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diff --git a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb index 68824e84..88fe84eb 100644 --- a/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb +++ b/Principles_And_Modern_Applications_Of_Mass_Transfer_Operations/Chapter8.ipynb @@ -1,432 +1,453 @@ { "metadata": { - "name": "Chapter_8" - }, - "nbformat": 2, + "name": "", + "signature": "sha256:1b9281c55596391056ad444a3bcc77ac238e8c52d286c4b6044d31a9fb1335cb" + }, + "nbformat": 3, + "nbformat_minor": 0, "worksheets": [ { "cells": [ { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h1>Chapter 8: FET Amplifiers<h1>" ] - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.1, Page Number: 253<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' Voltage gain'''", - "", - "# variable declaration", - "g_m=4.0*10**-3; #gm value", - "R_d=1.5*10**3; #resistance", - "", - "#calculation", - "A_v=g_m*R_d; #voltage gain", - "", - "#result", + "\n", + "# variable declaration\n", + "g_m=4.0*10**-3; #gm value\n", + "R_d=1.5*10**3; #resistance\n", + "\n", + "#calculation\n", + "A_v=g_m*R_d; #voltage gain\n", + "\n", + "#result\n", "print \"Voltage gain = %.2f\" %A_v" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ "Voltage gain = 6.00" ] } - ], + ], "prompt_number": 1 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.2, Page Number: 253<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' The Voltage gain'''", - "", - "# variable declaration", - "r_ds=10.0*10**3;", - "R_d=1.5*10**3; #from previous question", - "g_m=4.0*10**-3; #from previous question", - "", - "#calculation", - "A_v=g_m*((R_d*r_ds)/(R_d+r_ds)); #voltage gain", - "", - "#result", + "\n", + "\n", + "# variable declaration\n", + "r_ds=10.0*10**3;\n", + "R_d=1.5*10**3; #from previous question\n", + "g_m=4.0*10**-3; #from previous question\n", + "\n", + "#calculation\n", + "A_v=g_m*((R_d*r_ds)/(R_d+r_ds)); #voltage gain\n", + "\n", + "#result\n", "print \"Voltage gain = %.2f\" %A_v" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ "Voltage gain = 5.22" ] } - ], + ], "prompt_number": 2 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.3, Page Number:254<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' To find Voltage gain'''", - "", - "# variable declaration", - "R_s=560; #resistance in ohm", - "R_d=1.5*10**3; #resistance in ohm", - "g_m=4*10**-3; #g_m value", - "", - "#calculation", - "A_v=(g_m*R_d)/(1+(g_m*R_s)) #voltage gain", - "", - "#result", + "\n", + "\n", + "# variable declaration\n", + "R_s=560; #resistance in ohm\n", + "R_d=1.5*10**3; #resistance in ohm\n", + "g_m=4*10**-3; #g_m value\n", + "\n", + "#calculation\n", + "A_v=(g_m*R_d)/(1+(g_m*R_s)) #voltage gain\n", + "\n", + "#result\n", "print \"Voltage gain = %.2f\" %A_v" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ "Voltage gain = 1.85" ] } - ], + ], "prompt_number": 3 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.4, Page Number: 257<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "'''Unloaded amplifier'''", - "", - "#Variable declaration", - "vdd=12.0 #volts", - "Id=1.96*10**-3 #Amp", - "Rd=3.3*10**3 #ohm", - "Idss=12.0*10**-3 #Amp", - "Rs=910 # Ohm", - "vgsoff= 3 #v", - "vin=0.1 #V", - "", - "#calculation", - "vd=vdd-(Id*Rd)", - "vgs=-Id*Rs", - "gm0=2*Idss/(abs(vgsoff))", - "gm=0.00325 #mS", - "vout=gm*Rd*vin", - "vout=vout*2*1.414", - "#Result", + "\n", + "\n", + "#Variable declaration\n", + "vdd=12.0 #volts\n", + "Id=1.96*10**-3 #Amp\n", + "Rd=3.3*10**3 #ohm\n", + "Idss=12.0*10**-3 #Amp\n", + "Rs=910 # Ohm\n", + "vgsoff= 