+      "plot(V_CE,I_C);\n",
+      "title(\"DC load line\")\n",
+      "xlabel(\"V_CE in volts\")\n",
+      "ylabel(\"I_C in mA\")\n",
+      "text(8.55,2.15,'Q(8.55V,2.15mA)')\n",
+      "x1=arange(0,8.55,0.01)\n",
+      "y1=arange(0,2.15,0.01)\n",
+      "a=arange(0,8.55,0.01)\n",
+      "yd=2.15*(a/a)\n",
+      "plot(a,yd,'--r')\n",
+      "b=arange(-1,2.15,0.005)\n",
+      "xd=8.55*(b/b)\n",
+      "plot(xd,b,'--r')\n",
+      "show()\n",
+      "print \"DC load line shown in figure\"\n",
+      "print 'Operating point is ',V_CEQ,\" V and \",I_CQ*10**3,\" mA\""
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "metadata": {},
+       "output_type": "display_data",
+       "png": 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AMhERUQP06QOkpEjbZYeGAq+9BhQWyh2V8TEZEBkL9yZSLPZOYJmIyHh405nJ\nMIfeCexnQKRUTAYmxdR7J3DOgIioCVha7wSODIiMhSMDk/bDD8DrrwO5ucDGjdJoQclYJiJSKiYD\nk2dKvRNYJiJSKu5NZPLMuXcCRwZERI2UlSVtaXHpErBhAzB6tNwR/YZlIiIiI1Ni7wSWiYiIjMwc\neicwGRARNQFT753AMhERkQHI3TvB5MpESUlJ6NGjB7p164bVq1fLHQ6R4XBvIotiar0TZB0ZVFZW\nonv37jhy5AicnJwQFBSEuLg4eHp6/hYgRwZkLnifgcUqKAD+/GcgMRFYtQqYOlXaHM+QTGpkkJqa\niq5du8LV1RXNmzfHlClTcODAATlDIiJqch06AJ9+CuzfL218FxwsjRqURNZkcOPGDXTp0kX7tbOz\nM27cuCFjREREhqPk3gmyJgOVSiXn5YmIjE6pvROs5by4k5MTcnJytF/n5OTA2dm51uuWPjLxplar\noVarjRAdEZHh2NpKq4xeflladfTJJ0/XO0Gj0UCj0TQ6HlknkCsqKtC9e3ccPXoUnTt3Rp8+fTiB\nTOZr6VKuKCKdDNE7waQmkK2trbF582aMGDECXl5emDx5co1EQGRWmAioDkroncCbzoiIFKYpeidw\nozoiIjPwtL0TTKpMREREuhm7dwKTARGRgrVqJU03nTkDnD0rzSckJDT9dZgMiIyFE8j0FNzcgH37\ngI8+kvY5GjMGuHKl6c7POQMiY+HeRNREysulJjoffgi8+iqwZAnw7LM1X8M5AyIiM2eI3gkcGRAZ\nC0cGZCCP9k44cABo04ZLS4mUi8mADKiiAvjyS2D8eOl/NSYDIqViMiAj4pwBkVJFR8sdAVGdODIg\nIjJDHBkQEVGDMRkQERGTARERMRkQERGYDIiMh3sTkYJxNRGRsfA+AzIiriYiIqIGYzIgIiImAyIi\nYjIgIiIwGRAZD/cmIgXjaiIiIjPE1URERNRgTAZERMRkQERETAZERAQmAyLj4d5EpGCyJYO9e/fC\n29sbzZo1Q1pamlxhEBnPsmVyR0BUJ9mSga+vL/bt24dBgwbJFUKT0mg0coegF1OI0xRiBBhnU2Oc\n8pItGfTo0QMeHh5yXb7Jmcr/IKYQpynECDDOpsY45cU5AyIigrUhTz58+HDcvHmz1vEVK1YgLCzM\nkJcmIqKGEDJTq9Xi3LlzdT7v7u4uAPDBBx988NGAh7u7e4M+iw06MtCXqGf/jCtXrhgxEiIiyyTb\nnMG+fftwXOMYAAAJ20lEQVTQpUsXpKSkYPTo0Rg1apRcoRARWTzF71pKRESGp9jVRElJSejRowe6\ndeuG1atXyx2OTjk5OQgJCYG3tzd8fHywceNGuUOqV2VlJQICAhQ9ef/zzz9j0qRJ8PT0hJeXF1JS\nUuQOSaeVK1fC29sbvr6+iIyMxP/+9z+5QwIAzJo1C46OjvD19dUeKyoqwvDhw+Hh4YHnn38eP//8\ns4wRSnTFuWjRInh6esLPzw8TJkxAcXGxjBHqjrHa2rVrYWVlhaKiIhkiq6muODdt2gRPT0/4+Phg\n8eLFTz7RU83+GkhFRYVwd3cXWVlZory8XPj5+YnvvvtO7rBqyc/PF+np6UIIIe7duyc8PDwUGWe1\ntWvXisjISBEWFiZ3KHWaPn26+Oyzz4QQQjx8+FD8/PPPMkdUW1ZWlnBzcxMPHjwQQggRHh4uYmNj\nZY5KcuLECZGWliZ8fHy0xxYtWiRWr14thBBi1apVYvHixXKFp6UrzsOHD4vKykohhBCLFy+WPU5d\nMQohxPXr18WIESOEq6urKCwslCm63+iK89ixY2LYsGGivLxcCCFEQUHBE8+jyJFBamoqunbtCldX\nVzRv3hxTpkzBgQMH5A6rlo4dO8Lf3x8A0KZNG3h6eiIvL0/mqHTLzc1FYmIiXn75ZcU2CyouLsbJ\nkycxa9YsAIC1tTVsbGxkjqq2du3aoXnz5igrK0NFRQXKysrg5OQkd1gAgIEDB8LW1rbGsS+//BIz\nZswAAMyYMQP79++XI7QadMU5fPhwWFlJH0l9+/ZFbm6uHKFp6YoRABYuXIgPP/xQhoh00xXnli1b\nsGTJEjRv3hwA4ODg8MTzKDIZ3LhxA126dNF+7ezsjBs3bsgY0ZNlZ2cjPT0dffv2lTsUnd544w3E\nxMRof9mUKCsrCw4ODoiKikKvXr3whz/8AWVlZXKHVYudnR3efPNNuLi4oHPnzmjfvj2GDRsmd1h1\nunXrFhwdHQEAjo6OuHXrlswRPdm2bdsQGhoqdxi1HDhwAM7OzujZs6fcodTrp59+wokTJ9CvXz+o\n1WqcPXv2ie9R5CeDSqWSO4QGKS0txaRJk7Bhwwa0adNG7nBqiY+PR4cOHRAQEKDYUQEAVFRUIC0t\nDXPmzEFaWhqeffZZrFq1Su6wasnMzMT69euRnZ2NvLw8lJaW4osvvpA7LL2oVCrF/3598MEHaNGi\nBSIjI+UOpYaysjKsWLECyx7ZcFCpv08VFRW4e/cuUlJSEBMTg/Dw8Ce+R5HJwMnJCTk5Odqvc3Jy\n4OzsLGNEdXv48CEmTpyIqVOnYty4cXKHo1NycjK+/PJLuLm5ISIiAseOHcP06dPlDqsWZ2dnODs7\nIygoCAAwadIkRe5oe/bsWQwYMAD29vawtrbGhAkTkJycLHdYdXJ0dNTuBJCfn48OHTrIHFHdYmNj\nkZiYqMjkmpmZiezsbPj5+cHNzQ25ubkIDAxEQUGB3KHV4uzsjAkTJgAAgoKCYGVlhcLCwnrfo8hk\n0Lt3b/z000/Izs5GeXk5du/ejbFjx8odVi1CCMyePRteXl5YsGCB3OHUacWKFcjJyUFWVhZ27dqF\nIUOGYMeOHXKHVUvHjh3RpUsXXL58GQBw5MgReHt7yxxVbT169EBKSgp++eUXCCFw5MgReHl5yR1W\nncaOHYvPP/8cAPD5558r9o+WpKQkxMTE4MCBA2jZsqXc4dTi6+uLW7duISsrC1lZWXB2dkZaWpoi\nk+u4ceNw7NgxAMDly5dRXl4Oe3v7+t9kiNntppCYmCg8PDyEu7u7WLFihdzh6HTy5EmhUqmEn5+f\n8Pf3F/7+/uLgwYNyh1UvjUaj6NVE58+fF7179xY9e/YU48ePV+RqIiGEWL16tfDy8hI+Pj5i+vTp\n2lUbcpsyZYro1KmTaN68uXB2dhbbtm0ThYWFYujQoaJbt25i+PDh4u7du3KHWSvOzz77THTt2lW4\nuLhof5dee+01RcTYokUL7c/yUW5ubopYTaQrzvLycjF16lTh4+MjevXqJb7++usnnoc3nRERkTLL\nREREZFxMBkRExGRARERMBkREBCYDIiICkwEREYHJgIiIwGRAJmzIkCE4fPhwjWPr16/HnDlz6nzP\n5cuXERoaCg8PDwQGBmLy5MkoKCiARqOBjY0NAgICtI/qOzgfNXr0aJSUlDT591Jt5syZ+Ne//qX9\nXn755ReDXYvoUYrogUzUGBEREdi1axeef/557bHdu3cjJiZG5+sfPHiAMWPGYN26dRg9ejQA4Pjx\n47h9+zZUKhUGDRqEr776qt5rJiQkNN03oMOjG8lt2LAB06ZNQ6tWrQx6TSKAIwMyYRMnTkRCQgIq\nKioAQLuLaHBwsM7X79y5EwMGDNAmAgAYPHgwvL299d590tXVFUVFRcjOzoanpydeeeUV+Pj4YMSI\nEXjw4EGN1xYXF8PV1VX79f379+Hi4oLKykqcP38e/fr103b1erT7mBACmzZtQl5eHkJCQjB06FBU\nVVVh5syZ8PX1Rc+ePbF+/Xp9f0xEemEyIJNlZ2eHPn36IDExEQCwa