SEDRA/SMITH INSTRUCTOR’SSOLUTIONSMANUALFOR MicroelectronicCircuits EIGHTHEDITION AdelS.Sedra
UniversityofWaterloo
TonyChanCarusone
UniversityofToronto
VincentGaudet
UniversityofWaterloo
NewYorkOxford
OXFORDUNIVERSITYPRESS
OxfordUniversityPressisadepartmentoftheUniversityofOxford. ItfurtherstheUniversity’sobjectiveofexcellenceinresearch,scholarship, andeducationbypublishingworldwide.Oxfordisaregisteredtrademarkof OxfordUniversityPressintheUKandcertainothercountries.
PublishedintheUnitedStatesofAmericabyOxfordUniversityPress 198MadisonAvenue,NewYork,NY10016,UnitedStatesofAmerica.
Copyright©2020,2015,2010,2004,1998byOxfordUniversityPress; 1991,1987Holt,Rinehart,andWinston,Inc.;1982CBSCollegePublishing
FortitlescoveredbySection112oftheUSHigherEducation OpportunityAct,pleasevisitwww.oup.com/us/heforthe latestinformationaboutpricingandalternateformats.
Allrightsreserved.Nopartofthispublicationmaybereproduced,stored inaretrievalsystem,ortransmitted,inanyformorbyanymeans,without thepriorpermissioninwritingofOxfordUniversityPress,orasexpressly permittedbylaw,bylicense,orundertermsagreedwiththeappropriate reproductionrightsorganization.Inquiriesconcerningreproductionoutside thescopeoftheaboveshouldbesenttotheRightsDepartment,Oxford UniversityPress,attheaddressabove.
Youmustnotcirculatethisworkinanyotherform andyoumustimposethissameconditiononanyacquirer.
ISBN:978–0–19–085348–8
Printingnumber:987654321
PrintedbyBridgeportNationalBindery,Inc.,UnitedStatesofAmerica
Contents
ExerciseSolutions (Chapters1–18)
ProblemSolutions (Chapters1–18)
Preface ThisInstructor’sManual(ISM)containscompletesolutionsforabout450in-chapterexercisesand 1400end-of-chapterproblemsincludedinthebook MicroelectronicCircuits,EighthEdition
Thismanualhasbenefitedfromtheworkoftheaccuracycheckers,listedbelow.Wearegratefulto allofthem.Dr.AmirYazdaniofRyersonUniversitydeservesspecialmentionashehasdonea trulyoutstandingjobinensuringthatthismanualisasfreeoferrorsaspossible.However,despite allofourcombinedefforts,thereislittledoubtthatsomeerrorsremain.Wewillbemostgrateful toinstructorswhodiscovererrorsandpointthemouttous.Pleasesendcorrectionsandcomments byemailto:sedra@uwaterloo.ca.
AsshehasdoneforanumberofeditionsofthebookandtheISM,JenniferRodriguesmostably typedthemajorityofthesolutionsinthismanual.Weareverygratefulforherexcellentwork.
AtOUP,wewishtothankSeniorProductionEditorKeithFaivre,SeniorDevelopmentEditorEric Sinkins,andAssistantEditorMeganCarlson.
AdelSedra VincentGaudet Waterloo,Ontario,Canada TonyChanCarusone Toronto,Ontario,Canada
October2019
AccuracyCheckers • LeonidBelostotski,UniversityofCalgary
• DanieloRomero,UniversityofMaryland
• MuhammadUllah,FloridaPolytechnicUniversity
• DerekWright,UniversityofWaterloo
• CharlesWu,HardingUniversity
• AmirYazdani,RyersonUniversity
Chapter1
SolutionstoExerciseswithintheChapter
Ex:1.1 Whenoutputterminalsare open-circuited,asinFig.1.1a:
Forcircuita. v oc = v s (t )
Forcircuitb. v oc = i s (t ) × Rs
Whenoutputterminalsareshort-circuited,asin Fig.1.1b:
Forcircuita. i sc = v s (t ) Rs
Forcircuitb. i sc = i s (t )
Forequivalency
Rs i s (t ) = v s (t ) Rs a b vs (t) Figure1.1a is (t) a b Rs
Figure1.1b
Ex:1.2 voc vs Rs isc
v oc = 10mV
i sc = 10 µA
Rs = v oc i sc = 10mV 10 µA = 1k
Ex:1.3 Usingvoltagedivider:
v o (t ) = v s (t ) × R L Rs + R L
vs (t) vo Rs RL
Given v s (t ) = 10mV and Rs = 1k .
If R L = 100k
v o = 10mV × 100 100 + 1 = 9 9mV
If R L = 10k
v o = 10mV × 10 10 + 1 9 1mV
If R L = 1k
v o = 10mV × 1 1 + 1 = 5mV
If R L = 100
v o = 10mV × 100 100 + 1K 0.91mV
For v o = 0 8v s , R L R L + Rs = 0 8
Since Rs = 1k , R L = 4k
Ex:1.4 Usingcurrentdivider: Rs is 10 A RL io
i o = i s × Rs Rs + R L
Given i s = 10 µA, Rs = 100k .
For R L = 1k , i o = 10 µA × 100 100 + 1 = 9 9 µA
For R L = 10k , i o = 10 µA × 100 100 + 10 9 1 µA
For
R L = 100k , i o = 10 µA × 100 100 + 100 = 5 µA
For R L = 1M , i o = 10 µA × 100K 100K + 1M 0 9 µA
For i o = 0 8i s , 100 100 + R L = 0 8
⇒ R L = 25k
Ex:1.5 f = 1 T = 1 10 3 = 1000Hz
ω = 2π f = 2π × 103 rad/s
Ex:1.6 (a) T = 1 f = 1 60 s = 16.7ms
(b) T = 1 f = 1 10 3 = 1000s
(c) T = 1 f = 1 106 s = 1 µs
Ex:1.7 If6MHzisallocatedforeachchannel, then470MHzto806MHzwillaccommodate
806 470 6 = 56channels
Sincethebroadcastbandstartswithchannel14,it willgofromchannel14tochannel69.
Ex:1.8 P = 1 T T 0 v 2 R dt
= 1 T × V 2 R × T = V 2 R
Alternatively, P = P1 + P3 + P5 +··· = 4 V
Itcanbeshownbydirectcalculationthatthe infiniteseriesintheparentheseshasasumthat approaches π 2 /8; thus P becomes V 2 / R asfound fromdirectcalculation.
Fractionofenergyinfundamental
= 8/π 2 = 0 81
Fractionofenergyinfirstfiveharmonics
= 8 π 2 1 + 1 9 + 1 25 = 0 93
Fractionofenergyinfirstsevenharmonics
= 8 π 2 1 + 1 9 + 1 25 + 1 49 = 0.95
Fractionofenergyinfirstnineharmonics
= 8 π 2 1 + 1 9 + 1 25 + 1 49 + 1 81 = 0 96
Notethat90%oftheenergyofthesquarewaveis inthefirstthreeharmonics,thatis,inthe fundamentalandthethirdharmonic.
Ex:1.9 (a) D canrepresent15equally-spaced valuesbetween0and3.75V.Thus,thevaluesare spaced0.25Vapart.
v A = 0 V ⇒ D = 0000
v A = 0 25 V ⇒ D = 0000
v A = 1 V ⇒ D = 0000
v A = 3.75 V ⇒ D = 0000
(b)(i)1levelspacing: 20 ×+0 25 =+0 25 V
(ii)2levelspacings: 21 ×+0 25 =+0 5 V
(iii)4levelspacings: 22 ×+0 25 =+1 0 V
(iv)8levelspacings: 23 ×+0 25 =+2 0 V
(c)Theclosestdiscretevaluerepresentedby D is +1.25V;thus D = 0101.Theerroris-0.05V,or
0 05/1 3 × 100 =−4%
Ex:1.10
Voltagegain = 20 log 100 = 40dB
Currentgain = 20 log 1000 = 60dB
Powergain = 10 log A p = 10 log ( A v A i ) = 10 log 105 = 50dB
Ex:1.11 Pdc = 15 × 8 = 120mW
PL = (6/√2)2 1 = 18mW
Pdissipated = 120 18 = 102mW
η = PL Pdc × 100 = 18 120 × 100 = 15%
Ex:1.12 v o = 1 × 10 106 + 10 10 5 V = 10 µV
PL = v 2 o / R L = (10 × 10 6 )2 10 = 10 11 W
Withthebufferamplifier:
v o = 1 × Ri Ri + Rs × A v o × R L R L + Ro
= 1 × 1 1 + 1 × 1 × 10 10 + 10 = 0 25V
PL = v 2 o R L = 0 252 10 = 6 25mW
Voltagegain= v o v s = 0 25V 1V = 0 25V/V
=−12dB
Powergain ( A p ) ≡ PL Pi
where PL =6.25mWand Pi = v i i 1 ,
v i = 0 5Vand
i i = 1V 1M + 1M = 0 5 µA
Exercise1–3
ThisfigurebelongstoExercise1.15.
Stage 1
vs vi1 10vi1 100 k 1 M vi2 1 k
Thus,
Pi = 0 5 × 0 5 = 0 25 µW and
A p = 6.25 × 10 3
0 25 × 10 6 = 25 × 103
10log A p = 44dB
Ex:1.13 Open-circuit(noload)outputvoltage=
A v o v i
Outputvoltagewithloadconnected
= A v o v i R L R L + Ro
0 8 = 1 Ro + 1 ⇒ Ro = 0 25k = 250
Ex:1.14 A v o = 40dB = 100V/V
PL = v 2 o R L = A v o v i R L R L + Ro 2 R L
= v 2 i × 100 × 1 1 + 1 2 1000 = 2 5 v 2 i
Pi = v 2 i Ri = v 2 i 10,000
A p ≡ PL Pi = 2 5v 2 i 10 4 v 2 i = 2 5 × 104 W/W
10log Ap = 44dB
Ex:1.15 Withoutstage3(seefigureabove)
v L v s = 1M 100k + 1M (10) 100k 100k + 1k
×(100) 100 100 + 1k
v L v s = (0 909)(10)(0 9901)(100)(0 0909)
= 81 8V/V
Stage 2
k
vi2 vL 1 k
Ex:1.16 ReferthesolutiontoExample1.3inthe text.
v i 1 v s = 0 909V/V
v i 1 = 0 909 v s = 0 909 × 1 = 0 909mV
v i 2 v s = v i 2 v i 1 × v i 1 v s = 9 9 × 0 909 = 9V/V
v i 2 = 9 × v S = 9 × 1 = 9mV
v i 3 v s = v i 3 v i 2 × v i 2 v i 1 × v i 1 v s = 90 9 × 9 9 × 0 909
= 818V/V
v i 3 = 818 v s = 818 × 1 = 818mV
v L v s = v L v i 3 × v i 3 v i 2 × v i 2 v i 1 × v i 1 v s = 0 909 × 90 9 × 9 9 × 0 909 744V/V
v L = 744 × 1mV = 744mV
Ex:1.17 Usingvoltageamplifiermodel,the three-stageamplifiercanberepresentedas vi Ri Ro Avovi Ri = 1M Ro = 10
A v o = A v 1 × A v 2 × A v 3 = 9 9 × 90 9 × 1 = 900V/V
Theoverallvoltagegain
v o v s = Ri Ri + Rs × A v o × R L R L + Ro
For R L = 10
Overallvoltagegain
= 1M 1M + 100K × 900 × 10 10 + 10 = 409V/V
For R L = 1000
Overallvoltagegain
= 1M 1M + 100K × 900 × 1000 1000 + 10 = 810V/V
∴ Rangeofvoltagegainisfrom409 V/V to 810 V/V
Ex:1.18 ii io Ais ii RL Ro Rs Ri is
Ex:1.20 Usingthetransresistancecircuitmodel, thecircuitwillbe Ri Rs is ii Ro RL vo Rmii
i i = i s Rs Rs + Ri
i o = A is i i Ro Ro + R L = A is i s Rs Rs + Ri Ro R o + R L
Thus, i o i s = A is Rs Rs + Ri Ro R o + R L
Ex:1.19 Ri Ro Gmvi RL Ri vi vo
vs v i = v s Ri Ri + Rs
v o = G m v i ( Ro R L )
= G m v s Ri Ri + Rs ( Ro R L )
Thus,
v o v s = G m Ri Ri + Rs ( Ro R L )
i i i s = Rs Ri + Rs
v o = Rm i i × R L R L + Ro
v o i i = Rm R L R L + Ro
Now v o i s = v o i i × i i i s = Rm R L R L + Ro × Rs Ri + Rs = Rm Rs Rs + Ri × R L R L + Ro
Ex:1.21
v b = i b r π + (β + 1)i b Re = i b r π + (β + 1) Re But v b = v x and i b = i x , thus Rin ≡ v x i x = v b i b = r π + (β + 1) Re
Ex:1.22 f Gain 10Hz60dB 10kHz40dB 100kHz20dB 1MHz0dB
Gain (dB) 20 dB/decade
110 10 10 10 10 10 10 f (Hz)
Ex:1.23
RL Ro Ri Vi Vi Gm Vo CL
whichisoftheSTCLPtype.
ω0 = 1 C L ( R L Ro ) = 1 4.5 × 10 9 (103 Ro )
For ω0 tobeatleast wπ × 40 × 103 ,thehighest valueallowedfor Ro is Ro = 103 2π × 40 × 103 × 103 × 4 5 × 10 9 1 = 103 1 131 1 = 7 64 k
Thedcgainis
G m ( R L Ro )
Toensureadcgainofatleast40dB(i.e.,100), theminimumvalueof G m is
⇒ R L ≥ 100/(103 7 64 × 103 ) = 113 1 mA/V
Vo = G m Vi Ro R L C L = G m Vi 1 Ro + 1 R L + sC L
Thus, Vo Vi = G m 1 Ro + 1 R L × 1 1 + sC L 1 Ro + 1 R L
Vo Vi = G m ( R L Ro ) 1 + sC L ( R L Ro )
Ex:1.24 RefertoFig.E1.24
V2 Vs = Ri Rs + 1 sC + Ri = Ri Rs + Ri s s + 1 C ( Rs + Ri ) whichisanHPSTCfunction.
f 3dB = 1 2π C ( Rs + Ri ) ≤ 100Hz C ≥ 1 2π(1 + 9)103 × 100 = 0 16 µF
Chapter2
SolutionstoExerciseswithintheChapter
Ex:2.1 Theminimumnumberofterminals requiredbyasingleopampis5:twoinput terminals,oneoutputterminal,oneterminalfor positivepowersupply,andoneterminalfor negativepowersupply.
Theminimumnumberofterminalsrequiredbya quadopampis14:eachopamprequirestwo inputterminalsandoneoutputterminal (accountingfor12terminalsforthefourop amps).Inaddition,thefouropampscanallshare oneterminalforpositivepowersupplyandone terminalfornegativepowersupply.
Ex:2.2 Relevantequationsare:
v 3 = A (v 2 v 1 ); v Id = v 2 v 1 ,
v Ic m = 1 2 (v 1 + v 2 )
(a)
v 1 = v 2 v 3 A = 0 4 103 =−0 004V =−4mV
v Id = v 2 v 1 = 0 ( 0.004) =+0.004V = 4mV
v Ic m = 1 2 ( 4mV + 0) =−2mV
(b) 10 = 103 (2 v 1 ) ⇒ v 1 = 2 01V
v Id = v 2 v 1 = 2 2.01 =−0.01V =−10mV
v Ic m = 1 2 (v 1 + v 2 ) = 1 2 (2 01 + 2) = 2 005V 2V
(c)
v 3 = A (v 2 v 1 ) = 103 (1 998 2 002) =−4V
v Id = v 2 v 1 = 1 998 2 002 =−4mV
v Ic m = 1 2 (v 1 + v 2 ) = 1 2 (2 002 + 1 998) = 2V
(d)
1 2 = 103 v 2 ( 1 2) = 103 (v 2 + 1 2)
⇒ v 2 =−1 2012V
v Id = v 2 v 1 =−1.2012 ( 1.2) =−0 0012V =−1 2mV
v Ic m = 1 2 (v 1 + v 2 ) = 1 2 [ 1 2 + ( 1 2012)] −1 2V
Ex:2.3 FromFig.E2.3wehave: v 3 = μv d and
v d = ( G m v 2 G m v 1 ) R = G m R (v 2 v 1 )
Therefore:
v 3 = μ G m R (v 2 v 1 )
Thatis,theopen-loopgainoftheopampis
A = μ G m R .For G m = 20mA/V and
μ = 50,wehave:
A = 50 × 20 × 5 = 5000V/V,orequivalently, 74dB.
Ex:2.4 Thegainandinputresistanceofthe invertingamplifiercircuitshowninFig.2.5are R2 R1 and R1 ,respectively.Therefore,wehave:
R1 = 100k and R2 R1 =−10 ⇒ R2 = 10 R 1
Thus:
R2 = 10 × 100k = 1M
Ex:2.5 vi vo ii R 10 k
FromTable1.1wehave:
Rm = v o i i i o = 0 ;thatis,outputisopencircuit
Thenegativeinputterminaloftheopamp(i.e., v i )isavirtualground,thus v i = 0:
v o = v i Ri i = 0 Ri i =− Ri i Rm = v o i i i o 0 =− Ri i i i =− R ⇒ Rm =− R =−10k
Ri = v i i i and v i isavirtualground(v i = 0), thus Ri = 0 i i = 0 ⇒ Ri = 0
Sinceweareassumingthattheopampinthis transresistanceamplifierisideal,theopamphas zerooutputresistanceandthereforetheoutput resistanceofthistransresistanceamplifierisalso zero.Thatis Ro = 0
10 k
0.5 mA
10 k
vi vo
ConnectingthesignalsourceshowninFig.E2.5 totheinputofthisamplifier,wehave:
v i isavirtualgroundthatis v i = 0,thusthe currentflowingthroughthe 10-k resistor connectedbetween v i andgroundiszero. Therefore,
v o = v i R × 0 5mA = 0 10k × 0 5mA =−5V
Ex:2.6
Ex:2.7 R1 R2 v1 v2 vO Rf
Forthecircuitshownabovewehave:
v O =− R f R1 v 1 + R f R2 v 2
Sinceitisrequiredthat v O =−(v 1 + 4v 2 ), wewanttohave:
R f R1 = 1 and R f R2 = 4
Itisalsodesiredthatforamaximumoutput voltageof4V,thecurrentinthefeedback resistornotexceed1mA.
Therefore 4V R f ≤ 1mA ⇒ R f ≥ 4V 1mA ⇒ R f ≥ 4k
Letuschoose R f tobe 4k ,then R1 = R f = 4k and R2 = R f 4 = 1k
Ex:2.8
v 1 isavirtualground,thus v 1 = 0 V
i 1 = 2V v 1 R1 = 2 0 1k = 2mA
Assuminganidealopamp,thecurrentflowing intothenegativeinputterminaloftheopampis zero.Therefore, i 2 = i 1 ⇒ i 2 = 2mA
v O = v 1 i 2 R2 = 0 2mA × 5k =−10V
i L = v o R L = 10V 1k =−10mA
i O = i L i 2 =−10mA 2mA =−12mA
Voltagegain = v O 2V = 10V 1V =−5V/V or 14dB
Currentgain = i L i 1 = 10mA 2mA =−5A/A or 14dB
Powergain
= PL Pi = 10( 10mA) 2V × 2mA = 25W/W or14dB
NotethatpowergainindBis10log10 PL Pi
v O = Ra R1 Rc Rb v 1 + Ra R2 Rc Rb v 2 Rc R3 v 3
Wewanttodesignthecircuitsuchthat
v O = 2v 1 + v 2 4v 3
Thusweneedtohave
Ra R1 Rc Rb = 2, Ra R2 Rc Rb = 1,and Rc R3 = 4
Fromtheabovethreeequations,wehavetofind sixunknownresistors;therefore,wecan arbitrarilychoosethreeoftheseresistors.Letus choose Ra = Rb = Rc = 10k
Thenwehave
R3 = Rc 4 = 10 4 = 2 5k
Ra R1 Rc Rb = 2, ⇒ 10 R1 × 10 10 = 2
⇒ R1 = 5k
Ra R2 Rc Rb = 1 ⇒ 10 R2 × 10 10 = 1
⇒ R2 = 10k
Ex:2.9 Usingthesuperpositionprincipletofind thecontributionof v 1 totheoutputvoltage v O , weset v 2 = 0
Usingthesuperpositionprincipletofindthe contributionof v 1 to v O ,weset v 2 = v 3 = 0 Thenwehave(refertothesolutionofExercise 2.9): v O = 6v 1
Tofindthecontributionof v 2 to v O , weset
v 1 = v 3 = 0, then: v O = 2v 2
Tofindthecontributionof v 3 to v O weset
v 1 = v 2 = 0, then
v o =− 7k 1k v 3 =−7v 3
Combiningthecontributionsof v 1 , v 2 , and v 3 to v O wehave: v O = 6v 1 + 4v 2 7v 3 .
Ex:2.11 R1 R2 i vi vO
v O v i = 1 + R2 R1 = 2 ⇒ R2 R1 = 1 ⇒ R1 = R2
v + (thevoltageatthepositiveinputoftheopamp
is: v + = 3 1 + 3 v 1 = 0.75v 1
Thus v O = 1 + 7k 1k v + = 8 × 0 75v 1 = 6v 1
Tofindthecontributionof v 2 totheoutput voltage v O weset v 1 =0.
Then v + = 1 1 + 3 v 2 = 0 25v 2
Hence
v O = 1 + 7k 1k v + = 8 × 0.25v 2 = 2v 2
Combiningthecontributionsof v 1 and v 2 to v O ,wehave v O = 6v 1 + 2v 2
Ex:2.10 7 k 1 k 1 k 3 k v1 v3 vO v2
If v O = 10V,thenitisdesiredthat i = 10 µA
Thus, i = 10V R1 + R2 = 10 µA ⇒ R1 + R2 = 10V 10 µA R1 + R2 = 1M and R1 = R2 ⇒ R1 = R2 = 0.5M
Ex:2.12 (a) R1 vI vO v R2
v I v = v O / A ⇒ v = v I v O / A (1)
Butfromthevoltagedivideracross v O ,
v = v O R1 R1 + R2 (2)
EquatingEq.(1)andEq.(2)gives
v O R1 R1 + R2 = v I v O A
whichcanbemanipulatedtotheform
v O v I = 1 + ( R2 / R 1 ) 1 + 1 + ( R2 / R1 ) A
(b)For R1 = 1k and R2 = 9k theidealvalue fortheclosed-loopgainis 1 + 9 1 , thatis,10.The
actualclosed-loopgainis G = 10 1 + 10/ A .
If A = 103 ,then G = 9 901 and = G 10 10 × 100 =−0 99% −1%
For v I = 1V, v O = G × v I = 9 901V and
v O = A (v + v ) ⇒ v + v = v O A = 9.901 1000 9 9mV
If A = 104 ,then G = 9 99 and =−0 1%
For v I = 1V, v O = G × v I = 9.99V, therefore,
v + v = v O A = 9 99 104 = 0 999mV 1mV
If A = 105 ,then G = 9 999 and =−0 01%
For v I = 1V, v O = G × v I = 9 999 thus,
v + v = v O A = 9 999 105 = 0 09999mV 0 1mV
Ex:2.13
i I = 0A, v 1 = v I = 1V
i 1 = v 1 1k = 1V 1k = 1mA i 2 = i 1 = 1mA vO v1 iO i2 i1 iI iL 1 k 1 k 9 k vI 1 V
v O v i = 10V 1V = 10V/V or20dB
i L i I = 10mA 0 =∞
PL PI = v O × i L v I × i I = 10 × 10 1 × 0 =∞
Ex:2.14
(a)Loadvoltage = 1k 1k + 1M × 1V 1mV
(b) Loadvoltage = 1 V
Ex:2.15
v O = v 1 + i 2 × 9k = 1 + 1 × 9 = 10V
i L = v O 1k = 10V 1k = 10mA
i O = i L + i 2 = 11mA
(a) R1 = R3 = 2k , R2 = R4 = 200k
Since R4 / R3 = R2 / R1 wehave:
A d = v O v I 2 v I 1 = R2 R1 = 200 2 = 100V/V
(b) Rid = 2 R1 = 2 × 2k = 4k
Sinceweareassumingtheopampisideal,
Ro = 0
(c)
A cm ≡ v O v Ic m = R4 R3 + R4 1 R2 R1 R3 R4
Theworst-casecommon-modegain(i.e.,the largest A cm )occurswhentheresistortolerances aresuchthatthequantityinparenthesesis maximum.Thisinturnoccurswhen R2 and R3 areattheirhighestpossiblevalues(eachone percentabovenominal)and R1 and R4 areattheir lowestpossiblevalues(eachonepercentbelow nominal),resultingin
A cm = R4 R3 + R4 1 1.01 × 1.01 0 99 × 0 99
| A cm | R4 R3 + R4 ×0.04 200 202 ×0.04 0.04 V/V
ThecorrespondingCMRRis
CMRR = | A d | | A cm | = 100 0 04 = 2500 or68dB.
