{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Ch-6 : Bipolar Junction Transistor " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 151 Example 6.1." ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The emitter current is,\n", "IE = IB + IC\n", "10 = IB + 9.8\n", "Therefore, IB(mA) =0.20\n" ] } ], "source": [ "IE=10\n", "IC=9.8\n", "print \"The emitter current is,\"\n", "print \"IE = IB + IC\"\n", "print \"10 = IB + 9.8\"\n", "IB=IE-IC\n", "print \"Therefore, IB(mA) =%0.2f\"%IB" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 152 Example 6.2." ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The common-base d.c. current gain,\n", "alpha = IC/IE = 0.9873\n" ] } ], "source": [ "IE=6.28\n", "IC=6.20\n", "print \"The common-base d.c. current gain,\"\n", "alpha=IC/IE\n", "print \"alpha = IC/IE = %0.4f\"%alpha" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 155 Example 6.3." ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The common-base d.c. current gain (alpha) is,\n", "alpha = 0.967 = IC/IE = IC/10\n", "\n", "Therefore, IC = 9.67 mA\n", "The emitter current, IE = IB + IC\n", "Therefore, IB =0.33 mA\n" ] } ], "source": [ "alpha=0.967\n", "IE=10\n", "print \"The common-base d.c. current gain (alpha) is,\"\n", "print \"alpha = 0.967 = IC/IE = IC/10\"\n", "IC=alpha*IE\n", "print \"\\nTherefore, IC = %0.2f mA\"%IC\n", "print \"The emitter current, IE = IB + IC\"\n", "IB=IE-IC\n", "print \"Therefore, IB =%0.2f mA\"%IB" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 156 Example 6.4." ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The common-base d.c. current gain, alpha = IC/IE\n", "Therefore, IC =9.80 mA\n", "The emitter current, IE = IB + IC\n", "Therefore, IB =0.20 mA\n" ] } ], "source": [ "IE=10\n", "alpha=0.98\n", "print \"The common-base d.c. current gain, alpha = IC/IE\"\n", "IC=alpha*IE\n", "print \"Therefore, IC =%0.2f mA\"%IC\n", "print \"The emitter current, IE = IB + IC\"\n", "IB=IE-IC\n", "print \"Therefore, IB =%0.2f mA\"%IB" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 158 Example 6.5." ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "If alpha=0.97, beta = alpha/(1 - alpha)\n", "beta = 32.3333\n", "If beta=200, alpha = beta/(beta + 1)\n", "alpha =0.0000\n" ] } ], "source": [ "alpha=0.97\n", "print \"If alpha=0.97, beta = alpha/(1 - alpha)\"\n", "beta=alpha/(1-alpha)\n", "print \"beta = %0.4f\"%beta\n", "beta1=200\n", "print \"If beta=200, alpha = beta/(beta + 1)\"\n", "alpha1 =beta1/(beta1+1)\n", "print \"alpha =%0.4f\"%alpha1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 158 Example 6.6." ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "beta = 100 = IC / IB\n", "Therefore, IB =0.40 mA\n", "IE = IB + IC\n", "IE =40.40 mA\n" ] } ], "source": [ "beta=100.0\n", "IC=40\n", "print \"beta = 100 = IC / IB\"\n", "IB=IC/beta\n", "print \"Therefore, IB =%0.2f mA\"%IB\n", "print \"IE = IB + IC\"\n", "IE=IB+IC\n", "print \"IE =%0.2f mA\"%IE" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 159 Example 6.7." ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The common-base current gain, alpha = beta / (beta + 1) =0.9934\n", "Also, alpha = IC / IE\n", "Therefore, IC = 9.93 mA\n", "the emitter current, IE = IB + IC\n", "Therefore, IB =0.07 mA\n" ] } ], "source": [ "beta=150.\n", "IE=10\n", "alpha=beta/(beta+1)\n", "print \"The common-base current gain, alpha = beta / (beta + 1) =%0.4f\"%alpha\n", "print \"Also, alpha = IC / IE\"\n", "IC=alpha*IE\n", "print \"Therefore, IC = %0.2f mA\"%IC\n", "print \"the emitter current, IE = IB + IC\"\n", "IB=IE-IC\n", "print \"Therefore, IB =%0.2f mA\"%IB" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 160 Example 6.8." ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "We know that (beta), beta = 170 = IC / IB\n", "Therefore, IB =0.47 mA\n", "and IE = IB + IC =80.47 mA\n" ] } ], "source": [ "beta=170.