9cuTJ48uc7XZ2RkIDAwsM7n\nT548WaNMlJWVVes1j27/fOXKFcydOxfffvst2rdvry3vVLOxsYG/vz80Gg0AaSvxkSNHolmzZpg+\nfTpiYmJw4cIF+Pr61tgWWaVSYd68eejcuTM0Gg2OHj2K9PR05OXl4dKlS7h48SKioqL0+hkR6YvJ\ngExadakIkEpEERER9b6+vhHAwIEDkZ6ern24ubnVey43Nzdtk5PAwEBkZ2fXes3kyZOxe/duAL8l\nq+LiYhQXF2PgwIEApO5jJ06cqPda7u7uuHr1KubPn49Dhw6hXbt29b6eqKGYDMikjR07VvuXc1lZ\nGQICAup8rbe3N86dO9dk137mmWe0/27WrJm2XPWosLAwJCUl4e7du0hLS8OQIUNqvUafElX79u1x\n8eJFqNVqbN26FS+//PLTBU/0GCYDMmlt2rRBSEgIoqKintgZKzIyEsnJydqyEgCcOHECGRkZBo0v\nKCgI8+fPR1hYGFQqFWxsbGBra4tvvvkGAPCPf/wDarW61nvbtm2rXblUWFiIiooKTJgwAcuXL1dk\n0x8ybVxNRCYvIiICEyZMwJ49e+p9XcuWLREfH48FCxZgwYIFaN68Ofz8/LB+/XrcuXNHO2dQ7a9/\n/au2W1S1R+cMHm8fWVc7ycmTJyM8PFw7dwBITWZeffVVlJWVwd3dHdu3b6/1vldeeQUjR46Ek5MT\n1q1bh6ioKFRVVQGAItuBkmljPwMiImKZiIiIWCYiM3Tp0iVMnz69xrGWLVvi9OnTMkVEpHwsExER\nEctERETEZEBERGAyICIiMBkQERGYDIiICMD/A7QrYeqaDKtbAAAAAElFTkSuQmCC\n",
+       "text": [
+        "<matplotlib.figure.Figure at 0x7ff7b7899250>"
+       ]
+      },
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "DC load line shown in figure\n",
+        "Operating point is  8.55  V and  2.15  mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 10
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.10 page 220"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_CC= 10 # in V\n",
+      "V_BB= 3 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "V_T= 25*10**-3 # in V\n",
+      "bita=100 \n",
+      "RC= 3 # in k\u03a9\n",
+      "RC=RC*10**3 # in \u03a9\n",
+      "RB= 100 # in k\u03a9\n",
+      "RB=RB*10**3 # in \u03a9\n",
+      "I_B= (V_BB-V_BE)/RB # in V\n",
+      "I_C= bita*I_B # in A\n",
+      "V_C= V_CC-I_C*RC # in V\n",
+      "gm= I_C/V_T # in A/V\n",
+      "r_pi= bita/gm # in \u03a9\n",
+      "# v_be= r_pi/(RB+r_pi)*v_i\n",
+      "v_be_by_v_i= r_pi/(RB+r_pi) \n",
+      "# v_o= -gm*v_be*RC\n",
+      "v_o_by_v_i= -gm*v_be_by_v_i*RC # in V/V\n",
+      "Av= v_o_by_v_i # in V/V\n",
+      "print \"Voltage gain = %0.2f V/V  \" % (round(Av))\n",
+      "# Answer in the book is not accurate."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Voltage gain = -3.00 V/V  \n"
+       ]
+      }
+     ],
+     "prompt_number": 20
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.11 - page 253"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_B= 4 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "V_CC= 10 # in V\n",
+      "V_E= V_B-V_BE # in V\n",
+      "R_E= 3.3 # in k\u03a9\n",
+      "R_E=R_E*10**3 # in \u03a9\n",
+      "RC= 4.7 # in k\u03a9\n",
+      "RC=RC*10**3 # in \u03a9\n",
+      "I_E= V_E/R_E # in A\n",
+      "bita=100 \n",
+      "alpha= bita/(1+bita) \n",
+      "I_C= alpha*I_E #in A\n",
+      "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n",
+      "V_C= V_CC-I_C*RC # in V\n",
+      "print \"The value of V_C = %0.1f Volts\" %(V_C)\n",
+      "I_B= I_E/(1+bita) # in A\n",
+      "print \"The value of I_B = %0.2f mA\" %(I_B*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_C = 0.99 mA\n",
+        "The value of V_C = 5.3 Volts\n",
+        "The value of I_B = 0.01 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 21
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.12 - page 254"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "V_B= 5 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "V_CC= 10 # in V\n",
+      "bita=100 \n",
+      "R_B= 100 # in k\u03a9\n",
+      "R_C= 2 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "I_B= (V_B-V_BE)/R_B # in A\n",
+      "I_C= bita*I_B #in A\n",
+      "V_C= V_CC-I_C*R_C # in V\n",
+      "I_E= I_C # in A (approx)\n",
+      "print \"The value of I_B = %0.3f mA\" %(I_B*10**3)\n",
+      "print \"The value of I_C = %0.1f mA \" %(I_C*10**3)\n",
+      "print \"The value of V_C = %0.1f Volts\" %(V_C)\n",
+      "print \"The value of I_E = %0.1f mA\" %(I_E*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_B = 0.043 mA\n",
+        "The value of I_C = 4.3 mA \n",
+        "The value of V_C = 1.4 Volts\n",
+        "The value of I_E = 4.3 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 22
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.13 - page 255"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "from sympy import symbols, solve\n",
+      "V_B = symbols('V_B')\n",
+      "# Given data \n",
+      "V_EB= 0.7 # in V\n",
+      "V_E = 0.7 # in V\n",
+      "bita=100 \n",
+      "V_EC= 0.2 # in V\n",
+      "V_E= V_EB+V_B # in V\n",
+      "V_CC= 5 # in V\n",
+      "R_E= 1 # in k\u03a9\n",
+      "R_E=R_E*10**3 # in \u03a9\n",
+      "R_C= 10 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "R_B= 10 # in k\u03a9\n",
+      "R_B= R_B*10**3 # in \u03a9\n",
+      "V_E= V_B+V_EB       # (i)\n",
