{ "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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"text": [ "" ] }, { "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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KyjJKIiBSkqbonZCbm4sXXngBHh4e6Nq1KxYsWICHvy5dunTpEmbNmgUAuHPnDkaOHAl/\nf3/4+PggNjZW5/lmzpyJ5557DgEBAQgICMDFixcBABqNBjY2Ntrj77//PgAgJCQEhw8frnGO9evX\nY86cOTWO5eTkICQkBN7e3vDx8cHGjRt1Xv+HH35A//790bJlS6xdu7bGc66urujZsycCAgLQp0+f\nhv2gfrV//35YWVnhxx9/1B67desWQkNDG3U+qofQw/Dhw0Vpaak+L21yeoZIZFS5uUJERAjh4iLE\nnj1CVFU9+T1VVVUiKChIxMbGCiGEqKysFLNnzxaLFi0SQggxbdo0kZqaKoQQIjo6WrzzzjtCCCFu\n374t7OzsxMOHD2udc+bMmeJf//pXreNff/21CAsLq3X8k08+EVFRUTWO9evXT5w8ebLGsfz8fJGe\nni6EEOLevXvCw8NDfPfdd7XOV1BQIM6cOSPeffddsWbNmhrPubq6isLCQt0/DD2Fh4eLsLAwER0d\nXeN4ZGSkOHfu3FOd29w19LNTr5HBqlWr0L9/f/zxj3/EvHnzMG/ePMyfP9+wWYpIwZycpLuXd+wA\nli8Hhg2TJpzrc+zYMbRq1QozZswAAFhZWWHdunXYtm0bSkpKkJKSgqCgIABAp06dUFJSAkCaq7O3\nt4e1te57REUdpQBdxydOnIiEhARUVFQAALKzs5GXl4fg4OAar+vYsSP8/f0BAG3atIGnpyfy8vJq\nnc/BwQG9e/dG8+bN9Y5BrVZj4cKFCAoKgqenJ86cOaNdnfjXv/5V+7rS0lL897//xebNm7F79+4a\n5xg7dizi4uJ0XpMaR69k8Morr2DYsGHo168fevfujcDAQAQGBho6NiLFa0jvhIyMjFq/N23btoWL\niwsuXryI7t27a4//4Q9/QEZGBjp37gw/Pz9s2LChzhiWLFkCPz8/LFy4EOXl5QCkenFycjL8/PwQ\nGhqK737NVHZ2dujTpw8SExMBALt27cLkyZPr/R6zs7ORnp6Ovn37PvHn8SiVSoVhw4ahd+/e+Pvf\n/17j+DPPPIMzZ87gtddewwsvvICtW7fi22+/RWxsLO7evQsAOHDgAEaOHAkXFxc4ODggLS1Ne44+\nffrgxIkTDYqH6qdXMqisrMTf/vY3REVFYcaMGZg5c6b2rxsiS6dv7wSVSlXnOZKTk9GpUyft1ytW\nrIC/vz/y8vJw/vx5/OlPf8K9e/dqvW/lypW4fPkyzpw5g6KiIqxevRoA0KtXL+Tk5ODChQuYN28e\nxo0bp31PREQEdu3aBQDYvXs3IiIi6oyrtLQUkyZNwoYNG9CmTRu9fh7VTp06hfT0dBw8eBAfffQR\nTp48qX1u7NixAAAfHx/4+PjA0dERLVq0wHPPPYecnBwAQFxcHF588UUAwIsvvlhjJNCpUydkZ2c3\nKB6qn17JYNSoUfj444+Rn5+PoqIi7YOIftOhA/Dpp8D+/VLPhOBgadRQzcvLC+fOnavxnpKSEuTk\n5KBjx441SirJycnaD0J3d3e4ubnVmESt1vHXjZRatGiBqKgopKamApBGHK1btwYg/f4+fPhQ+zs7\nduxYHD16FOnp6SgrK0NAQIDO7+fhw4eYOHEipk6dWiOZ6Ks6uTk4OGD8+PHa2ADgmWeeASCVyqr/\nXf11RUUFioqK8PXXX2P27Nlwc3NDTEwM9uzZo32dEKLe5EoNp1cy2LlzJ1atWoUBAwZoS0S9e/c2\ndGxEJqmu3glDhw5FWVkZ/vGPfwCQRtxvvvkmIiMj0b179xo7Affo0QNHjhwBIK2e+fHHH/Hcc8/V\nulZ+fj4A6cNx37598PX11b6nOrmkpqZCCAE7OzsA0hxASEgIoqKiEBkZWeN8PXr00J5v9uzZ8PLy\nwoIFC574PT8+N1BWVqYdydy/fx+HDx/WxqbPuf75z39i+vTpyM7ORlZWFq5fvw43Nzft6CI/Px+/\n//3v9Tof6akpZ68NwQRCJKpTUZEQc+cK0aGDEFu2CJGdnSPGjh0runXrJtq3by/Cw8NFeXm5+OWX\nX0S3bt2077t9+7YYM2aM6Nmzp/Dx8RFffPGF9rnQ0FCRn58vhBBiyJAhwtfXV/j4+Ihp06aJ+/fv\nCyGE2Lx5s/D29hZ+fn6if//+4vTp0zXi2r9/v7CyshI//vhjjWt2795dCCHEyZMnhUqlEn5+fsLf\n31/4+/uLgwcPCiGE2Lp1q9i6dasQQlp15OzsLNq1ayfat28vunTpIu7duycyMzOFn5+f8PPzE97e\n3mLFihXa66jVau1KII1GU2PVk1qtFmfPnhUhISHi0KFDNWLeuHGjmDNnjhBCiLi4OPHWW2815j+J\nxWjoZ6fiP2khbRVT+/HYUjOt6Gi+nq9X3OvPnxdi4EAhAgKEOHVKOpacnCy8vb21SzZnzJghUlJS\ndF/HCOLj48WmTZtku35DREZGirS0NLnDULSGJgNZeyDn5ORg+vTpKCgogEqlwiuvvFJrySrvQCZz\nIYTUO+Htt3X3Tvj222+xdu1abN++Xb4gTUBBQQGioqKQkJAgdyiKZpDtKAzl5s2buHnzJvz9/VFa\nWorAwEDs378fnp6evwXIZEBm5t494P33gc8+A959F5g7F6hjmT5RoxlkOwoAuHHjBk6dOoUTJ07g\n+PHjTbLGV98bW4jMwtKlANg7gZRJr5HB4sWLsXv3bnh5eaFZs2ba41999VWTBZKdnY3BgwcjIyOj\nxnpmjgzIbKhUUq3oEeydQIbS0M/OensgV9u3bx9+/PHHGuuBm9KTbmxZ+utfVIB0K7tarTZIHETG\nVt07YcQIabTQq5eUGN58E2jZUu7oyJRoNBpoNJpGv1+vkcGoUaOwZ88etG3bttEXqsvDhw8xZswY\njBo1Sud6Zo4MyGzoGBk8jr0TqKkYZAJ5woQJuHDhAoYOHaodHahUqjq3tdWXEAIzZsyAvb091q1b\npztAJgMyF3okg2rsnUBPyyDJQNde6iqV6qn3J/rmm28waNAg9OzZU3tr+cqVKzFy5Mga12EyILPQ\ngGQAAOXlUiL48EPg1VeBJUuAZ581YHxkVkxqaak+mAzIbCxdql1R1BA3bgCLFgGnTkkTzJMmSXmF\nqD5NmgxefPFF7N27V+eeIiqVSttVyZCYDIgkx49Lu6M6OACbNgFeXnJHRErWpMkgLy8PnTt3rnOr\nWFdX14bG12BMBkS/qagAtmwB3nsPmDYNiI4GbGzkjoqUiGUiIgtQUAD8+c9AYiKwahUwdSpgpfct\npGQJmAyILEhqqrSdhbW11EOhVy+5IyKlMNh2FESkPHX1TiBqqHqTQUFBATIyMmodz8jIwO3btw0W\nFJFZasRKIn1YWUnJ4PvvpRGClxewdStQWWmQy5GZqjcZzJs3D3fu3Kl1vLCwEK+//rrBgiIyS8uW\nGfT0trbSKqPDh4GdO6W9jpKTDXpJMiP1zhkEBgbW6tlazdvbW+eooalxzoDMRgNvOnsaT+qdQOav\nSecMqnuY6vLw4UP9oyIio1KpgMhIqXTUsSPg4wOsWwfw15bqUm8y6Nq1q85uQomJiXB3dzdYUETU\nNNg7gfRVb5no8uXLGDNmDAYMGIDAwEAIIXDu3DkkJycjPj4e3bt3N3yALBORuTBimUgX9k6wLE1a\nJvLw8MDFixcxaNAgZGdn49q1axg8eDAuXbpklERAZFaio2W9fHXvhO++k1Yc9eoFfPAB8OCBrGGR\nQjTJTWf9+/fH6dOnmyKeWjgyIDIM9k4wb7LcdPaAf1oQmRw3N2DfPuCjj6SkMGYMcOWK3FGRXHgH\nMpGFGzFCGh0MGgT06wf85S/A/ftyR0XGxmRARGjRQron4cIF4OpVaU5h715Z57vJyJgMiEjLyUm6\ne3nHDmD5cmDYMGnCmcxfkySDHTt2NMVpiMybgfYmMoTBg4G0NGn10eDB0pxCcbHcUZEh1buaqE2b\nNtrexLXeqFKhpKTEYIE9eh2uJiKzIPN9Bo3F3gmmif0MiJTKRJNBNfZOMC3sZ0BEBsHeCeaNyYCI\n9MbeCeaLZSIiYzHxMpEuFy4A8+YBpaVS6WjAALkjomosExEplcx7ExmCnx9w/Djw1ltAeDgwYwZw\n86bcUVFjMBkQGYsJLS1tCPZOMA8sExFRk/rhB+D114HcXGDjRqnTGhkfl5YSkezYO0F+nDMgItmx\nd4LpkTUZzJo1C46OjvD19ZUzDCIykFatpKmSM2eAs2el+QQdnXRJAWRNBlFRUUhKSpIzBCLjMdMJ\nZH2wd4LyyZoMBg4cCFtbWzlDIDKeZcvkjkB27J2gXJwzICKjYu8EZbKWOwB9LH1keK1Wq6FWq2WL\nhYiaRnXvhOPHpbuYt24FNm2SkgM1nEajgUajafT7ZV9amp2djbCwMFy6dEnn81xaSmbDDLejaCoV\nFcCWLcB77wHTpkk3a9vYyB2VaePSUiIyOdbW0uggIwMoKQE8PaVua1VVckdmOWRNBhERERgwYAAu\nX76MLl26YPv27XKGQ2RYZrg3UVPr0AH49FNg/35p47vgYKnjGhme7GWiJ2GZiMgyVVUB27cD774L\njB8PvP8+YG8vd1Smg2UiIjIL7J1gXBwZEJFJYO+EhuFGdURktoQA4uKk+xSGDgVWr5a2zabaWCYi\nIrPF3gmGw2RAZCwWvDdRU2vbVhoVfPMNkJQE+PsDR4/KHZVpY5mIyFh405lBsHeCbiwTEZFFYe+E\npsFkQERmgb0Tng7LRETGwjKRUR06BMyfD3TrBqxfD3TtKndExsUyERER2DuhoZgMiIyFexMZHXsn\n6I9lIiKyGNW9ExwczL93AstERER1GDxY2gV13Djp3wsXAsXFckelDEwGRGRR2DtBN5aJiMiipaYC\nc+dKSWLzZuk+BXPAMhERUQP06QOkpEjbZYeGAq+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": [ "" ] }, { "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": {} } ] }