Ex:2.16 Wechoose R3 = R1 and R4 = R2
Thenforthecircuittobehaveasadifference
amplifierwithagainof10andaninputresistance of 20k ,werequire
A d = R2 R1 = 10 and
R Id = 2 R1 = 20k ⇒ R1 = 10k and
R2 = A d R1 = 10 × 10k = 100k
Therefore, R1 = R3 = 10k and
R2 = R4 = 100k
Ex:2.17 Given v Ic m =+5V
v Id = 10sin ω t mV
2 R1 = 1k , R2 = 0 5M
R3 = R4 = 10k
v I 1 = v Ic m 1 2 v Id = 5 1 2 × 0 01sin ω t
= 5 0 005sin ω t V
v I 2 = v Ic m + 1 2 v Id
= 5 + 0.005sin ω t V
v (opamp A 1 ) = v I 1 = 5 0 005sin ω t V
v (opamp A 2 ) = v I 2 = 5 + 0 005sin ω t V
v Id = v I 2 v I 1 = 0 01sin ω t
v O 1 = v I 1 R2 × v Id 2 R1
= 5 0 005sin ω t 500k × 0 01sin ω t 1k
= (5 5 005sin ω t ) V
v O 2 = v I 2 + R2 × v Id 2 R1
= (5 + 5 005sin ω t ) V
v + (opamp A 3 ) = v O 2 × R4 R3 + R4 = v O 2 10 10 + 10
= 1 2 v O 2 = 1 2 (5 + 5 005sin ω t )
= (2.5 + 2.5025sin ω t )V
v (opamp A 3 ) = v + (opamp A 3 )
= (2 5 + 2 5025sin ω t ) V
v O = R4 R3 1 + R2 R1 v Id
10k
10k 1 + 0 5M 0 5k × 0 01sin ω t
= 1(1 + 1000) × 0 01sin ω t
= 10 01sin ω t V
Ex:2.18
vi(t) vI (t ) dt C R 1 CR t 0 i i
vO (t)
Thesignalwaveformswillbeasshown. vI (t) vO (t) t t 2.5 V 2.5 V 2.5 V 2.5 V 0 2 s
When v I =+2 5 V,thecurrentthroughthe capacitorwillbeinthedirectionindicated, i = 2 5 V/ R ,andtheoutputvoltagewill decreaselinearlyfrom +2 5 Vto 2 5 V.Thusin ( T /2) seconds,thecapacitorvoltagechangesby 5V.Thechargeequilibriumequationcanbe expressedas
i ( T /2) = C × 5 V 2 5 R T 2 = 5C ⇒ CR = 2 5 T 10 = 1 4 × 2 × 10 6
= 0 5 µs
Ex:2.19 Vo Vi C R
Theinputresistanceofthisinvertingintegratoris R ;therefore, R = 10k
Sincethedesiredintegrationtimeconstant is 10 3 s, wehave: CR = 10 3 s ⇒
C = 10 3 s 10k = 0 1 µF
FromEq.(2.27)thetransferfunctionofthis integratoris:
Vo ( j ω)
Vi ( j ω) =− 1 j ω CR
For ω = 10 rad/s,theintegratortransferfunction hasmagnitude
Vo Vi = 1 1 × 10 3 = 100V/V andphase φ = 90◦
For ω = 1 rad/s,theintegratortransferfunction hasmagnitude
Vo Vi = 1 1 × 10 3 = 1000V/V andphase φ = 90◦
Thefrequencyatwhichtheintegratorgain magnitudeisunityis
ω int = 1 CR = 1 10 3 = 1000 rad/s
Ex:2.20
C = 0 01 µF istheinputcapacitanceofthis differentiator.Wewant CR = 10 2 s (thetime constantofthedifferentiator);thus,
R = 10 2 0 01 µF = 1M
FromEq.(2.33),thetransferfunctionofthe differentiatoris
Vo ( j ω)
Vi ( j ω) =− j ω CR
Thus,for ω = 10rad/s thedifferentiatortransfer functionhasmagnitude
Vo Vi = 10 × 10 2 = 0 1V/V
andphase φ =−90◦
For ω = 103 rad/s,thedifferentiatortransfer functionhasmagnitude
Vo Vi = 103 × 10 2 = 10V/V
andphase φ =−90◦
Ifweaddaresistorinserieswiththecapacitorto limitthehigh-frequencygainofthedifferentiator to100,thecircuitwouldbe:
Athighfrequenciesthecapacitor C actslikea shortcircuit.Therefore,thehigh-frequencygain ofthiscircuitis: R R1 .Tolimitthemagnitudeof thishigh-frequencygainto100,weshouldhave:
R R1 = 100 ⇒ R1 = R 100 = 1M 100 = 10k
Ex:2.21
RefertothemodelinFig.2.27andobservethat
v + v = V OS + v 2 v 1 = V OS + v Id
andsince v O = v 3 = A (v + v ),then
v O = A (v Id + V OS ) (1)
where A = 104 V/Vand V OS = 5 mV.From Eq.(1)weseethat v id = 0 resultsin v O = 50 V, whichisimpossible;thustheopampsaturates and v O =+10 V.Thissituationpertainsfor v Id ≥−4 mV.If v Id decreasesbelow 4 mV, theop-ampoutputdecreasescorrespondingly.
Forinstance, v Id =−4 5 mVresultsin v O =+5 V; v Id =−5 mVresultsin v O = 0 V; v Id =−5 5 mVresultsin v O =−5 V;and v Id =−6 mVresultsin v O =−10 V,atwhich pointtheopampsaturatesatthenegativelevelof 10 V.Furtherdecreasesin v Id havenoeffecton theoutputvoltage.Theresultisthetransfer characteristicsketchedinFig.E2.21.Observe thatthelinearrangeofthecharacteristicisnow centeredaround v Id =−5 mVratherthanthe idealsituationof v Id = 0;thisshiftisobviouslya resultoftheinputoffsetvoltage V OS
Ex:2.22 (a)Theinvertingamplifierof 1000 V/Vgainwillexhibitanoutputdcoffset voltageof ± V OS (1 + R2 / R1 ) = ±3 mV × (1 + 1000) =±3 03 V.Now,sincethe op-ampsaturationlevelsare ±10 V,theroomleft foroutputsignalswingisapproximately ±7 V. Thustoavoidop-ampsaturationandtheattendant clippingofthepeakoftheoutputsinusoid,we mustlimitthepeakamplitudeoftheinputsine wavetoapproximately 7 V/1000 = 7 mV.
(b)Ifatroomtemperature (25◦ C), V OS is trimmedtozeroand(i)thecircuitisoperatedata constanttemperature,thepeakoftheinputsine wavecanbeincreasedto10mV.(ii)However,if thecircuitistooperateinthetemperaturerange of 0◦ C to 75◦ C (i.e.,atatemperaturethatdeviates fromroomtemperaturebyamaximumof 50◦ C), theinputoffsetvoltagewilldriftfrombya maximumof 10 µV/◦ C × 50◦ C = 500 µVor0.5 mV.Thiswillreducetheallowedpeakamplitude oftheinputsinusoidto9.5mV.
Ex:2.23
(a)Iftheamplifieriscapacitivelycoupledinthe mannerofFig.2.32(a),thentheinputoffset voltage V OS willseeaunity-gainamplifier[Fig. 2.32(b)]andthedcoffsetvoltageattheoutput willbeequalto V OS ,thatis,3mV.Thus,almost theentireoutputrangeof ±10 Vwillbeavailable forsignalswing,allowingasine-waveinputof approximately10-mVpeakwithouttheriskof outputclipping.Obviously,inthiscasethereis noneedforoutputtrimming.
(b)Weneedtoselectavalueofthecoupling capacitor C thatwillplacethe3-dBfrequencyof theresultinghigh-passSTCcircuitat1000Hz, thus 1000 = 1 2π CR1 ⇒ C = 1 2π × 1000 × 1 × 103 = 0 16 µF
Ex:2.24 FromEq.(2.35)wehave:
V O = I B 1 R2 I B R2 = 100nA × 1M = 0 1V
FromEq.(2.37)thevalueofresistor R3 (placed inserieswithpositiveinputtominimizethe outputoffsetvoltage)is
R3 = R1 R2 =
9 9k
R3 = 9 9k 10k
Withthisvalueof R3 ,thenewvalueoftheoutput dcvoltage[usingEq.(2.38)]is:
V O = I OS R2 = 10nA × 10k = 0 01V
Ex:2.25 UsingEq.(2.39)wehave:
v O = V OS + V OS CR t ⇒ 12 = 2mV + 2mV 1ms t
⇒ t = 12V 2mV 2mV × 1ms 6s
Withthefeedbackresistor R F ,tohaveatleast ±10V ofoutputsignalswingavailable,wehave tomakesurethattheoutputvoltagedueto V OS hasamagnitudeofatmost2V.FromEq.(2.34), weknowthattheoutputdcvoltagedueto V OS is
V O = V OS 1 + R F R ⇒ 2V = 2mV 1 + R F 10k
1 + R F 10k = 1000 ⇒ R F 10M
ThecornerfrequencyoftheresultingSTC networkis ω 0 = 1 CR F
Weknow RC = 1ms and R = 10k ⇒ C = 0 1 µF
Thus ω 0 = 1 0 1 µF × 10M = 1rad/s
f 0 = ω 0 2π = 1 2π = 0 16Hz
Ex:2.26
20 log A 0 = 106 dB ⇒ A 0 = 200,000 V/V f t = 3 MHz
f b = f t / A 0 = 3 MHz 200,000 = 15 Hz
At f b ,theopen-loopgaindropsby3dBbelow itsvalueatdc;thusitbecomes103dB.
For f f b , | A | f t / f ;thus
At f = 300 Hz, | A |= 3 MHz 300 Hz = 104 V/V or80dB
At f = 3 kHz, | A |= 3 MHz 3 kHz = 103 V/V or60dB
At f = 12 kHz,whichistwooctaveshigherthan 3kHz,thegainwillbe 2 × 6 = 12 dBbelowits valueat3kHz;thatis, 60 12 = 48 dB.
At f = 60 kHz, | A |= 3 MHz 60 kHz = 50 V/V or34dB
Ex:2.27
A 0 = 106 V/V or120dB
Thegainfallsoffattherateof20dB/decade. Thus,itreaches40dBatafrequencyfour decadeshigherthan f b ,
104 f b = 100 kHz ⇒ f b = 10 Hz
Theunity-gainfrequency f t willbetwodecades higherthan100kHz,thatis,
f t = 100 × 100 kHz = 10 MHz
Alternatively,wecouldhavefound f t fromthe gain-bandwidthproduct
f t = A 0 f b = 106 × 10 Hz=10MHz
Atafrequency f f b , | A |= f t / f
For f = 10 kHz, | A |= 10 MHz 10 kHz = 103 V/Vor60dB
Ex:2.28
20 log A 0 = 106 dB ⇒ A 0 = 200,000 V/V
f t = 20 MHz
Foranoninvertingamplifierwithanominaldc gainof100, 1 + R2 R1 = 100
Sincethenominaldcgainismuchlowerthan A 0 ,
f 3dB f t 1 + R2 R1 = 20 MHz 100 = 200 kHz
Ex:2.29 Fortheinputvoltagestepofmagnitude V theoutputwaveformwillstillbegivenbythe exponentialwaveformofEq.(2.56)if ω t V ≤ SR
thatis, V ≤ SR ω t ⇒ V ≤ SR 2π f t resultingin V ≤ 0 16V
FromAppendixFweknowthatthe10%to90% risetimeoftheoutputwaveformoftheformof Eq.(2.56)is tr 2 2 × timeconstant = 2.2 ω t
Thus, tr 0 35 µs
Ifaninputstepofamplitude1.6V(10timesas largecomparedtothepreviouscase)isapplied, theoutputwillbeslew-ratelimitedandthus linearlyrisingwithaslopeequaltotheslewrate, asshowninthefollowingfigure.
SR
Ex:2.30 FromEq.(2.57)wehave: f M = SR 2π V O max = 318 kHz
UsingEq.(2.58),foraninputsinusoidwith frequency f = 5 f M , themaximumpossible amplitudethatcanbeaccommodatedatthe outputwithoutincurringSRdistortionis: V O = V O max f M 5 f M = 5 × 1 5 = 1V (peak)
Chapter3
SolutionstoExerciseswithintheChapter
Ex:3.1 T = 50K
n i = BT 3/2 e Eg /(2kT )
= 7 3 × 1015 (50)3/2 e 1 12/(2×8 62×10 5 ×50)
9 6 × 10 39 /cm3
T = 350K
n i = BT 3/2 e Eg /(2kT )
= 7 3 × 1015 (350)3/2 e 1 12/(2×8 62×10 5 ×350)
= 4 15 × 1011 /cm3
Ex:3.2 N D = 1017 /cm3
FromExercise3.1, n i at
T = 350K = 4 15 × 1011 /cm3
n n = N D = 1017 /cm3
pn ∼ = ni 2 N D
= (4.15 × 1011 )2 1017
= 1 72 × 106 /cm3
Ex:3.3 At300K, n i = 1 5 × 1010 /cm3
p p = N A
Wantelectronconcentration
= n p = 1 5 × 1010 106 = 1 5 × 104 /cm3
∴ N A = p p = ni 2 n p = (1 5 × 1010 )2 1 5 × 104 = 1 5 × 1016 /cm3
Ex:3.4 (a) νn drift =−μn E
Herenegativesignindicatesthatelectronsmove inadirectionoppositeto E
Weuse
νn -drift = 1350 × 1 2 × 10 4 ∵ 1 µm = 10 4 cm
= 6 75 × 106 cm/s = 6 75 × 104 m/s
(b)Timetakentocross 2-µm
length = 2 × 10 6 6 75 × 104 30ps
Exercise3–1
(c)In n -typesilicon,driftcurrentdensity Jn is Jn = qn μn E
= 1 6 × 10 19 × 1016 × 1350 × 1V 2 × 10 4
= 1 08 × 104 A/cm 2
(d)Driftcurrent In = AJn
= 0 25 × 10 8 × 1 08 × 104
= 27 µA
Theresistanceofthebaris
R = ρ × L A
= qn μn × L A
= 1 6 × 10 19 × 1016 × 1350 × 2 × 10 4 0 25 × 10 8
= 37 0 k
Alternatively,wemaysimplyusethepreceding resultforcurrentandwrite
R = V / In = 1 V/27 µA = 37 0 k
Notethat 0 25 µm2 = 0 25 × 10 8 cm2
Ex:3.5 Jn = qD n dn ( x ) dx
FromFig.E3.5,
n 0 = 1017 /cm3 = 105 /(µm)3
Dn = 35cm2 /s = 35 × (104 )2 (µm)2 /s
= 35 × 108 (µm)2 /s dn
dx = 105 0 0 5 = 2 × 105 µm 4
Jn = qDn dn ( x ) dx
= 1 6 × 10 19 × 35 × 108 × 2 × 105
= 112 × 10 6 A/µm2
= 112 µA/µm2
For In = 1mA = Jn × A
⇒ A = 1mA Jn = 103 µA 112 µA/(µm)2 9 µm2
Ex:3.6 UsingEq.(3.20), Dn
μn = D p μ p = VT
Dn = μn VT = 1350 × 25.9 × 10 3
∼ = 35cm2 /s
D p = μ p VT = 480 × 25 9 × 10 3
∼ = 12 4cm2 /s
Ex:3.7 Equation(3.25)
W = 2 s q 1 N A + 1 N D V0
= 2 s q N A + N D N A N D V0
W 2 = 2 s q N A + N D N A N D V0
V0 = 1 2 q s N A N D N A + N D W 2
Ex:3.8 Ina p + n diode NA ND
Equation(3.26) W = 2 s q 1 N A + 1 N D V0
Wecanneglecttheterm 1 N A ascomparedto 1 N D , thus
W 2 s qN D V0
Equation(3.26) x n = W N A N A + N D
W N A N A = W
Equation(3.28), x p = W N D N A + N D since NA ND
W N D N A = W N A N D
Equation(3.28), Q J = Aq N A N D N A + N D W
Aq N A N D N A W = AqN D W
Equation(3.29), Q J = A 2 s q N A N D N A + N D V0
A 2 s q N A N D N A V0 since NA ND = A 2 s qN D V0
Ex:3.9 InExample3.5, N A = 1018 /cm3 and N D = 1016 /cm3
Inthe n -regionofthis pn junction
n n = N D = 1016 /cm3
pn = n 2 i n n = (1 5 × 1010 )2 1016 = 2.25 × 104 /cm3
Asonecanseefromaboveequation,toincrease minority-carrierconcentration ( pn ) byafactorof 2,onemustlower ND (= nn )byafactorof2.
Ex:3.10
Equation(3.40) I S = Aqn 2 i D p L p N D + Dn L n N A
since D p L p and Dn L n haveapproximately
similarvalues,if NA N D ,thentheterm Dn L n N A canbeneglectedascomparedto D p L p N D
∴ I S ∼ = Aqn 2 i D p L p N D
Ex:3.11 I S = Aqn 2 i D p L p N D + Dn L n N A
= 10 4 × 1 6 × 10 19 × (1 5 × 1010 )2
10 5 × 10 4 × 1016 2 + 18 10 × 10 4 × 1018
= 1 46 × 10 14 A
I = I S (e V / V T 1)
I S e V / V T = 1 45 × 10 14 e 0 605/(25 9×10 3 )
= 0 2mA
Ex:3.12 W = 2 s q 1 N A + 1 N D ( V0 V F )
= 2 × 1 04 × 10 12 1 6 × 10 19 1 1018 + 1 1016 (0 814 0 605)
= 1 66 × 10 5 cm = 0 166 µm
Ex:3.13 W = 2 s q 1 N A + 1 N D ( V0 + V R )
= 2 × 1 04 × 10 12 1 6 × 10 19 1 1018 + 1 1016 (0 814 + 2)
= 6 08 × 10 5 cm = 0 608 µm
UsingEq.(3.28),
Q J = Aq N A N D N A + N D W
= 10 4 × 1 6 × 10 19 1018 × 1016 1018 + 1016 × 6 08 × 10 5 cm
= 9 63pC
Reversecurrent I = I S = Aqn 2 i D p L p N D + Dn L n N A
= 10 14 × 1.6 × 10 19 × (1.5 × 1010 )2 × 10 5 × 10 4 × 1016 + 18 10 × 10 4 × 1018
= 7 3 × 10 15 A
Ex:3.14 Equation(3.47),
C j 0 = A s q 2 N A N D N A + N D 1 V0
= 10 4 1 04 × 10 12 × 1 6 × 10 19 2 1018 × 1016 1018 + 1016 1 0 814 = 3 2pF
Equation(3.46),
C j = C j 0 1 + V R V0 = 3 2 × 10 12 1 + 2 0 814 = 1 72pF
Ex:3.15 C d = dQ dV = d dV (τ T I ) = d dV τ T × I S (e V / V T 1) = τ T I S d dV (e V / V T 1) = τ T I S 1 VT e V / V T
τ T VT × I S
V / V T ∼ = τ T VT I
Ex:3.16 Equation(3.49), τ p = L 2 p D p = (5 × 10 4 )2 10 = 25ns
Equation(3.57),
C d = τ T VT I
InExample3.6, N A = 1018 /cm3 , N D = 1016 /cm3
Assuming N A N D , τ T τ p = 25ns
∴ C d = 25 × 10 9 25 9 × 10 3 0 1 × 10 3 = 96 5pF
Exercise4–1 Chapter4
SolutionstoExerciseswithintheChapter
Ex:4.1 RefertoFig.4.3(a).For v I ≥ 0, the diodeconductsandpresentsazerovoltagedrop. Thus v O = v I .For v I < 0, thediodeiscutoff, zerocurrentflowsthrough R ,and v O = 0.The resultisthetransfercharacteristicinFig.E4.1.
Ex:4.2 SeeFig.4.3aand4.3b.Duringthe positivehalfofthesinusoid,thediodeisforward biased,soitconductsresultingin v D = 0.During thenegativehalfcycleoftheinputsignal v I , the diodeisreversebiased.Thediodedoesnot conduct,resultinginnocurrentflowinginthe circuit.So v O = 0 and v D = v I v O = v I .This resultsinthewaveformshowninFig.E4.2.
Ex:4.3 ˆ i
Ex:4.4 (a)
R = 3 133k
Ex:4.6 Themaximumcurrentariseswhen
|v I | = 20 V.Inthiscase,
i D = 20 5 R
Toensurethisis50mA, R = 20 5 50 = 0.3 k = 300
Ex:4.7Equation(4.5)
V2 V1 = 2 3 V T log I2 I1
Atroomtemperature VT = 25mV
V2 V1 = 2 3 × 25 × 10 3 × log 10 0 1 = 115mV
Ex:4.8 i = I S e v / VT (1)
1 (mA) = I S e 0 7/ VT (2)
Dividing(1)by(2),weobtain
i (mA) = e (v 0 7)/ VT
⇒ v = 0 7 + 0 025 ln(i ) where i isinmA.Thus, for i = 0 1 mA,
v = 0 7 + 0 025 ln(0 1) = 0 64 V andfor i = 10 mA,
v = 0 7 + 0 025 ln(10) = 0 76 V
Ex:4.9 i D = I S e v / VT
⇒ I S = i D e v / VT = 0 25 × e 300/25
= 1 23 × 106 m A = 1 23 × 10 9 A
Ex:4.10 T = 125 25 = 100◦ C
I S = 10 14 × 1 15 T = 1.17 × 10 8 A
Ex:4.11 At 20◦ C I = 1V 1M = 1 µA
Sincethereverseleakagecurrentdoublesfor every 10◦ C increase,at 40◦ C
I = 4 × 1 µA = 4 µA
⇒ V = 4 µA × 1M = 4 0V
@0◦ C I = 1 4 µA
⇒ V = 1 4 × 1 = 0 25V
Ex:4.12 a.Useiteration: Diodehas0.7Vdropat1mAcurrent.
Assume V D = 0 7V
I D = 5 0.7 10k = 0 43mA
UseEq.(4.5)andnotethat
V1 = 0 7V, I1 = 1mA ID VCC 5 V VD R 10 k
V2 V1 = 2 3 × VT log I2 I1
V2 = V1 + 2 3 × VT log I2 I1
Firstiteration
V2 = 0 7 + 2 3 × 25 × 10 3 log 0 43 1 = 0 679V
Seconditeration
I2 = 5 0 679 10k = 0 432mA
V2 = 0 7 + 2 3 × 25 3 × 10 3 log 0 432 1
= 0 679V 0 68V wegetalmostthesamevoltage.
∴ Theiterationyields
I D = 0.43mA, V D = 0.68V
b.Useconstantvoltagedropmodel:
V D = 0 7V constantvoltagedrop
I D = 5 0 7 10k = 0 43mA
Ex:4.13 R I 10 V
2.4 V
Diodeshave0.7Vdropat1mA
∴ 1mA = I S e 0 7/ VT (1)
Atacurrent I (mA),
I = I S e V D / VT (2)
Using(1)and(2),weobtain
I = e ( V D 0 7)/ VT
Foranoutputvoltageof2.4V,thevoltagedrop
acrosseachdiode = 2 4 3 = 0.8V
Now I,thecurrentthrougheachdiode,is
I = e (0 8 0 7)/0 025
= 54 6mA R = 10 2.4
6 × 10 3 = 139
Ex:4.15 Withareversevoltage,thediodein Fig.4.10willnotconduct.Thus,thevoltagedrop on R willbezero,andthereversevoltageonthe diodeis V D =− V DD .Toensurewerespectthe peakinversevoltage,werequire V DD = V D > 30 V.Hence,theminimum voltageon V DD is 30 V.
Ex:4.16 Whenconductingareversecurrentof 20mA,thereversevoltageis
V Z = V ZT + I Z r z = 3 5 V + (20 mA 10 mA)10 = 3 6 V
Themaximumcurrent, Imax ,isthecurrentat 200mWpowerdissipation.
200 mW = Imax (3 5 V + 10 ( Imax 10 mA))
0 2 = 3 5 Imax + 10 I 2 max 0 1 Imax
⇒ I 2 max + 0 34 Imax 0 02 = 0
⇒ Imax = 1 2 0 34 + 0 342 + 4 × 0 02 = 51 mA
Ex:4.17 r d = VT I D
I D = 0 1mA r d = 25 × 10 3 0 1 × 10 3 = 250
I D = 1mA r d = 25 × 10 3 1 × 10 3 = 25
I D = 10mA r d = 25 × 10 3 10 × 10 3 = 2 5
Ex:4.18 Forsmallsignalmodel,
i D = v D / r d (1)
where r d = VT I D Forexponentialmodel,
i D = I S e V / V T
i D 2 i D 1 = e ( V2 V1 ) / V T = e v D / V T i D = i D 2 i D 1 = i D 1 e v D / V T i D 1
= i D 1 e v D / V T 1 (2)
Inthisproblem, i D 1 = I D = 1mA.
UsingEqs.(1)and(2)with VT = 25mV,we obtain
v D (mV) i D (mA) i D (mA) smallexponential signalmodel
a 10 0.4 0.33
b 5 0.2 0.18
c + 5 + 0.2 + 0.22
d + 10 + 0.4 + 0.49
Ex:4.19
a.Inthisproblem, V O i L = 20mV 1mA = 20 .
∴ Totalsmall-signalresistanceofthefourdiodes = 20
∴ Foreachdiode, r d = 20 4 = 5
But r d = VT I D ⇒ 5 = 25mV I D
∴ I D = 5mA R IL VO 15 V and R = 15 3 5mA = 2.4k .
b.For V O = 3V, voltagedropacrosseachdiode
= 3 4 = 0 75V
i D = I S e V / VT I S = i D e V / VT = 5 × 10 3 e 0 75/0 025 = 4 7 × 10 16 A
c.If i D = 5 i L = 5 1 = 4mA
Acrosseachdiodethevoltagedropis
V D = VT ln I D I S = 25 × 10 3 × ln 4 × 10 3 4 7 × 10 16
= 0 7443V
Voltagedropacross4diodes
= 4 × 0 7443 = 2 977V
sochangein V O = 3 2.977 = 23mV.
Ex:4.20 Whenthediodecurrentishalved,the voltagechangesby
V Z = r z I Z = 80 × 5 mA 2 =−200 mV
⇒ V Z = 6 0 2 = 5 8 V
Whenthediodecurrentisdoubled,
V Z = r z I Z = 80 × 5 mA = 400 mV
⇒ V Z = 6 + 0.4 = 6.4 V
Finally,thevalueof V Z 0 isthatobtainedbyusing themodelatzerocurrent.
V Z = V Z 0 + r z I Z
⇒ V Z 0 = V Z r z I Z = 6 V 80 × 5 mA
= 5 6 V
Ex:4.21
vS VD 2 t Vs 12 2 0 u u
a.Thediodestartsconductionat
v S = V D = 0 7V
v S = Vs sin ω t, here Vs = 12√2
At ω t = θ ,
v S = Vs sin θ = V D = 0.7V
12√2sin θ = 0 7 θ = sin 1 0 7 12√2 2 4◦
Conductionstartsat θ andstopsat 180 θ.