+      "V_C= V_E-V_EC  # (ii)\n",
+      "I_E= (V_CC-V_E)/(R_E)*1000     # mA   (iii)\n",
+      "I_B= V_B/R_B            #         (iv)\n",
+      "I_C= (V_C+V_CC)/R_C   #     (v)\n",
+      "# By using relationship, I_E= I_B+I_C\n",
+      "expr = I_E*1000-(I_B*1000+I_C*1000)\n",
+      "V_B = solve(expr,V_B)\n",
+      "V_B= (9*V_CC-11*V_EB+V_EC)/12 # in V\n",
+      "V_E= V_B+V_EB # in V\n",
+      "V_C= V_B+V_EB-V_EC # in V\n",
+      "I_E= (V_CC-V_E)/R_E# in amp\n",
+      "I_C= (V_B+V_EB-V_EC+V_CC)/R_B # in amp\n",
+      "I_B= V_B/R_B # in amp\n",
+      "print \"The value of V_B = %0.2f Volts\" %V_B\n",
+      "print \"The value of V_E = %0.2f Volts\" %V_E\n",
+      "print \"The value of V_C = %0.2f Volts\" %V_C\n",
+      "print \"The value of I_E = %0.2f mA\" %(I_E*10**3)\n",
+      "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n",
+      "print \"The value of I_B = %0.2f mA\" %(I_B*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_B = 3.12 Volts\n",
+        "The value of V_E = 3.83 Volts\n",
+        "The value of V_C = 3.62 Volts\n",
+        "The value of I_E = 1.17 mA\n",
+        "The value of I_C = 0.86 mA\n",
+        "The value of I_B = 0.31 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 23
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.14 - page 257"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "bita=100 \n",
+      "hFE= 100 \n",
+      "VCEsat= 0.2 # in V\n",
+      "VBEsat= 0.8 # in V\n",
+      "VBEactive= 0.7 # in V\n",
+      "VBB= 5 # in V\n",
+      "VCC= 10 # in V\n",
+      "R_C= 3 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "R_B= 50 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "# Formula VCC= ICsat*R_C+VCEsat\n",
+      "ICsat= (VCC-VCEsat)/R_C #A\n",
+      "print \"The value of IC(sat) = %0.2f mA\" %(ICsat*10**3)\n",
+      "IBmin= ICsat/bita # in A\n",
+      "# Apply KVL to input circuit,  VBB= IB*R_B+VBEsat\n",
+      "IB= (VBB-VBEsat)/R_B # in A\n",
+      "print \"Actual base current = %0.f \u00b5A\" %(IB*10**6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of IC(sat) = 3.27 mA\n",
+        "Actual base current = 84 \u00b5A\n"
+       ]
+      }
+     ],
+     "prompt_number": 24
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.16 - page 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "# beta= alpha/(1-alpha)\n",
+      "# At alpha= 0.5\n",
+      "alpha= 0.5 \n",
+      "beta=  alpha/(1-alpha) \n",
+      "print \"At alpha=0.5, the value of beta = %0.f \" %beta\n",
+      "# At alpha= 0.9\n",
+      "alpha= 0.9 \n",
+      "beta = alpha/(1-alpha) \n",
+      "print \"At alpha=0.9, the value of beta is  %0.f \" %beta\n",
+      "# At alpha= 0.5\n",
+      "alpha= 0.999 \n",
+      "beta= alpha/(1-alpha) \n",
+      "print \"At alpha=0.999, the value of beta is %0.f \" %beta"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At alpha=0.5, the value of beta = 1 \n",
+        "At alpha=0.9, the value of beta is  9 \n",
+        "At alpha=0.999, the value of beta is 999 \n"
+       ]
+      }
+     ],
+     "prompt_number": 25
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.17 - page 259"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "# alpha= beta/(1-beta)\n",
+      "# At beta= 1\n",
+      "beta=1 \n",
+      "alpha= beta/(1+beta) \n",
+      "print \"At beta=1, the value of alpha is  %0.2f \" %alpha\n",
+      "# At beta= 2\n",
+      "beta=2 \n",
+      "alpha= beta/(1+beta) \n",
+      "print \"At beta=2, the value of alpha is  %0.2f \" %alpha\n",
+      "# At beta= 100\n",
+      "beta=100 \n",
+      "alpha= beta/(1+beta) \n",
+      "print \"At beta=100, the value of alpha is %0.2f  \" %alpha\n",
+      "# At beta= 200\n",
+      "beta=200 \n",
+      "alpha= beta/(1+beta) \n",
+      "print \"At beta=200, the value of alpha is %0.3f  \"%alpha"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At beta=1, the value of alpha is  0.50 \n",
+        "At beta=2, the value of alpha is  0.67 \n",
+        "At beta=100, the value of alpha is 0.99  \n",
+        "At beta=200, the value of alpha is 0.995  \n"
+       ]
+      }
+     ],
+     "prompt_number": 26
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.18 - page 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import exp, log\n",
+      "# Given data \n",
+      "VBE= 0.76 # in V\n",
+      "VT= 0.025 # in V\n",
+      "I_C= 10*10**-3 # in A\n",
+      "# Formula I_C= I_S*exp(VBE/VT)\n",
+      "I_S= I_C/(exp(VBE/VT)) # in A\n",
+      "print \"The value of I_S = %0.3e A\" %I_S\n",
+      "# Part(a) for VBE = 0.7 V\n",
+      "VBE= 0.7 # in V\n",
+      "I_C= I_S*exp(VBE/VT)\n",
