{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Chapter 5 Biasing Methods" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_1 PG 5.4" ] }, { "cell_type": "code", "execution_count": 1, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ " Refer to the figure-5.5 shown\n", "i) When Rb=300 kohm\n", " Base emitter junction is not reverse biased \n", " Assume transistor is operating in active region\n", " We know that Vcc=Ib*Rb+Vbe \n", "\n", " base current Ib is: 31.00 microA \n", "\n", "\n", " Collector current is 3.1 mA \n", "\n", " Applying KVL around collector loop ie Vcc=Ic*Rc+Vce\n", "\n", " Now Vce= 3.8 V \n", "\n", "\n", " Since Vce=3.8 V collector to base junction is reverse biased and we can say \n", " that our assumption that transistor is in active region is justified\n", "\n", "\n", " ii)When Rb=150 kohm\n", "\n", "\n", " base emitter junction is not reverse biased \n", " assume transistor is operating in active region\n", " Applying KVL around base loop ie Vcc=Ib*Rb+Vbe\n", "\n", " base current Ib is: 62 microA \n", "\n", "\n", " Collector current is 6.2 mA \n", "\n", " Applying KVL around collector loop ie Vcc=Ic*Rc+Vce\n", "\n", " Therefore Vce= -2.4 V \n", "\n", "\n", " Collector voltage Vce has to be +ve or zero but Vce=-2.4 V hence \n", " transistor is not in active region but it is in saturation region\n", "\n", " \n", " Applying KVL around base loop ie Vcc=Ib*Rb+Vbe_sat\n", "\n", " base current Ib is: 61 microA \n", "\n", " Applying KVL around collector loop ie Vcc=Ic*Rc+Vce_sat\n", "\n", " Collector current is 4.9 mA \n", "\n", " To justify transistor is in saturation then \n", " Ib must be greater than (Ic/Beta)\n", "\n", "\n", " Now Ib=61 microA \n", " \n", " (Ic/Beta)=49 microA \n", "\n", " Hence transistor in saturation region is satisfied \n" ] } ], "source": [ "print \" Refer to the figure-5.5 shown\"\n", "print \"i) When Rb=300 kohm\"\n", "print \" Base emitter junction is not reverse biased \"\n", "print \" Assume transistor is operating in active region\"\n", "Rb=300e3#\n", "Rc=2e3##collector resistance\n", "Vcc=10##supply voltage\n", "Vbe=0.7##base emitter voltage\n", "print \" We know that Vcc=Ib*Rb+Vbe \"\n", "Ib=(Vcc-Vbe)/Rb##since Vcc=Ib*Rb+Vbe\n", "print \"\\n base current Ib is: %.2f microA \\n\"%(Ib*1e6)\n", "Beta=100#\n", "Ic=Beta*Ib##colector current in active region\n", "print \"\\n Collector current is %.1f mA \\n\"%(Ic*1e3)\n", "print \" Applying KVL around collector loop ie Vcc=Ic*Rc+Vce\"\n", "Vce=Vcc-Ic*Rc##since Vcc=Ic*Rc+Vce\n", "print \"\\n Now Vce= %.1f V \\n\\n\"%(Vce)\n", "print \" Since Vce=3.8 V collector to base junction is reverse biased and we can say \\n that our assumption that transistor is in active region is justified\"\n", "print \"\\n\\n ii)When Rb=150 kohm\\n\\n\"\n", "print \" base emitter junction is not reverse biased \"\n", "Rb=150e3#\n", "print \" assume transistor is operating in active region\"\n", "print \" Applying KVL around base loop ie Vcc=Ib*Rb+Vbe\"\n", "Ib=(Vcc-Vbe)/Rb##since Vcc=Ib*Rb+Vbe\n", "print \"\\n