{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 08 : BJT specifications and performance" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.2, Page No 313" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "P2=25.0*10**-3#when frequency increase to 20KHz\n", "P1=50.0*10**-3#when signal frequency is 5KHz\n", "\n", "#Calculations\n", "Po=10*math.log((P2/P1),10)\n", "\n", "#Results\n", "print(\" output power change in decibels is %.2f dB \" %Po)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " output power change in decibels is -3.01 dB \n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.3, Page No 314" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#initialisation of variables\n", "v1=1# output voltage measured at 5KHz\n", "v2=.707# output voltage measure at 20kHz\n", "\n", "#Calculations\n", "Po=20*math.log((v2/v1),10)\n", "\n", "#Results\n", "print(\" output power change is %.2f dB \" %Po)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " output power change is -3.01 dB \n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.4 Page No 317" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#initialisation of variables\n", "Ic=1.0*10**-3\n", "hfe=50.0\n", "hie=1.3*10**3\n", "fT=250.0*10**6\n", "Cbc=5.0*10**-12\n", "Rc=8.2*10**3\n", "Rl=100.0*10**3\n", "\n", "#Calculations\n", "Ie=Ic\n", "Av=(hfe*((Rc*Rl)/(Rc+Rl)))/hie\n", "Cbe=(6.1*Ie)/fT\n", "Cin=(Cbe+(1+Av)*Cbc)*10**9\n", "\n", "#Results\n", "print(\" input capacitance when the circuit operated as CE is %.2fnF \"%Cin)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " input capacitance when the circuit operated as CE is 1.49nF \n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.5, Page No 319" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "R1=100*10**3\n", "R2=47.0*10**3\n", "Re=4.7*10**3\n", "Cbc=5.0*10**-12\n", "Cbe=24.4*10**-12\n", "hfe=50\n", "hie=1.3*10**3\n", "hib=24.5\n", "rs=hib\n", "rs=600.0\n", "\n", "#Calculations\n", "print(\" common emitter circuit\")\n", "Rb=(R1*R2)/(R1+R2)\n", "Zi=(Rb*hie)/(Rb+hie)\n", "Cin=1.48*10**-9\n", "f2=1/(2*3.14*Cin*((rs*Zi)/(rs+Zi)))\n", "print(\"input-capacitance upper cutoff frequency is %dHz \" %f2)\n", "print(\"common base circuit\")\n", "Zi=(Re*hib)/(Re+hib)\n", "Cin=(Cbe+Cbc)\n", "f2=(1/(2*3.14*Cin*((rs*Zi)/(rs+Zi))))*10**-6\n", "\n", "#Results\n", "print(\" input capacitance upper cutoff when operating as CB circuit with base bypassed to ground is %.2f MHz \" %f2)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " common emitter circuit\n", "input-capacitance upper cutoff frequency is 265447Hz \n", "common base circuit\n", " input capacitance upper cutoff when operating as CB circuit with base bypassed to ground is 231.25 MHz \n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.6 Page No 322" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "fT=50.0*10**6\n", "hfe=50.0\n", "f2o=60.0*10**3\n", "Rc=10.0*10**3\n", "\n", "#Calculations\n", "fae=fT/hfe\n", "C4=(1.0/(2*3.14*f2o*Rc))*10**12\n", "\n", "#Results\n", "print(\"capacitance required for C4 to give 60kHz upper cutoff frequency is %.2f pF \" %C4)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "capacitance required for C4 to give 60kHz upper cutoff frequency is 265.39 pF \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.8 Page No 326" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#initialisation of