+      "print \"For VBE = 0.7 V , The value of I_C = %0.3f mA\" %(I_C*10**3)\n",
+      "\n",
+      "# Part (b) for I_C= 10 \u00b5A\n",
+      "I_C= 10*10**-6 # in A\n",
+      "# Formula I_C= I_S*exp(VBE/VT)\n",
+      "VBE= VT*log(I_C/I_S) \n",
+      "print \"For I_C = 10 \u00b5A, The value of VBE = %0.3f V\" %VBE"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_S = 6.273e-16 A\n",
+        "For VBE = 0.7 V , The value of I_C = 0.907 mA\n",
+        "For I_C = 10 \u00b5A, The value of VBE = 0.587 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 27
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.19 - page 260"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from __future__ import division\n",
+      "# Given data \n",
+      "VBE= 0.7 # in V\n",
+      "VT= 0.025 # in V\n",
+      "I_B= 100 # in \u00b5A\n",
+      "I_B=I_B*10**-6 # in A\n",
+      "I_C= 10*10**-3 # in A\n",
+      "# Formula I_C= I_S*exp(VBE/VT)\n",
+      "I_S= I_C/(exp(VBE/VT)) # in A\n",
+      "alpha= I_C/(I_C+I_B) \n",
+      "beta= I_C/I_B \n",
+      "IS_by_alpha= I_S/alpha # in A\n",
+      "IS_by_beta= I_S/beta # in A\n",
+      "print \"The value of alpha is %0.2f \" %alpha\n",
+      "print \"The value of beta is  %0.2f \" %beta \n",
+      "print \"The value of Is/alpha = %0.2e A\" %IS_by_alpha\n",
+      "print \"The value of Is/beta = %0.2e A\" %IS_by_beta"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of alpha is 0.99 \n",
+        "The value of beta is  100.00 \n",
+        "The value of Is/alpha = 6.98e-15 A\n",
+        "The value of Is/beta = 6.91e-17 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 28
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.20 - page 261"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "VBE= 0.7 # in V\n",
+      "VCC= 10.7 # in V\n",
+      "R_C= 10 #in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "R_B= 10 #in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "I1= (VCC-VBE)/R_C # in A\n",
+      "print \"The value of I1 = %0.f mA\" %(I1*10**3)\n",
+      "# Part (b)\n",
+      "VC= -4 #in V\n",
+      "VB= -10 # in V\n",
+      "R_C= 5.6 #in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "R_B= 2.4 #in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "VCC=12 # V\n",
+      "I_C= (VC-VB)/R_B # in A\n",
+      "V2= VCC- (R_C*I_C) \n",
+      "print \"The value of V2 = %0.f Volt\" %V2\n",
+      "# Part (c)\n",
+      "VCC= 0 \n",
+      "VCE= -10 # in V\n",
+      "R_C= 10 #in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "I_C= (VCC-VCE)/R_C # in A\n",
+      "V4= 1 # in V\n",
+      "I3= I_C # in A (approx)\n",
+      "print \"The value of V4 = %0.f Volt\" %V4\n",
+      "print \"The value of I3 = %0.f mA\" %(I3*10**3)\n",
+      "# Part (d)\n",
+      "VBE= -10 # in V\n",
+      "VCC= 10 # in V\n",
+      "R_B= 5 #in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "R_C= 15 #in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "# I5= I_C and \n",
+      "# I5= (V6-0.7-VBE)/R_B and I_C= (VCC-V6)/R_C\n",
+      "V6= (VCC*R_B+R_C*(0.7+VBE))/(R_C+R_B) \n",
+      "print \"The value of V6 = %0.3f Volt\" %(V6)\n",
+      "I5= (V6-0.7-VBE)/R_B # in A\n",
+      "print \"The value of I5 = %0.3f mA\" %(I5*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I1 = 1 mA\n",
+        "The value of V2 = -2 Volt\n",
+        "The value of V4 = 1 Volt\n",
+        "The value of I3 = 1 mA\n",
+        "The value of V6 = -4.475 Volt\n",
+        "The value of I5 = 0.965 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 29
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.21 -page 264"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "# Part (a)\n",
+      "V_C= 2 # in V\n",
+      "R_C= 1 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "V_B= 4.3 # in V\n",
+      "R_B= 200 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "I_C= V_C/R_C # in A\n",
+      "I_B= V_B/R_B # in A\n",
+      "beta= I_C/I_B \n",
+      "print \"Part (a)\"\n",
+      "print \"Collector current = %0.f mA\" %(I_C*10**3)\n",
+      "print \"Base current = %0.1f \u00b5A\" %(I_B*10**6)\n",
+      "print \"The value of beta is %0.f \"%beta\n",
+      "\n",
+      "# Part (b)\n",
+      "V_C= 2.3 # in V\n",
+      "R_C= 230 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "V_B= 4.3 # in V\n",
+      "R_B= 20 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "I= V_C/R_C # current through 230\u03a9 resistro i.e. I_C + I_B in A\n",
+      "I_B= (V_B-V_C)/R_B # in A\n",
+      "I_C= I-I_B # in A\n",
+      "bita= abs(I_C/I_B) \n",
+      "print \"Part (b)\"\n",
+      "print \"Collector current = %0.2f mA\" %(I_C*10**3)\n",
+      "print \"Base current = %0.2f mA\" %(I_B*10**3)\n",