\n", "IC=80\n", "print \"We know that (beta), beta = 170 = IC / IB\"\n", "IB=IC/beta\n", "print \"Therefore, IB =%0.2f mA\"%IB\n", "IE=IB+IC\n", "print \"and IE = IB + IC =%0.2f mA\"%IE" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 161 Example 6.9." ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "beta = 200 = IC / IB\n", "Therefore, IC = 25.00 mA\n", "and IE = IB + IC =25.12 mA\n" ] } ], "source": [ "IB=0.125\n", "beta=200\n", "print \"beta = 200 = IC / IB\"\n", "IC=beta*IB\n", "print \"Therefore, IC = %0.2f mA\"%IC\n", "IE=IB+IC\n", "print \"and IE = IB + IC =%0.2f mA\"%IE" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 162 Example 6.10" ] }, { "cell_type": "code", "execution_count": 16, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "We know that base current, IB = IE / (1 + beta) =0.12 mA\n", "and collector current, IC = IE - IB =11.88 mA\n" ] } ], "source": [ "IE=12.\n", "beta=100\n", "IB=IE/(1+beta)\n", "print \"We know that base current, IB = IE / (1 + beta) =%0.2f mA\"%IB\n", "IC=IE-IB\n", "print \"and collector current, IC = IE - IB =%0.2f mA\"%IC" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 162 Example 6.11" ] }, { "cell_type": "code", "execution_count": 17, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(a) To find beta of the transistor \n", "beta = IC / IB = 20.0000\n", "(b) To find alpha of the transistor\n", "alpha = beta / (1+beta) = 0.9524\n", "(c) To find emitter current, IE\n", "IE = IB + IC = 2.10 mA\n", "(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA\n", "Therefore, IB = 125.00 uA\n", " IC = 2.60 mA\n", "New value of beta of the transistor,\n", "beta = IC / IB =20.8000\n" ] } ], "source": [ "IB=100*10**-6\n", "IC=2*10**-3\n", "beta=IC/IB\n", "print \"(a) To find beta of the transistor \"\n", "print \"beta = IC / IB = %0.4f\"%beta\n", "alpha=beta/(beta+1)\n", "print \"(b) To find alpha of the transistor\"\n", "print \"alpha = beta / (1+beta) = %0.4f\"%alpha\n", "IE=IB+IC\n", "IE1=IE*10**3\n", "print \"(c) To find emitter current, IE\"\n", "print \"IE = IB + IC = %0.2f mA\"%IE1\n", "# answer in the textbook is wrong\n", "print \"(d) To find the new value of beta when delta_IB = 25uA and delta_IC = 0.6mA\"\n", "delta_IB=25*10**-6\n", "delta_IC=0.6*10**-3\n", "IB1=IB+delta_IB\n", "IB11=IB1*10**6\n", "IC1=IC+delta_IC\n", "IC11=IC1*10**3\n", "print \"Therefore, IB = %0.2f uA\"%IB11\n", "print \" IC = %0.2f mA\"%IC11\n", "beta1=IC1/IB1\n", "print \"New value of beta of the transistor,\"\n", "print \"beta = IC / IB =%0.4f\"%beta1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 163 Example 6.12." ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The collector current is, IC = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha)) = 5.15 mA\n", "The emitter current is, IE = IB + IC =5.25 mA\n" ] } ], "source": [ "alpha=0.98\n", "ICO=5*10**-6\n", "ICBO=ICO\n", "IB=100*10**-6\n", "IC=((alpha*IB)/(1-alpha))+(ICO/(1-alpha))\n", "IC1=IC*10**3\n", "print \"The collector current is, IC = ((alpha*IB)/(1-alpha))+(ICO/(1-alpha)) = %0.2f mA\"%IC1\n", "IE=IB+IC\n", "IE1=IE*10**3\n", "print \"The emitter current is, IE = IB + IC =%0.2f mA\"%IE1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 164 Example 6.13." ] }, { "cell_type": "code", "execution_count": 20, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(a) To find the value of collector current when IB = 0.25mA\n", "IC = (beta*IB) + ((1+beta)*ICBO) = 13.01 mA\n", "(b) To find the value of new collector current if temperature rises to 50 C\n", "I''CBO(beta=50) = ICBO*(2**((T2-T1)/10)) =49.25 uA\n", "Therefore, the collector current at 50 C is\n", "IC = (beta*IB) + ((1+beta)*I''CBO) =15.01 mA\n" ] } ], "source": [ "ICBO=10*10**-6\n", "hFE=50\n", "beta=hFE\n", "IB=0.25*10**-3\n", "IC=(beta*IB)+((1+beta)*ICBO)\n", "IC1=IC*10**3\n", "print \"(a) To find the value of collector current when IB = 0.25mA\"\n", "print \"IC = (beta*IB) + ((1+beta)*ICBO) = %0.2f mA\"%IC1\n", "T1=27.