{ "metadata": { "name": "", "signature": "sha256:e29ad753b5f2886d343bb74ecb0ecc91fcb2a9898826cbfcabd7e953e57d2f63" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter04: Optical Detectors and Receivers" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.1:Pg-4.5" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Eg= 1.1 \n", "lamda_c = 1.24/Eg \n", "print \"The cut off wavelength in um= \",round(lamda_c,2) \n", "\n", "Eg_ger =0.67 \n", "lamda_ger= 1.24/Eg_ger \n", "print \" \\nThe cut off wavelength for Germanium in um= \",round(lamda_ger,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The cut off wavelength in um= 1.13\n", " \n", "The cut off wavelength for Germanium in um= 1.85\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.2:Pg-4.5" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given \n", "Eg = 1.43 \n", "lamda = 1.24/Eg \n", "lamda=lamda*1000 # converting in nm\n", "print \"The cut off wavelength in nm =\",round(lamda,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The cut off wavelength in nm = 867.13\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.3:Pg-4.5" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "P = 6*10**6 \n", "Eh_pair= 5.4*10**6 \n", "n= Eh_pair/P*100 \n", "print \" The quantum efficiency in % = \",n \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The quantum efficiency in % = 90.0\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.4:Pg-4.6" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "R= 0.65 \n", "P0= 10*10**-6 \n", "Ip= R*P0 \n", "Ip=Ip*10**6 # convertinf in uA...\n", "print \" The generated photocurrent in uA = \",Ip \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The generated photocurrent in uA = 6.5\n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.5:Pg-4.6" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "\n", "Ec= 1.2*10**11 \n", "P= 3*10**11 \n", "lamda = 0.85*10**-6 \n", "n= Ec/P*100 \n", "print \"The efficiency in % =\",n \n", "\n", "q= 1.602*10**-19 \n", "h= 6.625*10**-34 \n", "c= 3*10**8 \n", "n= n/100 \n", "R= n*q*lamda/(h*c) \n", "print \" \\n\\nThe Responsivity of the photodiode in A/W=\",round(R ,4)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The efficiency in % = 40.0\n", " \n", "\n", "The Responsivity of the photodiode in A/W= 0.2741\n" ] } ], "prompt_number": 12 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.6:Pg-4.7" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "\n", "n= 0.65 \n", "E= 1.5*10**-19 \n", "Ip= 2.5*10**-6 \n", "h= 6.625*10**-34 \n", "c= 3*10**8 \n", "lamda= h*c/E \n", "lamda=lamda*10**6 # converting in um for displaying...\n", "print \"The wavelength in um =\",lamda \n", "lamda=lamda*10**-6 \n", "q= 1.602*10**-19 \n", "R= n*q*lamda/(h*c) \n", "print \"\\nThe Responsivity in A/W =\",R \n", "Pin= Ip/R \n", "Pin=Pin*10**6 # converting in uW for displaying/..\n", "print \" \\nThe incidnt power in uW= \",round(Pin,1) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The wavelength in um = 1.325\n", "\n", "The Responsivity in A/W = 0.6942\n", " \n", "The incidnt power in uW= 3.6\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.7:Pg-4.8" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "Iin= 1 \n", "lamda= 1550*10**-9 \n", "q= 1.602*10**-19 \n", "h= 6.625*10**-34 \n", "c= 3*10**8 \n", "n=0.65 \n", "Ip=n*q*lamda*Iin/(h*c) \n", "Ip=Ip*1000 # converting in mA for displaying...\n", "print \" The average photon current in mA= \",int(Ip)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The average photon current in mA= 812\n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.1.8:Pg-4.9" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "\n", "n= 0.70 \n", "Ip= 4*10**-6 \n", "e= 1.602*10**-19 \n", "h= 6.625*10**-34 \n", "c= 3*10**8 \n", "E= 1.5*10**-19\n", "lamda = h*c/E \n", "lamda=lamda*10**6 # converting um for displaying...\n", "print \"The wavelength in um =\",round(lamda,3) \n", "R= n*e/E \n", "Po= Ip/R \n", "Po=Po*10**6 # converting um for displaying...\n", "print \" \\nIncident optical Power in uW =\",round(Po,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The wavelength in um = 1.325\n", " \n", "Incident optical Power in uW = 5.35\n" ] } ], "prompt_number": 14 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.2.1:Pg-4.14" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import math\n", "Ct= 7*10.0**-12\n", "Rt= 50*1*10.0**6/(50+(1*10**6))\n", "B= 1/(2*math.pi*Rt*Ct)\n", "B=B*10**-6 #converting in mHz for displaying...