+      "print \"The value of beta is %0.2f \"%beta\n",
+      "\n",
+      "# Part (c)\n",
+      "V_E= 10 # in V\n",
+      "R_E= 1 # in k\u03a9\n",
+      "R_E=R_E*10**3 # in \u03a9\n",
+      "V_1= 7 # in V\n",
+      "R_C= 1 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "V_B= 6.3 # in V\n",
+      "R_B= 100 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "I_E= (V_E-V_1)/R_C #in A\n",
+      "I_C=I_E # in A (approx)\n",
+      "V_C= I_C*R_C # in V\n",
+      "I_B= (V_B-V_C)/R_B # in A\n",
+      "beta= I_E/I_B-1 \n",
+      "print \"Part (c)\"\n",
+      "print \"Emitter current = %0.2f mA\" %(I_E*10**3)\n",
+      "print \"Base current = %0.2f \u00b5A\" %(I_B*10**6)\n",
+      "print \"Collector voltage = %0.2f Volts\" %(V_C)\n",
+      "print \"The value of beta is %0.2f \"%(beta)\n",
+      "\n",
+      "# Note : In the book the value of base current in the first part is wrong due to calculation error.\n",
+      "#In the part (b) the values of collector current and beta are wrong due to calculation error in the first line of part (b)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "Collector current = 2 mA\n",
+        "Base current = 21.5 \u00b5A\n",
+        "The value of beta is 93 \n",
+        "Part (b)\n",
+        "Collector current = -0.09 mA\n",
+        "Base current = 0.10 mA\n",
+        "The value of beta is 93.02 \n",
+        "Part (c)\n",
+        "Emitter current = 3.00 mA\n",
+        "Base current = 33.00 \u00b5A\n",
+        "Collector voltage = 3.00 Volts\n",
+        "The value of beta is 89.91 \n"
+       ]
+      }
+     ],
+     "prompt_number": 30
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.22 - page 266"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "# Part (a)\n",
+      "beta= 30 \n",
+      "R_C= 2.2 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "R_B= 2.2 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "VCC= 3 # in V\n",
+      "VCE= -3 # in V\n",
+      "VBE= 0.7 # in V\n",
+      "V_B= 0 # in V\n",
+      "V_E= V_B-VBE # in V\n",
+      "I_E= (V_E-VCE)/R_B # in A\n",
+      "I_C= I_E # in A\n",
+      "V_C= VCC-I_E*R_C # in V\n",
+      "I_B= I_C/beta # in A\n",
+      "print \"Part (a)\"\n",
+      "print \"The value of V_B = %0.2f V \" %V_B\n",
+      "print \"The value of V_E = %0.2f V\" %V_E\n",
+      "print \"The value of I_E = %0.2f mA\" %(I_E*10**3)\n",
+      "print \"The value of V_C = %0.3f V\" %(V_C)\n",
+      "print \"The value of I_B = %0.2f mA\" %(I_B*10**3)\n",
+      "# Part (b)\n",
+      "R_C= 560 # in \u03a9\n",
+      "R_B= 1.1 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "VCC= 9 # in V\n",
+      "VCE= 3 # in V\n",
+      "V_B= 3 # in V\n",
+      "V_E= V_B+VBE # in V\n",
+      "I_E= (VCC-V_E)/R_B # in A\n",
+      "alpha= beta/(1+beta) \n",
+      "I_C= I_E*alpha # in A\n",
+      "V_C= I_C*R_C # in V\n",
+      "I_B= I_C/beta # in A\n",
+      "print \"Part (b)\"\n",
+      "print \"The value of V_B = %0.2f V \" %V_B\n",
+      "print \"The value of V_E = %0.2f V\" %V_E\n",
+      "print \"The value of I_E = %0.2f mA\" %(I_C*10**3)\n",
+      "print \"The value of V_C = %0.2f V\" %(V_C)\n",
+      "print \"The value of I_B = %0.3f mA\" %(I_B*10**3)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "The value of V_B = 0.00 V \n",
+        "The value of V_E = -0.70 V\n",
+        "The value of I_E = 1.05 mA\n",
+        "The value of V_C = 0.700 V\n",
+        "The value of I_B = 0.03 mA\n",
+        "Part (b)\n",
+        "The value of V_B = 3.00 V \n",
+        "The value of V_E = 3.70 V\n",
+        "The value of I_E = 4.66 mA\n",
+        "The value of V_C = 2.61 V\n",
+        "The value of I_B = 0.155 mA\n"
+       ]
+      }
+     ],
+     "prompt_number": 31
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.23 - page 268"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "from numpy import inf\n",
+      "# Given data \n",
+      "VBE= 0.7 # in V\n",
+      "VCC= 9 # in V\n",
+      "VCE= -9 # in V\n",
+      "V_B= -1.5 # in V\n",
+      "R_C= 10 # in k\u03a9\n",
+      "R_C=R_C*10**3 # in \u03a9\n",
+      "R_B= 10 # in k\u03a9\n",
+      "R_B=R_B*10**3 # in \u03a9\n",
+      "I_B=  abs(V_B)/R_B # in A\n",
+      "V_E= V_B-VBE # in V\n",
+      "print \"The value of V_E = %0.2f Volt\" %V_E\n",
+      "I_E= (V_E-VCE)/R_B # in A\n",
+      "beta= I_E/I_B-1 \n",
+      "alpha= beta/(1+beta) \n",
+      "print \"The value of alpha = %0.2f Volt\" %alpha\n",
+      "print \"The value of beta = %0.2f Volt\" %beta\n",
+      "V_C= VCC-I_E*alpha*R_C # in V\n",
+      "print \"The value of V_C = %0.2f Volt\" %V_C\n",
+      "beta = inf\n",
+      "alpha= beta/(1+beta)\n",