∴ Totalconductionangle = 180 2θ = 175 2◦
b. v O ,avg = 1 2π (π θ ) θ ( Vs sin φ V D ) d φ = 1 2π [ Vs cos φ V D φ ]φ =π θ φ θ = 1 2π [ Vs cos θ Vs cos (π θ ) V D (π 2θ )]
But cos θ 1, cos (π θ ) − 1, and π 2θ π
v O ,avg = 2 Vs 2π V D 2 = Vs π V D 2
For Vs = 12√2and V D = 0 7V
v O ,avg = 12√2 π 0 7 2 = 5 05V
c.Thepeakdiodecurrentoccursatthepeak diodevoltage.
∴ ˆ i D = Vs V D R = 12√2 0 7 100 = 163mA
PIV =+ V S = 12√2 17V
Ex:4.22 () vS Vs VS VD output 0 input t
a.Asshowninthediagram,theoutputiszero between (π θ ) to (π + θ ) = 2θ
Here θ istheangleatwhichtheinputsignal reaches V D
∴ Vs sin θ = V D
θ = sin 1 V D Vs 2θ = 2sin 1 V D Vs
b.Averagevalueoftheoutputsignalisgivenby
V O = 1 2π ⎡ ⎣2 × (π θ ) θ ( Vs sin φ V D ) d φ ⎤ ⎦ = 1 π [ Vs cos φ V D φ ]π θ φ =θ 2 Vs π V D ,for θ small.
c.Peakcurrentoccurswhen φ = π 2
Peakcurrent = Vs sin (π/2) V D R = Vs V D R
If v S is12V(rms), then Vs = √2 × 12 = 12√2
Peakcurrent = 12√2 0 7 100 163mA
Nonzerooutputoccursforangle = 2 (π 2θ )
Thefractionofthecycleforwhich v O > 0 is = 2 (π 2θ ) 2π × 100
Averageoutputvoltage V O is
V O = 2 Vs π V D = 2 × 12√2 π 0 7 = 10 1V
Peakdiodecurrent ˆ i D is
ˆ
Ex:4.24 Full-wavepeakrectifier: RC vO vS vS D1 D2 Vp Vr assume ideal diodes t t { T 2
Theripplevoltageistheamountofvoltage reductionduringcapacitordischargethatoccurs whenthediodesarenotconducting.Theoutput voltageisgivenby
v O = V p e t / RC
V p Vr = V p e T /2 RC ← dischargeisonly halftheperiod.Wealsoassumed t T 2
Vr = V p 1 e T /2 RC e T /2 RC 1 T /2 RC ,for CR T/2 Thus Vr V p 1 1 + T /2 RC
Vr = V p 2 fRC (a) Q.E.D. Tofindtheaveragediodecurrent,notethatthe chargesuppliedto C duringconductionisequal tothechargelostduringdischarge.
Q SUPPLIED = Q LOST
i C av t = CV r SUB (a) i D ,av I L t = C V p 2 fRC = V p 2 fR = V p π ω R i D ,av = V p π ω tR + I L where ω t istheconductionangle.
Notethattheconductionanglehasthesame expressionasforthehalf-waverectifierandis givenbyEq.(4.30),
ω t ∼ = 2 Vr V p (b)
Substitutingfor ω t ,weget
⇒ i D ,av = π V p 2 Vr V p R + I L
Sincetheoutputisapproximatelyheldat Vp , V p R ≈ I L · Thus
⇒ i D av ∼ = π I L V p 2 Vr + I L
= I L 1 + π V p 2 Vr Q.E.D.
If t =0isatthepeak,themaximumdiodecurrent occursattheonsetofconductionorat t =−ω t .
Duringconduction,thediodecurrentisgivenby
i D = i C + i L
i D ,max = C d v S dt t =−ω t + i L assuming i L isconst. i L V p R = I L
= C d dt V p cos ω t + I L
=−C sin ω t × ω V p + I L
=−C sin( ω t ) × ω V p + I L
Forasmallconductionangle
sin( ω t ) ≈− ω t Thus
⇒ i D ,max = C ω t × ω V p + I L
Sub(b)toget
i D ,max = C 2 V r V p ω V p + I L
Substituting ω = 2π f andusing(a)togetherwith V p / R I L resultsin
i D max = I L 1 + 2π V p 2 Vr Q.E.D.
Ex:4.25
Theoutputvoltage, v O ,canbeexpressedas
Attheendofthedischargeinterval
v O = V p 2 V D Vr
Thedischargeoccursalmostoverhalfofthetime period T /2
Fortimeconstant RC T 2
e t / RC 1 T 2 × 1 RC
∴ V P 2 V D Vr = V p 2 V D 1 T 2 × 1 RC
⇒ Vr = V p 2 V D × T 2 RC
Here V p = 12√2and Vr = 1V
V D = 0 8V
T = 1 f = 1 60 s
1 = (12√2 2 × 0 8) × 1 2 × 60 × 100 × C
C = (12√2 1 6) 2 × 60 × 100 = 1281 µF
Withoutconsideringtheripplevoltage,thedc outputvoltage
= 12√2 2 × 0.8 = 15.4V
Ifripplevoltageisincluded,theoutputvoltageis
= 12√2 2 × 0 8 Vr 2 = 14 9V
I L = 14 9 100 0 15 A
v O = V p 2 V D e t / RC vS vO D2 D1 D3 D4 C R ac line voltage
Theconductionangle ω t canbeobtainedusing Eq.(4.30)butsubstituting V p = 12√2 2 × 0 8:
ω t = 2 Vr V p = 2 × 1 12√2 2 × 0 8
= 0 36rad = 20 7◦
Theaverageandpeakdiodecurrentscanbe calculatedusingEqs.(4.34)and(4.35):
i D av = I L 1 + π V p 2 Vr , where I L = 14 9V 100 ,
V p = 12√2 2 × 0.8, and Vr = 1V;thus
i D av = 1 45A
i D max = I 1 + 2π V p 2 Vr = 2 76A
PIVofthediodes
= VS V DO = 12√2 0.8 = 16.2V
Toprovideasafetymargin,selectadiodecapable ofapeakcurrentof3.5to4AandhavingaPIV ratingof20V.
Ex:4.26 vA iD vI i i vD vO 1 k iR
Thediodehas0.7Vdropat1mAcurrent.
i D = I S e v D / VT
i D 1mA = e (v D 0 7)/ V T
⇒ v D = VT ln i D 1mA + 0 7V
For v I = 10mV, v O = v I = 10mV
Itisanidealopamp,so i + = i = 0
∴ i D = i R = 10mV 1k = 10 µA
v D = 25 × 10 3 ln 10 µA 1mA + 0 7 = 0 58V
v A = v D + 10mV
= 0 58 + 0 01 = 0.59V
For v I = 1V
v O = v I = 1V
i D = v O 1k = 1 1k = 1mA
v D = 0 7V
V A = 0 7V + 1k × 1mA = 1 7V
For v I =−1V, thediodeiscutoff.
∴ v O = 0V
v A =−12V
Ex:4.27 vO R IL vI
v I > 0 ∼ diode iscutoff,loopisopen,andthe opampissaturated:
v O = 0V
v I < 0 ∼ diode conductsandclosesthenegative feedbackloop:
v O = v I
Ex:4.28 Reversingthedioderesultsinthepeak outputvoltagebeingclampedat0V: t vO 10 V
Herethedccomponentof v O = V O =−5V
Ex:4.29 Thecapacitorvoltageaccountsforthe shiftofthevoltagewaveformsfrom v I to v O Thus,fromFig.4.30,weseethecapacitorvoltage is V p + VCC = 8 V.Thediode’speakinverse voltagearisesatthepeaksof v O ,
PIV = v O max VCC
= VCC + 2 V p VCC
= 2 V p = 6 V
Ex:4.30 C j 0 = 100 fF, V0 = 3 V, m = 3.Using Equation(3.47),
at V R = 1 V: C j = 100 fF 1 + 1 3 3 = 42.2 fF,and
at V R = 3 V: C j = 100 fF 1 + 3 3 3 = 12 5 fF.
Ex:4.31 Thereversecurrentis
i D = I D + i P = I D + R × P
Atanincidentlightpowerof P = 1 mW,
i D = 10 4 + 0 5 × 1 = 0 5 mA
Atanincidentlightpowerof P = 1 µW,
i D = 10 4 + 0 5 × 10 3 = 6 × 10 4 mA = 0 6 µA
Ex:4.32 Neglectingdarkcurrent, i P = R × P
Thecapacitanceis10pFpermm2 or, equivalently,1nFpercm2 .Thus, C j = 1 × 3 33 = 3 33 nF
Ex:4.33 R = 9 3 × 1 8 20 = 0 18 k = 180
Ex:4.34 R = 9 3 × 2 2 20 = 0 12 k = 120
Exercise5–1
Chapter5
SolutionstoExerciseswithintheChapter
Ex:5.1
C ox = ox tox = 34 5pF/m 4nm = 8 625fF/µm2
μn = 450cm2 /V S
k n = μn C ox = 388 µA/V2
v OV = v G S Vt = 0 5V
g DS = 1 1k = k n W L v OV ⇒ W L = 5 15
L = 0 18 µm, so W = 0 93 µm
Ex:5.2
C ox = ox tox = 34.5pF/m 1 4nm = 24 6fF/µm2
μn = 216cm2 /V s
k n = μn C ox = 531 µA/V2
i D = 1 2 k n W L v 2 OV
50 = 1 2 × 531 × 10 × v 2 OV
∴ v OV = 0 14V
v G S = Vt + v OV = 0.49V
v DS , min = v OV = 0 14V,
Ex:5.3 i D = 1 2 k n W L v 2 OV insaturation
Changein i D is:
(a)double L ,0.5
(b)double W ,2
(c)double v OV , 22 = 4
(d)double v DS ,nochange(ignoringlength modulation)
(e)changes(a)–(d),4
Case(c)wouldcauseleavingsaturationif
v DS < 2v OV
Ex:5.4 Forsaturation v DS ≥ v OV , so v DS must bechangedto 2v OV i D = 1 2 k n W L v 2 OV , so i D increasesbyafactorof4.
Ex:5.5 v OV = 0 5V
g DS = k n W L v OV = 1 1k
∴ k n = k n W L = 1 1 × 0 5 = 2mA/V2
At v DS = 0 2V,v DS <v OV ,thusthetransistoris operatinginthetrioderegion,
i D = k n (v OV v DS 1 2 v 2 DS )
= 2(0 5 × 0 2 1 2 × 0 22 ) = 0 16mA
At v DS = 0 5V,v DS = v OV ,thusthetransistoris operatinginsaturation,
i D = 1 2 k n v 2 OV = 1 2 × 2 × 0 52 = 0 25mA
At v DS = 1V, v DS >v OV andthetransistoris operatinginsaturationwith i D = 0 25mA
Ex:5.6 V A = V A L = 5 × 0 8 = 4V
λ = 1 V A = 0 25V 1
v DS = 0 8V >v OV = 0 2V
⇒ Saturation: i D = 1 2 k n W L v 2 OV (1 + λv DS )
i D = 1 2 × 400 × 16 0 8 × 0 22 (1 + 0 25 × 0 8)
= 0 192mA
r o = V A i D = 4 0 16 = 25k where i D isthevalueof i D without channel-lengthmodulationtakenintoaccount.
r o = v DS i D ⇒ i D = 1V 25k = 0 04mA = 40 µA
Ex:5.7 vG vD 1.8 V iD
Vtp =−0 5V
k p = 100 µA/V2 W L = 10 ⇒ k p = 1mA/V2
(a)Conductionoccursfor VS G ≥| Vtp |= 0.5V
⇒ v G ≤ 1 8 0 5 =+1 3V
(b)Trioderegionoccursfor v D G ≥| Vtp |= 0 5V
⇒ v D ≥ v G + 0 5
(c)Conversely,forsaturation
v D G ≤| Vtp |= 0 5V
⇒ v D ≤ v G + 0 5
(d)Given λ ∼ = 0,
i D = 1 2 k p W L |v OV |2 = 50 µA
∴ |v OV |= 0 32V = v S G −| Vtp |
⇒ v S G =|v OV |+| Vtp |= 0 82V
v G = 1 8 v S G = 0 98V
v D ≤ v G + 0 5 = 1 48V
(e)For λ =−0.2V 1 and |v OV |= 0.32V,
i D = 50 µAand r o = 1 | λ |i D = 100k
(f)At v D =+1V,v SD = 0 8V,
i D = 1 2 k n W L |v OV |2 (1 +| λ ||v SD |)
= 50 µA (1 16) = 58 µA
At v D = 0V,v SD = 1 8V,
i D = 50 µA (1 36) = 68 µA
r o = v DS i D = 1V 10 µA = 100k
whichisthesamevaluefoundin(c).
Ex:5.8
R D = V DD v D I D = 1 0 2 0 1 = 8k
I D = 1 2 μn C ox W L V 2 OV ⇒
100 = 1 2 × 400 × 5 0 4 V 2 OV ⇒
V OV = 0.2V ⇒ VG S = V OV + Vt = 0.2 + 0.4 = 0 6V
VS =−0.6V ⇒ R S = VS VSS I D = 0.6 ( 1) 0 1
R S = 4k
Ex:5.9
Vtn = 0 5V
μn C ox = 0 4mA/V2
W L = 0.72 µm 0 18 µm = 4 0
λ = 0 VD R 1.8 V
Saturationmode (v G D = 0 < Vtn ) : I D = 1 2 μn C ox W L ( V D Vtn )2 = 0 032mA
V D = 0 7V = 1 8 I D R
∴ R = 1 8 0 7 0 032mA = 34 4k
Ex:5.10 R2 R Q2 Q1 1.8 V
Since Q 2 isidenticalto Q 1 andtheir VG S values arethesame,
I D 2 = I D 1 = 0 032mA
For Q 2 tooperateatthetriode–saturation boundary,wemusthave
V D 2 = V OV = 0 7 0 5 = 0 2V
∴ R2 = 1.8V 0.2V 0 032mA = 50k
Ex:5.11 R D = 6 55 × 2 = 13 1k
VG S = 2V, assumetrioderegion:
I D = k n W L ( VG S Vtn ) V DS 1 2 V 2 DS
I D = V DD V DS R ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ⇒ 2 V DS 13 1 = 2 × (2 0 5) V DS V 2 DS 2
⇒ V 2 DS 3 076 V DS + 0 15 = 0
⇒ V DS = 0 05V < V OV ⇒ trioderegion
I D = 2 0 05 13 1 = 0 15mA
Ex:5.12 AsindicatedinExample5.6,
V D ≥ VG Vt forthetransistortobeinthe saturationregion.
V D min = VG Vt = 5 1 = 4V
I D = 0 5mA ⇒ R D max = V DD V D min I D = 10 4 0 5 = 12k
Ex:5.13
I D = 0 32mA = 1 2 k n W L V 2 OV = 1 2 × 1 × V 2 OV
⇒ V OV = 0 8V
VG S = 0 8 + 1 = 1 8V
VG = VS + VG S = 1 6 + 1 8 = 3 4V
R G 2 = VG I = 3 4 1 µA = 3 4M
R G 1 = 5 3 4 1 µA = 1 6M
R S = VS 0 32 = 5k
V D = 3 4V, then R D = 5 3 4 0.32 = 5k
Ex:5.14 ID R 1.8 V
Vtp =−0 4V k p = 0 1mA/V2 W L = 10 µm 0 18 µm ⇒ k p = 5 56mA/V2
VS G =| Vtp |+| V OV |
= 0 4 + 0 6 = 1V
VS =+1V
Since V D G = 0,thetransistorisoperatingin saturation,and
I D = 1 2 k p V 2 OV = 1mA
∴ R = 1 8 VS I D = 1 8 1 1 = 0 8k = 800
Ex:5.15 V I = 0:sincethecircuitisperfectly symmetrical, V O = 0 andtherefore VG S = 0, whichimpliesthatthetransistorsareturnedoff and I DN = I DP = 0
V I = 2 5V:ifweassumethattheNMOSis turnedon,then V O wouldbelessthan2.5V,and thisimpliesthatPMOSisoff ( VS G P < 0)
I DN = 1 2 k n W L ( VG S Vt )2
I DN = 1 2 × 1(2 5 V O 1)2
I DN = 0 5(1 5 V O )2
Also: V O = R L I DN = 10 I DN
I DN = 0 5(1 5 10 I DN )2
⇒ 100 I 2 DN 32 I DN + 2 25 = 0 ⇒ I DN = 0 104mA I DP = 0
VO = 10 × 0 104 = 1 04V
Exercise5–4 V I =−2 5V:Againifweassumethat Q P is turnedon,then V O > 2 5Vand VG SN < 0, whichimpliesthattheNMOS Q N isturnedoff.
I DN = 0
Becauseofthesymmetry,
I DP = 0 104mA,
1 23 V
V O =− I DP × 10k =−1 04V Ex:5.16 Vt = 0 8 + 0 4
Ex:5.17 v DS min = v G S +| Vt | = 1 + 2 = 3 V
I D = 1 2 × 2 [1 ( 2)]2 = 9 mA
Chapter6
SolutionstoExerciseswithintheChapter
Ex:6.1 i C = I S e v BE / VT
v BE 2 v BE 1 = VT ln i C 2 i C 1
v BE 2 = 700 + 25ln 0 1 1 = 642mV
v BE 3 = 700 + 25ln 10 1 = 758mV
Ex:6.2 ∴ α = β β + 1
50 50 + 1 <α< 150 150 + 1 0 980 <α< 0 993
Ex:6.3 IC = I E I B
= 1 460mA 0 01446mA = 1 446mA
α = IC I E = 1 446 1 460 = 0 99
β = IC I B = 1 446 0.01446 = 100
IC = I S e v BE / VT
I S = IC e v BE / VT = 1 446 e 700/25 = 1 446 e 28 mA = 10 15 A
Ex:6.4 β = α 1 α and IC = 10mA
For α = 0 99,β = 0 99 1 0.99 = 99
I B = IC β = 10 99 = 0 1mA
For α = 0.98,β = 0 98 1 0 98 = 49
I B = IC β = 10 49 = 0 2mA
Ex:6.5 Given:
I S = 10 16 A,β = 100, I C = 1mA Wewrite
I SE = I S /α = I S × 1 + 1 β
= 10 16 × 1 01 = 1 01 × 10 16 A
Exercise6–1
I SB = I S β = 10 16 100 = 10 18 A
V BE = VT ln IC I S = 25ln 1mA 10 16 A
= 25 × 29 9336 = 748mV
Ex:6.6 VCC 5 V RC B 10 A bIB E C v BE
v BE = 690mV IC = 1mA
Foractiverange VC ≥ V B , RC max = VCC 0 690 IC
= 5 0 69 1 = 4 31k
Ex:6.7 I S = 10 15 A AreaC = 100 × Area E
I SC = 100 × I S = 10 13 A
Ex:6.8 i C = I S e v BE / VT I SC e v BC / VT for i C = 0
I S e v BE / VT = I SC e v BC / VT I SC I S = e v BE / VT e v BC / VT = e (v BE v BC )/ VT ∴ VCE = V BE V BC = VT ln I SC I S
ForcollectorArea=100 × Emitterarea
VCE = 25ln 100 1 = 115mV
Ex:6.9 IC = I S e V BE / VT I SC e V BC / VT
I B = I S β e V BE / VT + I SC e V BC / VT
β forced = IC I B sat <β = β I S e V BE / VT I SC e V BC / VT
I S e V BE / VT + β I SC e V BC / VT = β I S e ( V BE V BC )/ VT I SC
I S e ( V BE V BC )/ VT + β I SC = β e VCE sat / VT I SC / I S
e VCE sat / VT + β I SC / I S Q.E.D.
β forced = 100 e 200/25 100 e 200/25 + 100 × 100 = 100 × 0 2219 ≈ 22 2
Ex:6.10 IS /a IS ev EB /VT aiE 2 mA E B C 10 V
I E = I S α e V BE / VT
2mA = 51 50 10 14 e V BE / VT
V BE = 25ln 2 103 × 50 51 × 1014 = 650mV
IC = β β + 1 I E = 50 51 × 2 = 1 96mA
I B = IC β = 1.96 50 ⇒ 39 2 µA
Ex:6.11 IC = I S e V BE / VT = 1.5A
∴ V BE = VT ln 1 5/10 11 = 25 × 25 734 = 643mV
Ex:6.12
0 5 k = 500
Sinceat IC = 1 mA, V BE = 0 8 V,thenat IC = 2 mA,
V BE = 0.8 + 0.025 ln 2 1 = 0 8 + 0 017 = 0 817 V
V E =− V BE =−0 817 V I E = 2 mA α = 2 0 99 = 2 02 mA
I E = V E ( 1 5) R E Thus, R E = 0 817 + 1 5 2 02 = 0 338 k = 338
I E = V E ( 10) 10 = 0 7 + 10 10
= 0 93 mA
Assumingactive-modeoperation,
I B = I E β + 1 = 0 93 50 + 1 = 0 0182 mA
= 18 2 µA
IC = I E I B = 0 93 0 0182 = 0 91 mA
VC = 10 IC × 5
= 10 0 91 × 5 = 5 45 V
Since VC > V B ,thetransistorisoperatinginthe activemode,asassumed.
Ex:6.14
V B = 1 0 V
Thus,
I B = V B 100 k = 0 01 mA
V E =+1 7 V
Thus,
I E = 10 V E 5 k = 10 1 7 5 = 1 66 mA and β + 1 = I E I B = 1 66 0 01 = 166
⇒ β = 165
α = β β + 1 = 165 165 + 1 = 0 994
Assumingactive-modeoperation,
IC = α I E = 0 994 × 1 66 = 1 65 mA and VC =−10 + 1 65 × 5 =−1 75 V
Since VC < V B ,thetransistorisindeedoperating intheactivemode.
Ex:6.15 5 V IC VE VC RC 1 k 2 mA
Thetransistorisoperatingataconstantemitter current.Thus,achangeintemperatureof +30◦ C resultsinachangein V EB by
V EB =−2 mV × 30 =−60 mV
Thus,
V E =−60 mV
Sincethecollectorcurrentremainsunchangedat α I E ,thecollectorvoltagedoesnotchange:
VC = 0 V
Ex:6.16 RefertoFig.6.19(a):
i C = I S e v BE / VT + v CE r o (1) NowusingEqs.(6.21)and(6.22),wecanexpress r o as
r o = V A
I S e v BE / VT
SubstitutinginEq.(1),wehave
i C = I S e v BE / VT 1 + v CE V A whichisEq.(6.18).Q.E.D.
Ex:6.17 r o = V A IC = 100 IC
At IC = 0 1 mA, r o = 1 M
At IC = 1 mA, r o = 100 k
At IC = 10 mA, r o = 10 k
Ex:6.18 IC = VCE r o where
r o = V A IC = 100 1 = 100 k
IC = 11 1 100 = 0.1 mA
Thus, IC becomes1.1mA.
Ex:6.19 VCC 10 V IB IC VBB RC 10 k
RB 10 k
VBE 0.7 V VCE
(a)Foroperationintheactivemodewith
VCE = 5 V,
IC = VCC VC RC = 10 5 10 = 0 5 mA
I B = IC β = 0 5 50 = 0 01 mA
V BB = V BE + I B R B
= 0 7 + 0 01 × 10 = 0 8 V
(b)Foroperationattheedgeofsaturation,
VCE = 0 3 V
IC = VCC VCE RC = 10 0 3 10 = 0 97 mA
I B = IC β = 0 97 50 = 0 0194 mA
V BB = V B + I B R B
= 0 7 + 0 0194 × 10 = 0 894 V
(c)Foroperationdeepinsaturationwith βforced = 10, wehave
VCE 0 2 V
IC = 10 0 2 10 = 0 98 mA
I B = IC βforced = 0 98 10 = 0 098 mA
V BB = V B + I B R B
= 0 7 + 0 098 × 10 = 1 68 V
Ex:6.20 For V BB = 0 V, I B = 0 andthe transistoriscutoff.Thus,
IC = 0 and
VC = VCC =+10 V
Ex:6.21 RefertothecircuitinFig.6.22andlet V BB = 1 7 V.Thecurrent I B canbefoundfrom
I B = V BB V B R B = 1 7 0 7 10 = 0.1 mA
Assumingoperationintheactivemode,
IC = β I B = 50 × 0.1 = 5 mA
Thus,
VC = VCC RC IC
= 10 1 × 5 = 5 V whichisgreaterthan V B ,verifyingthatthe transistorisoperatingintheactivemode,as assumed.
(a)Toobtainoperationattheedgeofsaturation, RC mustbeincreasedtothevaluethatresultsin
VCE = 0 3 V:
RC = VCC 0 3 IC
= 10 0.3 5 = 1 94 k
(b)Furtherincreasing RC resultsinthetransistor operatinginsaturation.Toobtain saturation-modeoperationwith VCE = 0 2 Vand βforced = 10,weuse
IC = βforced × I B = 10 × 0 1 = 1 mA
Thevalueof RC requiredcanbefoundfrom
RC = VCC VCE
IC = 10 0 2 1 = 9 8 k
Ex:6.22 RefertothecircuitinFig.6.23(a)with thebasevoltageraisedfrom4Vto V B .Ifatthis valueof V B ,thetransistorisattheedgeof saturationthen,
VC = V B 0 4 V
Since IC I E ,wecanwrite 10 VC
RC = V E R E = V B 0 7 R E
Thus,
DividingEq.(1)byEq.(2),wehave
Ex:6.23
Ex:6.25 RefertothecircuitinFig.6.26(a).The largestvaluefor RC whiletheBJTremainsinthe activemodecorrespondsto
VC =+0 4 V
Sincetheemitterandcollectorcurrentsremain unchanged,thenfromFig.6.26(b)weobtain
Thus,
Toestablishareverse-biasvoltageof2Vacross theCBJ,
VC =+6 V
Fromthefigureweseethat RC
and
wherewehaveassumed α 1
Ex:6.24
Thefigureshowsthecircuitwiththebasevoltage at V B andtheBJToperatinginsaturationwith VCE = 0 2 Vand βforced = 5
Fora4-Vreverse-biasedvoltageacrosstheCBJ, VC =−4 V
Refertothefigure. IC = 1 mA = VC ( 10) RC
Ex:6.27 RefertothecircuitinFig.6.27:
I B = 5 0.7 100 = 0 043 mA
Toensurethatthetransistorremainsintheactive modefor β intherange50to150,weneedto select RC sothatforthehighestcollectorcurrent possible,theBJTreachestheedgeofsaturation, thatis, VCE = 0 3 V.Thus,
VCE = 0 3 = 10 RC IC max where
IC max = βmax I B
= 150 × 0 043 = 6 45 mA
Thus,
RC = 10 0 3 6 45 = 1 5 k
Forthelowest β ,
IC = βmin I B
= 50 × 0.043 = 2.15 mA
andthecorresponding VCE is
VCE = 10 RC IC = 10 1 5 × 2 15
= 6 775 V
Thus, VCE willrangefrom0.3Vto6.8V.