\n", "T2=50.\n", "I_CBO = ICBO * (2**((T2-T1)/10))\n", "I_CBO1=I_CBO*10**6\n", "print \"(b) To find the value of new collector current if temperature rises to 50 C\"\n", "print \"I''CBO(beta=50) = ICBO*(2**((T2-T1)/10)) =%0.2f uA\"%I_CBO1\n", "IC2=(beta*IB)+((1+beta)*I_CBO)\n", "IC3=IC2*10**3\n", "print \"Therefore, the collector current at 50 C is\"\n", "print \"IC = (beta*IB) + ((1+beta)*I''CBO) =%0.2f mA\"%IC3" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 165 Example 6.14." ] }, { "cell_type": "code", "execution_count": 21, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The current gain of the transistor is alpha = delta_IC/delta_IE = 0.9900\n" ] } ], "source": [ "delta_IC=0.99*10**-3\n", "delta_IE=1*10**-3\n", "alpha=delta_IC/delta_IE\n", "print \"The current gain of the transistor is alpha = delta_IC/delta_IE = %0.4f\"%alpha" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 165 Example 6.15" ] }, { "cell_type": "code", "execution_count": 23, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) =0.99\n" ] } ], "source": [ "beta_dc=100.\n", "alpha_dc=beta_dc/(1+beta_dc)\n", "print \"The d.c. current gain of the transistor in CB mode is, alpha_dc = beta_dc/(1+beta_dc) =%0.2f\"%alpha_dc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 165 Example 6.16." ] }, { "cell_type": "code", "execution_count": 24, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Common base current gain is, alpha = delta_IC/delta_IE =0.99\n", "Common-emitter current gain is beta = alpha / (1-alpha) =199.00\n" ] } ], "source": [ "delta_IC=0.995*10**-3\n", "delta_IE=1*10**-3\n", "alpha=delta_IC/delta_IE\n", "print \"Common base current gain is, alpha = delta_IC/delta_IE =%0.2f\"%alpha\n", "beta=alpha/(1-alpha)\n", "print \"Common-emitter current gain is beta = alpha / (1-alpha) =%0.2f\"%beta" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 166 Example 6.17." ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "We know that, alpha = beta/(1+beta)\n", "Therefore, the common base current gain is, alpha =0.98\n", "We also know that, alpha = IC / IE\n", "Therefore, IC = alpha * IE = 2.94 mA\n" ] } ], "source": [ "beta=49.\n", "alpha=beta/(1+beta)\n", "print \"We know that, alpha = beta/(1+beta)\"\n", "print \"Therefore, the common base current gain is, alpha =%0.2f\"%alpha\n", "print \"We also know that, alpha = IC / IE\"\n", "IE=3*10**-3\n", "IC=alpha*IE\n", "IC1=IC*10**3\n", "print \"Therefore, IC = alpha * IE = %0.2f mA\"%IC1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 166 Example 6.18." ] }, { "cell_type": "code", "execution_count": 32, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The collector current, IC(mA) = beta * IB =2.25 mA\n", "The emitter current, IE(mA) = IC + IB =2.26 mA\n", "Common-base current gain, alpha = beta/(1+beta) = 0.99\n" ] } ], "source": [ "IB=15*10**-6\n", "beta=150.\n", "IC=beta*IB\n", "IC1=IC*10**3\n", "print \"The collector current, IC(mA) = beta * IB =%0.2f mA\"%IC1\n", "IE=IC+IB\n", "IE1=IE*10**3\n", "print \"The emitter current, IE(mA) = IC + IB =%0.2f mA\"%IE1\n", "alpha=beta/(1+beta)\n", "print \"Common-base current gain, alpha = beta/(1+beta) = %0.2f\"%alpha" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 167 Example 6.19." ] }, { "cell_type": "code", "execution_count": 34, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Referring to fig.6.18, the base current is,\n", "IB = (VBB - VBE) / RB =16.50 uA\n", "The collector current is, IC = beta*IB =3.30 mA\n", "The emitter current is, IE = IC + IB =3.32 mA\n", "Therefore, VCE = VCC - IC*RC =3.40 V\n" ] } ], "source": [ "print \"Referring to fig.6.18, the base current is,\"\n", "VBB=4\n", "VBE=0.7\n", "RB=200*10**3\n", "IB=(VBB-VBE)/RB\n", "IB1=IB*10**6\n", "print \"IB = (VBB - VBE) / RB =%0.2f uA\"%IB1\n", "beta=200\n", "IC=beta*IB\n", "IC1=IC*10**3\n", "print \"The collector current is, IC = beta*IB =%0.2f mA\"%IC1\n", "IE=IC+IB\n", "IE1=IE*10**3\n", "print \"The emitter current is, IE = IC + IB =%0.2f mA\"%IE1\n", "VCC=10\n", "RC=2*10**3\n", "VCE=VCC-(IC*RC)\n", "print \"Therefore, VCE = VCC - IC*RC =%0.2f V\"%VCE" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 167 Example 6.20." ] }, { "cell_type": "code", "execution_count": 35, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2.48 IC = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) =2.48 mA\n", "Therefore, IE = IB + IC =2.50 mA\n" ] } ], "source": [ "alpha_dc=0.99\n", "ICBO=5*10**-6\n", "IB=20*10**-6\n", "IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))\n", "IC1=IC*10**3\n", "print IC1,\"IC = ((alpha_dc*IB)/(1-alpha_dc)) + (ICBO/(1-alpha_dc)) =%0.2f mA\"%IC1\n", "IE=IB+IC\n", "IE1=IE*10**3\n", "print \"Therefore, IE = IB + IC =%0.2f mA\"%IE1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 168 Example 6.21." ] }, { "cell_type": "code", "execution_count": 37, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The leakage current ICBO = 0.2 uA\n", " ICEO = 18 uA\n", "Assume that IB = 30 mA\n", "IE = IB + IC\n", "IC = IE - IB = (beta*IB)+((1+beta)*ICBO)\n", "We know that, ICEO = ICBO/(1-alpha) = (1+beta)*ICBO\n", "beta = (ICEO / ICBO)-1 =89.00\n", "IC = (beta*IB) + ((1+beta)*ICBO) =2.67 A\n", "alpha_dc = 1 - (ICBO / ICEO) =0.99\n", "beta_dc = (IC-ICBO) / (IB-ICEO) =89.05\n" ] } ], "source": [ "ICBO=0.2*10**-6\n", "ICEO=18*10**-6\n", "IB=30*10**-3\n", "print \"The leakage current ICBO = 0.2 uA\"\n", "print \" ICEO = 18 uA\"\n", "print \"Assume that IB = 30 mA\"\n", "print \"IE = IB + IC\"\n", "print \"IC = IE - IB = (beta*IB)+((1+beta)*ICBO)\"\n", "print \"We know that, ICEO = ICBO/(1-alpha) = (1+beta)*ICBO\"\n", "beta=(ICEO/ICBO)-1\n", "print \"beta = (ICEO / ICBO)-1 =%0.2f\"%beta\n", "IC=(beta*IB)+((1+beta)*ICBO)\n", "print \"IC = (beta*IB) + ((1+beta)*ICBO) =%0.2f A\"%IC\n", "alpha_dc=1-(ICBO/ICEO)\n", "print \"alpha_dc = 1 - (ICBO / ICEO) =%0.2f\"%alpha_dc\n", "beta_dc=(IC-ICBO)/(IB-ICEO)\n", "print \"beta_dc = (IC-ICBO) / (IB-ICEO) =%0.2f\"%beta_dc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 168 Example 6.22." ] }, { "cell_type": "code", "execution_count": 39, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Assume that, IB = 1 mA\n", "IC = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = 104.00 mA\n", "IE = IC + IB = 105.00 mA\n" ] } ], "source": [ "alpha_dc=0.99\n", "ICBO=50*10**-6\n", "IB=1*10**-3\n", "IC=((alpha_dc*IB)/(1-alpha_dc))+(ICBO/(1-alpha_dc))\n", "IC1=IC*10**3\n", "print \"Assume that, IB = 1 mA\"\n", "print \"IC = ((alpha_dc*IB) / (1-alpha_dc)) + (ICBO/(1-alpha_dc)) = %0.2f mA\"%IC1\n", "IE=IC+IB\n", "IE1=IE*10**3\n", "print \"IE = IC + IB = %0.2f mA\"%IE1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 169 Example 6.23." ] }, { "cell_type": "code", "execution_count": 43, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(i) DC load line:\n", "Maximum VCE = VCC = 24V\n", "Maximum IC = VCC / RC = 0.00 mA\n", "(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found\n", "Here, VCEQ(V) = VCC / 2 = 12.00 V\n", " ICQ = 1.5 mA\n", "\n", "(iii) AC load line\n", "AC load, R_a.c. = RC || RL = 6.00 kohm\n", "Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. = 21.00 \n", "This locates the point D(OD = 21V) on the VCE axis\n", "Maximum IC = ICQ + VCEQ/R_a.c. = 1.50 mA\n", "This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. \n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "%matplotlib inline\n", "from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend\n", "print \"(i) DC load line:\"\n", "print \"Maximum VCE = VCC = 24V\"\n", "IC=24/(8*10**3) #in Ampere\n", "x1=IC*10**3 #in mA\n", "print \"Maximum IC = VCC / RC = %0.2f mA\"%x1\n", "print \"(ii) For fixing the optimum operating point Q, mark the middle of the d.c. load line AB and the corresponding VCE and IC values can be found\"\n", "VCEQ=24./2\n", "print \"Here, VCEQ(V) = VCC / 2 = %0.2f V\"%VCEQ #in volts\n", "print \" ICQ = 1.5 mA\"\n", "print \"\"\n", "print \"(iii) AC load line\"\n", "Rac=(8*24.)