+      "I_B= 0 \n",
+      "V_B=0 \n",
+      "V_C= VCC-I_E*R_C # in volt\n",
+      "print \"The value of V_B = %0.2f V \" %V_B\n",
+      "print \"The value of V_E = %0.2f V\" %V_E\n",
+      "print \"The value of V_C = %0.2f V\" %(V_C)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_E = -2.20 Volt\n",
+        "The value of alpha = 0.78 Volt\n",
+        "The value of beta = 3.53 Volt\n",
+        "The value of V_C = 3.70 Volt\n",
+        "The value of V_B = 0.00 V \n",
+        "The value of V_E = -2.20 V\n",
+        "The value of V_C = 2.20 V\n"
+       ]
+      }
+     ],
+     "prompt_number": 32
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.24 - page 269"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "VBE_1= 0.7 # in V\n",
+      "VBE_2= 0.5 # in V\n",
+      "V_T= 0.025 # in V\n",
+      "I_C1= 10 # in mV\n",
+      "I_C1= I_C1*10**-3 # in A\n",
+      "#  I_C1= I_S*%e**(VBE_1/V_T)        (i)\n",
+      "#  I_C2= I_S*%e**(VBE_2/V_T)        (ii)\n",
+      "# Devide equation (ii) by (i)\n",
+      "I_C2= I_C1*exp((VBE_2-VBE_1)/V_T) # in A\n",
+      "print \"The value of I_C2 = %0.2f \u00b5A\" %(I_C2*10**6)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of I_C2 = 3.35 \u00b5A\n"
+       ]
+      }
+     ],
+     "prompt_number": 33
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.25 - page 270"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "R1= 10 # in k\u03a9\n",
+      "R1=R1*10**3 # in \u03a9\n",
+      "R2= 10 # in k\u03a9\n",
+      "R2=R2*10**3 # in \u03a9\n",
+      "I_C=.5 # mA\n",
+      "V_T= 0.025 #in V\n",
+      "I_C= I_C*10**-3 # in A\n",
+      "V= 10 # in V\n",
+      "Vth= V*R1/(R1+R2) # in V\n",
+      "Rth=  R1*R2/(R1+R2) #in \u03a9\n",
+      "vo= I_C*Rth # in V\n",
+      "vi=V_T # in V\n",
+      "vo_by_vi= vo/vi #in V/V\n",
+      "print \"The value of vo/vi = %0.f V/V \" %vo_by_vi"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of vo/vi = 100 V/V \n"
+       ]
+      }
+     ],
+     "prompt_number": 34
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.27 - page 272"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "V_B= 2 # in V\n",
+      "V_CC=5 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "R_E= 1*10**3 # in \u03a9\n",
+      "R_C= 1*10**3 # in \u03a9\n",
+      "V_E= V_B-V_BE # in V\n",
+      "I_E= V_E/R_E # in A\n",
+      "I_C= I_E # in A\n",
+      "V_C= V_CC-I_C*R_C #in V\n",
+      "print \"At V_B= +2 V\"\n",
+      "print \"The value of V_E = %0.2f Volts\" %V_E\n",
+      "print \"The value of V_C = %0.2f Volts\" %V_C\n",
+      "\n",
+      "# Part (b)\n",
+      "V_B= 0 #in V\n",
+      "V_E= 0 # in V\n",
+      "I_E=  0 # in A\n",
+      "V_C= 5 # in V\n",
+      "print \"At V_B= 0 V\"\n",
+      "print \"The value of V_E = %0.2f Volts\" %V_E\n",
+      "print \"The value of V_C = %0.2f Volts\" %V_C"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "At V_B= +2 V\n",
+        "The value of V_E = 1.30 Volts\n",
+        "The value of V_C = 3.70 Volts\n",
+        "At V_B= 0 V\n",
+        "The value of V_E = 0.00 Volts\n",
+        "The value of V_C = 5.00 Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 35
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.28 - page 273"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "V_B= 0 # in V\n",
+      "R_E=1*10**3 #in \u03a9\n",
+      "R_C=1*10**3 #in \u03a9\n",
+      "V_CC=5 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "V_E= V_B-V_BE # in V\n",
+      "I_E= (1+V_E)/R_E # in A\n",
+      "I_C= I_E # (approx) in A\n",
+      "V_C= V_CC-I_C*R_C #in V\n",
+      "print \"Part (a)\"\n",
+      "print \"The value of V_E = %0.2f Volts\" %V_E\n",
+      "print \"The value of V_C = %0.2f Volts\" %V_C\n",
+      "# For saturation \n",
+      "V_CE=0.2  # V\n",
+      "V_CB= -0.5 # in V\n",
+      "# I_C= 5-V_C/R_C and V_C= V_E-VCE, So\n",
+      "# I_C= (5.2-V_E)/R_C\n",
+      "# I_E= (V_E+1)/R_E and at the edge of saturation I_C=I_E,\n",
+      "V_E= 4.2/2 #/ in V\n",
+      "V_B= V_E+0.7 # in V\n",
+      "V_C= V_E+0.2 # in V\n",
+      "print \"Part (b) \"\n",
+      "print \"The value of V_E = %0.2f Volts\" %V_E\n",
+      "print \"The value of V_B = %0.2f Volts\" %V_B\n",
+      "print \"The value of V_C = %0.2f Volts\" %V_C\n",
+      "\n",
+      "# Note: In the book , there is a miss print in the last line of this question \n",
+      "#because V_E+0.2= 2.1+0.2 = 2.3  (not 2.8) , so answer in the book is wrong "
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a)\n",
+        "The value of V_E = -0.70 Volts\n",
+        "The value of V_C = 4.70 Volts\n",
+        "Part (b) \n",
+        "The value of V_E = 2.10 Volts\n",