Ex:6.28 RefertothesolutionofExample6.10.
I E = V BB V BE R E + R BB /(β + 1)
= 5 0 7 3 + (33.3/51) = 1 177 mA
IC = α I E = 0 98 × 1 177 = 1 15 mA
Thusthecurrentisreducedby
IC = 1 28 1 15 = 0 13 mA whichisa 10% change.
Ex:6.29 RefertothecircuitinFig.6.30(b).The totalcurrentdrawnfromthepowersupplyis
I = 0 103 + 1 252 + 2 78 = 4 135 mA
Thus,thepowerdissipatedinthecircuitis
P = 15 V × 4 135 mA = 62 mW
Ex:6.30 15 V
Fromthefigureweseethat
V E 3 = IC 3 α × 0 47
VC 2 = V E 3 + 0 7 = IC 3 α × 0 47 + 0 7 (1)
Anodeequationatthecollectorof Q 2 yields
2 75 = VC 2 2 7 + IC 3 β
Substitutingfor VC 2 fromEq.(1),weobtain
2 75 = (0 47 IC 3 /α) + 0 7 2.7 + IC 3 β
Substituting α = 0 99 and β = 100 andsolving for IC 3 resultsin
IC 3 = 13 4 mA
Now, V E 3 and VC 2 canbedetermined:
V E 3 = IC 3 α × 0 47 = 13 4 0 99 × 0 47 =+6 36 V
VC 2 = V E 3 + 0 7 =+7 06 V
Ex:6.31
V
V
Fromthefigureweseethat Q 1 willbeoffand Q 2 willbeon.Sincethebaseof Q 2 willbeata voltagehigherthan 5 V,transistor Q 2 willbe operatingintheactivemode.Wecanwritealoop equationfortheloopcontainingthe10-k resistor,theEBJof Q 2 andthe1-k resistor:
I E × 1 0 7 I B × 10 =−5
Substituting I B = I E /(β + 1) = I E /101 and rearranginggives
I
Thus,
V E =−3 9 V
V B 2 =−4 6 V
I B = 0 039 mA
Ex:6.32 Withtheinputat + 10V,thereisa strongpossibilitythattheconductingtransistor
ThisfigurebelongstoExercise6.32.
Q 1 willbesaturated.Assumingthistobethe case,theanalysisstepswillbeasfollows:
VCE sat | Q 1 = 0 2 V V E = 5 V VCE sat =+4 8 V I E 1 = 4 8 V 1 k = 4 8 mA
whichislowerthan βmin ,verifyingthat Q 1 is indeedsaturated.
Finally,since Q 2 isoff,
IC 2 = 0
Ex:6.33 V O =+10 BV BCO = 10 70 =−60 V
Chapter7
SolutionstoExerciseswithintheChapter
Ex:7.1 RefertoFig.7.2(a)and7.2(b).
CoordinatesofpointA: Vt and V DD ;thus0.4V and1.8V.Todeterminethecoordinatesof pointB,weuseEqs.(7.7)and(7.8)asfollows:
V OV B = √2k n R D V DD + 1 1 k n R D
= √2 × 4 × 17 5 × 1 8 + 1 1 4 × 17 5
= 0 213 V
Thus,
VG S B = Vt + V OV B = 0 4 + 0 213 = 0 613 V and
V DS B = V OV B = 0 213 V
Thus,coordinatesofBare0.613Vand0.213V. AtpointC,theMOSFETisoperatinginthe trioderegion,thus
i D = k n (v G S C Vt )v DS C 1 2 v 2 DS C
If v DS C isverysmall,
i D k n (v G S C Vt )v DS C
= 4(1 8 0 4)v DS C
= 5 6v DS C , mA
But
i D = V DD v DS C R D V DD R D = 1.8 17 5 = 0 1 mA
Thus, v DS C = 0 1 5 6 = 0 018V = 18mV,which isindeedverysmall,asassumed.
Ex:7.2 RefertoExample7.1andFig.7.4(a).
Design1:
V OV = 0 2 V, VG S = 0 6 V
I D = 0 08 mA
Now,
A v =−k n V OV R D
Thus, 10 =−0.4 × 10 × 0.2 × R D
⇒ R D = 12 5 k
V DS = V DD R D I D = 1 8 12 5 × 0 08 = 0 8 V
Design2:
R D = 17 5 k
A v =−k n V OV R D 10 =−0 4 × 10 × V OV × 17 5
Thus,
V OV = 0.14 V
VG S = Vt + V OV = 0 4 + 0 14 = 0 54 V
I D = 1 2 k n W L V 2 OV
= 1 2 × 0 4 × 10 × 0 142 = 0 04 mA
R D = 17.5 k
V DS = V DD R D I D
= 1 8 17 5 × 0 04 = 1 1 V
Ex:7.3 A v =− IC RC VT
320 =− 1 × RC 0 025 ⇒ RC = 8 k
VC = VCC IC RC
= 10 1 × 8 = 2 V
Sincethecollectorvoltageisallowedtodecrease to +0.3V,thelargestnegativeswingallowedat theoutputis 2 0 3 = 1 7V.Thecorresponding inputsignalamplitudecanbefoundbydividing 1.7Vbythegainmagnitude(320V/V),resulting in5.3mV.
Ex:7.4 iD VDD RD vGS vgs vDS VGS
V DD = 5V
VG S = 2V
Vt = 1V
λ = 0
k n = 20 µA/V2
R D = 10k
W L = 20
(a) VG S = 2V ⇒ V OV = 1V
I D = 1 2 k n W L V 2 OV = 200 µA = 0 2mA
V DS = V DD I D R D =+3V
(b) gm = k n W L V OV = 400 µA/V = 0 4mA/V
(c) A v = v ds v gs =− gm R D =−4V/V
(d) v gs = 0 2sin ω t V
v ds =−0 8sin ω t V
v DS = V DS + v ds ⇒ 2.2V ≤ v DS ≤ 3.8V
(e)UsingEq.(7.28),weobtain
i D = 1 2 k n ( VG S Vt )2
+ k n ( VG S Vt )v gs + 1 2 k n v 2 gs
i D = 200 + 80sin ω t
+ 8sin2 ω t , µA
= [200 + 80sin ω t + (4 4cos2ω t )]
= 204 + 80sin ω t 4cos2ω t , µA
Thus, I D shiftsby 4 µA and
2HD = ˆ i 2ω ˆ i ω = 4 µA 80 µA = 0 05 (5%)
Ex:7.5
(a) VG S = 1 5V ⇒ V OV = 1.5 1 = 0.5V
gm = 2 I D V OV
I D = 1 2 k n W L V 2 OV = 1 2 × 60 × 40 × 0 52
I D = 300 µA = 0 3mA
gm = 2 × 0 3 0 5 = 1 2mA/V
r o = V A I D = 15 0 3 = 50k
(b) I D = 0 5mA ⇒ gm = 2 μn C ox W L I D
= 2 × 60 × 40 × 0 5 × 103
gm = 1 55mA/V
r o = V A I D = 15 0.5 = 30k
Ex:7.6
I D = 0 1mA, gm = 1mA/V, k n = 500 µA/V2
gm = 2 I D V OV ⇒ V OV = 2 × 0 1 1 = 0 2V
I D = 1 2 k n W L V 2 OV ⇒ W L = 2 I D k n V 2 OV
= 2 × 0 1
500
1000 × 0 22 = 10
Ex:7.7
gm = μn C ox W L V OV
Samebiasconditions,sosame V OV andalsosame L and gm forbothPMOSandNMOS.
μn C ox Wn = μ p C ox W p ⇒ μ p μn = 0 4 = Wn W p
⇒ W p Wn = 2.5
Ex:7.8
I D = 1 2 k p W L ( VS G −| Vt |)2
= 1 2 × 60 × 16 0 8 × (1 6 1)2
I D = 216 µA
gm = 2 I D | V OV | = 2 × 216 1 6 1 = 720 µA/V
= 0 72mA/V
|λ|= 0 04 ⇒| V A |= 1 |λ| = 1 0 04 = 25V/µm
r o = | V A |× L I D = 25 × 0.8 0 216 = 92 6k
Ex:7.9
gm r o = 2 I D V OV × V A I D = 2 V A V OV
V A = V A × L = 6 × 3 × 0 18 = 3 24V
gm r o = 2 × 3 24 0 2 = 32.4V/V
Ex:7.10
RefertothesolutionofExample7.3.From Eq.(7.47), A v ≡ v o v i =− gm R D (notethat R L isabsent).
Thus,
gm R D = 25
5 V RD vo vi RG
Substitutingfor gm = k n V OV ,wehave
k n V OV R D = 25
where k n = 1mA/V2 ,thus
V OV R D = 25 (1)
Next,considerthebiasequation
VG S = V DS = V DD R D I D
Thus,
Vt + V OV = V DD R D I D
Substituting Vt = 0 7V, V DD = 5V,and
I D = 1 2 k n V 2 OV = 1 2 × 1 × V 2 OV = 1 2 V 2 OV
weobtain
0 7 + V OV = 5 1 2 V 2 OV R D (2)
Equations(1)and(2)canbesolvedtoobtain
V OV = 0 319 V and
R D = 78 5k
Thedccurrent I D canbenowfoundas
I D = 1 2 k n V 2 OV = 50.9 µA
Todeterminetherequiredvalueof R G weuseEq. (7.48),againnotingthat R L isabsent:
Rin = R G 1 + gm R D
0 5 M = R G 1 + 25
⇒ R G = 13 M
Finally,themaximumallowableinputsignal ˆ v i canbefoundasfollows:
ˆ v i = Vt | A v |+ 1 = 0 7 V 25 + 1 = 27 mV
Ex:7.11 D 0 Req G S i i it ro vt 1 gm i t = v t r o + i = v t r o + gm v t
Req = v t i t = r o 1 gm
Ex:7.12
Given: gm = ∂ i C ∂v BE i C = IC where IC = I S e v BE / VT
∂ i C ∂v BE i C = IC = I S e v BE / VT VT = IC VT
Thus, gm = IC VT
Ex:7.13
gm = IC VT = 0 5mA 25mV = 20mA/V
Ex:7.14
IC = 0 5mA ( constant)
For β = 50: gm = IC VT = 0 5mA 25mV = 20mA/V
I B = IC β = 0.5 50 = 10 µA
r π = β gm = 50 20 = 2 5k
For β = 200, gm = IC VT = 20mA/V
I B = IC β = 0.5mA 200 = 2 5 µA
r π = β gm = 200 20 = 10k
Ex:7.15
β = 100 IC = 1mA
gm = 1mA 25mV = 40mA/V
r e = VT I E = α VT IC 25mV 1mA = 25
r π = β gm = 100 40 = 2.5k
Ex:7.16
gm = IC VT = 1mA 25mV = 40mA/V
A v = v ce v be =− gm RC
=−40 × 10
=−400V/V
VC = VCC IC RC
= 15 1 × 10 = 5V
v C (t ) = VC + v c (t )
= VC + A v v be (t )
= 5 400 × 0 005sin ω t
= 5 2sin ω t
i B (t ) = I B + i b (t ) where
I B = IC β = 1mA 100 = 10 µA and i b (t ) = gm v be (t ) β = 40 × 0 005sin ω t 100 = 2sin ω t , µA
Thus,
i B (t ) = 10 + 2sin ω t , µA
Ex:7.17
Exercise7–5
I E = 10 0 7 10 = 0.93mA
IC = α I E = 0 99 × 0 93 = 0 92mA
VC =−10 + IC RC
=−10 + 0 92 × 7 5 =−3 1V
A v = v o v i = α RC r e
where r e = 25mV 0 93mA = 26.9
A v = 0 99 × 7 5 × 103 26 9 = 276 2V/V
For v i = 10mV, v o = 276 2 × 10 = 2 76V
Ex:7.20 10V 8 k 10 k Y Z X I 1 mA
I E = 1mA
IC = 100 101 × 1 = 0 99mA
I B = 1 101 × 1 = 0 0099mA
(a) VC = 10 8 × 0.99 = 2.08 2.1V
V B =−10 × 0 0099 =−0 099 −0 1V
ThisfigurebelongstoExercise7.20c.
V E =−0 1 0 7 =−0 8V
(b) gm = IC VT = 0.99 0 025 40mA/V
r π = β gm = 100 40 2 5k
r o = V A IC = 100 0 99 = 101 100k (c)Seefigurebelow.
Rsig = 2k R B = 10k r π = 2 5k
gm = 40mA/V
RC = 8k R L = 8k r o = 100k
V y Vsig = Vπ Vsig × V y Vπ = R B r π ( R B r π ) + Rsig ×− gm ( RC R L r o )
= 10 2.5 (10 2 5) + 2 ×−40(8 8 100)
=−0 5 × 40 × 3 846 =−77V/V
If r o isnegelected, V y Vsig =−80,foranerror of3.9%.
Ex:7.21
gm = 2 I D V OV = 2 × 0 25 0.25 = 2mA/V
Rin =∞
Av o =− gm R D =−2 × 20 =−40V/V
Ro = R D = 20k
A v = Av o R L R L + Ro =−40 × 20 20 + 20 =−20V/V
G v = A v =−20V/V
ˆ v i = 0 1 × 2 V OV = 0 1 × 2 × 0 25 = 0 05V
ˆ v o = 0 05 × 20 = 1V
Ex:7.22
IC = 0 5mA
gm = IC VT = 0 5mA 0 025V = 20mA/V
r π = β gm = 100 20 = 5k
Rin = r π = 5k
Av o =− gm RC =−20 × 10 =−200V/V
Ro = RC = 10k
A v = Av o R L R L + Ro =−200 × 5 5 + 10
=−66 7V/V
G v = Rin Rin + Rsig A v = 5 5 + 5 ×−66 7
=−33 3V/V
v π = 5mV ⇒ˆ v sig = 2 × 5 = 10mV
ˆ v o = 10 × 33 3 = 0 33V
Althoughalargerfractionoftheinputsignal reachestheamplifierinput,linearity considerationscausetheoutputsignaltobein factsmallerthanintheoriginaldesign!
Ex:7.23 RefertothesolutiontoExercise7.21.
If v sig = 0 2V andwewishtokeep
ˆ v gs = 50mV,thenweneedtoconnecta
resistance Rs = 3 gm inthesourcelead.Thus,
Rs = 3 2mA/V = 1 5k
G v = A v =− R D R L 1 gm + Rs =− 20 20
0 5 + 1 5 =−5V/V
ˆ v o = Gv ˆ v sig = 5 × 0 2 = 1 V(unchanged)
Ex:7.24
Fromthefollowingfigureweseethat
ˆ v sig = ˆ i b Rsig +ˆ v π + ˆ i e Re
= ˆ i e β + 1 Rsig +ˆ v π + ˆ i e Re
= v π (β + 1)r e Rsig +ˆ v π + v π r e Re
ˆ v sig =ˆ v π 1 + Re r e + Rsig r π Q E D
For IC = 0 5mA and β = 100,
r e = VT I E = α VT IC = 0 99 × 25 0 5 50
r π = (β + 1)r e 5k
For ˆ v sig = 100mV, Rsig = 10k andwith ˆ v π limitedto10mV,thevalueof Re requiredcanbe foundfrom
100 = 10 1 + Re 50 + 10 5
⇒ Re = 350
Rin = (β + 1)(r e + Re ) = 101 × (50 + 350)
= 40 4k
G v =−β RC R L Rsig + (β + 1)(r e + Re )
=−100 10 10 + 101 × 0.4 =−19 8V/V
Ex:7.25 1
gm = Rsig = 100
⇒ gm = 1 0 1k = 10mA/V
But
gm = 2 I D V OV
Thus, 10 = 2 I D 0 2
⇒ I D = 1mA
G v = Rin Rin + Rsig × gm R D
= 0 5 × 10 × 2 = 10V/V
Ex:7.26
IC = 1mA
r e = VT I E VT IC = 25mV 1mA = 25
Rin = r e = 25
Av o = gm RC = 40 × 5 = 200V/V
Ro = RC = 5k
A v = Av o R L R L + Ro = 200 × 5 5 + 5 = 100V/V
G v = Rin Rin + Rsig × A v = 25 25 + 5000 × 100 = 0 5V/V
Ex:7.27
Rin = r e = 50
⇒ I E = VT r e = 25mV 50 = 0 5mA
IC I E = 0 5mA
G v = RC R L r e + Rsig
40 = RC R L (50 + 50)
RC R L = 4k
Ex:7.28 RefertoFig.7.42(c).
Ro = 100
Thus, 1 gm = 100 ⇒ gm = 10mA/V
But
gm = 2 I D V OV
Thus,
I D = 10 × 0 25 2 = 1 25mA
ˆ v o =ˆ v i × R L R L + Ro = 1 × 1 1 + 0 1 = 0 91V
ˆ v gs =ˆ v i 1 gm 1 gm + R L = 1 × 0 1 0 1 + 1 = 91mV
Ex:7.29
Ro = 200
1 gm = 200
⇒ gm = 5mA/V
But
g m = k n W L V OV
Thus, 5 = 0 4 × W L × 0 25
⇒ W L = 50
I D = 1 2 k n W L V 2 OV
= 1 2 × 0 4 × 50 × 0 252
= 0 625mA
R L = 1k to10k
Correspondingly,
G v = R L R L + Ro = R L R L + 0.2 willrangefrom
G v = 1 1 + 0 2 = 0 83V/V to
G v = 10 10 + 0 2 = 0.98V/V
Ex:7.30 IC = 5mA
r e = VT I E VT IC = 25mV 5mA = 5
Rsig = 10k R L = 1k
Rin = (β + 1)(r e + R L ) = 101 × (0 005 + 1) = 101 5k
Gv o = 1V/V
Rout = r e + Rsig β + 1
= 5 + 10,000 101 = 104
G v = R L R L + r e + Rsig β + 1 = R L R L + Rout
= 1 1 + 0 104 = 0 91V/V
v π = v sig r e r e + R L + Rsig β + 1
v sig =ˆ v π 1 + R L r e + Rsig (β + 1) r e
v sig = 5 1 + 1000 5 + 10,000 101 × 5 = 1 1V/V
Correspondingly,
ˆ v o = G v × 1 1 = 0 91 × 1 1 = 1V
Ex:7.31 Rsig Rin vsig (b2 1) (re2 RL) vo RL Rout re1 Q1 Q2
Fromthefigurewecanwrite
Rin = (β1 + 1) [r e 1 + (β2 + 1)(r e 2 + R L )]
Rout = R L r e 2 + r e 1 + Rsig /(β1 + 1) β2 + 1
v o v sig = R L R L + r e 2 + r e 1 + Rsig /(β1 + 1) β2 + 1
For I E 2 = 5 mA, β1 = β2 = 100, R L = 1 k , and Rsig = 100 k ,weobtain
r e 2 = 25 mV 5 mA = 5
I E 1 = 5 β2 + 1 = 5 101 0 05 mA
r e 1 = 25 mV 0 05 mA = 500
Rin = 101 × (0 5 + 101 × 1 005) = 10 3 M
Rout = 1 0 005 + 0 5 + (100/101) 101 20
v o v sig = 1 1 + 0 005 + 0.5 + (100/101) 101 = 0 98 V/V
Ex:7.32 I D = 1 2 k n W L ( VG S Vt )2
Exercise7–8
0.5 = 1 2 × 1( VG S 1)2
⇒ VG S = 2V
If Vt = 1 5V,then
I D = 1 2 × 1 × (2 1 5)2 = 0 125mA
⇒ I D I D = 0 125 0 5 0 5 =−0 75 =−75%
Ex:7.33
R D = V DD V D I D = 5 2 0 5 = 6k
→ R D = 6 2k
I D = 1 2 k n W L V 2 OV ⇒ 0 5 = 1 2 × 1 × V 2 OV
⇒ V OV = 1V
⇒ VG S = V OV + Vt = 1 + 1 = 2V
⇒ VS =−2V
R S = VS VSS I D = 2 ( 5) 0 5 = 6k
→ R S = 6 2k
Ifwechoose R D = R S = 6 2k ,then I D will changeslightly:
I D = 1 2 × 1 × ( VG S 1)2
Also,
VG S =− VS = 5 R S I D
Thus,
2 I D = (4 6 2 I D )2
⇒ 38 44 I 2 D 51 6 I 2 D + 16 = 0
⇒ I D = 0.49mA, 0.86mA
I D = 0 86 resultsin VS > 0 or VS > VG ,whichis notacceptable.Therefore I D = 0 49mA and VS =−5 + 6 2 × 0 49 =−1 96V
V D = 5 6 2 × 0 49 =+1 96V
R G shouldbeselectedintherangeof 1M to 10M
Ex:7.34
I D = 0 5mA = 1 2 k n W L V 2 OV
⇒ V 2 OV = 0.5 × 2 1 = 1
⇒ V OV = 1V ⇒ VG S = 1 + 1 = 2V
= V D ⇒ R D = 5 2 0 5 = 6k
⇒ R D = 6 2k (standardvalue).Forthis R D
wehavetorecalculate I D :
I D = 1 2 × 1 × ( VG S 1)2
= 1 2 ( V DD R D I D 1)2
( VG S = V D = V DD R D I D )
I D = 1 2 (4 6 2 I D )2 ⇒ I D ∼ = 0 49mA
V D = 5 6 2 × 0 49 = 1 96V
Ex:7.35 RefertoExample7.12.
(a)Fordesign1, R E = 3k , R1 = 80k ,and R2 = 40k .Thus, V BB = 4V.
I E = V BB V BE R E + R1 R2 β + 1
Forthenominalcase, β = 100 and
I E = 4 0.7 3 + 40 80 101 = 1 01 1mA
For β = 50,
I E = 4 0 7 3 + 40 80 51 = 0 94mA
For β = 150,
I E = 4 0 7 3 + 40 80 151 = 1 04mA
Thus, I E variesoverarangeapproximately10% ofthenominalvalueof1mA.
(b)Fordesign2, R E = 3 3k , R1 = 8k ,and R2 = 4k .Thus, V BB = 4V.Forthenominal case, β = 100 and
I E = 4 0 7
3 3 + 4 8 101 = 0 99 1mA
For β = 50,
I E = 4 0 7
3.3 + 4 8 51 = 0 984mA
For β = 150,
I E = 4 0 7
3 3 + 4 8 151 = 0.995mA
Thus, I E variesoverarangeof1.1%ofthe nominalvalueof1mA.Notethatlowering theresistancesofthevoltagedividerconsiderably decreasesthedependenceonthevalueof β ,a
highlydesirableresultobtainedattheexpense ofincreasedcurrentandhencepower dissipation.
Ex:7.36 RefertoFig.7.55.Sincethecircuitisto beusedasacommon-baseamplifier,wecan dispensewith R B altogetherandgroundthebase; thus R B = 0.Thecircuittakestheformshownin thefigurebelow.
10 V
RC
5 V vo vi
RE
Toestablish I E = 1mA,
I E = 5 V BE R E
1mA = 5 0.7 R E
⇒ R E = 4 3k
Thevoltagegain v o v i = gm RC ,where gm = IC VT = 40mA/V
Tomaximizethevoltagegain,weselect RC as largeaspossible,consistentwithobtaininga ±2-Vsignalswingatthecollector.Tomaintain active-modeoperationatalltimes,thecollector voltageshouldnotbeallowedtofallbelowthe valuethatcausestheCBJtobecomeforward biased,namely, 0.4V.Thus,the lowestpossibledcvoltageatthe collectoris 0 4V + 2V =+1 6V Correspondingly,
RC = 10 1 6 IC 10 1 6 1mA = 8 4k
Ex:7.37 RefertoFig.7.56.For I E = 1mA and VC = 2 3V,
I E = VCC VC RC
1 = 10 2.3 RC
⇒ RC = 7 7k
Now,usingEq.(7.147),weobtain
I E = VCC V BE
RC + R B β + 1
1 = 10 0 7
7 7 + R B 101
⇒ R B = 162k
Selectingstandard5%resistors(AppendixJ),we use
R B = 160k and RC = 7 5k
Theresultingvalueof I E isfoundas
I E = 10 0 7
7 5 + 160 101 = 1.02mA
andthecollectorvoltagewillbe
VC = VCC I E RC = 2 3V
Ex:7.38 RefertoFig.7.57(b).
VS = 3 5Vand I D = 0 5mA; thus
R S = VS I D = 3 5 0 5 = 7k
V DD = 15Vand V D = 6V; thus
R D = V DD V D I D = 15 6 0 5mA = 18k
Toobtain V OV ,weuse
I D = 1 2 k n V 2 OV
0 5 = 1 2 × 4 V 2 OV
⇒ V OV = 0 5V
Thus,
VG S = Vt + V OV = 1 + 0 5 = 1 5V
Wenowcanobtainthedcvoltagerequiredatthe gate,
VG = VS + VG S = 3 5 + 1 5 = 5V
Usingacurrentof 2 µA inthevoltagedivider,we have
R G 2 = 5V 2 µA = 2 5M
Thevoltagedropacross R G 1 is10V,thus
R G 1 = 10V 2 µA = 5M
Thiscompletesthebiasdesign.Toobtain gm and r o ,weuse
gm = 2 I D V OV = 2 × 0 5 0 5 = 2mA/V
r o = V A I D = 100 0 5 = 200k
Ex:7.39 RefertoFig.7.57(a)and(c)andtothe valuesfoundinthesolutiontoExercise7.38 above.
