{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 28 : Bipolar Junction Transistors" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_1 Page No. 910" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Emitter Current Ie = 5.00 Amps\n" ] } ], "source": [ "# A transistor has the following currents: Ib is\u0002 20 mA and Ic is 4.98 A. Calculate Ie.\n", "\n", "# Given data\n", "\n", "Ib = 20*10**-3# # Base current=20 mAmps\n", "Ic = 4.98# # Collector current=4.98 Amps\n", "\n", "Ie = Ic+Ib#\n", "print 'The Emitter Current Ie = %0.2f Amps'%Ie" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_2 Page No. 912" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Collector Current Ic = 0.09804 Amps\n", "i.e 98.04 mAmps\n" ] } ], "source": [ "# A transistor has the following currents: Ie is\u0002 100 mA, Ib is\u0002 1.96 mA. Calculate Ic.\n", "\n", "# Given data\n", "\n", "Ie = 100.0*10**-3# # Emitter current=100 mAmps\n", "Ib = 1.96*10**-3# # Base current=4.98 Amps\n", "\n", "Ic = Ie-Ib#\n", "print 'The Collector Current Ic = %0.5f Amps'%Ic\n", "print 'i.e 98.04 mAmps'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_3 Page No. 913" ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Base Current Ib = 0.0010 Amps\n", "i.e 1 mAmps\n" ] } ], "source": [ "# A transistor has the following currents: Ie is\u0002 50 mA, Ic is\u0002 49 mA. Calculate Ib.\n", "\n", "# Given data\n", "\n", "Ie = 50.0*10**-3# # Emitter current=50 mAmps\n", "Ic = 49.0*10**-3# # Collector current=20 mAmps\n", "\n", "Ib = Ie-Ic#\n", "print 'The Base Current Ib = %0.4f Amps'%Ib\n", "print 'i.e 1 mAmps'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_4 Page No. 914" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Value of Alpha(dc) = 0.9960\n" ] } ], "source": [ "# A transistor has the following currents: Ie is\u0002 15 mA, Ib is\u0002 60 u\u0004A. Calculate \u0002Alpha(dc).\n", "\n", "# Given data\n", "\n", "Ie = 15.*10**-3# # Emitter current=15 mAmps\n", "Ib = 60.*10**-6# # Base current=60 uAmps\n", "\n", "Ic = Ie-Ib#\n", "\n", "Adc = Ic/Ie#\n", "print 'The Value of Alpha(dc) = %0.4f'%Adc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_5 Page No. 916" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Value of Beta(dc) = 200\n" ] } ], "source": [ "# A transistor has the following currents: Ic is\u0002 10 mA and Ib is 50 uA. Calculate Beta(dc).\n", "\n", "# Given data\n", "\n", "Ic = 10.*10**-3# # Collector current=10 mAmps\n", "Ib = 50.*10**-6# # Base current=50 uAmps\n", "\n", "Bdc = Ic/Ib#\n", "print 'The Value of Beta(dc) = %0.f'%Bdc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_6 Page No. 918" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Collector Current Ic = 0.01125 Amps\n", "i.e 11.25 mAmps\n" ] } ], "source": [ "# A transistor has Beta(dc) of 150 and Ib of 75 uAmps. Calculate Ic.\n", "\n", "# Given data\n", "\n", "Bdc = 150.# # Beta(dc)=150\n", "Ib = 75.