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-{
- "metadata": {
- "name": "",
- "signature": "sha256:bace9e702292e3d3629b1af1b04f785ea7b14f0684491da077287b62be815475"
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 8: FET Amplifiers<h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.1, Page Number: 253<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage gain = 6.00"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.2, Page Number: 253<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage gain = 5.22"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.3, Page Number:254<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Voltage gain = 1.85"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.4, Page Number: 257<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Total output ac voltage(peak-to-peak) = 3.033030 V \n",
- "ridig on DC value of 5.532000V "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.5, Page Number: 258<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Output voltage rms value = 0.63 Volts"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.6, Page Number: 259<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "Input resistance as seen by signal source = 9708737.86 ohm"
- ]
- }
- ],
- "prompt_number": 6
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.7, Page Number: 260<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "DC output voltage = 8.40 Volts\n",
- "AC output voltage = 3.29 volts"
- ]
- }
- ],
- "prompt_number": 7
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8.8, Page Number: 262<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\n",
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "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",
- "metadata": {},
- "source": [
- "<h3>Example 8.9, Page Number:263 <h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "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",
- "metadata": {},
- "source": [
- "<h3>Example 8.10, Page Number: 266<h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "\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",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "minimum voltage gain = 10.00\n",
- "Input resistance seen from signal source = 451.92 ohms"
- ]
- }
- ],
- "prompt_number": 10
- }
- ],
- "metadata": {}
- }
- ]
-} \ No newline at end of file
generated by cgit (git 2.34.1) at 2025-04-01 15:19:26 +0000