Rin = R G 1 R G 2 = 5 2 5 = 1 67M
Ro = R D r o = 18 200 = 16 5k
G v =− Rin Rin + Rsig gm (r o R D R L )
=− 1 67 1 67 + 0 1 × 2 × (200 18 20) =−17 1V/V
Ex:7.40 Toreduce v gs tohalfitsvalue,the unbypassed Rs isgivenby
Rs = 1 gm
FromthesolutiontoExercise7.38above, gm = 2mA/V.Thus
Rs = 1 2 = 0 5k
Neglecting r o , G v isgivenby
G v =− Rin Rin + Rsig ×− R D R L 1 gm + Rs
=− 1 67 1.67 + 0.1 × 18 20 0.5 + 0.5
=−8.9V/V
Ex:7.41 RefertoFig.7.58(a).For V B = 5V and 50-µA currentthrough R B 2 ,wehave
R B 2 = 5V 0 05mA = 100k
Thebasecurrentis
I B = I E β + 1 0.5mA 100 = 5 µA
Thecurrentthrough R B 1 is
I R B 1 = I B + I R B 2 = 5 + 50 = 55 µA
Sincethevoltagedropacross R B 1 is VCC V B = 10V,thevalueof R B 1 canbe foundfrom
R B 1 = 10V 0 055 µA = 182k
Thevalueof R E canbefoundfrom
I E = V B V BE R E
⇒ R E = 5 0 7 0 5 = 8 6k
Thevalueof RC canbefoundfrom VC = VCC IC RC
6 = 15 0 99 × 0 5 × RC
RC 18k
Thiscompletesthebiasdesign.Thevaluesof gm , r π ,and r o canbefoundasfollows:
gm = IC VT 0 5mA 0 025V = 20mA/V
r π = β gm = 100 20 = 5k
r o = V A IC 100 0 5 = 200k
Ex:7.42 RefertoFig.7.58(b)andtothesolution ofExercise7.41above.
Rin = R B 1 R B 2 r π = 182 100 5 = 4 64k
Ro = RC r o = 18 200 = 16 51k
G v =− Rin Rin + Rsig gm ( RC R L r o )
G v =− 4.64 4 64 + 10 × 20 × (18 20 200)
=−57 3V/V
Ex:7.43 RefertothesolutionsofExercises7.41 and7.42above.With Re included(i.e.,left unbypassed),theinputresistancebecomes[refer toFig.7.59(b)]
Rin = R B 1 R B 2 [(β + 1)(r e + Re )]
Thus,
10 = 182 100 [101(0 05 + Re )]
wherewehavesubstituted r e = VT I E =
25 0 5 = 50 .Thevalueof Re isfoundfromthe equationabovetobe
Re = 67 2
Theoverallvoltagegaincanbefoundfrom
G v =−α Rin Rin + Rsig RC R L r e + Re
G v =−0 99 × 10 10 + 10 18 20 0 05 + 0 0672 =−40V/V
Ex:7.44 RefertoFig.7.60(a).
Rin = 50 = r e R E r e
r e = 50 = VT I E
⇒ I E = 0 5mA
IC = α I E I E = 0 5mA
VC = VCC RC IC
For VC = 1V and VCC = 5V,wehave
1 = 5 RC × 0.5
⇒ RC = 8k
Toobtaintherequiredvalueof R E ,wenotethat thevoltagedropacrossitis
( V EE V BE ) = 4 3V.Thus,
R E = 4 3 0 5 = 8 6k
G v = Rin Rin + Rsig gm ( RC R L ) = 50 50 + 50 × 20(8 8)
= 40V/V
ˆ v o = 40 ˆ v sig = 40 × 10mV = 0 4V
Ex:7.45 RefertoFig.7.61.Considerfirst thebiasdesignofthecircuitinFig.7.61(a).
Sincetherequired I E = 1mA,thebasecurrent
I B = I E β + 1 = 1 101 0 01mA.Foradcvoltage
dropacross R B of1V,weobtain
R B = 1V 0 01mA = 100k
Theresultisabasevoltageof–1Vandan emittervoltageof–1.7V.Therequiredvalueof R E cannowbedeterminedas
R E = 1.7 ( 5) I E = 3.3 1mA = 3 3k
Rin = R B (β + 1)r e + ( R E r o R L )
where r o = V A IC = 100V 1mA = 100k
Rin = 100 (100 + 1)0 025 + (3 3 100 1)
= 44 3k
v i v sig = Rin Rin + Rsig = 44 3 44 3 + 50 = 0 469V/V
v o v i = R E r o R L
r e + ( R E r o R L ) = 0 968V/V
G v ≡ v o v sig = 0.469 × 0.968 = 0.454V/V
Rout = r o R E r e + R B Rsig β + 1
= 100 3 3 0 025 + 100 50 101
= 320
Chapter8
SolutionstoExerciseswithintheChapter
Ex:8.1 InthecurrentsourceofExample8.1
(Fig.8.1)wehave I O = 100 µA andwewantto reducethechangeinoutputcurrent,
I O , correspondingtoa1-Vchangeinoutput voltage, V O , to1% of I O
Thatis, I O = V O r o2 = 0 01 I O ⇒ 1V r o2
= 0 01 × 100 µA = 1 µA
r o2 = 1V 1 µA = 1M
r o2 = V A × L I O ⇒ 1M = 20 × L 100 µA
⇒ L = 100V 20V/ µm = 5 µm
Tokeep VOV ofthematchedtransistorsthesame asthatinExample8.1, W L ofthetransistor shouldremainthesame.Therefore,
W 5 µm = 10 µm 1 µm ⇒ W = 50 µm
Sothedimensionsofthematchedtransistors Q1 and Q2 shouldbechangedto W = 50 µmand L = 5 µm
Ex:8.2 ForthecircuitofFig.8.4wehave
I2 = IREF ( W/ L )2 ( W/ L )1 , I3 = IREF ( W/ L )3 ( W/ L )1 and I5 = I4 ( W/ L )5 ( W/ L )4
Sinceallchannellengthsareequal,thatis,
L 1 = L 2 =···= L 5 = 1 µm and
IREF = 10 µA, I2 = 60 µA, I3 = 20 µA, I4 = I3 = 20 µA, and I5 = 80 µA, wehave
I2 = IREF W2 W1 ⇒ W2 W1 = I2 IREF = 60 10 = 6
I3 = IREF W3 W1 ⇒ W3 W1 = I3 IREF = 20 10 = 2
I5 = I4 W5 W4 ⇒ W5 W4 = I5 I4 = 80 20 = 4
Toallowthevoltageatthedrainof Q2 togo downtowithin0.2Vofthenegativesupply voltage,weneed V OV 2 = 0 2V: I2 = 1 2 μn C ox W L 2 V 2 OV 2 = 1 2 k n W L 2 V 2 OV 2
60 µA = 1 2 200 µA V2 W L 2 (0 2)2 ⇒ W L 2 = 120 200 × (0 2)2 = 15 ⇒ W2 = 15 × L 2 W2 = 15 µm, W2 W1 = 6 ⇒ W1 = W2 6 = 2 5 µm W3 W1 = 2 ⇒ W3 = 2 × W1 = 5 µm
Toallowthevoltageatthedrainof Q5 togoupto within0.2Vofpositivesupply,weneed V OV 5 = 0 2V:
I5 = 1 2 k p W L 5 V 2 OV 5
80 µA = 1 2 80 µA V2 W L 5 (0 2)2 ⇒ W L 5 = 2 × 80 80 × (0 2)2 = 50 ⇒ W5 = 50 L5
W5 = 50 µm W5 W4 = 4 ⇒ W4 = 50 µm 4 = 12 5 µm
Thus:
W1 = 2.5 µm, W2 = 15 µm, W3 = 5 µm W4 = 12 5 µm, and W5 = 50 µm
Ex:8.3 V BE = VT ln( I REF / I S ) = 0 025ln(10 3 /10 15 ) = 0 691V
FromEq.(8.21)wehave
I O = IREF ⎛ ⎜ ⎜ ⎝ m 1 + m + 1 β ⎞ ⎟ ⎟ ⎠ 1 + V O VBE V A 2
I O = 1mA⎛ ⎜ ⎝ 1 1 + 1 + 1 100 ⎞ ⎟ ⎠ 1 + 5 0 691 100 = 1 02mA
I O = 1 02mA Ro = ro2 = V A I O = 100V 1 02mA = 98k 100k
Ex:8.4
FromEq.(8.23),wehave
I O = IREF 1 + (2/β ) 1 + V O VBE V A
Q1 R Q2 VCC IREF IO
where V BE = VT ln I O I S = 0 025ln 0 5 × 10 3 10 15 = 0 673V
0 5mA = IREF 1 + (2/100) 1 + 2 0 673 50 ⇒
IREF = 0 5mA 1 02 1 0265 = 0 497mA
IREF = VCC VBE R ⇒ R = VCC VBE IREF
R = 5 0 673 0 497mA = 8 71k
V O min = VCE sat = 0 3V
For V O = 5 V,FromEq.(8.23)wehave
I O = IREF 1 + (2/β ) 1 + V O VBE V A
I O = 0 497 1 + (2/100) 1 + 5 0 673V 50 = 0 53mA
Ex:8.5 I1 = I2 =···= I N = IC | Q REF
Attheinputnode,
IREF = IC | Q REF + I B | Q REF + I B 1 +···+ I BN
= IC | Q REF + ( N + 1) I B | Q REF
= IC | Q REF + ( N + 1) β IC | Q REF ⇒ IC | Q REF = IREF 1 + N + 1 β
Thus, I1 = I2 =···= I N = IREF 1 + N + 1 β Q.E.D For β = 100,tolimittheerrorto10%, 0 1 = N + 1 β = N + 1 100
⇒ N = 9
Ex:8.6 Rin 1 gm 1
Now, Rin = 1 k ,thus gm 1 = 1 mA/V
But
gm 1 = 2(μn C ox ) W L 1 I D 1
1 = 2 × 0 4 × W L 1 × 0 1 ⇒ W L 1 = 12.5
Toobtain
A is ideal = 5 5 = ( W/ L )2 ( W/ L )1 ⇒ W L 2 = 5 × 12 5 = 62 5
Ro = r o2 = V A 2 I D 2 = V A 2 5 I D 1
Thus,
40 k = V A 2 5 × 0 1 ⇒ V A 2 = 20 V
But
V A 2 = V A 2 L 2 20 = 20 × L 2 ⇒ L 2 = 1 µm
Selecting L 1 = L 2 ,then L 1 = L 2 = 1 µm W1 = 12.5 µm W2 = 62 5 µm
Theactualshort-circuitcurrent-transferratiois givenbyEq.(8.31),
A is = 5 1 + 1 gm 1 r o1
Thusthepercentageerroris
Error = 1/ gm 1 r o1 1 + (1/ gm 1 r o1 ) × 100%
Substituting, gm 1 = 1 mA/V
r o1 = V A 1 I D 1 = V A L 1 I D 1 = 20 × 1 0 1 = 200 k
Error = 1/200 1 + (1/200) × 100 =−0 5%
Ex:8.7
UsingEq.(8.42):
gm = 2μn C ox W L I D
For I D = 10 µA,wehave
gm = 2(387 µA/V2 )(10)(10 µA)
= 0.28mA/V
UsingEq.(8.46):
A 0 = V A 2μn C ox ( W/ L ) √ I D
= 5V/µm 2(387 µA/V2 )(10)(0 36)2 √10 µA
A 0 = 50V/V
Since gm varieswith I D and A0 with 1 √ I D , for I D = 100 µA ⇒ gm = 0 28mA/V 100 10 1/2
= 0 88mA/V
A 0 = 50 10 100 1/2 = 15 8V/V
For I D = 1mA,wehave
gm = 0 28mA/V 1 0.010 1/2 = 2 8mA/V
A 0 = 50 0 010 1 1/2 = 5V/V
Ex:8.8
Sincealltransistorshavethesame
W L = 7 2 µm 0 36 µm , wehave
IREF = I D 3 = I D 2 = I D 1 = 100 µA gm 1 = 2μn C ox W L 1 I D 1 = 2 387 µA/V2 7 2 0 36 (100 µA)
= 1 24mA/V
r o1 = V An L 1 I D 1 = 5V/µm (0 36 µm) 0 1mA = 18k
r o2 = | V Ap | L 2 I D 2 = 6V/µm (0 36 µm) 0 1mA = 21 6k
Voltagegainis
A v =− gm 1 (r o1 r o2 )
A v =− (1 24mA/V)(18k 21 6k ) =−12.2V/V
Ex:8.9 VCC vo vi Rin VBIAS Q1 Q2 I
IC 1 = I = 100 µA = 0.1mA
gm 1 = IC 1 VT = 0 1mA 25mV = 4mA/V
Rin = r π 1 = β 1 gm 1 = 100 4mA/V = 25k
r o1 = V A I = 50V 0 1mA = 500k
r o2 = | V A | I = 50V 0 1mA = 500k
A 0 = gm 1 r o1 = (4mA/V)(500k ) = 2000V/V
A v =− gm 1 (r o1 r o2 ) =−(4mA/V) × (500k 500k ) =−1000V/V
Ex:8.10 RefertoFig.1, vsig Rsig RL i i Rin vo
vo = iR L v sig = i ( Rs + Rin )
Thus,
vo v sig = R L Rs + Rin Q.E.D
Ex:8.11 Since gm r o 1,weuseEq.(8.52),
Rin 1 gm + R L gm r o
R L 0 r o ( gm r o )r o ∞
Rin 1 gm 2 gm r o ∞
Ex:8.12 For gm r o 1,weuseEq.(8.57),
Rout r o + ( gm r o ) Rs toobtain
Rs 0 r o ( gm r o )r o ∞
Rout r o ( gm r o )r o ( gm r o )2 r o ∞
Ex:8.13 Substituting Rs = r o inEq.(8.53)gives
A is = 1 + 1 gm r o 1 + 2 gm r o
Substitutingthegivenvaluesof ( gm r o ) weobtain
gm r o 20 50 100
A is (A/A) 0.95 0.98 0.99
Ex:8.14 A v o remainsunchangedat gm r o .Witha loadresistance R L connected,
A v = A v o R L R L + Ro = ( gm r o ) R L R L + (1 + gm Rs )r o
Ex:8.15 Since gm r o 1 weuseEq.(8.62),
Rin r π 1 gm + R L gm r o For R L = 0,
Rin r π 1 gm = r e
For R L = r o ,
Rin = r π ( 1 gm + 1 gm ) = r π 2 gm 2r e
For R L = β r o ,
Rin = r π ( 1 gm + β r o gm r o ) = r π ( 1 gm + r π )
r π r π = 1 2 r π
For R L =∞,
Rin = r π
Summary:
R L 0 r o β r o ∞
Rin r e 2r e 1 2 r π r π
Ex:8.16 UsingEq.(8.69),
Rout r o + ( gm r o )( Re r π ) weobtain
Re 0 r e r π r o ∞
Rout r o 2r o β 2 + 1 r o (β + 1)r o (β + 1)r o
Ex:8.17 UsingEq.(8.65)with Re = r o ,
A is = 1 + 1 gm r o 1 + 1 gm r o + 1 gm (r o r π ) = 1 + 1
gm r o 1 + 2 gm r o + 1 gm r π = 1 + 1 gm r o 1 + 2 gm r o + 1 β
For β = 100 and gm r o = 100,
A is = 1 + 1 100 1 + 2 100 + 1 100 = 0 98 A/A
For β = 100 and gm r o = 1000,
A is = 1 + 1 1000 1 + 2 1000 + 1 100 = 0 99 A/A
Ex:8.18 Ro = 1 + gm ( Re r π )r o where
gm = 4 mA/V, r π = β gm = 25 k ,
Re = 1 k , and r o = V A IC = 20 0 1 = 200 k
Thus,
Ro = [1 + 4(1 25)] × 200 1 M
Withoutemitterdegeneration,
Ro = r o = 200 k
Ex:8.19 If L ishalved L = 0 55 µm 2 and | V A | = V A · L ,weobtain
| V A |= 5V/µm 0 55 µm 2 = 1 375V
Ro = | V A | | VOV |/2 · | V A | I D = 2 (1 375V)2 (0 3V)(100 µA) = 126k
Since I D = 1 2 μ p C ox W L | VOV |2 1 + VSD | V A |
W L = 2 (100 µA) 90 µA/V2 (0.3V)2 1 + 0 3V 1 375V W L = 20 3
Ex:8.20 VG4 1.1 V VDD 1.8 V
Ifalltransistorsarematchedandareobviously operatingatthesame I D ,thenall | VOV | willbe
equalandequaltothatof Q 1 ,namely, | V OV | = 0 7 0 5 = 0 2V
Tokeep Q 2 insaturation,
v O min = VG 2 −| Vt |= 0 5V
Tokeep Q 3 insaturation,
v O max = VG 3 +| Vt |= 1 3V
Thus,theallowablerangeof v O isfrom0.5Vto 1.3V.
Ex:8.21 RefertoFig.8.31.
gm 1 = gm 2 = gm 3 = gm 4 = 2 I D | VOV | = 2 × 0 2 0 2 = 2 mA/V
r o1 = r o2 = r o3 = r o4 = | V A | I D = 2 0 2 = 10 k
Ron = ( gm 2 r o2 )r o1 = (2 × 10) × 10 = 200 k
Rop = ( gm 3 r o3 )r o4 = (2 × 10) × 10 = 200 k
Ro = Ron Rop = 200 200 = 100 k
A v =− gm 1 Ro =−2 × 100 =−200 V/V
Ex:8.22 gm 1 = gm 2 = gm = I D V OV 2 = 0 1mA (0 2/2) V = 1mA/V
r o1 = r o2 = r o
= V A I D = 2V 0.1mA = 20k
so, gm r o = 1mA/V (20k ) = 20
(a)For R L = 20k ,
Rin2 = R L + r o2 1 + gm 2 r o2 = 20k + 20k 1 + 20 = 1 9k
∴ A v 1 =− gm 1 (r o1 Rin2 ) =−1mA/V (20 1 9) =−1 74V/V or IfweusetheapproximationofEq.(8.84),
Rin2 ≈ R L gm 2 r o2 + 1 gm 2 = 20k 20 + 1 1mA/V = 2k then
A v1 =−1mA/V (20k 2k ) =−1 82V/V Continuing,fromEq.(8.81),
A v =− gm 1 ( gm 2 r o2 r o1 ) R L
A v =−1mA/V {[(20)(20k )] 20k }
=−19 0V/V
A v 2 = A v A v 1 = 19 0 1 82 = 10 5V/V
(b)Now,for R L = 400k ,
Rin2 R L gm 2 r o2 + 1 gm 2 = 400k 20 + 1 1mA/V
= 21k
A v 1 =−1mA/V (20k 21k ) =−10 2V/V
A v =−1mA/V [(20)(20k )] 400k
=−200V/V
A v 2 = A v A v 1 = 200 10 2 = 19 6V/V
Ex:8.23 ReferringtoFig.8.34,
Rop = ( gm 3 r o3 )(r o4 r π 3 ) and
Ron = ( gm 2 r o2 )(r o1 r π 2 )
Themaximumvaluesoftheseresistancesare obtainedwhen r o r π andaregivenby
Ron max = ( gm 2 r o2 ) r π 2
Rop max = ( gm 3 r o3 ) r π 3
Since gm r π = β,
Ron max = β 2 r o2
Rop max = β 3 r o3
Since A v =− gm 1 Ron Rop ,
| A v max | = gm 1 β 2 r o2 β 3 r o3
Ex:8.24 Forthe npn transistors,
gm 1 = gm 2 = | IC | | VT | = 0 2mA 25mV = 8mA/V
r π 1 = r π 2 = β gm = 100 8mA/V = 12 5k
r o1 = r o2 = | V A | | IC | = 5V 0 2mA = 25k
FromFig.8.34,
Ron = ( gm 2 r o2 )(r o1 r π 2 )
= (8mA/V)(25k )(25k 12 5k )
Ron = 1 67M
Forthe pnp transistors,
gm 3 = gm 4 = | IC | VT = 0 2mA 25mV = 8mA/V
r π 3 = r π 4 = β gm = 50 8mA/V = 6 25k
r o3 = r o4 = | V A | | IC | = 4V 0 2mA = 20k
Rop = ( gm 3 r o3 )(r o4 r π 3 )
= (8mA/V)(20k )(20k 6.25k )
Rop = 762k
A v =− gm 1 Ron Rop
=− (8mA/V)(1 67M 762k )
A v =−4186V/V
A v max occurswhen r o1 and r o4 are r π
Then
Ron = ( gm 2 r o2 ) r π 2 = β 2 r o2
Ron = 100 (25k ) = 2 5M
Rop = ( gm 3 r o3 ) r π 3 = β 3 r o3
Rop = 50 (20k ) = 1M
Finally,
A v max =−(8mA/V)(2 5M 1 0M )
A v max =−5714V/V
Ex:8.25
gm = 2 I D VOV = 2 × 0 2 0 2 = 2 mA/V
gm b = χ gm = 0 2 × 2 = 0 4 mA/V
r o1 = r o3 = V A I D = 5 0 2 = 25 k
R L = r o1 r o3 1 gm b = 25 25 2 5 k
= 2.083 k
v o v i = R L R L + 1 gm = 2 083 2 083 + 1 2 = 0 81 V/V
Toobtain Rout weuseEq.(8.95),
Rout = 1 gm 1 gm b r o1 r o3 = 0 5 2 5 25 25 = 0 403 k = 403
Ex:8.26 RefertothecircuitinFig.8.36.All transistorsareoperatingat I D = IREF = 100 µA andequal VOV ,foundfrom
I D = 1 2 μn C ox W L V 2 OV
100 = 1 2 × 387 × 3 6 0 36 × V 2 OV
⇒ VOV = 0 227 V
VGS = 0 227 + 0 5 = 0 727 V
V O min = VG 3 Vt 3
= VG S 4 + VG S 1 Vt 3
Thus,
V O min = 2 VGS Vt
= Vt + 2 VOV
= 0 5 + 2 × 0 227 = 0 95 V
gm = 2 I D VOV = 2 × 0 1 0 227 = 0.88 mA/V
r o = V A I D = V A L I D = 5 × 0 36 0 1 = 18 k
Ro = ( gm 3 r o3 )r o2 = (0.88 × 18) × 18 = 285 k
Ex:8.27 FortheWilsonmirrorfromEq.(8.97), wehave
I O IREF 1 1 + 2 β 2 = 0 9998
Thus | I O IREF | IREF × 100 =0.02%
whereasforthesimplemirrorfromEq.(8.18)we have
I O IREF = 1 1 + 2 β = 0 98
Hence | I O IREF | IREF × 100 = 2%
FortheWilsoncurrentmirror,wehave
Ro = β r o 2 = 100 × 100k 2 = 5M
andforthesimplemirror, Ro = r o = 100k .
Ex:8.28 Forthetwocurrentsourcesdesignedin
Example8.5,wehave
gm = IC VT = 10 µA 25mV = 0 4 mA V
r o = V A IC = 100V 10 µA = 10M ,
r n = β gm = 250k
ForthecurrentsourceinFig.8.40(a),wehave
Ro = r o2 = r o = 10M
ForthecurrentsourceinFig.8.40(b),from Eq.(8.105),wehave
Rout 1 + gm ( R E r n ) r o
FromExample8.5, R E = R3 = 11 5k ; therefore,
Rout 1 + 0 4 mA V (11 5k 250k ) 10M
∴ Rout = 54M
Chapter9
SolutionstoExerciseswithintheChapter
Ex:9.1 ReferringtoFig.9.3, If R D isdoubledto5k ,
V D 1 = V D 2 = V DD I 2 R D
= 1 5 0 4mA 2 (5k ) = 0 5V
VCMmax = Vt + V D = 0.5 + 0.5 =+1.0V
Sincethecurrents I D 1 ,and I D 2 arestill0.2mA each,
VG S = 0 82V
So, VCMmin = VSS + VCS + VG S =−1 5V + 0 4V + 0 82V =−0 28V
So,thecommon-moderangeis 0 28V to +1.0V
Ex:9.2 (a)Thevalueof v id thatcauses Q 1 to conducttheentirecurrentis √2 V OV
→ √2 × 0 316 = 0 45V then, V D 1 = V DD I × R D = 1.5 0.4 × 2.5 = 0.5V
V D 2 = V DD =+1 5V
v O = V D 2 V D 1 =+1V
(b)For Q 2 toconducttheentirecurrent:
v id =−√2 V OV =−0 45V then,
V D 1 = V DD =+1 5V
V D 2 = 1 5 0 4 × 2 5 = 0 5V
v O = V D 2 V D 1 =−1V
(c)Thusthedifferentialoutputrangeis +1Vto 1V.