variables\n", "ton=100.0*10**-9\n", "Rs=600.0\n", "Rb=4.7*10**3\n", "\n", "#Calculations\n", "C1=(ton/Rs)*10**12\n", "print(\" suitable speed up capacitor is %dpF \" %C1)\n", "C1=160*10**-12#standard value\n", "PWmin=(5*Rs*C1)\n", "SWmin=5*Rb*C1\n", "fmax=1/(PWmin+SWmin)\n", "\n", "#Results\n", "print(\"maximum signal frequency is %.2f Hz \" %(fmax/1000))" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " suitable speed up capacitor is 166pF \n", "maximum signal frequency is 235.85 Hz \n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.9, Page No 330" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "R1=30.0*10**3\n", "R2=30.0*10**3\n", "rs=30.0*10**3\n", "f2=40.0*10**3\n", "f1=100.0\n", "k=1.37*10**-23\n", "R=10.0*10**3\n", "Av=600.0\n", "Ri=3.0*10**3\n", "\n", "#Calculations\n", "Rb=(R1*R2)/(R1+R2)\n", "Rg=(rs*Rb)/(rs+Rb)\n", "T=(273+25)\n", "B=f2-f1\n", "en=math.sqrt(4*k*T*B*R)\n", "eni=en*((Ri/(Ri+Rg)))\n", "eno=(Av*eni)*10**6\n", "\n", "#Results\n", "print(\"noise output voltage is %.2f uV \" %eno)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "noise output voltage is 353.44 uV \n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.10 Page No 331" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math \n", "\n", "#initialisation of variables\n", "Ic=30.0*10**-6\n", "Vce=5.0\n", "eno=354.0*10**-6\n", "NF=10.0\n", "F=2.51#F=antilog(NF/10)\n", "\n", "#Calculations\n", "Vn=((math.sqrt(F))*eno)*10**6\n", "\n", "#Results\n", "print(\"total noise output volateg for amplifier is %.2f uV \" %Vn)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "total noise output volateg for amplifier is 560.84 uV \n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.11 Page No 333" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "#initialisation of variables\n", "Pd25=625.0*10**-3\n", "D=5.0*10**-3\n", "Vce=10.0\n", "T2=55.0\n", "\n", "#Calculations\n", "Pdt2=Pd25-D*(T2-25)\n", "Pd=Pdt2\n", "Ic=Pd/Vce\n", "\n", "#Results\n", "print(\" maximum Ic level is %.2fA \" %(Ic*1000))\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ " maximum Ic level is 47.50A \n" ] } ], "prompt_number": 9 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.13 Page No 335" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#initialisation of variables\n", "Pd=80.0\n", "Vce=60.0\n", "\n", "#Calculations\n", "Ic=Pd/Vce\n", "print(\"point 1 Vce=60 and Ic= %.2f A\" %Ic)\n", "Vce=40.0\n", "Ic=Pd/Vce\n", "print(\"point 2 Vce=40 and Ic= %.2f A\" %Ic)\n", "Vce=20.0\n", "Ic=Pd/Vce\n", "print(\" point 3 Vce=20 and Ic= %.2f A\" %Ic)\n", "Vce=10.0\n", "Ic=Pd/Vce\n", "\n", "#Results\n", "print(\" point 4 Vce=10 and Ic= %.2f A\" %Ic)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "point 1 Vce=60 and Ic= 1.33 A\n", "point 2 Vce=40 and Ic= 2.00 A\n", " point 3 Vce=20 and Ic= 4.00 A\n", " point 4 Vce=10 and Ic= 8.00 A\n" ] } ], "prompt_number": 10 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 8.14, Page No 339" ] }, { "cell_type": "code", "collapsed": false, "input": [ "import math\n", "\n", "#initialisation of variables\n", "Vce=20.0\n", "Ic=1.0\n", "T2=90.0\n", "T1=25.0\n", "\n", "#Calculations\n", "Q=Vce*Ic\n", "Qcs=.4\n", "Qjc=1#from table\n", "Qsa=((T2-T1)/Q)-(Qjc+Qcs)\n", "\n", "#Results\n", "print(\"Qsa= %.2f \" %Qsa)" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Qsa= 1.85 \n" ] } ], "prompt_number": 11 } ], "metadata": {} } ] }