*10**-6# # Base current=75 uAmps\n", "\n", "Ic = Bdc*Ib#\n", "print 'The Collector Current Ic = %0.5f Amps'%Ic\n", "print 'i.e 11.25 mAmps'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_7 Page No. 920" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Value of Alpha(dc) = 0.9901\n" ] } ], "source": [ "# A transistor has Beta(dc) of 100. Calculate Alpha(dc).\n", "\n", "# Given data\n", "\n", "Bdc = 100.0# # Beta(dc)=100\n", "\n", "Adc = Bdc/(1+Bdc)#\n", "print 'The Value of Alpha(dc) = %0.4f'%Adc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_8 Page No. 922" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Value of Beta(dc) =199.00\n" ] } ], "source": [ "# A transistor has Alpha(dc) of 0.995. Calculate Beta(dc).\n", "\n", "# Given data\n", "\n", "Adc = 0.995# # Alpha(dc)=100\n", "\n", "Bdc = Adc/(1-Adc)#\n", "print 'The Value of Beta(dc) =%0.2f'%Bdc" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_9 Page No. 922" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Power Dissipation = 0.05 Watts\n", "i.e 50 mWatts\n" ] } ], "source": [ "# Calculate Pd if Vcc is 10 V and Ib is 50 uAmps. Assume Beta(dc) is 100.\n", "\n", "# Given data\n", "\n", "Bdc = 100.# # Beta(dc)=100\n", "Ib = 50.*10**-6# # Base current=50 uAmps\n", "Vcc = 10.# # Supply voltage=10 Volts\n", "\n", "Vce = Vcc\n", "\n", "Ic = Bdc*Ib#\n", "\n", "Pd = Vce*Ic#\n", "print 'The Power Dissipation = %0.2f Watts'%Pd\n", "print 'i.e 50 mWatts'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_10 Page No. 922" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Maximum Allowable Collector Current Ic(max) = 0.025 Amps\n", "i.e 25 mAmps\n" ] } ], "source": [ "# The transistor has a power rating of 0.5 W. If Vce is\u0002 20 V, calculate the maximum allowable collector current, Ic, that can exist without exceeding the transistor’s power rating.\n", "\n", "# Given data\n", "\n", "Pdmax = 0.5# # Power dissipation(max)=0.5 Watts\n", "Vce = 20.# # Voltage (collector to emitter)=20 Volts\n", "\n", "Ic = Pdmax/Vce#\n", "print 'The Maximum Allowable Collector Current Ic(max) = %0.3f Amps'%Ic\n", "print 'i.e 25 mAmps'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_11 Page No. 923" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Power Rating at 50°C = 0.28 Watts\n", "i.e 280 mWatts\n" ] } ], "source": [ "# Assume that a transistor has a power rating Pd(max) of 350 mW at an ambient temperature Ta of 25°C. The derating factor is 2.8 mW/°C. Calculate the power rating at 50°C.\n", "\n", "# Given data\n", "\n", "f = 2.8*10**-3# # Derating factor=2.8 mW/°C\n", "Pd = 350.*10**-3# # Power dissipation(max)=350 mWatts\n", "Ta = 25.# # Ambient Temperature=25°C\n", "Tp = 50.# # Power rating at 50°C\n", "\n", "delT = Tp-Ta# # Difference between max and min temp\n", "\n", "delPd = delT*f#\n", "\n", "Prat = Pd-delPd#\n", "print 'The Power Rating at 50°C = %0.2f Watts'%Prat\n", "print 'i.e 280 mWatts'" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_12 Page No. 923" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Base Current = 0.0000 Amps.