base current Ib is: %.0f microA \\n\"%(Ib*1e6)\n", "Ic=Beta*Ib##colector current in active region\n", "print \"\\n Collector current is %.1f mA \\n\"%(Ic*1e3)\n", "print \" Applying KVL around collector loop ie Vcc=Ic*Rc+Vce\"\n", "Vce=Vcc-Ic*Rc##since Vcc=Ic*Rc+Vce\n", "print \"\\n Therefore Vce= %.1f V \\n\\n\"%(Vce)\n", "print \" Collector voltage Vce has to be +ve or zero but Vce=-2.4 V hence \\n transistor is not in active region but it is in saturation region\\n\\n \"\n", "Vbe_sat=0.8##base saturation voltage\n", "Vce_sat=0.2##collector saturation voltage\n", "print \" Applying KVL around base loop ie Vcc=Ib*Rb+Vbe_sat\"\n", "Ib=(Vcc-Vbe_sat)/Rb##since Vcc=Ib*Rb+Vbe_sat\n", "print \"\\n base current Ib is: %.0f microA \\n\"%(Ib*1e6)\n", "print \" Applying KVL around collector loop ie Vcc=Ic*Rc+Vce_sat\"\n", "Ic=(Vcc-Vce_sat)/Rc#since Vcc=Ic*Rc+Vce_sat\n", "print \"\\n Collector current is %.1f mA \\n\"%(Ic*1e3)\n", "print \" To justify transistor is in saturation then \\n Ib must be greater than (Ic/Beta)\"\n", "x=Ic/Beta\n", "print \"\\n\\n Now Ib=%.0f microA \\n \\n (Ic/Beta)=%.0f microA \\n\"%(Ib*1e6,x*1e6)\n", "if (Ib>x) :\n", " x=(Ic/Beta)\n", " print \" Hence transistor in saturation region is satisfied \" " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_2 PG-5.6" ] }, { "cell_type": "code", "execution_count": 2, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "\n", " base current is 0.00007533 A \n", "\n", "\n", " for hFE_min=50\n", "\n", " Ic=3.767 A \n", "\n", " Vce=4.4667 V \n", "\n", "\n", "\n", " for hFE_max=60\n", "\n", " Ic=4.52 A \n", "\n", " Vce=3 V \n", "\n" ] } ], "source": [ "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=12##supply voltage\n", "Rb=150e3#\n", "Rc=2e3\n", "hFE_min=50##minimum voltage gain\n", "hFE_max=60##maximum voltage gain\n", "Ib=(Vcc-Vbe)/Rb##since Vcc=Ib*Rb+Vbe\n", "print \"\\n base current is %.8f A \\n\"%(Ib)\n", "print \"\\n for hFE_min=50\"\n", "Ic=hFE_min*Ib\n", "print \"\\n Ic=%.3f A \\n\"%(Ic*1e3)\n", "Vce=Vcc-Ic*Rc\n", "print \" Vce=%.4f V \\n\"%(Vce)\n", "print \"\\n\\n for hFE_max=60\"\n", "Ic=hFE_max*Ib\n", "print \"\\n Ic=%.2f A \\n\"%(Ic*1e3)\n", "Vce=Vcc-Ic*Rc\n", "print \" Vce=%.0f V \\n\"%(Vce)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_3 PG5.8 " ] }, { "cell_type": "code", "execution_count": 3, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.8 shown\n", "\n", " base current is 10.18 microA \n", "\n", "\n", " Ic=1.018 mA \n", "\n", "\n", " Vce=1.7180 V \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.8 shown\"\n", "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=12##supply voltage\n", "Rb=100e3#\n", "Rc=10e3#\n", "Beta=100##voltage gain\n", "Ib=(Vcc-Vbe)/((1+Beta)*Rc+Rb)##since Vcc=Ib*Rb+Vbe\n", "print \"\\n base current is %.2f microA \\n\"%(Ib*1e6)\n", "Ic=Beta*Ib\n", "print \"\\n Ic=%.3f mA \\n\"%(Ic*1e3)\n", "Vce=Vcc-(Ib+Ic)*Rc\n", "print \"\\n Vce=%.4f V \\n\"%(Vce)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_4 PG-5.9" ] }, { "cell_type": "code", "execution_count": 4, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "i) for hFE_min=50\n", "\n", " base current is 44.84 microA \n", "\n", " Ic=2.242 mA \n", "\n", " Vce=7.426 V \n", "\n", "\n", " for hFE_max=60\n", "\n", " base current is 41.54 microA \n", "\n", " Ic=2.493 mA \n", "\n", " Vce=6.932 V \n", "\n" ] } ], "source": [ "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=12##supply voltage\n", "Rb=150e3#\n", "Rc=2e3\n", "hFE_min=50#\n", "hFE_max=60#\n", "print \"i) for hFE_min=50\"\n", "Beta=hFE_min##minimum voltage gain\n", "Ib=(Vcc-Vbe)/((1+Beta)*Rc+Rb)##since Vcc=Ib*Rb+Vbe\n", "print \"\\n base current is %.2f microA \\n\"%(Ib*1e6)\n", "Ic=Beta*Ib\n", "print \" Ic=%.3f mA \\n\"%(Ic*1e3)\n", "Vce=Vcc-(Ib+Ic)*Rc\n", "print \" Vce=%.3f V \\n\"%(Vce)\n", "print \"\\n for hFE_max=60\"\n", "Beta=hFE_max##maximum voltage gain\n", "Ib=(Vcc-Vbe)/((1+Beta)*Rc+Rb)##since Vcc=Ib*Rb+Vbe\n", "print \"\\n base current is %.2f microA \\n\"%(Ib*1e6)\n", "Ic=Beta*Ib\n", "print \" Ic=%.3f mA \\n\"%(Ic*1e3)\n", "Vce=Vcc-(Ib+Ic)*Rc\n", "print \" Vce=%.3f V \\n\"%(Vce)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_5 PG-5.11" ] }, { "cell_type": "code", "execution_count": 5, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.13 shown\n", "\n", " Vb=3.33 V\n", "\n", "\n", " Ve=2.63 V \n", "\n", "\n", " Ie=5.27 mA \n", "\n", "\n", " Ib=52.08 microA \n", "\n", "\n", " Ic=5.208 mA \n", "\n", "We apply KVL to the collector circuit\n", "Vcc-Ic*Rc-Vce-Ie*Re=0\n", "\n", " Vce=2.16 V \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.13 shown\"\n", "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=10##supply voltage\n", "R1=10e3#\n", "Rc=1e3#\n", "R2=5e3#\n", "Re=500#\n", "Beta=100##voltage gain\n", "Vb=R2*Vcc/(R1+R2)##base voltage\n", "print \"\\n Vb=%.2f V\\n\"%(Vb)\n", "Ve=Vb-Vbe##emitter voltage\n", "print \"\\n Ve=%.2f V \\n\"%(Ve)\n", "Ie=Ve/Re#\n", "print \"\\n Ie=%.2f mA \\n\"%(Ie*1e3)\n", "Ib=5.26e-3/(1+Beta)##Ie=0.00526 A=5.26 mA\n", "print \"\\n Ib=%.2f microA \\n\"%(Ib*1e6)\n", "Ic=Beta*Ib#\n", "print \"\\n Ic=%.3f mA \\n\"%(Ic*1e3)\n", "print \"We apply KVL to the collector circuit\"\n", "print \"Vcc-Ic*Rc-Vce-Ie*Re=0\"\n", "Vce=Vcc-Ic*Rc-Ie*Re##since Vcc-Ic*Rc-Vce-Ie*Re=0\n", "print \"\\n Vce=%.2f V \\n\"%(Vce)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_6 PG-5.12" ] }, { "cell_type": "code", "execution_count": 6, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.13 shown in the question no 5.5\n", "\n", "\n", " Vt=3.33 V\n", "\n", "\n", " Rb=3333 ohm \n", "\n", "\n", " Ib=48.916 microA \n", "\n", "\n", " Ic=4.89 mA \n", "\n", "\n", " Vce=2.638 V \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.13 shown in the question no 5.5\\n\"\n", "#We must find the value of Ic,Ve,Vce using exact analysis\n", "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=10##supply voltage\n", "R1=10e3#\n", "Rc=1e3#\n", "R2=5e3#\n", "Re=500#\n", "Beta=100##voltage gain\n", "Vt=R2*Vcc/(R1+R2)##thevenin's voltage\n", "print \"\\n Vt=%.2f V\\n\"%(Vt)\n", "Rb=R1*R2/(R1+R2)#\n", "print \"\\n Rb=%.0f ohm \\n\"%(Rb)\n", "Ib=(Vt-Vbe)/(Rb+(1+Beta)*Re)#\n", "print \"\\n Ib=%.3f microA \\n\"%(Ib*1e6)\n", "Ic=Beta*Ib#\n", "print \"\\n Ic=%.2f mA \\n\"%(Ic*1e3)\n", "Vce=Vcc-Ic*Rc-(Ic+Ib)*Re##since Vcc-Ic*Rc-Vce-Ie*Re=0\n", "print \"\\n Vce=%.3f V \\n\"%(Vce)\n", "#the ans for Ve in the book is 2.648V whereas in output it is 2.638V because\n", "# in the book the values has been rounded off so that the final answer is \n", "#2.648V same is the case for Rb,Ib andIc " ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_7 PG-5.13" ] }, { "cell_type": "code", "execution_count": 7, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "\n", " Vb=2 V\n", "\n", "\n", " Ve=1.3 V \n", "\n", "\n", " Ie=2.77 mA \n", "\n", "\n", "i) for hFE_min=50\n", "\n", " Ic=2.712 mA \n", "\n", " Vce=5.277 V \n", "\n", "ii)for hFE_max=60\n", " Ic=2.721 mA \n", "\n", " Vce=5.26 V \n", "\n" ] } ], "source": [ "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=12##supply voltage\n", "R1=10e3#\n", "R2=2e3\n", "Re=470#\n", "Rc=2e3\n", "hFE_min=50#\n", "hFE_max=60#\n", "Vb=R2*Vcc/(R1+R2)##base voltage\n", "print \"\\n Vb=%.0f V\\n\"%(Vb)\n", "Ve=Vb-Vbe##emitter voltage\n", "print \"\\n Ve=%.1f V \\n\"%(Ve)\n", "Ie=Ve/Re#\n", "print \"\\n Ie=%.2f mA \\n\"%(Ie*1e3)\n", "print \"\\ni) for hFE_min=50\"\n", "Beta=hFE_min#\n", "Ib=Ie/(Beta+1)#\n", "Ic=Beta*Ib#\n", "print \"\\n Ic=%.3f mA \\n\"%(Ic*1e3)\n", "Vce=Vcc-Ic*Rc-Ve#\n", "print \" Vce=%.3f V \\n\"%(Vce)\n", "print \"ii)for hFE_max=60\"\n", "Beta=hFE_max#\n", "Ib=Ie/(Beta+1)#\n", "Ic=Beta*Ib#\n", "print \" Ic=%.3f mA \\n\"%(Ic*1e3)\n", "Vce=Vcc-Ic*Rc-Ve#\n", "print \" Vce=%.2f V \\n\"%(Vce)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_8 PG-5.16" ] }, { "cell_type": "code", "execution_count": 8, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "refer to the figure-5.17 shown\n", "We apply KVL to the collector circuit ie Vcc-Vce-Ic*Rc=0\n", "\n", " Rc=1 kohm \n", "\n", "\n", " Ib=50.00 microA \n", "\n", "We apply KVL to the base circuit ie Vcc-Vbe-Ib*Rb=0\n", "\n", " Rb=186 kohm \n", "\n", "the standard value of Rb=200k ohm\n" ] } ], "source": [ "print \"refer to the figure-5.17 shown\"\n", "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=10##supply voltage\n", "Beta=100##voltage gain\n", "Vce=5##colector to emitter voltage \n", "Ic=5e-3##collector current\n", "print \"We apply KVL to the collector circuit ie Vcc-Vce-Ic*Rc=0\"\n", "Rc=(Vcc-Vce)/Ic##since Vcc-Vce-Ic*Rc=0\n", "print \"\\n Rc=%.0f kohm \\n\"%(Rc*1e-3)\n", "Ib=Ic/Beta##base current\n", "print \"\\n Ib=%.2f microA \\n\"%(Ib*1e6)\n", "print \"We apply KVL to the base circuit ie Vcc-Vbe-Ib*Rb=0\"\n", "Rb=(Vcc-Vbe)/Ib##since Vcc-Vbe-Ib*Rb=0\n", "print \"\\n Rb=%.0f kohm \\n\"%(Rb*1e-3)\n", "print \"the standard value of Rb=200k ohm\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_9 PG-5.17" ] }, { "cell_type": "code", "execution_count": 9, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "refer to the figure-5.19 shown\n", "\n", " Ib=50 microA \n", "\n", "\n", " Rc=2.178 kohm \n", "\n", "Rc=2 kohm