/(8+24) #in k-ohm\n", "print \"AC load, R_a.c. = RC || RL = %0.2f kohm\"%Rac\n", "VCE=12+((1.5*10**-3)*(6*10**3)) #in Volts\n", "print \"Therefore, maximum VCE(V) = VCEQ + ICQ*R_a.c. = %0.2f \"%VCE\n", "print \"This locates the point D(OD = 21V) on the VCE axis\"\n", "IC=(1.5*10**-3)+(12/(6*10**3)) #in Ampere\n", "x3=IC*10**3 #in mA\n", "print \"Maximum IC = ICQ + VCEQ/R_a.c. = %0.2f mA\"%x3\n", "print \"This locates the point C(OC = 3.5mA) on the IC axis. By joining points C and D a.c. load line CD is constructed. \"\n", "x=[24,0]\n", "y=[0,3]\n", "plot(x,y)\n", "x1=[21,0]\n", "y1=[0,3.5]\n", "plot(x1,y1)\n", "title(\"Fig.6.22(b)\")\n", "xlabel(\"VCE(V)\")\n", "ylabel(\"IC(mA)\")\n", "show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 170 Example 6.24." ] }, { "cell_type": "code", "execution_count": 45, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(i) DC load line:\n", "Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line\n", "Maximum IC = VCC / (RC+RE) = 0.00 mA\n", "(ii) Operating point Q\n", "Therefore, V2 = 0.00 V\n", " V2 = VBE + IE*RE\n", "Therefore, IE = V2-VBE / RE = 3.30 mA\n", " IC = IE =3.30 mA\n", "VCE = VCC - IC(RC+RE) = 5.40 V\n", "(iii) AC load line\n", "AC load, Ra.c.(k-ohm) = RC || RL = 0.60 kohm\n", "Therefore, maximum VCE = VCEQ + ICQ*Ra.c. = 7.38 V\n", "Maximum IC(mA) = ICQ + VCEQ/Ra.c. = 12.30 mA\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "%matplotlib inline\n", "from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend\n", "print \"(i) DC load line:\"\n", "print \"Maximum VCE = VCC = 12V, which locates the point B(OB = 12V) of the d.c. load line\"\n", "IC=12/(2*10**3) #in Ampere\n", "x1=IC*10**3 #in mA\n", "print \"Maximum IC = VCC / (RC+RE) = %0.2f mA\"%x1\n", "\n", "print \"(ii) Operating point Q\"\n", "V2=((4*10**3)/(12*10**3))*12 #in V\n", "print \"Therefore, V2 = %0.2f V\"%V2\n", "print \" V2 = VBE + IE*RE\"\n", "IE=(4-0.7)/(1*10**3) #in Ampere\n", "x2=IE*10**3 #in mA\n", "print \"Therefore, IE = V2-VBE / RE = %0.2f mA\"%x2\n", "IC=x2 #in mA\n", "print \" IC = IE =%0.2f mA\"%IC\n", "VCE=12-((3.3*10**-3)*(2*10**3)) #in volts\n", "print \"VCE = VCC - IC(RC+RE) = %0.2f V\"%VCE\n", "print \"(iii) AC load line\"\n", "Rac=1.5/2.5 #in k-ohm\n", "print \"AC load, Ra.c.(k-ohm) = RC || RL = %0.2f kohm\"%Rac\n", "VCE=5.4+((3.3*10**-3)*(0.6*10**3)) #in Volts\n", "print \"Therefore, maximum VCE = VCEQ + ICQ*Ra.c. = %0.2f V\"%VCE\n", "IC=(3.3*10**-3)+(5.4/(0.6*10**3)) #in Ampere\n", "x3=IC*10**3 #in mA\n", "print \"Maximum IC(mA) = ICQ + VCEQ/Ra.c. = %0.2f mA\"%x3\n", "x=[7.38,0]\n", "y=[0,12.3]\n", "plot(x,y)\n", "x1=[12,0]\n", "y1=[0,6]\n", "plot(x1,y1)\n", "title(\"Fig.6.23(b)\")\n", "xlabel(\"VCE(V) -->\")\n", "ylabel(\"IC(mA) -->\")\n", "show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 171 Example 6.25." ] }, { "cell_type": "code", "execution_count": 49, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The collector resistance is, RC = (VCC - VCEQ) / ICQ =4.00 kohm\n", "The base current is, IBQ = ICQ / beta =10.00 uA\n", "The base resistance is, RB = (VCC - VBE(on)) / IBQ =0.93 Mohm\n" ] } ], "source": [ "ICQ=1*10**-3\n", "VCEQ=6.\n", "VCC=10.\n", "beta=100.