{ "metadata": { "name": "", "signature": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter6 - Oscillators" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.1 - page 438" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "Vf= 0.0125 # in volt\n", "Vo= 0.5 # in volt\n", "Beta= Vf/Vo \n", "# For oscillator A*Beta= 1\n", "A= 1/Beta \n", "print \"Amplifier Should have a minimum gain of\",A,\"to provide oscillation\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Amplifier Should have a minimum gain of 40.0 to provide oscillation\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.2 - page 439" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import pi, sqrt\n", "# Given data\n", "R1= 50 # in kohm\n", "R1=R1*10**3 # in ohm\n", "R2=R1 # in ohm\n", "R3=R2 # in ohm\n", "C1= 60 # in pF\n", "C1= C1*10**-12 # in F\n", "C2=C1 # in F\n", "C3=C2 # in F\n", "f= 1/(2*pi*R1*C1*sqrt(6)) \n", "print \"Frequency of oscilltions = %0.2f kHz\" %( f*10**-3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Frequency of oscilltions = 21.66 kHz\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.3 - page 445" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from numpy import pi\n", "# Given data\n", "f=2 # in kHz\n", "f=f*10**3 # in Hz\n", "# Let\n", "R= 10 # in kohm (As R should be greater than 1 kohm)\n", "R=R*10**3 # in ohm\n", "# Formula f= 1/(2*pi*R*C)\n", "C= 1/(2*pi*f*R) # in F\n", "C= C*10**9 # in nF\n", "# For Bita to be 1/3, Choose\n", "R4= R # in ohm\n", "R3= 2*R4 # in ohm\n", "print \"Value of C = %0.2f nF\" %C\n", "print \"Value of R3 = %0.f kohm\" %(R3*10**-3)\n", "print \"Value of R4 = %0.f kohm\" %(R4*10**-3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Value of C = 7.96 nF\n", "Value of R3 = 20 kohm\n", "Value of R4 = 10 kohm\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.4 - page 445" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "R1= 200 # in kohm\n", "R1=R1*10**3 # in ohm\n", "R2=R1 # in ohm\n", "C1= 200 # in pF\n", "C1= C1*10**-12 # in F\n", "C2=C1 # in F\n", "f= 1/(2*pi*R1*C1) # in Hz\n", "print \"Frequency of oscilltions = %0.2f kHz\" %(f*10**-3)\n", "\n", "# Note: Calculation to find the value of f in the book is wrong, so answer in the book is wrong" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Frequency of oscilltions = 3.98 kHz\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.5 - page 460" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "L= 100 # in \u00b5H\n", "L= L*10**-6 # in H\n", "C1= .001 # in \u00b5F\n", "C1= C1*10**-6 # in F\n", "C2= .01 # in \u00b5F\n", "C2= C2*10**-6 # in F\n", "C= C1*C2/(C1+C2) # in F\n", "# (i)\n", "f= 1/(2*pi*sqrt(L*C)) # in Hz\n", "print \"Operating frequency = %0.f kHz\" %(round(f*10**-3))\n", "# (ii)\n", "Beta= C1/C2 \n", "print \"Feedback fraction = %0.1f \" %Beta\n", "# (iii)\n", "# A*Bita >=1, so Amin*Bita= 1\n", "Amin= 1/Beta \n", "print \"Minimum gain to substain oscillations is\",Amin" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Operating frequency = 528 kHz\n", "Feedback fraction = 0.1 \n", "Minimum gain to substain oscillations is 10.0\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.6 - page 460" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "L= 15 # in \u00b5H\n", "L= L*10**-6 # in H\n", "C1= .004 # in \u00b5F\n", "C1= C1*10**-6 # in F\n", "C2= .04 # in \u00b5F\n", "C2= C2*10**-6 # in F\n", "C= C1*C2/(C1+C2) # in F\n", "f= 1/(2*pi*sqrt(L*C)) # in Hz\n", "print \"Frequency of oscilltions = %0.1f kHz\" %(f*10**-3)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Frequency of oscilltions = 681.5 kHz\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.7 - page 461" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "L= 0.01 # in H\n", "C= 10 # in pF\n", "C= C*10**-12 # in F\n", "f= 