+        "The value of V_B = 2.80 Volts\n",
+        "The value of V_C = 2.30 Volts\n"
+       ]
+      }
+     ],
+     "prompt_number": 36
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.29 - page 275"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data \n",
+      "V_CC=5 # in V\n",
+      "V_E= 1 # in V\n",
+      "V_BE= 0.7 # in V\n",
+      "R_E=5*10**3 #in \u03a9\n",
+      "R_C=5*10**3 #in \u03a9\n",
+      "R_B= 20*10**3 # in \u03a9\n",
+      "I_E= (V_CC-V_E)/R_E #  in A\n",
+      "# For pnp transistor V_BE= V_E-V_B\n",
+      "V_B= V_E-V_BE # in V\n",
+      "I_B= V_B/R_B # in A\n",
+      "I_C= I_E-I_B # in A\n",
+      "V_C= I_C*R_C-V_CC # in V\n",
+      "beta= I_C/I_B \n",
+      "alpha= I_C/I_E \n",
+      "print \"The value of V_B = %0.1f Volts\" %V_B\n",
+      "print \"The value of I_B = %0.3f mA\" %(I_B*10**3)\n",
+      "print \"The value of I_E = %0.1f mA\" %(I_E*10**3)\n",
+      "print \"The value of I_C = %0.3f mA\" %(I_C*10**3)\n",
+      "print \"The value of V_C = %0.3f Volts\" %V_C \n",
+      "print \"The value of beta is %0.1f \"%beta \n",
+      "print \"The value of alpha is %0.2f \"%alpha"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of V_B = 0.3 Volts\n",
+        "The value of I_B = 0.015 mA\n",
+        "The value of I_E = 0.8 mA\n",
+        "The value of I_C = 0.785 mA\n",
+        "The value of V_C = -1.075 Volts\n",
+        "The value of beta is 52.3 \n",
+        "The value of alpha is 0.98 \n"
+       ]
+      }
+     ],
+     "prompt_number": 37
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.30 - page 276"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_CC=5 # in V\n",
+      "V_T= 0.025 # in V\n",
+      "R_C=7.5*10**3 #in \u03a9\n",
+      "I_C= 0.5 # in mA\n",
+      "I_C= I_C*10**-3 # in A\n",
+      "I_E=I_C # (approx) in A\n",
+      "V_C= V_CC-I_C*R_C # in V\n",
+      "print \"dc voltage at the collector = %0.2f Volt\" %V_C\n",
+      "gm= I_C/V_T # in A/V\n",
+      "print \"The value of gm = %0.f mA/V\" %(gm*10**3)\n",
+      "# v_be= -v_i\n",
+      "# v_c= -gm*v_be*R_C\n",
+      "vcbyvi= gm*R_C # in V/V\n",
+      "print \"The value of vc/vi = %0.f V/V\" %(vcbyvi)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "dc voltage at the collector = 1.25 Volt\n",
+        "The value of gm = 20 mA/V\n",
+        "The value of vc/vi = 150 V/V\n"
+       ]
+      }
+     ],
+     "prompt_number": 38
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.31 - page 277"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_T= 0.025 # in V\n",
+      "I_E= 0.5 # in mA\n",
+      "I_E= I_E*10**-3 # in mA\n",
+      "Rsig= 50 # in \u03a9\n",
+      "R_C= 5*10**3 # in \u03a9\n",
+      "re= V_T/I_E # in ohm\n",
+      "Rin= Rsig+re # in ohm\n",
+      "print \"Input resistance = %0.f \u03a9\" %Rin\n",
+      "# Part(b)\n",
+      "# vo= -0.99*ie*R_C and ie= -v_sig/Rin\n",
+      "vo_by_v_sig= 0.99*R_C/Rin # in V/V\n",
+      "print \"The value of vo/vsig = %0.1f V/V\" %(vo_by_v_sig)"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Input resistance = 100 \u03a9\n",
+        "The value of vo/vsig = 49.5 V/V\n"
+       ]
+      }
+     ],
+     "prompt_number": 39
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.32 - page 278"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "beta= 200 \n",
+      "alpha= beta/(1+beta) \n",
+      "R_C= 100 # in \u03a9\n",
+      "R_B= 10 # in k\u03a9\n",
+      "Rsig= 1 # in k\u03a9\n",
+      "Rsig= Rsig*10**3 # in \u03a9\n",
+      "R_B= R_B*10**3 # in \u03a9\n",
+      "V_T= 25*10**-3 \n",
+      "V=1.5 # in V\n",
+      "I_E= 10 # in mA\n",
+      "I_E= I_E*10**-3 # in A\n",
+      "I_C= alpha*I_E # in A\n",
+      "V_C= I_C*R_C # in V\n",
+      "I_B= I_C/beta # in A\n",
+      "V_B= V-(R_B*I_B)\n",
+      "gm= I_C/V_T # in A/V\n",
+      "rpi= beta/gm # in \u03a9\n",
+      "Rib= rpi # in \u03a9\n",
+      "print \"The value of Rib = %0.2f \u03a9 \" %Rib\n",
+      "Rin= R_B*rpi/(R_B+rpi) # in \u03a9\n",
+      "print \"The value of Rin = %0.2f \u03a9\" %Rin\n",
+      "# vbe= v_sig*Rin/(Rsig+Rin) \n",
+      "vbe_by_vsig= Rin/(Rsig+Rin) \n",
+      "# vo= -gm*vbe*R_C and = -gm*v_sig*Rin/(Rsig+Rin)\n",
+      "vo_by_vsig= -gm*R_C*vbe_by_vsig # in V/V\n",
+      "print \"Overall voltage gain = %0.2f V/V\" %vo_by_vsig\n",
+      "# if \n",
+      "vo= 0.4 #(\u00b1) in V\n",
+      "vs= vo/abs(vo_by_vsig) # in V\n",
+      "vbe= vbe_by_vsig*vs # in V\n",
+      "print \"The value of v_sig = %0.2f mV\" %(vs*10**3)\n",
+      "print \"The value of v_be = %0.2f mV\" %(vbe*10**3)\n",
+      "\n",
+      "# Note: There is some difference between in this coding and book solution. But Coding is correct."