standard value\n", "We apply KVL to the input circuit ie Vce-Vbe-Ib*Rb=0\n", "\n", " Rb=86 kohm \n", "\n", "the standard value of Rb=91 kohm\n" ] } ], "source": [ "print \"refer to the figure-5.19 shown\"\n", "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=16##supply voltage\n", "Beta=100##voltage gain\n", "Vce=5##colector to emitter voltage \n", "Ic=5e-3##collector current\n", "Ib=Ic/Beta##base current\n", "print \"\\n Ib=%.0f microA \\n\"%(Ib*1e6)\n", "Rc=(Vcc-Vce)/(Ic+Ib)##since Vcc-Vce-Ic*Rc=0\n", "print \"\\n Rc=%.3f kohm \\n\"%(Rc*1e-3)\n", "print \"Rc=2 kohm standard value\"\n", "print \"We apply KVL to the input circuit ie Vce-Vbe-Ib*Rb=0\"\n", "Rb=(Vce-Vbe)/Ib##since Vce-Vbe-Ib*Rb=0\n", "print \"\\n Rb=%.0f kohm \\n\"%(Rb*1e-3)\n", "print \"the standard value of Rb=91 kohm\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_10 PG-5.18" ] }, { "cell_type": "code", "execution_count": 10, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.20 shown\n", "\n", " Ib=30 microA \n", "\n", "\n", " Ie=3.03 mA \n", "\n", "\n", " Re=990 ohm \n", "\n", " the standard value of Re=910 ohm\n", "\n", "\n", " Ve=2.757 V \n", "\n", "\n", " Rc=1414 ohm \n", "\n", " the lower side standard value is selected to reduce Ic*Rc and increase Vce \n", "\n", "\n", " Therefore Vb=3.45730 V \n", "\n", "\n", " I=0.3 mA \n", "\n", "\n", " R2=11524 ohm \n", "\n", " the standard value of R2=11 kohm\n", "\n", "the lower side standard is selected to satisfy I>=10*Ib\n", "\n", " I=0.3143 mA \n", "\n", "\n", " R1=24.812 kohm \n", "\n", " the standard value of R1=22kohm\n", "\n", "The lowest standard value is selected to satisfy I>=10*Ib\n" ] } ], "source": [ "print \"Refer to the figure-5.20 shown\"\n", "Vbe=0.7##base emitter voltage for silicon\n", "Vcc=12##supply voltage\n", "Beta=100##voltage gain\n", "Vce=5##colector to emitter voltage \n", "Ve=3##assumption\n", "Ic=3e-3##collector current\n", "Ib=Ic/Beta##base current\n", "print \"\\n Ib=%.0f microA \\n\"%(Ib*1e6)\n", "Ie=Ic+Ib##emitter current\n", "print \"\\n Ie=%.2f mA \\n\"%(Ie*1e3)\n", "Re=Ve/Ie#\n", "print \"\\n Re=%.0f ohm \\n\"%(Re)\n", "print \" the standard value of Re=910 ohm\"\n", "Re=910##standard value\n", "Ve=Ie*Re#\n", "print \"\\n\\n Ve=%.3f V \\n\"%(Ve)\n", "Rc=(Vcc-Ve-Vce)/Ic\n", "print \"\\n Rc=%.0f ohm \\n\"%(Rc)\n", "print \" the lower side standard value is selected to reduce Ic*Rc and increase Vce \"\n", "Vb=Ve+Vbe\n", "print \"\\n\\n Therefore Vb=%.5f V \\n\"%(Vb)\n", "I=10*Ib\n", "print \"\\n I=%.1f mA \\n\"%(I*1e3)\n", "R2=Vb/I#\n", "print \"\\n R2=%.0f ohm \\n\"%(R2)\n", "print \" the standard value of R2=11 kohm\\n\"\n", "print \"the lower side standard is selected to satisfy I>=10*Ib\"\n", "R2=11e3#\n", "I=Vb/R2#\n", "print \"\\n I=%.4f mA \\n\"%(I*1e3)\n", "R1=(Vcc-Vb)/(I+Ib)\n", "print \"\\n R1=%.3f kohm \\n\"%(R1*1e-3)\n", "print \" the standard value of R1=22kohm\\n\"\n", "print \"The lowest standard value is selected to satisfy I>=10*Ib\"" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_11 PG-5.26" ] }, { "cell_type": "code", "execution_count": 11, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "\n", " For base bias: stability factor=101 \n", "\n", "\n", " For collector to base bias: stability factor=10 \n", "\n", "\n", " For voltage divider bias: stability factor=7.19 \n", "\n" ] } ], "source": [ "#for base bias take the figure as shown the Example5.1\n", "Beta=100#\n", "S=1+Beta##stability factor\n", "print \"\\n For base bias: stability factor=%.0f \\n\"%(S)\n", "#for collector to base bias the figure as shown the Example5.3\n", "Beta=100#\n", "Rc=10e3#\n", "Rb=100e3\n", "S=(1+Beta)/(1+Beta*(Rc/(Rc+Rb)))#\n", "print \"\\n For collector to base bias: stability factor=%.0f \\n\"%(S)\n", "#for voltage divider bias take the figure as shown the Example5.5\n", "Re=500#\n", "R1=10e3#\n", "R2=5e3#\n", "Rb=R1*R2/(R1+R2)##R1 and R2 are in parallel\n", "S=(1+Beta)/(1+Beta*(Re/(Re+Rb)))#\n", "print \"\\n For voltage divider bias: stability factor=%.2f \\n\"%(S)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_12 PG-5.28" ] }, { "cell_type": "code", "execution_count": 12, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.25 shown\n", "Icbo doubles for every 10 degree Celsius\n", "\n", " Therefore Icbo2=64.00 microA \n", "\n", "Apply KVL to the base circuit Vbb=Vbe+Icbo2*Rb we get\n", "\n", " Rb=76.563 kohm \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.25 shown\"\n", "Icbo1=2e-6##at a temperature T1=25 degree celsius\n", "Vbb=5#\n", "Vbe=0.1\n", "print \"Icbo doubles for every 10 degree Celsius\"\n", "T1=25##temperature in degree celsius\n", "T2=80##temperature in degree celsius\n", "Icbo2=Icbo1*2**((T2-T1)/10)##at a temperature T2=80 degree celsius\n", "print \"\\n Therefore Icbo2=%.2f microA \\n\"%(Icbo2*1e6)\n", "print \"Apply KVL to the base circuit Vbb=Vbe+Icbo2*Rb we get\"\n", "Rb=(Vbb-Vbe)/Icbo2#\n", "print \"\\n Rb=%.3f kohm \\n\"%(Rb*1e-3)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_13 PG-5.29" ] }, { "cell_type": "code", "execution_count": 13, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.26 shown\n", "\n", " Apply KVL to the collector side Vcc=Ic*R3+Vce+Ie*R4\n", "\n", " Therefore R3=5.394 kohm \n", "\n", "\n", " Apply KVL to the base side I*R2=Vbe+Ie*R4\n", "\n", " Therefore I=0.1812 mA \n", "\n", "\n", " Apply KVl to the potential divider side we get Vcc=(I+Ib)*R1+I*R2\n", "\n", " therefore R1=65.547 kohm \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.26 shown\"\n", "Vcc=15##supply voltage\n", "Beta=100##voltage gain\n", "Vbe=0.6##base emitter voltage \n", "Ic=2e-3\n", "Vce=3#\n", "R4=600#\n", "R2=10e3#\n", "Ib=Ic/Beta#\n", "Ie=Ic+Ib##collector current\n", "print \"\\n Apply KVL to the collector side Vcc=Ic*R3+Vce+Ie*R4\"\n", "R3=(Vcc-Vce-Ie*R4)/Ic##since Vcc=Ic*R3+Vce+Ie*R4\n", "print \"\\n Therefore R3=%.3f kohm \\n\"%(R3*1e-3)\n", "print \"\\n Apply KVL to the base side I*R2=Vbe+Ie*R4\"\n", "I=(Vbe+Ie*R4)/R2##since I*R2=Vbe+Ie*R4\n", "print \"\\n Therefore I=%.4f mA \\n\"%(I*1e3)\n", "print \"\\n Apply KVl to the potential divider side we get