1/(2*pi*sqrt(L*C)) # in Hz\n", "print \"Frequency of oscilltions = %0.2f kHz\" %(f*10**-3)\n", "# Note: Calculation to find the value of f in the book is wrong, so answer in the book is wrong" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Frequency of oscilltions = 503.29 kHz\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.8 - page 463" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "L= 0.8 # in H\n", "\n", "C= .08 # in pF\n", "C= C*10**-12 # in F\n", "C_M= 1.9 # in pF\n", "C_M= C_M*10**-12 # in F\n", "C_T= C*C_M/(C+C_M) # in F\n", "R=5 # in kohm\n", "f_s= 1/(2*pi*sqrt(L*C)) # in Hz\n", "print \"Series resonant frequency = %0.f kHz\" %(f_s*10**-3)\n", "# (ii)\n", "f_p= 1/(2*pi*sqrt(L*C_T)) # in Hz\n", "print \"Parallel resonant frequency = %0.f kHz\" %(f_p*10**-3)\n", "# Note: Calculation to find the value of parallel resonant frequency in the book is wrong, so answer in the book is wrong" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Series resonant frequency = 629 kHz\n", "Parallel resonant frequency = 642 kHz\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.10 - page 466" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "R1= 220 # in kohm\n", "R1=R1*10**3 # in ohm\n", "R2=R1 # in ohm\n", "C1= 250 # in pF\n", "C1= C1*10**-12 # in F\n", "C2=C1 # in F\n", "f= 1/(2*pi*R1*C1) \n", "print \"Frequency of oscilltions = %0.2f Hz\" %f" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Frequency of oscilltions = 2893.73 Hz\n" ] } ], "prompt_number": 11 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.11 - page 467" ] }, { "cell_type": "code", "collapsed": false, "input": [ "from math import tan\n", "# Given data\n", "R= 10 # in kohm\n", "R=R*10**3 # in ohm\n", "f=1000 \n", "fie= 60 # in \u00b0\n", "# The impedence of given circuit , Z= R+j*1/(omega*C)\n", "# the phase shift, tan(fie)= imaginary part/ Real part\n", "# tand(fie) = 1/(omega*R*C)\n", "C= 1/(2*pi*R*tan(fie*pi/180)) \n", "print \"The value of C = %0.2f pF\" %(C*10**12)\n", "# Note : There is an calculation error to evaluate the value of C, So the answer in the book is wrong" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The value of C = 9188814.92 pF\n" ] } ], "prompt_number": 17 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.12 - page 467" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "L= 50 # in \u00b5H\n", "L= L*10**-6 # in H\n", "C1= 300 # in pF\n", "C1= C1*10**-12 # in F\n", "C2= 100 # in pF\n", "C2= C2*10**-12 # in F\n", "C_eq= C1*C2/(C1+C2) # in F\n", "f= 1/(2*pi*sqrt(L*C_eq)) # in Hz\n", "print \"Frequency of oscillations = %0.1f MHz\" %(f*10**-6)\n", "Beta= C2/C1 \n", "# (iii)\n", "# A*Beta >=1, so A*Bita= 1 (for sustained oscillations)\n", "Amin= 1/Beta \n", "print \"Minimum gain to substain oscillations is\",Amin" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Frequency of oscillations = 2.6 MHz\n", "Minimum gain to substain oscillations is 3.0\n" ] } ], "prompt_number": 18 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Exa 6.14 - page 469" ] }, { "cell_type": "code", "collapsed": false, "input": [ "# Given data\n", "L1= 2 # in mH\n", "L1= L1*10**-3 # in H\n", "L2= 1.5 # in mH\n", "L2= L2*10**-3 # in H\n", "# Formula f= 1/(2*pi*sqrt((L1+L2)*C)\n", "# For f= 1000 kHz, C will be maximum\n", "f=1000 # in kHz\n", "f=f*10**3 # in Hz\n", "Cmax= 1/((2*pi*f)**2*(L1+L2)) # in F\n", "# For f= 2000 kHz, C will be maximum\n", "f=2000 # in kHz\n", "f=f*10**3 # in Hz\n", "Cmin= 1/((2*pi*f)**2*(L1+L2)) # in F\n", "print \"Maximum Capacitance = %0.1f pF\" %(Cmax*10**12)\n", "print \"Minimum Capacitance = %0.1f pF\" %(Cmin*10**12)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Maximum Capacitance = 7.2 pF\n", "Minimum Capacitance = 1.8 pF\n" ] } ], "prompt_number": 19 } ], "metadata": {} } ] }