\n", "Approx 28.97 mAmps\n", "The Collector Current = 0.0043 Amps\n", "Approx 4.35 mAmps\n", "The Voltage Collector-Emitter = 5.48 Volts\n", "Q(5.480769,0.004346)\n", "\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "%matplotlib inline\n", "from matplotlib.pyplot import plot,xlabel,ylabel,show,title\n", "# Solve for Ib, Ic, Vce. Also, Construct a dc load line showing the valuse of Ic(sat), Vce(off), Icq, Vceq.\n", "\n", "# Given data\n", "Vcc = 12.# # Supply voltage=12 Volts\n", "Vbe = 0.7# # Base-Emitter Voltage=0.7 Volts\n", "Rb = 390.*10**3# # Base Resistor=390K Ohms\n", "Rc = 1.5*10**3# # Collector Resistor=1.5K Ohms\n", "B = 150.# # Beta(dc)=150\n", "\n", "Ib = (Vcc-Vbe)/Rb#\n", "print 'The Base Current = %0.4f Amps.'%Ib\n", "print 'Approx 28.97 mAmps'\n", "\n", "Icq = B*Ib#\n", "print 'The Collector Current = %0.4f Amps'%Icq\n", "print 'Approx 4.35 mAmps'\n", "\n", "Vceq = Vcc-(Icq*Rc)#\n", "print 'The Voltage Collector-Emitter = %0.2f Volts'%Vceq\n", "\n", "# For DC load line\n", "\n", "Icsat = (Vcc/Rc)#\n", "Vceoff = Vcc#\n", "\n", "Vce1=[Vceoff, Vceq ,0]\n", "Ic1=[0 ,Icq ,Icsat]\n", "\n", "#To plot DC load line\n", "\n", "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", "plot(Vce1, Ic1)\n", "plot(Vceq,Icq)\n", "plot(0,Icq)\n", "plot(Vceq,0)\n", "plot(0,Icsat)\n", "plot(Vceoff,0)\n", "xlabel(\"Vce in volt\")\n", "ylabel(\"Ic in Ampere\")\n", "title(\"DC Load-line for Base-Biased Transistor Circuit\")\n", "show() " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_13 Page No. 924" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Base Voltage = 2.61 Volts\n", "The Emmiter Voltage = 1.91 Volts\n", "The Collector Voltage = 10.65 Volts\n", "Approx 10.65 Volts\n", "The Collector-Emitter Voltage = 8.74 Volts\n", "Approx 8.74 Volts\n", "The Current Ic(sat) = 0.01 Amps\n", "i.e 9.52 mAmps\n", "The Voltage Vce(off) = 18.00 Volts\n", "Q(8.737067,0.004901)\n", "\n" ] }, { "data": { "image/png": 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"text/plain": [ "" ] }, "metadata": {}, "output_type": "display_data" } ], "source": [ "%matplotlib inline\n", "from matplotlib.pyplot import plot,xlabel,ylabel,show,title\n", "\n", "# Solve for Vb, Ve, Ic, Vc, and Vce. Also, calculate Ic(sat) and Vce(off). Finally, construct a dc load line showing the values of Ic(sat), Vce(off), Icq, and Vceq.\n", "\n", "# Given data\n", "\n", "R1 = 33.*10**3# # Resistor 1=33 kOhms\n", "R2 = 5.6*10**3# # Resistor 2=5.6 kOhms\n", "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", "Re = 390.# # Emitter resistance=390 Ohms\n", "Bdc = 200.# # Beta(dc)= 200\n", "Vcc = 18.# # Supply voltage = 18 Volts\n", "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", "\n", "Vb = Vcc*(R2/(R1+R2))#\n", "print 'The Base Voltage = %0.2f Volts'%Vb\n", "\n", "Ve = Vb-Vbe#\n", "print 'The Emmiter Voltage = %0.2f Volts'%Ve\n", "\n", "Ie = Ve/Re# # Emitter current\n", "\n", "Ic = Ie#\n", "\n", "Vc = Vcc-(Ic*Rc)#\n", "print 'The Collector Voltage = %0.2f Volts'%Vc\n", "print 'Approx 10.65 Volts'\n", "\n", "Vce = Vcc-(Ic*(Rc+Re))#\n", "print 'The Collector-Emitter Voltage = %0.2f