\n", "print \"The bandwidth of photodetector in MHz =\",round(B,2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The bandwidth of photodetector in MHz = 454.75\n" ] } ], "prompt_number": 43 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.2.2:Pg-4.15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import math\n", "W= 25*10**-6 \n", "Vd= 3*10**4 \n", "Bm= Vd/(2*math.pi*W) \n", "RT= 1/Bm \n", "RT=RT*10**9 # converting ns for displaying...\n", "print \" The maximum response time in ns =\",round(RT,2) " ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The maximum response time in ns = 5.24\n" ] } ], "prompt_number": 45 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "\n", "Ex4.2.3:Pg-4.15" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "e= 1.602*10**-19 \n", "h= 6.625*10**-34 \n", "v= 3*10**8 \n", "n=0.65 \n", "I= 10*10**-6 \n", "lamda= 900*10**-9 \n", "R= n*e*lamda/(h*v) \n", "Po= 0.5*10**-6 \n", "Ip= Po*R \n", "M= I/Ip \n", "print \" The multiplication factor =\",round(M,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The multiplication factor = 42.41\n" ] } ], "prompt_number": 48 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.3.1:Pg-4.18" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "n=0.65 \n", "lamda = 900*10**-9 \n", "Pin= 0.5*10**-6 \n", "Im= 10*10**-6 \n", "q= 1.602*10**-19 \n", "h= 6.625*10**-34 \n", "c= 3*10**8 \n", "R= n*q*lamda/(h*c) \n", "Ip= R*Pin \n", "M= Im/Ip \n", "print \" The multiplication factor =\",round(M,2)\n", "print \"\\n***NOTE-Answer wrong in textbook...\"" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The multiplication factor = 42.41\n", "\n", "***NOTE-Answer wrong in textbook...\n" ] } ], "prompt_number": 51 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.6.1:Pg-4.34" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import math\n", "lamda = 1300*10**-9 \n", "Id= 4*10**-9 \n", "n=0.9 \n", "Rl= 1000 \n", "Pincident= 300*10**-9 \n", "BW= 20*10**6 \n", "q= 1.602*10**-19 \n", "h= 6.625*10**-34 \n", "v= 3*10**8 \n", "Iq= math.sqrt((q*Pincident*n*lamda)/(h*v)) \n", "Iq= math.sqrt(Iq) \n", "Iq=Iq*100 # converting in proper format for displaying...\n", "print \"Mean square quantum noise current in Amp*10^11 =\",round(Iq,2)\n", "I_dark= 2*q*BW*Id \n", "I_dark=I_dark*10**19 # converting in proper format for displaying...\n", "print \" \\nMean square dark current in Amp*10^-19 =\",round(I_dark,3) \n", "k= 1.38*10**-23 \n", "T= 25+273 \n", "It= 4*k*T*BW/Rl \n", "It=It*10**16 # converting in proper format for displaying...\n", "print \" \\nMean square thermal nise current in Amp*10^-16 =\",round(It,2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Mean square quantum noise current in Amp*10^11 = 2.31\n", " \n", "Mean square dark current in Amp*10^-19 = 0.256\n", " \n", "Mean square thermal nise current in Amp*10^-16 = 3.29\n" ] } ], "prompt_number": 61 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.8.1:Pg-4.39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "\n", "lamda = 850*10**-9 # meters\n", "BER= 1*10**-9 \n", "N_bar = 9*log(10) \n", "h= 6.625*10**-34 # joules-sec\n", "v= 3*10**8 # meters/sec\n", "n= 0.65 # assumption\n", "E=N_bar*h*v/(n*lamda) \n", "E=E*10**18 # /converting in proper format for displaying...\n", "print \" The Energy received in Joules*10^-18 =\",round(E,2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The Energy received in Joules*10^-18 = 7.45\n" ] } ], "prompt_number": 64 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.8.2:Pg-4.39" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "lamda = 850*10**-9 \n", "BER = 1*10**-9 \n", "BT=10*10**6 \n", "h= 6.625*10**-34 \n", "c= 3*10**8 \n", "Ps= 36*h*c*BT/lamda \n", "Ps=Ps*10**12 # /converting in proper format for displaying...\n", "print \"The minimum incidental optical power required id in pW =\",round(Ps,2) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "The minimum incidental optical power required id in pW = 84.18\n" ] } ], "prompt_number": 67 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Ex4.8.3:Pg-4.40" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "import math\n", "C= 5*10**-12 \n", "B =50*10**6 \n", "Ip= 1*10**-7 \n", "e= 1.602*10**-19 \n", "k= 1.38*10**-23 \n", "T= 18+273 \n", "M= 1 \n", "Rl= 1/(2*math.pi*C*B) \n", "S_N= Ip**2/((2*e*B*Ip)+(4*k*T*B/Rl)) \n", "S_N = 10*math.log10(S_N) # in db\n", "print \" The S/N ratio in dB =\",round(S_N,2) \n", "M=41.54 \n", "S_N_new= (M**2*Ip**2)/((2*e*B*Ip*M**2.3)+(4*k*T*B/Rl)) \n", "S_N_new = 10*math.log10(S_N_new) # in db\n", "print \" \\n\\nThe new S/N ratio in dB =\",round(S_N_new,2)\n", "print \" \\n\\nImprovement over M=1 in dB =\",round(S_N_new-S_N,1) \n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " The S/N ratio in dB = 8.99\n", " \n", "\n", "The new S/N ratio in dB = 32.49\n", " \n", "\n", "Improvement over M=1 in dB = 23.5\n" ] } ], "prompt_number": 70 } ], "metadata": {} } ] }