\n", "VBE=0.7\n", "RC=(VCC-VCEQ)/ICQ\n", "RC1=RC*10**-3\n", "RC2=round(RC1)\n", "print \"The collector resistance is, RC = (VCC - VCEQ) / ICQ =%0.2f kohm\"%RC2\n", "IBQ=ICQ/beta\n", "IBQ1=IBQ*10**6\n", "print \"The base current is, IBQ = ICQ / beta =%0.2f uA\"%IBQ1\n", "RB=(VCC-VBE)/IBQ\n", "RB1=RB*10**-6\n", "print \"The base resistance is, RB = (VCC - VBE(on)) / IBQ =%0.2f Mohm\"%RB1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 171 Example 6.26." ] }, { "cell_type": "code", "execution_count": 50, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "VBB = IB*RB + VBE(on) + IE*RE\n", "Also, IE = IB + IC = IB + beta*IB = (1 + beta)*IB\n", "The base current, IB = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) =53.35 uA\n", "Therefore, IC = beta*IB =5.33 mA\n", "IE = IC + IB = 5.39 mA\n", "VCE = VCC - (IC*RC) - (IE*RE) =4.63 V\n", "The Q point is at\n", "VCEQ = 4.63 V\n", "and ICQ(mA) = 5.33 mA\n" ] } ], "source": [ "beta=100\n", "VBE=0.7\n", "VCC=10\n", "RB=20*10**3\n", "RC=0.4*10**3\n", "RE=0.6*10**3\n", "VBB=5\n", "print \"VBB = IB*RB + VBE(on) + IE*RE\"\n", "print \"Also, IE = IB + IC = IB + beta*IB = (1 + beta)*IB\"\n", "IB=(VBB-VBE)/(RB+((1+beta)*RE))\n", "IB1=IB*10**6\n", "print \"The base current, IB = (VBB - VBE(on)) / (RB + ((1+beta)*RE)) =%0.2f uA\"%IB1\n", "IC=beta*IB\n", "IC1=IC*10**3\n", "print \"Therefore, IC = beta*IB =%0.2f mA\"%IC1\n", "IE=IC+IB\n", "IE1=IE*10**3\n", "print \"IE = IC + IB = %0.2f mA\"%IE1\n", "VCE=VCC-(IC*RC)-(IE*RE)\n", "print \"VCE = VCC - (IC*RC) - (IE*RE) =%0.2f V\"%VCE\n", "print \"The Q point is at\"\n", "print \"VCEQ = %0.2f V\"%VCE\n", "print \"and ICQ(mA) = %0.2f mA\"%IC1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 172 Example 6.27. " ] }, { "cell_type": "code", "execution_count": 54, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(i) DC load line:\n", "When VCE = 0, IC = VCC/RC = 0.00 mA\n", "\n", "(ii) Operating point Q:\n", "Therefore, IB = VCC-VBE / RB = 10.00 uA\n", "Therefore, IC(mA) = beta*IB = 1.00 mA\n", " VCE = VCC - IC*RC = 03 V\n", "Therefore operating point is VCEQ = 3 V and ICQ = 1 mA\n", "\n", "(iii) Stability factor: S = 1 + beta = 1 + 100 = 101\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "%matplotlib inline\n", "from matplotlib.pyplot import plot,title,xlabel,ylabel,show,legend\n", "print \"(i) DC load line:\"\n", "IC=6/(3*10**3) #in Ampere\n", "x1=IC*10**3 #in mA\n", "print \"When VCE = 0, IC = VCC/RC = %0.2f mA\"%x1\n", "print \"\"\n", "print \"(ii) Operating point Q:\"\n", "IB=(6-0.7)/(530*10**3)\n", "x2=IB*10**6\n", "print \"Therefore, IB = VCC-VBE / RB = %0.2f uA\"%x2\n", "IC=100*10*10**-6 # in Ampere\n", "x3=IC*10**3 # in mA\n", "print \"Therefore, IC(mA) = beta*IB = %0.2f mA\"%x3\n", "VCE=6-((1*10**-3)*(3*10**3)) # in volts\n", "print \" VCE = VCC - IC*RC = %02.f V\"%VCE\n", "print \"Therefore operating point is VCEQ = 3 V and ICQ = 1 mA\"\n", "print \"\"\n", "print \"(iii) Stability factor: S = 1 + beta = 1 + 100 = 101\"\n", "x=[6,0]\n", "y=[0,2]\n", "plot(x,y)\n", "title(\"DC load line\")\n", "xlabel(\"VCE (V) --->\")\n", "ylabel(\"IC (mA) --->\")\n", "show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 172 Example 6.28." ] }, { "cell_type": "code", "execution_count": 55, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(a) To determine RB :\n", "RC = (VCC - VCE) / IC =5.00 kohm\n", "IB = IC / beta = 10.00 uA\n", "RB = (VCC - VBE - (IC*RC)) / IB = 630.00 kohm\n", "(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) =56.51\n", "(c) VCC = (beta*IB*RC) + (IB*RB) + VBE\n", " = IB * ((beta*RC) + RB) + VBE\n", "IB = (VCC-VBE) / ((beta*RC)+RB) =12.84 uA\n", "IC = beta*IB =0.64 mA\n", "VCE = VCC - (IC*RC) =8.79 V\n", "Therefore the coordinates of new operating point are :\n", "VCEQ(V) =8.79 V\n", "ICQ =0.64 mA\n" ] } ], "source": [ "VCC=12.\n", "beta=100.