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "The value of Rib = 502.50 \u03a9 \n",
+        "The value of Rin = 478.46 \u03a9\n",
+        "Overall voltage gain = -12.88 V/V\n",
+        "The value of v_sig = 31.06 mV\n",
+        "The value of v_be = 10.05 mV\n"
+       ]
+      }
+     ],
+     "prompt_number": 40
+    },
+    {
+     "cell_type": "heading",
+     "level": 2,
+     "metadata": {},
+     "source": [
+      "Exa 3.33 - page 280"
+     ]
+    },
+    {
+     "cell_type": "code",
+     "collapsed": false,
+     "input": [
+      "# Given data\n",
+      "V_T= 0.025 # in V\n",
+      "# Part(a)\n",
+      "print \"Part (a) : \"\n",
+      "V_BE= 690 # in mV\n",
+      "V_BE=V_BE*10**-3 # in V\n",
+      "I_C= 1 # in mA\n",
+      "I_B= 50 # in \u00b5A\n",
+      "I_C=I_C*10**-3 # in A\n",
+      "I_B=I_B*10**-6 # in A\n",
+      "beta= I_C/I_B \n",
+      "alpha= beta/(1+beta) \n",
+      "I_E= I_C/alpha # in A\n",
+      "# I_C= I_S*exp(V_BE/V_T)\n",
+      "I_S= I_C/(exp(V_BE/V_T)) \n",
+      "print \"The value of beta is %0.1f \" %beta\n",
+      "print \"The value of alpha is %0.4f \"%alpha\n",
+      "print \"The value of I_E = %0.2f mA\" %(I_E*10**3)\n",
+      "print \"The value of I_S = %0.2e A\" %I_S\n",
+      "\n",
+      "# Part(b)\n",
+      "print \"Part (b) : \"\n",
+      "V_BE= 690 # in mV\n",
+      "V_BE=V_BE*10**-3 # in V\n",
+      "I_C= 1 # in mA\n",
+      "I_C=I_C*10**-3 # in A\n",
+      "I_E= 1.070 # in mA\n",
+      "I_E=I_E*10**-3 # in A\n",
+      "beta= I_C/I_B \n",
+      "alpha= I_C/I_E \n",
+      "beta= alpha/(1-alpha) \n",
+      "I_B= I_C/beta # in A\n",
+      "# I_C= I_S*exp(V_BE/V_T)\n",
+      "I_S= I_C/(exp(V_BE/V_T)) \n",
+      "print \"The value of beta is %0.3f \" %beta\n",
+      "print \"The value of alpha is %0.4f \"%alpha\n",
+      "print \"The value of I_B = %0.2f \u00b5A\" %(I_B*10**6)\n",
+      "print \"The value of I_S = %0.2e A\" %I_S\n",
+      "\n",
+      "# Part(c)\n",
+      "print \"Part (C) : \"\n",
+      "V_BE= 580 # in mV\n",
+      "V_BE=V_BE*10**-3 # in V\n",
+      "I_E= 0.137 # in mA\n",
+      "I_B= 7 # in \u00b5A\n",
+      "I_E=I_E*10**-3 # in A\n",
+      "I_B=I_B*10**-6 # in A\n",
+      "#  I_C= alpha*I_E  = bita*I_B\n",
+      "beta= I_E/I_B-1 \n",
+      "alpha= beta/(1+beta) \n",
+      "I_C= beta*I_B # in A\n",
+      "# I_C= I_S*exp(V_BE/V_T)\n",
+      "I_S= I_C/(exp(V_BE/V_T)) \n",
+      "print \"The value of beta is %0.3f \" %beta\n",
+      "print \"The value of alpha is %0.4f \"%alpha\n",
+      "print \"The value of I_C = %0.3f mA\" %(I_C*10**3)\n",
+      "print \"The value of I_S = %0.3e A\" %I_S\n",
+      "\n",
+      "# Part(d)\n",
+      "print \"Part (d) : \"\n",
+      "V_BE= 780 # in mV\n",
+      "V_BE=V_BE*10**-3 # in V\n",
+      "I_C= 10.10 # in mA\n",
+      "I_B= 120 # in \u00b5A\n",
+      "I_C=I_C*10**-3 # in A\n",
+      "I_B=I_B*10**-6 # in A\n",
+      "beta= I_C/I_B \n",
+      "alpha= beta/(1+beta) \n",
+      "I_E= I_C/alpha # in A\n",
+      "# I_C= I_S*%e**(V_BE/V_T)\n",
+      "I_S= I_C/(exp(V_BE/V_T)) \n",
+      "print \"The value of beta is %0.3f \" %beta\n",
+      "print \"The value of alpha is %0.4f \"%alpha\n",
+      "print \"The value of I_E = %0.2f mA\" %(I_E*10**3)\n",
+      "print \"The value of I_S = %0.4e A\" %I_S\n",
+      "\n",
+      "# Part(e)\n",
+      "print \"Part (e) : \"\n",
+      "V_BE= 820 # in mV\n",
+      "V_BE=V_BE*10**-3 # in V\n",
+      "I_E= 75 # in mA\n",
+      "I_B= 1050 # in \u00b5A\n",
+      "I_E=I_E*10**-3 # in A\n",
+      "I_B=I_B*10**-6 # in A\n",
+      "#  I_C= alpha*I_E  = bita*I_B\n",
+      "beta= I_E/I_B-1 \n",
+      "alpha= beta/(1+beta) \n",
+      "I_C= beta*I_B # in A\n",
+      "# I_C= I_S*exp(V_BE/V_T)\n",
+      "I_S= I_C/(exp(V_BE/V_T)) \n",
+      "print \"The value of beta is %0.3f \" %beta\n",
+      "print \"The value of alpha is %0.3f \"%alpha\n",
+      "print \"The value of I_C = %0.2f mA\" %(I_C*10**3)\n",
+      "print \"The value of I_S = %0.3e A\" %I_S"
+     ],
+     "language": "python",
+     "metadata": {},
+     "outputs": [
+      {
+       "output_type": "stream",
+       "stream": "stdout",
+       "text": [
+        "Part (a) : \n",
+        "The value of beta is 20.0 \n",
+        "The value of alpha is 0.9524 \n",
+        "The value of I_E = 1.05 mA\n",
+        "The value of I_S = 1.03e-15 A\n",
+        "Part (b) : \n",
+        "The value of beta is 14.286 \n",
+        "The value of alpha is 0.9346 \n",
+        "The value of I_B = 70.00 \u00b5A\n",
+        "The value of I_S = 1.03e-15 A\n",
+        "Part (C) : \n",
+        "The value of beta is 18.571 \n",
+        "The value of alpha is 0.9489 \n",
+        "The value of I_C = 0.130 mA\n",
+        "The value of I_S = 1.092e-14 A\n",
+        "Part (d) : \n",
+        "The value of beta is 84.167 \n",
+        "The value of alpha is 0.9883 \n",
+        "The value of I_E = 10.22 mA\n",
+        "The value of I_S = 2.8466e-16 A\n",
+        "Part (e) : \n",
+        "The value of beta is 70.429 \n",
+        "The value of alpha is 0.986 \n",
+        "The value of I_C = 73.95 mA\n",
+        "The value of I_S = 4.208e-16 A\n"
+       ]
+      }
+     ],
+     "prompt_number": 41
+    }
+   ],
+   "metadata": {}
+  }
+ ]
+}
\ No newline at end of file