Volts'%Vce\n", "print 'Approx 8.74 Volts'\n", "\n", "Icsat = Vcc/(Rc+Re)#\n", "print 'The Current Ic(sat) = %0.2f Amps'%Icsat\n", "print 'i.e 9.52 mAmps'\n", "\n", "Vceoff = Vcc#\n", "print 'The Voltage Vce(off) = %0.2f Volts'%Vceoff\n", "\n", "Icq = Ic\n", "Vceq = Vce\n", "\n", "Vce1=[Vcc, Vceq, 0]\n", "Ic1=[0, Icq, Icsat]\n", "\n", "#To plot DC load line\n", "\n", "print \"Q(%f,%f)\\n\"%(Vceq,Icq)\n", "plot(Vce1, Ic1)\n", "plot(Vceq,Icq)\n", "plot(0,Icq)\n", "plot(Vceq,0)\n", "plot(0,Icsat)\n", "plot(Vceoff,0)\n", "xlabel(\"Vce in Volt\")\n", "ylabel(\"Ic in mAmps\")\n", "title(\"DC Load-line for Voltage Divider-Biased Transistor Circuit\")\n", "show()" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_14 Page No. 925" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Base Voltage = -1.90 Volts\n", "Approx -1.9 Volts\n", "The Emitter Voltage = -1.20 Volts\n", "Approx -1.2 Volts\n", "The Collector Current = 0.00 Amps\n", "Approx 2.4 mAmps\n", "The Collector Voltage = -7.21 Volts\n", "The Collector-Emitter Voltage = -6.01 Volts\n" ] } ], "source": [ "# For the pnp transistor, solve for Vb, Ve, Ic, Vc, and Vce.\n", "\n", "# Given data\n", "\n", "R1 = 33.*10**3# # Resistor1=33 kOhms\n", "R2 = 6.2*10**3# # Resistor2=6.2 kOhms\n", "Rc = 2.*10**3# # Collector resistance=2 kOhms\n", "Re = 500.# # Emitter resistance=500 Ohms\n", "Vcc = 12.# # Supply voltage=12 Volts\n", "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", "\n", "\n", "Vb = -Vcc*(R2/(R1+R2))#\n", "print 'The Base Voltage = %0.2f Volts'%Vb\n", "print 'Approx -1.9 Volts'\n", "\n", "Ve = Vb-(-Vbe)#\n", "print 'The Emitter Voltage = %0.2f Volts'%Ve\n", "print 'Approx -1.2 Volts'\n", "\n", "Ic = -(Ve/Re)# # Ic =~ Ie\n", "print 'The Collector Current = %0.2f Amps'%Ic\n", "print 'Approx 2.4 mAmps'\n", "\n", "Vc = -Vcc+(Ic*Rc)\n", "print 'The Collector Voltage = %0.2f Volts'%Vc\n", "\n", "Vce = -Vcc+(Ic*(Rc+Re))#\n", "print 'The Collector-Emitter Voltage = %0.2f Volts'%Vce" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Example No. 28_15 Page No. 926" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "The Emitter current = 0.0053 Amps\n", "i.e 5.3 mAmps\n", "The Collector voltage = 7.05 Volts\n" ] } ], "source": [ "# Calculate Ie and Vc\n", "\n", "# Given data\n", "\n", "Vee = 6.# # Supply voltage at emitter=6 Volts\n", "Vcc = 15.# # Supply voltage at collector=15 Volts\n", "Vbe = 0.7# # Base-Emmiter Voltage=0.7 Volts\n", "Rc = 1.5*10**3# # Collector resistance=1.5 kOhms\n", "Re = 1.*10**3# # Emitter resistance=1 kOhms\n", "\n", "Ie = (Vee-Vbe)/Re#\n", "print 'The Emitter current = %0.4f Amps'%Ie\n", "print 'i.e 5.3 mAmps'\n", "\n", "Ic = Ie# # Ic =~ Ie\n", "\n", "Vc = Vcc-Ic*Rc#\n", "print 'The Collector voltage = %0.2f Volts'%Vc" ] } ], "metadata": { "kernelspec": { "display_name": "Python 2", "language": "python", "name": "python2" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 2 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython2", "version": "2.7.9" } }, "nbformat": 4, "nbformat_minor": 0 }