\n", "VBE=0.7\n", "print \"(a) To determine RB :\"\n", "VCE=7\n", "IC=1*10**-3\n", "RC=(VCC-VCE)/IC\n", "RC1=RC*10**-3\n", "print \"RC = (VCC - VCE) / IC =%0.2f kohm\"%RC1\n", "IB=IC/beta\n", "IB1=IB*10**6\n", "print \"IB = IC / beta = %0.2f uA\"%IB1\n", "RB=(VCC-VBE-(IC*RC))/IB\n", "RB1=RB*10**-3\n", "print \"RB = (VCC - VBE - (IC*RC)) / IB = %0.2f kohm\"%RB1\n", "S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n", "print \"(b) Stability factor, S =(1 + beta) / (1 + (beta*(RC / (RC+RB)))) =%0.2f\"%S\n", "beta1=50\n", "print \"(c) VCC = (beta*IB*RC) + (IB*RB) + VBE\"\n", "print \" = IB * ((beta*RC) + RB) + VBE\"\n", "IB=(VCC-VBE)/((beta1*RC)+RB)\n", "IB1=IB*10**6\n", "print \"IB = (VCC-VBE) / ((beta*RC)+RB) =%0.2f uA\"%IB1\n", "IC=beta1*IB\n", "IC1=IC*10**3\n", "print \"IC = beta*IB =%0.2f mA\"%IC1\n", "VCE=VCC-(IC*RC)\n", "print \"VCE = VCC - (IC*RC) =%0.2f V\"%VCE\n", "print \"Therefore the coordinates of new operating point are :\"\n", "print \"VCEQ(V) =%0.2f V\"%VCE\n", "print \"ICQ =%0.2f mA\"%IC1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 173 Example 6.29." ] }, { "cell_type": "code", "execution_count": 56, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "RB = VCEQ / IB =32.00 kohm\n", "Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) =56.90\n" ] } ], "source": [ "VCC=12.\n", "RC=250.\n", "IB=0.25*10**-3\n", "beta=100.\n", "VCEQ=8.\n", "RB=VCEQ/IB\n", "RB1=RB*10**-3\n", "print \"RB = VCEQ / IB =%0.2f kohm\"%RB1\n", "S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n", "print \"Stability factor, S = (1+beta) / 1 + (beta*(RC/RC+RB)) =%0.2f\"%S" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 173 Example 6.30." ] }, { "cell_type": "code", "execution_count": 59, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "For a germanium transistor, VBE=0.3V. As alpha=0.985\n", "beta = alpha / ( 1 - alpha) =66.00\n", "(a) To find the coordinates of the operating point\n", "Referring to fig. 6.29,\n", "Thevenin voltage, VT = (R2 / (R1+R2)) * VCC =0.00 V\n", "Thevenin resistance, RB = (R1 * R2) / (R1 + R2) = 14.74 kohm\n", "Therefore, IC =-0.13 mA\n", "Since IB is very small, IC ~ IE = 1.73 mA\n", "Therefore, VCE = VCC - (IC*RC) - (IE*RE) =16.67 V\n", "Therefore, the coordinates of the operating point are :\n", "IC = -0.13 mA\n", "VCE =16.67 V\n", "(b) To find the stability factor S\n", "S =7.24\n" ] } ], "source": [ "VCC=16\n", "RC=3*10**3\n", "RE=2*10**3\n", "R1=56*10**3\n", "R2=20*10**3\n", "alpha=0.985\n", "VBE=0.3\n", "print \"For a germanium transistor, VBE=0.3V. As alpha=0.985\"\n", "beta=alpha/(1-alpha)\n", "beta1=round(beta)\n", "print \"beta = alpha / ( 1 - alpha) =%0.2f\"%beta1\n", "print \"(a) To find the coordinates of the operating point\"\n", "print \"Referring to fig. 6.29,\"\n", "VT=(R2/(R1+R2))*VCC\n", "print \"Thevenin voltage, VT = (R2 / (R1+R2)) * VCC =%0.2f V\"%VT\n", "RB=(R1*R2)/(R1+R2)\n", "RB1=RB*10**-3\n", "print \"Thevenin resistance, RB = (R1 * R2) / (R1 + R2) = %0.2f kohm\"%RB1\n", "IC=(VT-VBE)/((RB/beta)+(RE/beta)+RE)\n", "IC1=IC*10**3\n", "print \"Therefore, IC =%0.2f mA\"%IC1\n", "print \"Since IB is very small, IC ~ IE = 1.73 mA\"\n", "IE=IC\n", "VCE=VCC-(IC*RC)-(IE*RE)\n", "print \"Therefore, VCE = VCC - (IC*RC) - (IE*RE) =%0.2f V\"%VCE\n", "print \"Therefore, the coordinates of the operating point are :\"\n", "print \"IC = %0.2f mA\"%IC1\n", "print \"VCE =%0.2f V\"%VCE\n", "print \"(b) To find the stability factor S\"\n", "S = (1+beta)*((1+(RB/RE))/(1+beta+(RB/RE)))\n", "print \"S =%0.2f\"%S" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 174 Example 6.31." ] }, { "cell_type": "code", "execution_count": 60, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "(a) To determine RE,\n", "VCE = VCC - (IC*RC) - (IE*RE)\n", "Therefore, RE = 1.30 kohm\n", "(b) To determine R1 and R2,\n", "Therefore, RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE = 5.92 kohm\n", "Therefore, R2 = 6.50 kohm\n", "Therefore, R1 = R2 / ((R2/RB)-1)= 64.00 kohm\n" ] } ], "source": [ "VCE=12\n", "IC=2*10**-3\n", "VCC=24\n", "VBE=0.7\n", "beta=50\n", "RC=4.7*10**3\n", "S=5.1\n", "print \"(a) To determine RE,\"\n", "print \"VCE = VCC - (IC*RC) - (IE*RE)\"\n", "RE = (VCC - (IC*RC) - VCE)/IC #IC=IE\n", "RE1=RE*10**-3\n", "print \"Therefore, RE = %0.2f kohm\"%RE1\n", "\n", "print \"(b) To determine R1 and R2,\"\n", "RB=((RE*beta)/(((1+beta)/S)-1))-RE\n", "RB1=(RB*10**-3)\n", "print \"Therefore, RB(k-ohm) = ((RE*beta) / (((1+beta)/S)-1)) - RE = %0.2f kohm\"%RB1\n", "R2=0.1*beta*RE\n", "R2_1=R2*10**-3\n", "print \"Therefore, R2 = %0.2f kohm\"%R2_1\n", "R1=(5.9*10**3*R2)/(R2-(5.9*10**3)) #RB=5.9\n", "R1_1=round(R1*10**-3)\n", "print \"Therefore, R1 = R2 / ((R2/RB)-1)= %0.2f kohm\"%R1_1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 174 Example 6.32." ] }, { "cell_type": "code", "execution_count": 61, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "RTH = R1 || R2 = 10.00 kohm\n", "VTH = (R2 / (R1+R2)) * VCC = 1.79 V\n", "IBQ = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) =15.46 uA\n", "Therefore, ICQ = beta * IBQ =2.32 mA\n", "IEQ = IBQ + ICQ = 2.33 mA\n", "VCEQ = VCC - (ICQ*RC) - (IEQ*RE) = 4.43 V\n", "The Q point is at :\n", "VCEQ = 4.43 V\n", "ICQ = 2.32 mA\n" ] } ], "source": [ "R1=56*10**3\n", "R2=12.2*10**3\n", "RC=2*10**3\n", "RE=400\n", "VCC=10\n", "VBE=0.7\n", "beta=150\n", "RTH=(R1*R2)/(R1+R2)\n", "RTH1=round(RTH*10**-3)\n", "print \"RTH = R1 || R2 = %0.2f kohm\"%RTH1\n", "VTH=(R2/(R1+R2))*VCC\n", "print \"VTH = (R2 / (R1+R2)) * VCC = %0.2f V\"%VTH\n", "IBQ=(VTH-VBE)/(RTH+((1+beta)*RE))\n", "IBQ1=IBQ*10**6\n", "print \"IBQ = (VTH-VBE(on)) / (RTH + ((1+beta)*RE)) =%0.2f uA\"%IBQ1\n", "ICQ=beta*IBQ\n", "ICQ1=ICQ*10**3\n", "print \"Therefore, ICQ = beta * IBQ =%0.2f mA\"%ICQ1\n", "IEQ=IBQ+ICQ\n", "IEQ1=IEQ*10**3\n", "print \"IEQ = IBQ + ICQ = %0.2f mA\"%IEQ1\n", "VCEQ=VCC-(ICQ*RC)-(IEQ*RE)\n", "print \"VCEQ = VCC - (ICQ*RC) - (IEQ*RE) = %0.2f V\"%VCEQ\n", "print \"The Q point is at :\"\n", "print \"VCEQ = %0.2f V\"%VCEQ\n", "print \"ICQ = %0.2f mA\"%ICQ1" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 174 Example 6.33. " ] }, { "cell_type": "code", "execution_count": 64, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "IB = 128.90 uA\n", "IC = 7.73 mA \n", "VCE = 5.75 V\n", "To find stability factor, (S):\n", "Stability factor for voltage divider bias is\n", "S =(1+beta)/(1+(beta*(RE/(RE+RB)))) = where RB = R1 || R2 = 61\n" ] } ], "source": [ "VCC=22\n", "RC=2*10**3\n", "beta=60\n", "VBE=0.6\n", "R1=100*10**3\n", "R2=5*10**3\n", "RE=100\n", "a=VCC-VBE-((R1*VCC)/(R1+R2))\n", "c=(((R1*R2)/(R1+R2))+((1+beta)*RE))\n", "IB=a/c\n", "IB1=IB*10**6\n", "print \"IB = %0.2f uA\"%IB1\n", "IC=beta*IB\n", "IC1=IC*10**3\n", "print \"IC = %0.2f mA \"%IC1\n", "VCE = VCC - (IC*RC) - ((1+beta)*IB*RE)\n", "print \"VCE = %0.2f V\"%VCE\n", "print \"To find stability factor, (S):\"\n", "print \"Stability factor for voltage divider bias is\"\n", "RB=(R1*R2)/(R1+R2)\n", "S=(1+beta)/(1+(beta*(RE/(RE+RB))))\n", "print \"S =(1+beta)/(1+(beta*(RE/(RE+RB)))) = where RB = R1 || R2 = %0.f\"%S" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Page No. 175 Example 6.34." ] }, { "cell_type": "code", "execution_count": 68, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "IB = 46.50 uA\n", "Hence, IC = beta * IB =2.33 mA\n", "VCE = VCC - IC*RC = 5.35 V\n", "Therefore,the co-ordinates of the new operating point are:\n", "VCEQ = 5.35 V\n", "ICQ = 2.33 mA\n", "To find the stability factor S\n", "S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) =51.00\n" ] } ], "source": [ "VCC=10\n", "RC=2*10**3\n", "beta=50\n", "RB=100*10**3\n", "VBE=0.7 #collector to base resistor\n", "IB=(VCC-VBE)/(RB+(beta*RC))\n", "IB1=IB*10**6\n", "print \"IB = %0.2f uA\"%IB1\n", "IC=beta*IB\n", "IC1=IC*10**3\n", "print \"Hence, IC = beta * IB =%0.2f mA\"%IC1\n", "VCE=VCC-(IC*RC)\n", "print \"VCE = VCC - IC*RC = %0.2f V\"%VCE\n", "print \"Therefore,the co-ordinates of the new operating point are:\"\n", "print \"VCEQ = %0.2f V\"%VCE\n", "print \"ICQ = %0.2f mA\"%IC1\n", "print \"To find the stability factor S\"\n", "S=(1+beta)/(1+(beta*(RC/(RC+RB))))\n", "print \"S = (1+beta) / (1 + (beta*[RC/(RC+RB)])) =%0.2f\"%S" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.9" } }, "nbformat": 4, "nbformat_minor": 0 }