Ex:9.3 RefertoanswertableforExercise9.3 wherevalueswereobtainedinthefollowingway:
I 2 = 1 2 k n W L V 2 OV ⇒ W L = I k n V 2 OV = 2 V 2 OV
gm = 2( I /2) V OV = I V OV = 0 4 V OV mA/V
v id /2 V OV 2 = 0 1 → v id = 2 V OV √0 1 = 0 632 V OV
Ex:9.4 I D = I 2 = 0 8mA 2 = 0 4mA
I D = 1 2 k n W L ( V OV )2 RD 5 k RD 5 k I 0.8 mA VSS VDD ID ID
Thus, V OV = 2 I D k n W L = 2 (0 4mA) 0.2 mA/V2 (100) = 0 2V
gm = I D V OV /2 = 0 4mA × 2 0 2V = 4mA/V
r o = V A I D = 20V 0.4mA = 50k
A d = gm ( R D r o )
A d = (4mA/V)(5k 50k ) = 18 2V/V
Ex:9.5 With I = 200 µA,foralltransistors,
I D = I 2 = 200 µA 2 = 100 µA
L = 2(0 18 µm) = 0 36 µm
r o1 = r o2 = r o3 = r o4 = V A L I D
= (10V/ µm)(0 36 µm) 0 1mA = 36k
Since I D 1 = I D 2 = 1 2 μn C ox W L V 2 OV , W L 1 = W L 2 = 2 I D μn C ox V 2 OV
2(100 µA)
400 µA/V2 (0 2V)2 = 12.5
W
L 3 = W L 4 = 2 I D μ p C ox | V OV |2
2(100 µA)
100 µA/V2 (0 2)2 = 50
gm = I D V OV /2 = (100 µA)(2) 0 2V = 1mA/V,
so
A d = gm 1 (r o1 r o3 ) = 1(mA/V)(36k 36k ) = 18V/V
Ex:9.6 L = 2 (0 18 µm) = 0 36 µm
All r o = V A · L I D
Thedraincurrentforalltransistorsis
I D = I 2 = 200 µA 2 = 100 µA
r o = (10V/ µm)(0.36 µm) 0 1mA = 36k
ReferringtoFig.9.13(a),
Since I D = 1 2 μn C ox W L V 2 OV forallNMOS transistors,
W L 1 = W L 2 = W L 3 = W L 4
= 2 I D μn C ox V 2 OV = 2(100 µA) 400 µA/V2 (0 2V)2 = 12 5
W L 5 = W L 6 = W L 7 = W L 8
= 2 I D μ p C ox V 2 OV = 2(100 µA) 100 µA/V2 (0.2V)2 = 50
Foralltransistors,
gm = | I D | | V OV | /2 = (0.1mA)(2) (0 2V) = 1mA/V
FromFig.9.13(b),
Ron = ( gm 3 r o3 ) r o1 = (1 × 36) × 36 = 1.296M
Rop = ( gm 5 r o5 )r o7 = (1 × 36) × 36
= 1 296M
A d = gm 1 Ron Rop
= (1mA/V)(1 296M 1 296M ) = 648V/V
Ex:9.7
Ex:9.8
IC2
2.5 V RC 5 k IC1 RC
2.5 V VEE VCC I 0.4 mA
IC 1 = IC 2 I E 1 = I E 2 = I 2 = 0 4mA 2 = 0 2mA
VCM max VC + 0 4V
= VCC IC RC + 0 4V
= 2 5 0 2mA (5k ) + 0 4V =+1 9V
VCM min =− V EE + VCS + V BE
VCM min =−2 5V + 0 3V + 0 7V =−1 5V
Inputcommon-moderangeis 1 5V to +1 9V
Ex:9.9 Substituting i E 1 + i E 2 = I inEq.(9.45) yields
i E 1 = I 1 + e (v B 2 v B 1 )/ VT
0 99 I = I 1 + e (v B 2 v B 1 )/ VT
v B 1 v B 2 =− VT ln 1 0 99 1 =−25ln (1/99)
= 25ln (99) = 115mV
Ex:9.10 (a)TheDCcurrentineachtransisteris
0.5mA.Thus V BE foreachwillbe
V BE = 0.7 + 0.025ln 0 5 1
= 0.683V
⇒ v E = 5 0 683 =+4 317V
(b) gm = IC VT = 0 5 0 025 = 20 mA V
(c) i C 1 = 0 5 + gm 1 v BE 1
= 0 5 + 20 × 0 005sin (2π × 1000t )
= 0 5 + 0 1sin(2π × 1000t ), mA
i C 2 = 0 5 0 1sin(2π × 1000t ), mA
(d) v C 1 = ( VCC IC RC ) 0 1
× RC sin (2π × 1000t )
= (15 0 5 × 10) 0 1 × 10sin (2π × 1000t )
= 10 1sin(2π × 1000t ) , V
v C 2 = 10 + 1sin(2π × 1000t ) , V
(e) v C 2 v C 1 = 2 sin(2π × 1000t ) , V
(f)Voltagegain ≡ v C 2 v C 1 v B 1 v B 2
= 2Vpeak 0 01Vpeak = 200 V/V
Ex:9.11 Thetransconductanceforeach transistoris
gm = 2μn C ox ( W/ L ) I D
I D = I 2 = 0 8mA 2 = 0 4mA
Thus,
gm = √2 × 0 2 × 100 × 0 4 = 4mA/V
Thedifferentialgainformatched
R D valuesis A d = v O 2 v O 1 v id = gm R D
Ifweignorethe1%here,thenweobtain
A d = gm R D = (4mA/V)(5k ) = 20V/V
| A cm |= R D 2 R SS R D R D = 5 2 × 25 (0 01) = 0 001V/V
CMRR (dB) = 20log | A d | | A CM | = 20log 20 0.001 = 86dB
Ex:9.12 FromExercise9.11,
gm = 4mA/V
UsingEq.(9.85)andthefactthat R SS = 25k , weobtain
CMRR = (2gm R SS ) gm gm = 2(4mA/V)(25k ) 0.01
= 20, 000
CMRR (dB) = 20log10 (20, 000) = 86dB
Ex:9.13 IftheoutputofaMOSdifferential amplifieristakensingle-endedly,then
| A d |= 1 2 gm R D
(thatis,halfthegainobtainedwiththeoutput takendifferentially),andfromEq.(9.74)wehave
| A cm | R D 2 R SS
Thus,
CMRR ≡ | A d | | A cm | = gm R SS
Substituting gm = 4 mA/V, R D = 5 k ,and R SS = 25 k , | A d |= 10V/V
| A cm |= 0 1V/V
CMRR = 100 or 40 dB
Ex:9.14 VCC
i1 VB3 VB5 VEE Q5 Q1 Q2 Q3 Q4 VB4 vo2 vi2 I 200 A vo1 I = 200 µA Since β 1, IC 1 ≈ IC 2 ≈ I 2 = 200 µA 2 = 100 µA
gm 1 = gm 2 = gm = IC VT = 100 µA 25mV = 4mA/V RC 1 = RC 2 = RC = r o = | V A | IC = 10V 100 µA = 100k
r o1 = r o2 = V A I /2 = 10 0 1 = 100k
r e 1 = r e 2 = r e = VT I E = 25mV 0 1mA = 0 25k
| A d |= RC r o r e = 100k 100k 0 25k = 200V/V
Exercise9–4
Rid = 2r π , r π = β gm = 100 4mA/V = 25k
Rid = 2(25k ) = 50k
R EE = V A I = 10V 200 µA = 50k
Ifthetotalloadresistanceisassumedtobe mismatchedby1%,thenwehave
| A cm |= RC 2 R EE RC RC = 100 2 × 50 × 0.01 = 0.01V/V
CMRR (dB) = 20log10 A d A cm = 20log10 200 0 01 = 86dB
UsingEq.(9.96),weobtain
Ricm = β R EE 1 + RC β r o 1 + RC + 2 R EE r o = 100 × 50 × 1 + 100 100 × 100 1 + 100 + 2 × 50 100
Ricm 1 68M
Ex:9.15 FromExercise9.4:
V OV = 0 2V
UsingEq.(9.97)weobtain V OS dueto R D / R D as:
V OS = V OV 2 · R D R D
= 0.2 2 × 0 02 = 0 002V i.e2mV
Toobtain V OS dueto k n , useEq.(9.98),
V OS = V OV 2 k n k n
⇒ V OS = 0 2 2 × 0 02 = 0 002
⇒ 2mV
Theoffsetvoltagearisingfrom Vt isobtained fromEq.(9.99):
V OS = Vt = 2mV
Finally,fromEq.(9.100)thetotalinputoffsetis
V O S
3 × 2 × 10 3 2 = 3 5mV
Ex:9.16 FromEq.(9.103),weget V OS = VT RC RC 2 + I S I S 2
= 25 (0 02)2 + (0 1)2 = 2 55mV I B = 100 2(β + 1) = 100 2 × 101 ∼ = 0 5 µA
I OS = I B β β
= 0 5 × 0 1 µA = 50nA
Ex:9.17 I D = 1 2 I = 0 4 mA
I D = 1 2 μn C ox W L n V 2 OV 0 4 = 1 2 × 0 2 × 100 × V 2 OV ⇒ V OV = 0 2 V
gm 1,2 = 2 I D V OV = 2 × 0 4 0 2 = 4 mA/V G m d = gm 1,2 = 4 mA/V
r o2 = V An I D = 20 0 4 = 50 k
r o4 = | V Ap | I D = 20 0 4 = 50 k
Ro = r o2 r o4 = 50 50 = 25 k
A d = G m Ro = 4 × 25 = 100 V/V
Ex:9.18 G m d = gm 1,2 I /2 VT = 0 4 mA 0 025 V = 16 mA/V
r o2 = r o4 = V A IC = V A I /2 = 100 0 4 = 250 k
Ro = r o2 r o4 = 250 250 = 125 k
A d = G m Ro = 16 × 125 = 2000 V/V
Rid = 2r π = 2 × β gm 1,2 = 2 × 160 16 = 20 k
Ex:9.19 FromExercise9.17,weget
I D = 0.4 mA
V OV = 0 2 V gm 1,2 = 4 mA/V
G m d = 4 mA/V A d = 100 V/V
Now,
R SS = 25 k
gm 3 = 2μ p C ox W L p I D
= √2 × 0 1 × 200 × 0 4 = 4 mA/V
| A cm |= 1 2 gm 3 R SS = 1 2 × 4 × 25 = 0 005 V/V
CMRR = | A d | | A cm | = 100 0 005
= 20,000 or 20 log 20,000 = 86 dB
Ex:9.20 RefertoFig.(9.37).
(a)UsingEq.(9.142),weobtain
I6 = ( W/ L )6 ( W/ L )4 ( I /2)
⇒ 128 = ( W/ L )6 11 1 × 64
thus, ( W/ L )6 = 22 2
UsingEq.(9.143),weget
I7 = ( W/ L )7 ( W/ L )5 I
⇒ 128 = ( W/ L )7 88 8 × 128
thus, ( W/ L )7 = 88 8
(b)For Q 1 ,
I 2 = 1 2 μ p C ox W L 1 V 2 OV1
⇒ V OV 1 = 64 1 2 × 128 × 44.4 = 0 15V
Similarlyfor Q 2 , V OV 2 = 0 15V
For Q 6 ,
128 = 1 2 × 512 × 22 2 V 2 OV 6
⇒ V OV 6 = 0 15V
(c) gm = 2 I D V OV
I D V OV gm
Q 1 64 µA0.15V0.853mA/V
Q 2 64 µA0.15V0.853mA/V
Q 6 128 µA0.15V1.71mA/V
(d)
r o2 = 3.2/0.064 = 50k
r o4 = 3 2/0 064 = 50k
r o6 = 3 2/0 128 = 25k
r o7 = 3 2/0 128 = 25k
(e)Eq.(9.140):
A 1 =− gm 1 (r o2 r o4 )
=−0 853 (50 50) =−21 3 V V
Eq.(9.141):
A 2 =− gm 6 (r o6 r o7 )
=−21.3V/V
Overallvoltagegainis
A 1 × A 2 =−21 3 ×− 21 3 = 454V/V
Ex:9.21 Rid = 20 2k
A v o = 8513V/V
Ro = 152
With R S = 10k and R L = 1k ,
G v = 20 2 20 2 + 10 × 8513 × 1 (1 + 0 152)
= 4943V/V
Ex:9.22 i e 8 i b8 = β 8 + 1 = 101A/A
i b8 i c 7 = R5 R5 + Ri 4 = 15 7 15 7 + 303 5 = 0 0492A/A i c 7 i b7 = β 7 = 100A/A
i b7 i c 5 = R3 R3 + Ri 3 = 3 3 + 234 8 = 0 0126A/A
i c 5 i b5 = β 5 = 100A/A
i b5 i c 2 = R1 + R2 R1 + R 2 + Ri 2 = 40 40 + 5 05 = 0 8879A/A
i c 2 i 1 = β 2 = 100A/A
Thustheoverallcurrentgainis
i e 8 i 1 = 101 × 0 0492 × 100 × 0 0126 × 100
× 0 8879 × 100
= 55,599A/A andtheoverallvoltagegainis
v o v id = R6 Ri 1 i e 8 i 1 = 3 20.2 × 55599 = 8257V/V
Exercise10–1
Chapter10
SolutionstoExerciseswithintheChapter
Ex:10.1 C ox = ox tox = 3 45 × 10 11 F/m 3 × 10 9 m
= 11 5 × 10 3 F/m2
= 11 5 fF/µm2
C ov = WL ov C ox
= 1 5 × 0 03 × 11 5 = 0 52 fF
C gs = 2 3 WLC ox + C ov
= 2 3 × 1.5 × 0.15 × 11.5 + 0.52
= 2 25 fF
C gd = C ov = 0 52 fF
C sb = C sb0 1 + VSB V0 = 1 1 + 0 2 0 9 = 0 9 fF
C db = C db0 1 + V DB V0 = 1 1 + 0 8 + 0 2 0 9 = 0 7 fF
Ex:10.2 gm = 2k n ( W/ L ) I D
= 2 × 0 5 × (1 5/0 15) × 0 1 = 1 mA/V
f T = gm
2π(C gs + C gd )
= 1 × 10 3
2π(2 25 + 0 52) × 10 15
f T = 57 5 GHz
Ex:10.3 C de = τ F gm where
τ F = 20 ps
gm = IC VT = 1 mA 0 025 V = 40 mA/V
Thus,
C de = 20 × 10 12 × 40 × 10 3 = 0 8 pF
C je 2C je 0
= 2 × 20 = 40 fF
C π = C de + C je
= 0.8 + 0.04 = 0.84 pF
C μ = C μ0 1 + VCB V0c m = 20 1 + 2 0 5 0 33 = 12 fF
f T = gm 2π(C π + C μ ) = 40 × 10 3 2π(0 84 + 0 012) × 10 12 = 7 47 GHz
Ex:10.4 |β |= 10 at f = 50 MHz
Thus,
f T = 10 × 50 = 500 MHz
C π + C μ = gm 2π f T
= 40 × 10 3
2π × 500 × 106 = 12.7 pF
C π = 12 7 2 = 10 7 pF
Ex:10.5 C π = C de + C je
10 7 = C de + 2
⇒ C de = 8.7 pF
Since C de isproportionalto gm andhence IC ,at
IC = 0 1 mA,
C de = 0 87 pF and
C π = 0 87 + 2 = 2 87 pF
f T = 4 × 10 3 2π(2 87 + 2) × 10 12 = 130 7 MHz
Ex:10.6 A M =− gm R L where
gm = 2 mA/V
R L = R L r o = 10 20 = 6 67 k
Thus,
A M =−2 × 6 67 =−13 3 V/V
C eq = (1 + gm R L )C gd
= (1 + 13 3) × 5 = 71 5 fF
C in = C gs + C eq = 20 + 71 5 = 91 5
f H = 1 2π C in Rsig
= 1 2π × 91 5 × 10 15 × 20 × 103 = 86 9 MHz
Ex:10.7 f H = 1 2π C in Rsig
For f H ≥ 100 MHz,
C in ≤ 1 2π × 100 × 106 × 20 × 103 = 79 6 fF
But,
C in = C gs + C eq = 20 + C eq
Thus,
C eq ≤ 59 6 fF
C gd (1 + gm R L ) ≤ 59 6
C gd ≤ 59 6 1 + 20 = 2 8 fF
Ex:10.8 RefertoExample10.2.
Tomove f t from10.6GHzto3GHz, f H mustbe movedfrom530.5MHzto
f H = 530 5 × 3 10.6 = 150 1 MHz
Since,
f H = 1 2π(C L + C gd ) R L then
C L + C gd = 1 2π × 150 1 × 106 × 10 × 103 = 106 fF
C L = 106 5 = 101 fF
Ex:10.9 Toreducethemidbandgaintohalfthe valuefound,wereduce R L bythesamefactor, thus
R L = 4 76 2 = 2 38 k
But,
R L = R L r o
2 38 = R L 100
⇒ R L = 2 44 k
C in = C π + C μ (1 + gm R L )
= 7 + 1(1 + 40 × 2 38)
= 103 2 pF
f H = 1 2π C in Rsig
= 1 2π × 103.2 × 10 12 × 1.67 × 103 = 923 MHz
Thus,byacceptingareductioningainbyafactor of2,thebandwidthisincreasedbyafactorof 923/480 = 1.9,approximathelythesamefactoras thereductioningain.
Ex:10.10 T (s ) = 1000 1 + s 2π × 105
GB = 1000 × 100 × 103 = 108 Hz
Ex:10.11 | A M |= gm R L = 2 × 10 = 20 V/V
GB =| A M | f H
= 20 × 56 8 = 1 14 GHz
Ex:10.12 | A M |= 1 2 × 20 = 10 V/V
R gs = 20 k
R gd = Rsig (1 + gm R L ) + R L
= 20(1 + 10) + 5 = 225 k
RC L = R L = 5 k
τ gs = C gs R gs = 20 × 10 15 × 20 × 103 = 400 ps
τ gd = C gd R gd = 5 × 10 15 × 225 × 103 = 1125 ps
τC L = C L RC L = 25 × 10 15 × 5 × 103 = 125 ps
τ H = τ gs + τ gd + τC L
= 400 + 1125 + 125
= 1650 ps
f H = 1 2πτ H
= 1 2π × 1650 × 10 12 = 96 5 MHz
GB = 10 × 96.5 = 0.965 GHz
Ex:10.13 gm = 2μn C ox W L I D
Since I D isincreasedbyafactorof4, gm doubles:
gm = 2 × 2 = 4 mA/V
Since R L is r o /2,increasing I D byafactoroffour resultsin r o andhence R L decreasingbyafactor of4,thus
R L = 1 4 × 10 = 2 5 k
| A M |= gm R L = 4 × 2 5 = 10 V/V
R gs = Rsig = 20 k
R gd = Rsig (1 + gm R L ) + R L
= 20(1 + 10) + 2 5
= 222 5 k
RC L = R L = 2 5 k
τ H = τ gs + τ gd + τC L
= C gs Rsig + C gd R gd + C L RC L
= 20 × 10 15 × 20 × 103 + 5 × 10 15 × 222 5
× 103 + 25 × 10 15 × 2.5 × 103
= 400 + 1112 5 + 62 5
= 1575 ps
f H = 1 2π × 1575 × 10 12 = 101 MHz
GB =| A M | f H
= 10 × 101 = 1 01 GHz
Ex:10.14 (a) gm = 40 mA/V
r π = 200 40 = 5 k
r on = V An I = 130 1 = 130 k
r op = | V Ap | I = 50 1 = 50 k
R L = r on r op = 130 50 = 36.1 k
A M =− r π r π + Rsig gm R L
=− 5 5 + 36 × 40 × 36 1
=−176 V/V
(b) C in = C π + C μ (1 + gm R L )
= 16 + 0 3(1 + 40 × 36 1)
= 450 pF
Rsig = r π Rsig
= 5 36 = 4 39 k
f H = 1 2π C in Rsig
= 1 2π × 450 × 10 12 × 4.39 × 103
= 80 6 kHz
(c) Rπ = Rsig = 4 39 k
Rμ = Rsig (1 + gm R L ) + R L
= 4.39(1 + 40 × 36.1) + 36.1
= 6.38 M
RCL = R L = 36 1 k
τ H = C π + C μ Rμ + C L RCL
= 16 × 4 39 + 0 3 × 6 38 × 103 + 5 × 36 1
= 70 2 + 1914 + 180 5
= 2164 7 ns
f H = 1 2π × 2164 7 × 10 9
= 73.5 kHz
(d) f Z = gm 2π C μ
= 40 × 10 3
2π × 0 3 × 10 12 = 21 2 GHz
(e)GB = 175 × 73 5 = 12 9 MHz
Ex:10.15 Rin = R L + r o 1 + gm r o
= 800 + 20 1 + 40 20 k
G v = R L Rsig + Rin = 800 20 + 20 = 20 V/V
R gs = Rsig Rin = 20 20 = 10 k
R gd = R L Ro
= 800 840 = 410 k
τ H = C gs R gs + (C gd + C L ) R gd
= 20 × 10 15 × 10 × 103 + (5 + 25) × 10 15
× 410 × 103
= 200 + 12,300
= 12,500 ps
f H = 1 2π × 12,500 × 10 12 = 12 7 MHz
Thus,whilethemidbandgainhasbeenincreased substantially(byafactorof21),thebandwidth hasbeensubstantiallylowered(byafactorof 20.7).Thus,thehigh-frequencyadvantageofthe CGamplifieriscompletelylost!
Ex:10.16 (a) A CS =− gm ( R L r o )
=− gm (r o r o ) =− 1 2 gm r o
=− 1 2 × 40 =−20 V/V
A cascode =− gm ( R L Ro )
=− gm (r o gm r o r o )
− gm r o =−40 V/V
Thus,
A cascode A CS = 2 (b)FortheCSamplifier,
τ H = C gs R gs + C gd R gd where
R gs = Rsig
R gd = Rsig (1 + gm R L ) + R L Rsig (1 + gm R L )
= Rsig 1 + 1 2 gm r o
= Rsig 1 + 1 2 × 40 = 21 Rsig
τ H = C gs Rsig + C gd × 21 Rsig
= C gs Rsig + 0.25C gs × 21 Rsig
= 6 25C gs Rsig
f H = 1 2π × 6 25C gs Rsig
Forthecascodeamplifier,
τ H Rsig C gs 1 + C gd 1 (1 + gm 1 Rd 1 )
where
Rd 1 = r o1 Rin2 = r o r o + r o gm r o
= r o 2 gm = 2 gm r o 2 gm + r o
= 2r o 2 + gm r o = 2r o 2 + 40 = r o 21
τ H = C gs Rsig 1 + 0.25 1 + gm r o 21
= C gs Rsig 1 + 0 25 1 + 40 21
= 1 73C gs Rsig
f H = 1 2π × 1 73 C gs Rsig
Thus,
f H (cascode)
f H (CS) = 6 25 1 73 = 3 6 (c) f t (cascode)
f t (CS) = 2 × 3 6 = 7 2
Ex:10.17 gm = 40 mA/V
r π = β gm = 200 40 = 5 k
Rin = r π = 5 k
A 0 = gm r o = 40 × 130 = 5200
Ro1 = r o1 = 130 k
Rin2 = r π 2 r o2 + R L gm 2 r o2 = 5 130 + 50 5200
= 35
Ro β2 r o2 = 200 × 130 = 26 M
A M =− r π r π + Rsig gm ( Ro R L )
=− 5 5 + 36 40(26,000 50)
Exercise10–4
A M =−244 V/V
Rsig = r π 1 Rsig
= 5 36 = 4 39 k
Rπ 1 = Rsig = 4 39 k
Rc 1 = r o1 Rin2
= 130 k 35 35
Rμ1 = Rsig (1 + gm 1 Rc 1 ) + Rc 1
= 4 39(1 + 40 × 0 035) + 0 035
= 10 6 k
τ H = C π 1 Rπ 1 + C μ1 Rμ1 + C π 2 Rc 1 + (C L + C μ2 )( R L Ro )
= 16 × 4 39 + 0 3 × 10 6 + 16 × 0 035
+ (5 + 0 3)(50 26,000)
= 70 24 + 3 18 + 0 56 + 264 5
= 338 5 ns
f H = 1 2π × 338 5 × 10 9 = 470 kHz
f t =| A M | f H = 244 × 470 = 113.8 MHz
Thus,incomparisontotheCEamplifierof Exercise10.19,weseethat | A M | hasincreased from175V/Vto242V/V, f H hasincreasedfrom 73.5kHzto470kHz,and f t hasincreasedfrom 12.9MHzto113.8MHz.
Tohave f H equalto1MHz,
τ H = 1 2π f H = 1 2π × 1 × 106 = 159.2 ns
Thus, 159 2 = 70 24 + 3 18 + 0 56
+ (C L + C μ )(50 26,000)
⇒ C L + C μ = 1.71 pF
Thus, C L mustbereducedto1.41pF.
Ex:10.18 FromEq.(10.103),weobtain
R gs = Rsig gm R L + 1 + R L gm R L + 1 = Rsig + R L gm R L + 1
R gd = Rsig
RC L = R L gm R L + 1
Ex:10.19 FromExample10.8,weget
τ H = b1 = 104 ps
f H = 1
2πτ H
= 1 2π × 104 × 10 12 = 1 53 GHz
Thisislowerthantheexactvaluefoundin Example10.8(i.e.,1.86GHz)byabout18%,still notabadestimate!
Ex:10.20 gm = 40 mA/V
r e = 25
r π = β gm = 100 40 = 2 5 k
Rsig = 1 k
R L = R L r o = 1 100 = 0.99 k
A M = R L R L + r e + Rsig β + 1
= 0 99 0 99 + 0 025 + (1/101) = 0 97 V/V
C π + C μ = gm 2π f T = 40 × 10 3 2π × 400 × 106
= 15 9 pF
C μ = 2 pF
C π = 13.9 pF f Z = 1 2π C π r e = 1 2π × 13 9 × 10 12 × 25
= 458 MHz
b1 = C π + C μ 1 + R L r e Rsig + C π + C L 1 + Rsig r π R L 1 + R L r e + Rsig r π = 13 9 + 2 1 + 0.99 0 025 × 1 + (13 9 + 0)0 99 1 + 0 99 0 025 + 1 2 5
= 2 66 × 10 9 s
b2 = C π C μ R L Rsig 1 + R L r e + Rsig r π = 13 9 × 2 × 0 99 × 1 1 + 0.99 0 025 + 1 2 5
= 0.671 × 10 18
ω P 1 and ω P 2 aretherootsoftheequation
1 + b1 s + b2 s 2 = 0
Solvingweobtain,
f P 1 = 67 2 MHz
f P 2 = 562 MHz
Since f P 1 f P 2 ,
f H f P 1 = 67.2 MHz
Ex:10.21 (a) I D 1,2 = 1 2 μn C ox W L 1,2 V 2 OV
0 4 = 1 2 × 0 2 × 100 V 2 OV
⇒ V OV = 0.2 V
gm = 2 I D V OV = 2 × 0 4 0 2 = 4 mA/V
(b) A d = gm ( R D r o ) where
r o = V A I D = 20 0 4 = 50 k
A d = 4(5 50) = 4 × 4 545 = 18.2 V/V
(c) f H = 1 2π(C L + C gd + C db )( R D r o ) = 1 2π(100 + 10 + 10) × 10 15 × 4 545 × 103
= 292 MHz
(d) τ gs = C gs Rsig = 50 × 10 = 500 ps
τ gd = C gd R gd = C gd Rsig (1 + gm R L ) + R L
= 10 [10(1 + 18 2) + 4 545]
= 1965 5 ps
τC L = (C L + C db ) R L = 110 × 4 545 = 500 ps
τ H = τ gs + τ gd + τC L
= 500 + 1965 5 + 500 = 2965 5 ps
f H = 1 2π × 2965 5 × 10 12
= 53 7 MHz
Ex:10.22 f Z = 1 2π R SS C SS
= 1 2π × 75 × 103 × 0 4 × 10 12
= 5 3 MHz
Thus,the3-dBfrequencyoftheCMRRis 5.3MHz.