Vcc=(I+Ib)*R1+I*R2\"\n", "R1=(Vcc-I*R2)/(I+Ib)##since Vcc=(I+Ib)*R1+I*R2\n", "print \"\\n therefore R1=%.3f kohm \\n\"%(R1*1e-3)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_14 PG-5.30" ] }, { "cell_type": "code", "execution_count": 14, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.27 shown\n", "apply KVl to the potential divider side we get Vcc-(I+Ib)*R1-I*R2=0 we get\n", "\n", " Re=148.51 ohm \n", "\n", "Apply KVL to the collector side Vcc-Ic*Rc-Vce+Ie*Re=0\n", "\n", " Therefore Vce=7.7 V \n", "\n", "\n", " stability factor=24.256 \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.27 shown\"\n", "Vcc=12##supply voltage\n", "Beta=100##voltage gain\n", "Vbe=0.7##base emitter voltage \n", "Ic=2e-3#\n", "Ib=Ic/Beta#\n", "R1=50e3#\n", "R2=5e3#\n", "Rc=2e3#\n", "print \"apply KVl to the potential divider side we get Vcc-(I+Ib)*R1-I*R2=0 we get\"\n", "I=(Vcc-R1*Ib)/(R1+R2)##since Vcc-(I+Ib)*R1-I*R2=0\n", "Vb=R2*I#\n", "Ie=Ib+Ic#\n", "Re=(Vb-Vbe)/(Ib+Ic)##Vb=Vbe+Re*Ie\n", "print \"\\n Re=%.2f ohm \\n\"%(Re)\n", "print \"Apply KVL to the collector side Vcc-Ic*Rc-Vce+Ie*Re=0\"\n", "Vce=Vcc-Ic*Rc-Ie*Re##since Vcc-Ic*Rc-Vce+Ie*Re=0\n", "print \"\\n Therefore Vce=%.1f V \\n\"%(Vce)\n", "Rb=R1*R2/(R1+R2)##R1 and R2 are in parallel\n", "S=(1+Beta)/(1+Beta*(Re/(Re+Rb)))#\n", "print \"\\n stability factor=%.3f \\n\"%(S)" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "## Ex 5_15 PG-5.31" ] }, { "cell_type": "code", "execution_count": 15, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "Refer to the figure-5.28 shown\n", "at T1=25 degree celsius, applying KVL to the base circuit we get\n", "Vcc-Ib*Rb-Vbe=0\n", "\n", " Therefore Ic=11.3 mA \n", "\n", "Apply KVL to the collector side Vcc-Ic*Rc-Vce=0\n", "\n", " Therefore Vce=5.22 V \n", "\n", "\n", " At 75 degree celsius Vce=3.525\n", "\n", "\n", " change in Ic=25 % (an increase)\n", "\n", "\n", " change in Vce=-32.47 % (a decrease) \n", "\n" ] } ], "source": [ "print \"Refer to the figure-5.28 shown\"\n", "Vbe=0.7#\n", "Rb=100e3#\n", "Rc=600#\n", "Vcc=12##supply voltage\n", "T1=25##temperature in degree celsius\n", "T2=75##temperature in degree celsius\n", "print \"at T1=25 degree celsius, applying KVL to the base circuit we get\"\n", "print \"Vcc-Ib*Rb-Vbe=0\"\n", "Beta=100##voltage gain at T1=25 degree celsius\n", "Ib=(Vcc-Vbe)/Rb##since Vcc-Ib*Rb-Vbe=0\n", "Ic=Beta*Ib#\n", "print \"\\n Therefore Ic=%.1f mA \\n\"%(Ic*1e3)\n", "print \"Apply KVL to the collector side Vcc-Ic*Rc-Vce=0\"\n", "Vce=Vcc-Ic*Rc##since Vcc-Ic*Rc-Vce=0\n", "print \"\\n Therefore Vce=%.2f V \\n\"%(Vce)\n", "Beta=125##voltage gain at T1=25 degree celsius\n", "Ic1=Beta*Ib#\n", "Vce1=Vcc-Ic1*Rc##since Vcc-Ic*Rc-Vce=0\n", "print \"\\n At 75 degree celsius Vce=%.3f\\n\"%(Vce1)\n", "Ic_change=(Ic1-Ic)*100/Ic##percentage in Ic \n", "print \"\\n change in Ic=%.0f %% (an increase)\\n\"%(Ic_change)\n", "Vce_change=(Vce1-Vce)*100/Vce##percentage in Vce\n", "print \"\\n change in Vce=%.2f %% (a decrease) \\n\"%(Vce_change)" ] } ], "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 }