3 #v\n", + "vin=0.1 #V\n", + "\n", + "#calculation\n", + "vd=vdd-(Id*Rd)\n", + "vgs=-Id*Rs\n", + "gm0=2*Idss/(abs(vgsoff))\n", + "gm=0.00325 #mS\n", + "vout=gm*Rd*vin\n", + "vout=vout*2*1.414\n", + "#Result\n", "print\"Total output ac voltage(peak-to-peak) = %f V \\nridig on DC value of %fV \"%(vout,vd)" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ - "Total output ac voltage(peak-to-peak) = 3.033030 V ", + "Total output ac voltage(peak-to-peak) = 3.033030 V \n", "ridig on DC value of 5.532000V " ] } - ], + ], "prompt_number": 4 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.5, Page Number: 258<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' RMS voltage'''", - "", - "# variable declaration", - "R_D=3.3*10**3; #resistance in ohm", - "R_L=4.7*10**3; #load resistance in ohm", - "g_m=3.25*10**-3; #from previous question", - "V_in=100.0*10**-3; #previous question", - "", - "#calculation", - "R_d=(R_D*R_L)/(R_D+R_L); #Equivalent drain resistance", - "V_out=g_m*R_d*V_in; #output RMS voltage in volt", - "", - "#result", + "\n", + "# variable declaration\n", + "R_D=3.3*10**3; #resistance in ohm\n", + "R_L=4.7*10**3; #load resistance in ohm\n", + "g_m=3.25*10**-3; #from previous question\n", + "V_in=100.0*10**-3; #previous question\n", + "\n", + "#calculation\n", + "R_d=(R_D*R_L)/(R_D+R_L); #Equivalent drain resistance\n", + "V_out=g_m*R_d*V_in; #output RMS voltage in volt\n", + "\n", + "#result\n", "print \"Output voltage rms value = %.2f Volts\" %V_out" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ "Output voltage rms value = 0.63 Volts" ] } - ], + ], "prompt_number": 5 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.6, Page Number: 259<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' Input resistance wrt signal source '''", - "", - "# variable declaration", - "I_GSS=30.0*10**-9; #current in ampere", - "V_GS=10.0; #ground-source voltage", - "R_G=10.0*10**6; #resistance in ohm", - "", - "#calculation", - "R_IN_gate=V_GS/I_GSS; #gate input resistance", - "R_in=(R_IN_gate*R_G)/(R_IN_gate+R_G); #parallel combination", - "", - "#result", + "\n", + "# variable declaration\n", + "I_GSS=30.0*10**-9; #current in ampere\n", + "V_GS=10.0; #ground-source voltage\n", + "R_G=10.0*10**6; #resistance in ohm\n", + "\n", + "#calculation\n", + "R_IN_gate=V_GS/I_GSS; #gate input resistance\n", + "R_in=(R_IN_gate*R_G)/(R_IN_gate+R_G); #parallel combination\n", + "\n", + "#result\n", "print \"Input resistance as seen by signal source = %.2f ohm\" %R_in" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ "Input resistance as seen by signal source = 9708737.86 ohm" ] } - ], + ], "prompt_number": 6 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.7, Page Number: 260<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' AC and DC output voltage '''", - "", - "# variable declaration", - "I_DSS=200.0*10**-3;", - "g_m=200.0*10**-3;", - "V_in=500.0*10**-3;", - "V_DD=15.0;", - "R_D=33.0;", - "R_L=8.2*10**3;", - "", - "#calculation", - "I_D=I_DSS; #Amplifier is zero biased", - "V_D=V_DD-I_D*R_D;", - "R_d=(R_D*R_L)/(R_D+R_L);", - "V_out=g_m*R_d*V_in;", - "", - "#result", - "print \"DC output voltage = %.2f Volts\" %V_D", + "\n", + "\n", + "# variable declaration\n", + "I_DSS=200.0*10**-3;\n", + "g_m=200.0*10**-3;\n", + "V_in=500.0*10**-3;\n", + "V_DD=15.0;\n", + "R_D=33.0;\n", + "R_L=8.2*10**3;\n", + "\n", + "#calculation\n", + "I_D=I_DSS; #Amplifier is zero biased\n", + "V_D=V_DD-I_D*R_D;\n", + "R_d=(R_D*R_L)/(R_D+R_L);\n", + "V_out=g_m*R_d*V_in;\n", + "\n", + "#result\n", + "print \"DC output voltage = %.2f Volts\" %V_D\n", "print \"AC output voltage = %.2f volts\" %V_out" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ - "DC output voltage = 8.40 Volts", + "DC output voltage = 8.40 Volts\n", "AC output voltage = 3.29 volts" ] } - ], + ], "prompt_number": 7 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.8, Page Number: 262<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' Drain Current '''", - "", - "# Theoretical example", - "# result", - "", - "print \"Part A:\\nQ point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,\"", - "print \"At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is\" ", - "print \"the difference of the two drain currents=1.6mA\"", - "print \"\\nPart B:\\nQ point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,\"", - "print \"At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is\"", - "print\" the difference of the two drain currents=2.8mA\"", - "print \"\\nPart C:\\nQ point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,\"", - "print \" At V_GS=7V, I_D=1.7mA. So peak to peak drain current is\"", + "\n", + "\n", + "# Theoretical example\n", + "# result\n", + "\n", + "print \"Part A:\\nQ point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,\"\n", + "print \"At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is\" \n", + "print \"the difference of the two drain currents=1.6mA\"\n", + "print \"\\nPart B:\\nQ point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,\"\n", + "print \"At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is\"\n", + "print\" the difference of the two drain currents=2.8mA\"\n", + "print \"\\nPart C:\\nQ point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,\"\n", + "print \" At V_GS=7V, I_D=1.7mA. So peak to peak drain current is\"\n", "print \" the difference of the two drain currents=2.2mA\"" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ - "Part A:", - "Q point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,", - "At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is", - "the difference of the two drain currents=1.6mA", - "", - "Part B:", - "Q point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,", - "At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is", - " the difference of the two drain currents=2.8mA", - "", - "Part C:", - "Q point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,", - " At V_GS=7V, I_D=1.7mA. So peak to peak drain current is", + "Part A:\n", + "Q point: V_GS=-2V I_D=2.5mA. At V_GS=-1V, I_D=3.4mA,\n", + "At V_GS=-3V, I_D=1.8mA. So peak to peak drain current is\n", + "the difference of the two drain currents=1.6mA\n", + "\n", + "Part B:\n", + "Q point: V_GS=0V I_D=4mA. At V_GS=1V, I_D=5.3mA,\n", + "At V_GS=-1V, I_D=2.5mA. So peak to peak drain current is\n", + " the difference of the two drain currents=2.8mA\n", + "\n", + "Part C:\n", + "Q point: V_GS=8V I_D=2.5mA. At V_GS=9V, I_D=3.9mA,\n", + " At V_GS=7V, I_D=1.7mA. So peak to peak drain current is\n", " the difference of the two drain currents=2.2mA" ] } - ], + ], "prompt_number": 8 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.9, Page Number:263 <h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' To find Voltage gain'''", - "", - "# variable declaration", - "R_1=47.0*10**3;", - "R_2=8.2*10**3;", - "R_D=3.3*10**3;", - "R_L=33.0*10**3;", - "I_D_on=200.0*10**-3;", - "V_GS=4.0;", - "V_GS_th=2.0;", - "g_m=23*10**-3;", - "V_in=25*10**-3;", - "V_DD=15.0;", - "", - "#calculation", - "V_GSnew=(R_2/(R_1+R_2))*V_DD;", - "K=I_D_on/((V_GS-V_GS_th)**2)", - "#K=value_of_K(200*10**-3,4,2);", - "K=K*1000;", - "I_D=K*((V_GSnew-V_GS_th)**2);", - "V_DS=V_DD-I_D*R_D/1000;", - "R_d=(R_D*R_L)/(R_D+R_L);", - "V_out=g_m*V_in*R_d;", - "", - "#result", - "print \"Drain to source voltage = %.2f volts\" %V_GSnew", - "print \"Drain current = %.2f mA\" %I_D", - "print \"Gate to source voltage = %.2f volts\" %V_DS", - "print \"AC output voltage = %.2f volts\" %V_out", + "\n", + "\n", + "# variable declaration\n", + "R_1=47.0*10**3;\n", + "R_2=8.2*10**3;\n", + "R_D=3.3*10**3;\n", + "R_L=33.0*10**3;\n", + "I_D_on=200.0*10**-3;\n", + "V_GS=4.0;\n", + "V_GS_th=2.0;\n", + "g_m=23*10**-3;\n", + "V_in=25*10**-3;\n", + "V_DD=15.0;\n", + "\n", + "#calculation\n", + "V_GSnew=(R_2/(R_1+R_2))*V_DD;\n", + "K=I_D_on/((V_GS-V_GS_th)**2)\n", + "#K=value_of_K(200*10**-3,4,2);\n", + "K=K*1000;\n", + "I_D=K*((V_GSnew-V_GS_th)**2);\n", + "V_DS=V_DD-I_D*R_D/1000;\n", + "R_d=(R_D*R_L)/(R_D+R_L);\n", + "V_out=g_m*V_in*R_d;\n", + "\n", + "#result\n", + "print \"Drain to source voltage = %.2f volts\" %V_GSnew\n", + "print \"Drain current = %.2f mA\" %I_D\n", + "print \"Gate to source voltage = %.2f volts\" %V_DS\n", + "print \"AC output voltage = %.2f volts\" %V_out\n", "print \"Answer in textbook are approximated\"" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ - "Drain to source voltage = 2.23 volts", - "Drain current = 2.61 mA", - "Gate to source voltage = 6.40 volts", - "AC output voltage = 1.72 volts", + "Drain to source voltage = 2.23 volts\n", + "Drain current = 2.61 mA\n", + "Gate to source voltage = 6.40 volts\n", + "AC output voltage = 1.72 volts\n", "Answer in textbook are approximated" ] } - ], + ], "prompt_number": 9 - }, + }, { - "cell_type": "markdown", + "cell_type": "markdown", + "metadata": {}, "source": [ "<h3>Example 8.10, Page Number: 266<h3>" ] - }, + }, { - "cell_type": "code", - "collapsed": false, + "cell_type": "code", + "collapsed": false, "input": [ - "''' AC and DC output voltage''' ", - "", - "# variable declaration", - "V_DD=-15.0; #p=channel MOSFET", - "g_m=2000.0*10**-6; #minimum value from datasheets", - "R_D=10.0*10**3;", - "R_L=10.0*10**3;", - "R_S=4.7*10**3;", - "", - "#calculation", - "R_d=(R_D*R_L)/(R_D+R_L); #effective drain resistance", - "A_v=g_m*R_d;", - "R_in_source=1.0/g_m;", - "#signal souce sees R_S in parallel with ip rest at source terminal(R_in_source)", - "R_in=(R_in_source*R_S)/(R_in_source+R_S); ", - "", - "#result ", - "print \"minimum voltage gain = %.2f\" %A_v", + "\n", + "\n", + "# variable declaration\n", + "V_DD=-15.0; #p=channel MOSFET\n", + "g_m=2000.0*10**-6; #minimum value from datasheets\n", + "R_D=10.0*10**3;\n", + "R_L=10.0*10**3;\n", + "R_S=4.7*10**3;\n", + "\n", + "#calculation\n", + "R_d=(R_D*R_L)/(R_D+R_L); #effective drain resistance\n", + "A_v=g_m*R_d;\n", + "R_in_source=1.0/g_m;\n", + "#signal souce sees R_S in parallel with ip rest at source terminal(R_in_source)\n", + "R_in=(R_in_source*R_S)/(R_in_source+R_S); \n", + "\n", + "#result \n", + "print \"minimum voltage gain = %.2f\" %A_v\n", "print \"Input resistance seen from signal source = %.2f ohms\" %R_in" - ], - "language": "python", + ], + "language": "python", + "metadata": {}, "outputs": [ { - "output_type": "stream", - "stream": "stdout", + "output_type": "stream", + "stream": "stdout", "text": [ - "minimum voltage gain = 10.00", + "minimum voltage gain = 10.00\n", "Input resistance seen from signal source = 451.92 ohms" ] } - ], + ], "prompt_number": 10 } - ] + ], + "metadata": {} } ] }
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