Ex:10.23 A d = gm 1,2 (r o2 r o4 ) where
gm 1,2 = 0 5 0 025 = 20 mA/V
r o2 = r o4 = 100 0 5 = 200 k
A d = 20(200 200) = 2000 V/V
Thedominanthigh-frequencypoleisthat introducedattheoutputnode,
f H = 1
2π C L (r o2 r o4 )
= 1 2π × 2 × 10 12 × 100 × 103
= 0 8 MHz
Ex:10.24 (a) A M =− gm R L
where
R L = R L r o = 20 20 = 10 k
A M =−2 × 10 =−20 V/V
τ H = C gs R gs + C gd R gd + C L R L
= C gs Rsig + C gd Rsig (1 + gm R L ) + R L + C L R L
= 20 × 20 + 5 [20(1 + 20) + 10] + 5 × 10
= 400 + 2150 + 50
= 2600 ps
f H = 1 2πτ H
= 1 2π × 2600 × 10 12
= 61 2 MHz
GB =| A M | f H
= 20 × 61 2
= 1 22 GHz
(b) G m = gm 1 + gm Rs = 2 1 + 2 = 0 67 mA/V
Ro r o (1 + gm Rs )
= 20 × 3 = 60 k
R L = R L Ro = 20 60 = 15 k
A M =− G m R L
=−0 67 × 15 =−10 V/V
R gd = Rsig (1 + G m R L ) + R L
= 20(1 + 10) + 15
= 235 k
RC L = R L = 15 k
R gs = Rsig + Rs + Rsig Rs /(r o + R L )
1 + gm Rs
r o r o + R L
where Rs = 2 gm = 1 k
R gs = 20 + 1 + 20 × 1 20 + 20 1 + 2 × 20 20 + 20
= 10 75 k
τ H = C gs R gs + C gd R gd + C L RC L
= 20 × 10.75 + 5 × 235 + 5 × 15
= 215 + 1175 + 75 = 1465 ps
f H = 1 2π × 1465 × 10 12 = 109 MHz
GB = 10 × 109 = 1 1 GHz
Ex:10.25 RefertoFig.10.35(b).
A M = 2r π 2r π + Rsig × 1 2 × gm R L
where
gm = 20 mA/V
r π = 100 20 = 5 k
A M = 10 10 + 10 × 1 2 × 20 × 10 = 50 V/V f P 1 = 1 2π C π 2 + C μ (2r π Rsig ) = 1 2π 6 2 + 2 × 10 12 (10 10) × 103
= 6 4 MHz
f P 2 = 1 2π C μ R L = 1 2π × 2 × 10 12 × 10 × 103
= 8 MHz
T (s ) = 50 1 + s ω P 1 1 + s ω P 2 | T ( j ω)|= 50 1 + ω ω P 1 2 1 + ω ω P 2 2
At ω = ω H , | T |= 50/√2,thus
2 = 1 + ω H ω P 1 2 1 + ω H ω P 2 2
1 +
ω 4 H
ω 2 P 1 ω 2 P 2 + ω 2 H 1 ω 2 P 1 + 1 ω 2 P 2 1 = 0
f 4 H
f 2 P 1 f 2 P 2 + f 2 H 1 f 2 P 1 + 1 f 2 P 2 1 = 0
f 4 H 6.42 × 82 + f 2 H 1 6.42 + 1 82 1 = 0
⇒ f H = 4 6 MHz(Exactvalue)
P 2
UsingEq.(10.150),anapproximathevaluefor f H canbeobtained:
f H 1/ 1 f 2 P 1 + 1 f 2 P 2
= 1/ 1 6.42 + 1 82 = 5 MHz
Ex:10.26 A M =− R G R G + Rsig gm ( R D R L )
=− 10 10 + 0 1 × 2(10 10)
=−9 9 V/V
f P 1 = 1 2π C C 1 ( Rsig + R G )
= 1 2π × 1 × 10 6 (0 1 + 10) × 106
= 0 016 Hz
f P 2 = gm + 1/ R S 2π C S
= (2 + 0 1) × 10 3
2π × 1 × 10 6 = 334 2 Hz
f P 3 = 1 2π C C 2 ( R D + R L ) = 1 2π × 1 × 10 6 (10 + 10) × 103
= 8 Hz
f Z = 1
2π C S R S
= 1
2π × 1 × 10 6 × 10 × 103
= 15 9 Hz
Sincethehighest-frequencypoleis f P 2 = 334 2 andthenexthighest-frequencysingularityis f Z at15.9Hz,thelower3-dBfrequency f L willbe
f L f P 2 = 334.2 Hz
Ex:10.27 RefertoFig.10.42.
τC 1 = C C 1 Rsig + ( R B r π )
= 1 × 10 6 [5 + (100 2 5)] × 103
= 7 44 ms
τCE = C E R E r e + R B Rsig β + 1
β = gm r π = 40 × 2 5 = 100
r e 1/ gm = 25
τCE = 1 × 10 6 5 0 025 + 100 5 101 × 103
τCE = 0 071 ms
τC 2 = C C 2 ( RC + R L )
= 1 × 10 6 (8 + 5) × 103
= 13 ms
f L = 1 2π 1 τC 1 + 1 τCE + 1 τC 2
= 1 2π 1 7 44 + 1 0 071 + 1 13 × 103
= 2 28 kHz
f Z = 1
2π C E R E = 1
2π × 1 × 10 6 × 5 × 103 = 31 8 Hz
Since f Z ismuchlowerthan f L itwillhavea negligibleeffecton f L
Chapter11
SolutionstoExerciseswithintheChapter
Ex:11.1 (b)
β = 1/ A f ideal = 1 10 = 0 1 V/V
0 1 = R1 R1 + R2
⇒ R2 R1 = 9
(c)
A = 100 β = 0 1
Aβ = 10
1 + Aβ = 11 or 20 8 dB
A f = A 1 + Aβ = 100 11 = 9 091 V/V
A f differsfromtheidealvalueof10V/Vby 9 1%.
(d)
10 = A 1 + Aβ = 100 1 + 100β
⇒ β = 0 09 V/V
0 09 = R1 R1 + R2
⇒ R2 R1 = 10 11
(e)
Vo = A f Vs = 10 × 1 = 10 V
V f = β Vo = 0 09 × 10 = 0 9 V
Vi = Vs V f = 1 0 9 = 0 1 V
(f)
A = 0 8 × 100 = 80
A f = 80 1 + 80 × 0 09 = 9 76 V/V whichisa 2 44%decrease.
Ex.11.2 (b)
β = 1/ A f ideal = 1/103
β = 0.001 V/V
0 001 = R1 R1 + R2
⇒ R2 R1 = 999
(c)
Aβ = 104 × 10 3 = 10
1 + Aβ = 11 or 20 8 dB
A f = A 1 + Aβ = 104 11 = 909 1 V/V
(d)
1000 = 104 1 + 104 β
⇒ β = 0 0009
0 0009 = R1 R1 + R2
⇒ R2 R1 = 1110 1
(e)
Vo = A f Vs = 1000 × 0 01 = 10 V
V f = β Vo = 0.0009 × 10 = 0.009 V
Vi = Vs V f = 0 01 0 009 = 0 001 V
(f)
A = 0 8 × 104
A f = 0 8 × 104 1 + 0 8 × 104 × 0 0009 = 975 6 V/V
whichisa 2 44%decrease.
Ex.11.3 Toconstrainthecorrespondingchange in A f to0.1%,weneedanamount-of-feedback ofatleast
1 + Aβ = 10% 0 1% = 100
Thusthelargestobtainableclosed-loopgainwill be
A f = A 1 + Aβ = 1000 100 = 10 V/V
Eachamplifierinthecascadewillhaveanominal gainof10V/Vandamaximumvariabilityof 0.1%;thustheoverallvoltagegainwillbe
(10)3 = 1000 V/Vandthemaximumvariability willbe0.3%.
Ex.11.4
β = R1 R1 + R2 = 1 1 + 9 = 0 1
Aβ = 104 × 0 1 = 1000
1 + Aβ = 1001
A f = A 1 + Aβ
A f = 104 1 + 104 × 0 1 = 9 99 V/V
f Hf = f H (1 + Aβ) = 100 × 1001 = 100 1 kHz
Ex.11.5 (a)RefertoFig.11.8(c).
β = R1 R1 + R2 (b) RD R2 R1 Vt Vr Vd
Figure1showsthecircuitpreparedfor determiningtheloopgain Aβ .Observethatwe haveeliminatedtheinputsignal Vs ,andopened theloopatthegateof Q wheretheinput impedanceisinfiniteobviatingtheneedfora terminationresistanceattheright-handsideof thebreak.Nowweneedtoanalyzethecircuitto determine
Aβ ≡− Vr Vt
First,wewriteforthegainoftheCSamplifier Q ,
Vd Vt =− gm R D ( R1 + R2 )(1) thenweusethevoltage-dividerruletofind Vr , Vr
Vd = R1 R1 + R2 (2)
CombiningEqs.(1)and(2)gives
Aβ ≡− Vr Vt = gm R D ( R1 + R2 ) R1 R1 + R2 whichcanbesimplifiedto
β
(c) A = Aβ β = gm R D ( R1 + R2 ) R D + R1 + R2 (d) β = R1 R1 + R2 = 20 20 + 80 = 0.2 V/V
Aβ = 4 10 × 20 10 + 20 + 80 = 7 27
A = 7 27 0 2 = 36 36 V/V
A f = A 1 + Aβ = 36 36 1 + 36 36 × 0 2 = 4 4 V/V
If Aβ were 1,then
A f 1 β = 1 0 2 = 5 V/V
Ex.11.6 FromthesolutionofExample11.5, 1 + Aβ = 60 1
Thus,
f Hf = (1 + Aβ) f H = 60 1 × 1 = 60 1 kHz
Ex.11.7 RefertoFig.E11.7.The1-mAbias currentwillsplitequallybetweentheemittersof Q 1 and Q 2 ,thus
I E 1 = I E 2 = 0 5 mA
Transistor Q 3 willbeoperatingatanemitter current
I E 3 = 5 mA determinedbythe5-mAcurrentsource.Sincethe dccomponentof Vs = 0,thenegativefeedback willforcethedcvoltageattheoutputtobe approximatelyzero.
The β circuitisshowninFig.1(onnextpage) togetherwiththedeterminationof β andofthe loadingeffectsofthe β circuitonthe A circuit,
β = R1 R1 + R2 = 1 1 + 9 = 0 1 V/V
R11 = R1 R2 = 1 9 = 0 9 k
R22 = R1 + R2 = 1 + 9 = 10 k
The A circuitisshowninFig.2(onnextpage).
r e 1 = r e 2 = VT I E 1,2 = 25 mV 0 5 mA = 50
r e 3 = VT I E 3 = 25 mV 5 mA = 5
i e = Vi
r e 1 + r e 2 + Rs + R11 β + 1
ThesefiguresbelongtoExercise11.7.
i e = Vi
0 05 + 0 05 + 10 + 0 9 101
⇒ i e = 4 81 Vi (1)
Rb3 = (β + 1)[r e 3 + ( R22 R L )] = 101[0 005 + (10 2)] = 168 84 k
i b3 = α i e RC RC + Rb3
= 0 99 i e 20 20 + 168 84
Figure2
⇒ i b3 = 0 105i e (2) Vo = i e 3 ( R22 R L ) = (β + 1)i b3 ( R22 R L )
= i b3 × 101(10 2)
⇒ Vo = 168 33i b3 (3)
Combining(1)–(3),weobtain
A ≡ Vo Vi = 85 V/V
β = 0 1 V/V
Aβ = 8 5
1 + Aβ = 9 5
A f = 85 9 5 = 8 95 V/V
Fromthe A circuit,wehave
Ri = Rs + R11 + (β + 1)(r e 1 + r e 2 )
= 10 + 0 9 + 101 × 0 1
= 21 k
Rif = Ri (1 + Aβ)
= 21 × 9 5 = 199 5 k
Rin = Rif Rs = 199 5 10 = 189 5 k
Fromthe A circuit,wehave
Ro = R L R22 r e 5 + RC β + 1
= 2 10 0 005 + 20 101
= 181
Rof = Ro 1 + Aβ
= 181 9 5 = 19 1
Rof = R L Rout
19 1 = 2 k Rout
⇒ Rout = 19 2
Ex.11.8 Figure1showsthe β circuittogether withthedeterminationof β , R11 and R22
β = R1 R1 + R2
ThisfigurebelongstoExercise11.8.
Exercise11–4 R11 = R1 R2
R22 = R1 + R2
Figure2showsthe A circuit.Wecanwrite
Vo = gm ( R D R22 ) Vi
Thus, A ≡ Vo Vi = gm R D ( R1 + R2 )
f = A 1 + Aβ
From A circuit,wehave Ri = 1 gm
Figure1
Ro = R D R22
Rin = Rif = Ri (1 + Aβ)
Rin = 1 gm (1 + Aβ)
Rout = Rof = Ro 1 + Aβ
Rout = R D ( R1 + R2 ) 1 + Aβ
ComparisonwiththeresultsofExercise11.6
showsthattheexpressionsfor A and β are identical.However, Rin and Rout cannotbe determinedusingthemethodofExercise11.6.
Ex.11.9 A f ideal = 10 A/A
β = 1 A f ideal = 0 1 A/A
A = 1000 A/A
1 + Aβ = 1 + 1000 × 0 1 = 101
A f = A 1 + Aβ = 1000 101 = 9.9 A/A
Rif = Ri 1 + Aβ = 1000 101 = 9 9
Rof = Ro (1 + Aβ) = 100 k × 101 = 10 1 M
Ex.11.10 RefertothesolutionofExample11.7.
If μ isreducedfrom1000V/Vto100V/V, A willbereducedbythesamefactor(10)tobecome
A = 64 9 mA/V
andtheloopgainwillbecorrespondingly reduced,
Aβ = 64 9
Thus,theamountoffeedbackbecomes
1 + Aβ = 65 9
andtheclosed-loopgain A f becomes
A f = A 1 + Aβ = 64 9 65 9 = 0 985 mA/V
Theinputresistancebecomes
Rin = Rif = (1 + Aβ) Ri = 65 9 × 101 = 6 7 M
Theoutputresistancebecomes
Rout = Rof = (1 + Aβ) Ro = 65 9 × 61 = 4 M
Ex.11.11 RefertothesolutionofExample11.7. Todouble A f ideal weneedtoreduce β tohalfits
value.Since β = R F ,thefeedbackresistance R F becomes
R F = 500
Now,since A isgivenby
A = Rid Rid + R F μ 1 gm + ( R F r o2 ) r o2 R F + r o2
thevalueof A becomes
A = 100 100 + 0 5 1000 0 5 + (0 5 20) 20 0 5 + 20 = 982 7 mA/V
Thus, A f becomes
A f = A 1 + Aβ = 982 7 1 + 982 7 × 0 5 = 1 996 mA/V
Ex.11.12 Toobtain A f ideal = 100 mA/V,the feedbackfactor β mustbe
β = 1 100 = 0 01 V/mA = 10
But,
β = R E 1 R E 2 R E 1 + R E 2 + R F
For R E 1 = R E 2 = 100 ,
10 = 100 × 100 100 + 100 + R F
⇒ R F = 800
Sincetheidealvalueof ( Io / Vs ) is100mA/V,the idealvalueofthevoltagegainis,
Vo Vs =− Io RC Vs =−100 × 0 6
=−60 V/V
Ex.11.13 Seefigureonnextpage.Figure1 showsthecircuitpreparedforthedetermination oftheloopgain,
Aβ ≡− Vr Vt
Wewilltracethesignalaroundtheloopas follows:
Ic 2 = gm 2 Vt (1)
Ib3 = Ic 2 RC 2 RC 2 + Ri 3 (2) where
Ri 3 = (β + 1) r e 3 + R E 2 ( R F + ( R E 1 r e 1 )) (3) Ie 3 = (β + 1) Ib3 (4)
ThisfigurebelongstoExercise11.13.
I f = Ie 3 R E 2 R E 2 + R F + ( R E 1 r e 1 ) (5)
Ie 1 = I f R E 1 R E 1 + r e 1 (6)
Ic 1 = α Ie 1 (7)
Vr =− Ic 1 ( RC 1 r π 2 )(8)
Combining(1)–(8)gives Vr intermsof Vt and hence Aβ ≡− Vr / Vt .Weshalldothis numericallyusingthevaluesinExample11.8:
gm 2 = 40 mA/V, RC 2 = 5 k ,β = 100,
r e 3 = 6 25 , R E 1 = R E 2 = 100 , R F = 640 ,
r e 1 = 41 7 ,α1 = 0 99, RC 1 = 9 k , and r π 2 = 2 5 k
Ri 3 = 101 0 00625 + 0 1 (0 64 + (0 1 0 0417))
= 9 42 k
Ic 2 = 40 Vt (9)
Ib3 = 0 347 Ic 2 (10)
Ie 3 = 101 Ib3 (11)
I f = 0 13 Ie 3 (12)
Ie 1 = 0 706 I f (13)
Ic 1 = 0 99 Ie 1 (14)
Vr =−1 957 Ic 1 (15)
Combining(9)–(15),weobtain
Aβ =− Vr Vt = 249 2
Ex.11.14 mVt RF ro Rs Rid RL Rid VrVt
Figure1
Figure1showsthecircuitpreparedfor determiningtheloopgain
Aβ ≡− Vr Vt
Usingthevoltage-dividerrule,wecanwriteby inspection
Vr = μ Vt R L [ R F + ( Rs Rid )]
r o + R L [ R F + ( Rs Rid )] ( Rs Rid ) R F + ( Rs Rid )
Vr = μ Vt R L ( Rs Rid )
r o [ R L + R F + ( Rs Rid )] + R L [ R F + ( Rs Rid )]
Thus, Aβ =− Vr Vt = μ R L ( Rid Rs )
r o [ R L + R F + ( Rid Rs )] + R L [ R F + ( Rid Rs )]
Q.E.D.
UsingthenumericalvaluesinExample11.9,we get
Aβ = 1035 2
Figure1
ThisfigurebelongstoExercise11.15.
Ex.11.15 Seefigureonnextpage.Figure1(a) showsthefeedbackamplifiercircuit.The β circuitisshowninFig.1(b),andthe determinationof β isshowninFig.1(c),
Finally,the A circuitisshowninFig.1(f).We canwritebyinspection
Ri = Rs R11 = Rs R F Ro = r o R22 = r o R F Vgs = Ii Ri Vo =− gm Vgs (r o R22 )
(b)Thedeterminationof R11 and R22 is illustratedinFigs.1(d)and(e),respectively: R11 = R22 = R F
Figure1
A f = gm ( Rs R F )(r o R F )
1 + gm ( Rs R F )(r o R F )/ R F Q E D
(c) Rif = Ri 1 + Aβ
Rif = Rs R F
1 + gm ( Rs R F )(r o R F )/ R F 1 Rif = 1 Rs + 1 R F + gm (r o R F ) R F
But, 1 Rif = 1 Rs + 1 Rin thus, 1 Rin = 1 R F 1 + gm (r o R F )
⇒ Rin = R F 1 + gm (r o R F ) Q E D
(d) Rout = Rof = Ro 1 + Aβ
= r o R F
1 + gm ( Rs R F )(r o R F )/ R F 1 Rout = 1 r o + 1 R F + gm ( Rs R F ) R F
⇒ Rout = r o R F 1 + gm ( Rs R F ) Q E D
(e) A =−5(1 10)(20 10)
A =−30 3 k β =− 1 R F =− 1 10 =−0 1 mA/V
Aβ = 3 03
1 + Aβ = 4 03
A f = A 1 + Aβ =− 30 3 4 03 =−7 52 k
(Comparetotheidealvalueof 10 k ).
Ri = Rs R F = 1 10 = 909
Ro = r o R F = 20 10 = 6.67 k
Rif = Ri 1 + Aβ = 909 4 03 = 226
Rin = 1 1 Rif 1 Rs = 291
Rof = Ro 1 + Aβ = 6 67 4 03 = 1 66 k
Rout = Rof = 1 66 k
Ex.11.16 RefertothesolutionofExample
11.10.
A =−μ Ri 1/ gm + ( R1 R
=−100 100 1 5 + (10 90 20) 20 20 + (10 90)
=−1076.4 A/A
β =−0 1
A f = A 1 + Aβ =− 1076 4 1 + 1076 4 × 0 1
=− 1076 4 1 + 107 64 =− 1076 4 108 64
=−9 91 A/A
Ri = 100 k
Rin = Rif = Ri 1 + Aβ = 100 108.64 = 921
Rout = Rof = Ro (1 + Aβ) = 929 × 108 64 = 101 M
Ex.11.17 RefertothesolutionofExample
11.10.With R2 = 0, β =− 1 1 + 0 =−1
A f ideal = 1 β =−1 A/A
Ri = Rs Rid ( R1 + R2 )
⇒ Ri = R1
Ro = r o2 + ( R1 R2 ) + ( gm r o2 )( R1 R2 )
⇒ Ro = r o2
A =−μ Ri 1/ gm + ( R1 R2 r o2 ) r o2 r o2 + ( R1 R2 )
⇒ A =−μgm R1
Aβ = μgm R1
A f = A 1 + Aβ =− μgm R1 1 + μgm R1
Rin = Rif = Ri 1 + Aβ
Rin = R1 1 + μgm R1 1 μgm
Rout = Rof = Ro (1 + Aβ)
= r o2 (1 + μgm R1 )
⇒ Rout μ( gm r o2 ) R1
Ex.11.18 Thefeedbackshiftsthepolebya factorequaltotheamountoffeedback:
1 + A 0 β = 1 + 105 × 0 01 = 1001
Thepolewillbeshiftedtoafrequency
f Pf = f P (1 + A 0 β)
= 100 × 1001 = 100.1 kHz
If β ischangedtoavaluethatresultsinanominal closed-loopgainof1,thenweobtain
β = 1
and
1 + A 0 β = 1 + 105 × 1 105 thenthepolewillbeshiftedtoafrequency
f Pf = 105 × 100 = 10 MHz
Ex.11.19 FromEq.(11.40),weseethatthe polescoincidewhen
(ω P 1 + ω P 2 )2 = 4(1 + A 0 β)ω P 1 ω P 2
(104 + 106 )2 = 4(1 + 100β) × 104 × 106
⇒ 1 + 100β = 25 5
⇒ β = 0 245
Thecorrespondingvalueof Q = 0 5.Thiscan alsobeverifiedbysubstitutinginEq.(11.42).
Amaximallyflatresponseisobtainedwhen Q = 1/√2.SubstitutinginEq.(11.42),weobtain
1 √2 = (1 + 100β) × 104 × 106 104 + 106
⇒ β = 0 5
Inthiscase,thelow-frequencyclosed-loopgainis
A f (0) = A 0 1 + A 0 β = 100 1 + 100 × 0 5 = 1 96 V/V
Ex.11.20 Theclosed-looppolesaretherootsof thecharacteristicequation
1 + A (s )β = 0
1 + ⎛ ⎜ ⎝ 10 1 + s 104 ⎞ ⎟ ⎠ 3 β = 0
Tosimplifymatters,wenormalize s bythe factor 104 ,thusobtainingthenormalized
complex-frequencyvariable S = s /104 ,andthe characteristicequationbecomes
( S + 1)3 + 103 β = 0 (1)
Thisequationhasthreeroots,arealoneanda pairthatcanbecomplexconjugate.Therealpole canbefoundfrom
( S + 1)3 =−103 β
Dividingthecharacteristicpolynomialin(1)by S + 1 + 10β 1/3 givesaquadraticwhosetwo rootsaretheremainingpolesofthefeedback amplifier.Aftersomestraightforwardbut somewhattediousalgebra,weobtain
S 2 + 10β 1/3 2 S + 1 + 100β 2/3 10β 1/3 = 0 (3)
Thepairofpolescannowbeobtainedas
S = 1 + 5β 1/3 ± j 5√3 β 1/3 (4)
Equations(1)and(3)describethethreepoles showninFig.E11.20.
FromEq.(2)weseethatthepairofcomplex poleslieonthe j ω axisforthevalueof β that makesthecoefficientof S equaltozero,thus
βcr = 2 10 3 = 0 008
Ex.11.21
A (s) 99R R
Figure1
FromFig.1,wecaneasilyobtaintheloopgainas
Aβ = A (s ) × 0 01 = 105 1 + s 2π × 10 × 0 01 = 1000 1 + s 2π × 10
Fromthissingle-poleresponse(low-passSTC response)wecanfindtheunity-gainfrequency byinspectionas
f 1 = f P × 1000 = 104 Hz
Thephaseangleat f 1 willbe 90◦ andthusthe phasemarginis 90◦
Ex.11.22 FromEq.(11.48),weobtain
| A f ( j ω1 )| 1/β = 1/|1 + e j θ |
= 1/|1 + cos θ j sin θ |
(a)ForPM = 30◦ , θ = 180 30 = 150◦ ,thus
| A f ( j ω1 )| 1/β = 1/|1 + cos 150◦ j sin 150◦ | = 1 93
(b)ForPM = 60◦ , θ = 180 60 = 120◦ ,thus
| A f ( j ω1 )| 1/β = 1/|1 + cos 120◦ j sin 120◦ | = 1
(c)ForPM = 90◦ , θ = 180 90 = 90◦ ,thus
| A f ( j ω1 )| 1/β = 1/|1 + cos 90◦ j sin 90◦ |
= 1/√2 = 0 707
Ex.11.23 Foraphasemarginof 45◦ , φ =−135◦ .Locatingthepoint φ =−135◦ on thephaseplotinFig.11.38,wefindthe correspondingpointonthegainplotat80dB, thatis,
20 log(1/β) = 80 dB
⇒ β = 10 4 A f = A 1 + Aβ = 105 1 + 105 × 10 4 = 9.09 × 103 = 79 2 dB 80 dB
Ex.11.24 Toobtainstableperformancefor closed-loopgainsaslowas20dB(whichis80 dBbelow A 0 ,orequivalently 104 below A 0 ), wemustplacethenewdominantpoleat 1MHz/104 = 100 Hz.
Ex.11.25 Thefrequencyofthefirstpolemustbe loweredfrom1MHztoanewfrequency f D = 10 MHz 104 = 1000 Hz
thatis,byafactorof1000.Thus,thecapacitance atthecontrollingnodemustbeincreasedbya factorof1000.
Chapter12
SolutionstoExerciseswithintheChapter
Ex:12.1 Toallowfor v O toreach
VCC + VCE sat =−15 + 0.2 =−14.8 V,with Q 1 justcuttingoff(i.e. i E 1 = 0),
I = 14.8 V R L = 14.8 1 k = 14 8 mA
Thevalueof R cannowbefoundfrom
I = V R R = VCC V D R
14 8 = 15 0 7 R ⇒ R = 14.3 14 8 = 0 97 k
Theresultingoutputsignalswingwillbe 14 8
Vto +14 8 V.Theminimumcurrentin Q 1 = 0
Themaximumcurrentin
Q 1 = 14 8 + 14 8 = 29 6 mA
Ex:12.2 At v O =−10 V,wehave
i L = 10 1 =−10 mA
i E 1 = I + i L = 14 8 10 = 4 8 mA
v BE 1 = 0 6 + 0 025 ln 4 8 1
= 0 64 V
v I = v O + v BE 1
=−10 + 0 64 =−9 36 V
At v O = 0 V,wehave
i L = 0 mA
i E 1 = I = 14 8 mA
v BE 1 = 0 6 + 0 025 ln 14 8 1
= 0 67 V
v I = v O + v BE 1
= 0 + 0 67 = 0 67 V
At v O = 10 V
i L = 10 1 = 10 mA
i E 1 = I + i L = 14 8 + 10 = 24 8 mA
v BE 1 = 0 6 + 0 025 ln 24 8 1
= 0 68 V
v I = v O + v BE 1
= 10 + 0 68 = 10 68 V
At v O =−10 V,wehave
i E 1 = 4 8 mA
r e 1 = 25 mV 4 8 mA = 5 2
v o v i = R L R L + r e 1 = 1 1 + 0 0052 = 0 995 V/V
At v O = 0 V,
i E 1 = 14 8 mA
r e 1 = 25 mV 14 8 mA = 1 7
v o v i = R L R L + r e 1 = 1 1 + 0 0017 = 0 998 V/V
At v O =+10 V,
i E 1 = 24 8 mA
r e 1 = 25 mV 24.8 mA = 1 0
v o v i = R L R L + r e 1 = 1 1 + 0.001 = 0 999 V/V
Ex:12.3 a. PL = Vo /√2 2 R L = 8/√2 2 100 = 0.32W
PS = 2 VCC × I = 2 × 10 × 100 × 10 3
= 2W
Efficiency η = PL PS × 100
= 0 32 2 × 100 = 16%
Ex:12.4 (a) PL = 1 2 ˆ V 2 o R L
= 1 2 (4 5)2 4 = 2 53W
(b) PS + = PS = VCC × 1 π ˆ Vo R L
= 6 × 1 π × 4 5 4 = 2 15W
(c) η = PL PS × 100 = 2 53 2 × 2 15 × 100 = 59%
(d)Peakinputcurrents = 1 β + 1
ˆ Vo R L
= 1 51 × 4 5 4
= 22 1mA
(e)UsingEq.(12.22),weobtain
PDN max = PDP max = V 2 CC π 2 R L
= 62 π 2 × 4 = 0.91W
Ex:12.5 (a)Thequiescentpowerdissipatedin
eachtransistor = I Q × VCC
Totalpowerdissipatedinthetwotransistors
= 2 I Q × VCC
= 2 × 2 × 10 3 × 15
= 60mW
(b) I Q isincreasedto10mA
At v O = 0,wehave i N = i P = 10mA
FromEq.(12.31),weobtain
Rout = VT i P + i N = 25 10 + 10 = 1 25
v o v i = R L R L + Rout = 100 100 + 1.25
v o v i = 0 988at v O = 0V
At v O = 10V,wehave
i L = 10V 100 = 0 1A = 100mA
UseEq.(12.27)tocalculate i N :
i 2 N i N i L I 2 Q =0
i 2 N 100 i N 102 = 0
⇒ i N = 101 0mA
UsingEq.(12.26),weobtain
i P = I 2 Q i N 1mA
Rout = VT i N + i P = 25 101.0 + 1 0 2451
v o v i = R L R L + Rout = 100 100 + 0 2451 1
%change = 1 0 988 1 × 100 = 1 2%
InExample12.3, I Q =2mA,andfor v O = 0
Rout = VT
i N + i P = 25 2 + 2 = 6 25
v o v i = R L R L + Rout = 100 100 + 6 25 = 0 94
v O = 10V
i L = 10V 100 = 100mA
Againcalculate i N (for I Q = 2mA)using Eq.(12.27) (i N =100.04mA):
i P = I 2 Q I N = 22 100.04 = 0 04mA
Rout = VT i N + i P = 25 100 04 + 0 04 = 0.25
v o v i = R L R L + Rout 1
%Change = 1 0 94 1 × 100 = 6%
For I Q =10mA,changeis1.2%
For I Q =2mA,changeis6%
(c)Thequiesentpowerdissipatedineach transistor = I Q × VCC
Totalpowerdissipated = 2 × 10 × 10 3 × 15 = 300mW
Ex:12.6 FromExample12.4,wehave VCC =15
V, R L = 100
Q N and Q P matchedand I S = 10 13 Aand β = 50, IBias = 3mA
For v O =10V,wehave i L = 10 100 = 0 1A
Asafirstapproximation, i N 0.1A,
i P = 0, i BN 0 1A 50 + 1 2mA
i D = IBias i BN = 3 2 = 1mA
V BB = 2 VT ln⎛ ⎜ ⎝ 10 3 1 3 × 10 13 ⎞ ⎟ ⎠ (1)
This 1 3 isbecausebiasingdiodeshave 1 3 areaof theoutputdevices.
But V BB = V BEN + V BEP
= VT ln i N I S + VT ln i N i L I S
= V T ln i N (i N i L ) I 2 S (2)
EquatingEqs.1and2,weobtain
2 VT ln⎛ ⎜ ⎝ 10 3 1 3 × 10 13 ⎞ ⎟ ⎠ = VT ln i N (i N i L ) I 2 S
⎜
IBias
iN iD iP
10 3 1 3 × 10 13
⎟ ⎠ 2 = i N (i N 0 1) 10 13 2
i N (i N 0 1) = 9 × 10 6
If i N isinmA,then
i N (i N 100) = 9
i 2 N 100 i N 9 = 0
⇒ i N = 100 1mA
i P = i N i L = 0 1mA
For v O =−10V and i L = 10 100 =−0 1A
=−100mA:
Asafirstapproximationassume i P ∼ = 100mA, i N 0 Since i N = 0,currentthrough diodes = 3mA
∴ V BB = 2 VT ln⎛ ⎜ ⎝ 3 × 10 3 1 3 × 10 13 ⎞ ⎟ ⎠ (3)
But V BB = VT ln i N 10 13 + VT ln i P 10 13 = V T ln i P + i L 10 13 + VT ln i P 10 13 (4)
Here i L =−0 1A
EquatingEqs.(3)and(4),weobtain 2 VT ln⎛ ⎜ ⎝ 3 × 10 3 1 3 × 10 13 ⎞ ⎟ ⎠ =
VT ln i P 0 1 10 13 + VT ln i P 10 13 ⎛ ⎜ ⎝ 3 × 10 3 1 3 × 10 13 ⎞ ⎟ ⎠ 2 = i P (i P 0 1) 10 13 2
i P (i P 0.1) = 81 × 10 6
ExpressingcurrentsinmA,wehave
i P (i P 100) = 81
i 2 P 100 i P 81 = 0
⇒ i P = 100 8mA
i N = i P + i L = 0.8mA
Ex:12.7 IC = gm × 2mV/◦ C × 5 ◦ C, mA
where gm isinmA/mV
gm = 10mA 25mV = 0 4mA/mV
Thus, IC = 0 4 × 2 × 5 = 4mA
Ex:12.8 RefertoFig.12.14.
(a)Toobtainaterminalvoltageof1.2V,and since β 1 isverylarge,itfollowsthat
V R 1 = V R 2 = 0 6V
Thus IC 1 = 1mA
I R = 1 2V R1 + R2 = 1 2 2 4 = 0 5mA
Thus, I = IC 1 + I R = 1 5mA
(b)For V BB =+50mV:
V BB = 1 25V I R = 1 25 2.4 = 0 52mA
V BE = 1 25 2 = 0 625V
IC 1 = 1 × e V BE / VT = e 0 025/0 025
= 2 72mA
I = 2 72 + 0 52 = 3 24mA
For V BB =+100mV,wehave
V BB = 1 3V, I R = 1 3 2.4 = 0 54mA
V BE = 1 3 2 = 0 65V
IC 1 = 1 × e V BE / VT = 1 × e 0 05/0 025
= 7 39mA
I = 7 39 + 0 54 = 7 93mA
For V BB =+200mV:
V BB = 1.4V, I R = 1 4 2 4 = 0.58mA
V BE = 0 7V
IC 1 = 1 × e 0 1/0 025 = 54 60mA
I = 54.60 + 0.58 = 55.18mA
For V BB =−50mV:
V BB = 1 15V, I R = 1 15 2 4 = 0 48mA
V BE = 1 15 2
= 0 575
IC 1 = 1 × e 0 025/0 025 = 0 37mA
I = 0.48 + 0.37 = 0.85mA
For V BB =−100mV:
V BB = 1 1V I R = 1 1 2 4 = 0 46mA
V BE = 0 55V
IC 1 = 1 × e 0 05/0 025 = 0 13mA
I = 0 46 + 0 13 = 0 59mA
For V BB =−200mV:
V BB = 1 0V I R = 1 2 4 = 0 417mA
V BE = 0 5V
IC 1 = 1 × e 0 1/0 025 = 0 018mA
I = 0 43mA
Ex:12.9 (a)Fromsymmetryweseethatall transistorswillconductequalcurrentsandhave equal V BE ’s.Thus,
v O = 0V
If V BE 0 7V, then
V E 1 = 0 7Vand I1 = 15 0 7 5 = 2 86mA
Ifweneglect I B 3 ,then
IC 1 2 86mA
Atthiscurrent, | V BE | isgivenby
| V BE |= 0 025ln 2 86 × 10 3 3 3 × 10 14 0 63V
Thus V E 1 = 0 63Vand I1 = 2 87mA
Nomoreiterationsarerequiredand
i C 1 = i C 2 = i C 3 = i C 4 2 87mA
(b)For v I =+10V
Tostarttheiterations,let V BE 1 0 7V
E 1 = 10 7V
I1 = 15 10 7 5 = 0 86mA
Neglecting I B 3 ,weobtain IC 1 I E 1 I1 = 0 86mA Butatthiscurrent | V BE 1 |= VT ln IC 1 I S = 0 025ln 0 86 × 10 3 3 3 × 10 14 = 0 6V
Thus, V E 1 =+10 6Vand I1 = 0 88mA No furtheriterationsarerequiredand
IC 1 0 88mA
Tofind IC 2 ,weuseanidenticalprocedure:
V BE 2 0 7V
V E 2 = 10 0 7 =+9 3V
I2 = 9 3 ( 15) 5 = 4 86mA
V BE 2 = 0 025ln 4.86 × 10 3 3 3 × 10 14
= 0 643V
V E 2 = 10 0 643 =+9 357
I2 = 4 87mA
IC 2 4.87mA
Finally,
IC 3 = IC 4 = 3 3 × 10 14 e V BE / VT where
V BE = V E 1 V E 2 2 = 0 62V
Thus, IC 3 = IC 4 = 1 95mA
Thesymmetryofthecircuitenablesustofindthe valuesfor v I =−10V asfollows:
IC 1 = 4 87mA IC 2 = 0 88mA
IC 3 = IC 4 = 1 95mA
For v I =+10V,wehave v O = V E 1 V BE 3
= 10 6 0 62 =+9 98V
For v I =−10V,wehave v O = V E 1 V BE 3
=−9 357 0 62 =−9 98V
(c)For v I =+10V,wehave
v O 10V I L 100mA
IC 3 100mA
I B 3 = 100 201 0 5mA
Assumingthat | V BE 1 | hasnotchangedmuchfrom 0.6V,then
V E 1 10.6V
I1 = 15 10 6 5 = 0 88mA
I E 1 = I1 I B 3 = 0 88 0 5 = 0 38mA
IC 1 0 38mA
| V BE 1 |= 0 025ln 0 38 × 10 3 3 3 × 10 14
vI = 0 58V
V E 1 = 10 58V
I1 = 15 10 58 5 = 0 88mA
Thus, IC 1 0 38mA
Nowfor Q 2 wehave
V BE 2 = 0 643V
V E 2 = 10 0 643 = 9 357
I2 = 4 87mA
I B 4 0
IC 2 4 87mA (asin(b))
Assumingthat IC 3 100mA, wehave
V BE 3 = 0 025ln 100 × 10 3 3 3 × 10 14
= 0 72V
Thus, v O = V E 1 V BE 3
= 10 58 0 72 =+9 86V
| V BE 4 |= v O V E 2 9 86 9 36 = 0 5V
Thus, IC 4 = 3 3 × 10 14 e 0 5/0 025 0 02mA
Fromsymmetrywefindthevaluesforthecase
v I =−10Vas:
IC 1 = 4 87mA, IC 2 = 0 38mA
IC 3 = 0 02mA, IC 4 = 100mA
v O =−9 86V
Ex:12.10 For Q 1 :
i C 1 = I SP e v EB / VT
i C β N + 1 = I SP e v EB / VT
i C β N I SP e v EB / VT
Thus,effectivescale current = β N I SP
Ex:12.12
vI vO
(b)Effectivecurrentgain ≡ i C i B = β P β N
= 20 × 50 = 1000
100 × 10 3 = 50 × 10 14 e v EB /0 025
v EB = 0 025ln 2 × 1011 = 0 651V
Ex:12.11
v O = 0 V ⇒ I D 1 = 10 mA
⇒ VG S 1 = Vtn + V OV 1
= v tn + 2 I D 1 k n
= 0 5 V + 20 mA 50 mA/V2 = 1 13 V
⇒ v I = 1 13 V
FromEq.(12.38),
R L ≥ VSS V OV 2 I = 2 5 V 0 63 V 10 mA = 187
When v O = 0 V,
I D 1 = 50 mA
gm 1 = 2k n I D 1 = √2 × 80 × 50 = 89 mA/V
A v = R L R L + 1/ gm 1 = 100 100 + 1/0 089 = 0 899 V/V
V OV 2 = 2 I k n = 2 × 50 80 = 1 12 V
⇒ v O min =−2 5 + 1 12
=−1 38 V
WemustalsoensureEquation(12.38)issatisfied.
I ≥ VSS V OV 2 R L
50 mA ≥ 1 38 V 100 = 13 8 mA
Thus,when v O = v O min :
I D 1 = 50 mA 1 38 V 100 = 36.2 mA
⇒ gm 1 = 2k n I D 1
= √2 × 80 × 36 2
= 76 1 mA/V
⇒ A v = R L R L + 1/ gm 1
= 100 100 + 1/0 0761 = 0 884 V/V
Thepercentagechangeistherefore, 0 884 0 899
0.899 × 100% = 1 7%
Exercise12–7 Ex:12.13 AccordingtoEquation(12.44),
Rout ∼ = 1 μgm p
⇒ μ ∼ = 1 gm p Rout
= 1
0 05 × 1
= 20 V/V
gm p = 2 I V OV 1
⇒ I = gm p V OV 1 2 = 50 × 0.3 2 = 7 5 mA
Ex:12.14 I Q = 0 1 × 2 V 100 = 2 mA
gm n = gm p = 2 I Q V OV 1,2 = 2 × 2 0 3 = 13 33 mA/V
AccordingtoEquation(12.45), Rout = 1 μ( gm p + gm n )
= 1 10 × (13 3 + 13 3) = 3 75
Ex:12.15 SeeFig.1.
Figure1
Ex:12.16 f s = 10 × highestfrequencyinaudiosignal
= 10 × 20 = 200 kHz
Since f s isadecadehigherthan f P ,thegainwill havefallenby40dB.ThusthePWMcomponent at f s willbeattenuatedby40dB.
Ex:12.17 Maximumpeakamplitude = V DD
Maximumpowerdeliveredto R L = ( V DD /√2)2 R L = V 2 DD 2 R L
Figure1 continued
For V DD = 35 Vand R L = 8 :
Peakamplitude = 35 V
Maximumpower = 352 2 × 8 = 76 6 W
Exercise12–8 Powerdeliveredbypowersupplies = PL η = 76 6 0 9 = 85 1 W
Chapter13
SolutionstoExerciseswithintheChapter
Ex:13.1 UsingEq.(13.2),weobtain
V ICM min =− VSS + Vtn + V OV 3 −| Vtp |
=−1 65 + 0 5 + 0 3 0 5
=−1 35 V
UsingEq.(13.3),weget
V ICM max = V DD −| V OV 3 |−| Vtp |−| V OV 1 |
= 1 65 0 3 0 5 0 3
=+0 55 V
Thus,
1 35 V ≤ V ICM ≤+0 55 V
UsingEq.(13.5),weobtain
VSS + V OV 6 ≤ v O ≤ V DD −| V OV 7 |
Thus
1 65 + 0 5 ≤ v O ≤ 1 65 0 5
⇒−1 15 V ≤ v O ≤+1 15 V
Inaunity-gainconfiguration,theoutputis connectedtothegateof Q 1 andmust,therefore, respectbothlimits.Theoverlapis
1 15 V ≤ v O ≤+0 55 V
Ex:13.2 Foralldevices,wehave
| V An | = 5 V/ µm × 0.5 µm = 2.5 V
V Ap = 6 V/ µm × 0 5 µm = 3 V
UsingEq.(13.11),weget
A 1 =− 2 | V OV 1 | 1 | V A 2 | + 1 V A 4 =− 2 0 2 V 1 3 V + 1 2 5 V =−13 6 V/V
UsingEq.(13.16),weobtain A 2 =− 2 | V OV 6 | 1 V A 6 + 1 | V A 4 | =− 2 0 3 V 1 2 5 V + 1 3 V =−9 1 V/V
Thetotalgainis
A v = A 1 A 2 =−13 6 ×−9 1
= 123 8 V/V
Wefindtheop-ampoutputresistanceasfollows,
r o6 = V An I D 6 = 2 5 V 0 1 mA = 25 k
r o7 = V Ap I D 6 = 3 V 0 1 mA = 30 k
Ro = r o6 r o7 = 25 30 = 13 6 k
Ex:13.3 Thefeedbackisofthevoltagesampling type(i.e.,theconnectionattheoutputisashunt one),thus
Rout = Rof = Ro 1 + Aβ where
Ro = r o6 r o7
A = gm 1 (r o2 r o4 ) gm 6 (r o6 r o7 )
β = 1
Thus, Rout = r o6 r o7 1 + gm 1 (r o2 r o4 ) gm 6 (r o6 r o7 )
Usually,
A 1
Thus,
Rout 1 gm 6 [ gm 1 (r o2 r o4 )]
Ex:13.4 UsingEq.(13.31),weget
f t = G m 1 2π C C
⇒ C C = G m 1 2π f t
= 0 35 × 10 3
2π × 20 × 106
= 2 8 pF
FromEq.(13.26),wehave
f Z = G m 2 2π C C = 1 0 × 10 3
2π × 2 8 × 10 12 57 MHz
FromEq.(13.30),wehave
f P 2 = G m 2 2π C 2
= 1.0 × 10 3 2π × 2 × 10 12 = 80 MHz
Thus, f t islowerthan f Z and f P 2
Exercise13–2
Ex:13.5 (a)UsingEq.(13.36),wehave
f t = G m 1 2π C C
⇒ C C = G m 1 2π f t = 1 × 10 3
2π × 100 × 106 = 1 6 pF
A 0 = G m 1 (r o2 r o4 ) G m 2 (r o6 r o7 )
= 1(100 100) × 2(40 40)
= 50 × 2 × 20 = 2000 V/V
f 3dB = f t A 0 = 100 × 106 2000 = 50 kHz
(b)FromEq.(13.34),wehave
R = 1 G m 2 = 1 2 × 10 3 = 500 (c)FromEq.(13.35),wehave
f P 2 = G m 2 2π C 2 = 2 × 10 3 2π × 1 × 10 12 = 318 MHz
φ P 2 =−tan 1 f t f P 2 =−tan 1 100 318 =−17 4◦
PM = 90 17 4 = 72 6◦
Ex:13.6 UsingEq.(13.47),weobtain
SR = V OV 1 ωt = 0 2 × 2π × 100 × 106 = 126 V/µs
UsingEq.(13.45),
SR = I C C
⇒ I = 126 × 106 × 1 6 × 10 12 = 200 µA
Ex:13.7
G m 2 = gm 6 = 2 I D 6 V OV = 2 × 0 5 0 25 = 4 mA/V
f P 2 = G m 2 2π × C 2 = 4 2π × 15 × 10 12 = 42 MHz
G m 1 = gm 1 = 2( I /2) V OV = 2 × 0 3 0 25 = 1.2 mA/V
f t = G m 1 2π × C C ⇒ C C = G m 1 2π × f t = 1.2 × 10 3 2π × 15 3 × 106 = 12 5 pF
Ex:13.8 I REF = 0 2 mA 10 = 20 µA = I D 8
I D 8 = 1 2 k p W L 8 V 2 OV ⇒ W L 8 = 2 I D 8 k p × V 2 OV = 2 × 20 × 10 6 86 × 10 6 × 0 52 = 1 86
W8 = 1 86 × L 8 = 1 86 × 0 6 = 1 1 µm
W5 W8 = I D 5 I D 8 ⇒ W5 = 200 20 × 1 1 = 11 µm
W6 W8 = I D 6 I D 8 ⇒ W6 = 400 20 × 1.1 = 22 µm
| VG S 8 |=| Vtp |+| V OV |= 0 5 + 0 5 = 1 0 V
Therequiredresistorvalueis
R = 1 V 1 V ( 1 V) I REF = 1 V 20µA = 50 k
Ex:13.9 Totalbiascurrent = 300 µA = 2 I B
⇒ I B = 150 µA
I B = I D 1 + I D 3 150 = I D 1 + 0 25 I D 1
⇒ I D 1 = 120 µA
I = I D 1 + I D 2
= 120 + 120 = 240 µA
I D 3,4 = 0 25 I D 1,2 = 0 25 × 120 = 30 µA
Ex:13.10 UsingEq.(13.50),weget
V ICM max = V DD −| V OV 9 |+ Vtn
= 1.65 0.3 + 0.5 =+1.85 V UsingEq.(13.51),weobtain
V ICM min =− VSS + V OV 11 + V OV 1 + Vtn
V ICM min =−1 65 + 0 3 + 0 3 + 0 5
tan 1 f t f P 2 = 20◦ ⇒ f t = f P 2 × tan20◦ = 15 3 MHz
=−0 55 V
Thus,
0 55 V ≤ V ICM ≤+1 85 V
UsingEq.(13.54),weget
V O max = V DD −| V OV 10 |−| V OV 4 |
= 1 65 0 3 0 3 =+1 05 V
UsingEq.(13.55),weobtain
V O min =− VSS + V OV 7 + V OV 5 + Vtn
=−1 65 + 0 3 + 0 3 + 0 5
=−0 55 V
Thus,
0 55 V ≤ v O ≤+1 05 V
Ex:13.11 G m = gm 1 = gm 2
G m = 2( I /2) V OV 1 = I V OV 1
= 0 24 0 2 = 1 2 mA/V
r o2 = | V A | I /2 = 20 0 12 = 166 7 k
r o4 = | V A | I D 4 = | V A | I B I 2 = 20 0 150 0 120
= 20 0 03 = 666 7 k
r o10 = | V A | I B = 20 0 15 = 133 3 k
gm 4 = 2 I D 4 | V OV | = 2 × 0 03 0 2 = 0 3 mA/V
Ro4 = ( gm 4 r o4 )(r o2 r o10 )
= (0 3 × 666 7)(166 7 133 3)
= 14 8 M
gm 6 = 2 I D 6 V OV = 2 × 0 03 0 2 = 0.3 mA/V
r o6 = r o8 = | V A | I D 6,8 = 20 0 03 = 666 7 k
Ro6 = gm 6 r o6 r o8
= 0.3 × 666.7 × 666.7 = 133.3 M
Ro = Ro4 Ro6 = 14 8 133 3 = 13 3 M
A v = G m Ro = 1 2 × 13 3 × 103 = 16,000 V/V
Ex:13.12 (a)TheNMOSinputstageoperates overthefollowinginputcommon-moderange:
VSS + 2 V OV + Vtn ≤ V ICM ≤ V DD −| V OV |+ Vtn thatis,
( 2 5 + 0 6 + 0 7) ≤ V ICM ≤ (2 5 0 3 + 0 7)
1 2 V ≤ V ICM ≤+2 9 V
(b)ThePMOSinputstageoperatesoverthe followinginputcommon-moderange: VSS + V OV −| Vtp |≤ V ICM ≤ V DD 2| V OV |−| Vtp | thatis,
( 2 5 + 0 3 0 7) ≤ V ICM ≤ (2 5 0 6 0 7) 2 9 V ≤ V ICM ≤+
(c)Theoverlaprangeis
(d)Theinputcommon-moderangeis 2 9 V ≤ V ICM ≤+2 9 V
Ex:13.13 Denotethe ( W/ L ) ofthetransistorsin thewide-swingmirrorby ( W/ L ) M .Transistor Q 4 has ( W/ L )5 = 1 4 ( W/ L ) M IREF = 1 2
Thus, V OV 5 2 = V OV where V OV istheoverdrivevoltageforeachof themirrortransistors.Thus, V5 = Vtn + 2 V OV whichisthevalueof VBIAS neededinthecircuit ofFig.13.13(b).
Ex:13.14 I S 2 = 2 I S 1 UsingEq.(13.72),weobtain I = VT R2 ln I S 2 I S 1 0 01 = 0 025 R2 ln2 ⇒ R2 = 1 73 k
R3 = R4 = 0 2 V 0 01 mA = 20 k
Ex:13.15 Toobtain I5 = 10 µA,transistor Q 5 musthavethesameemitterareaas Q 2 and
R5 = R2 = 1.73 k
Toobtain I6 = 5 µA,transistor Q 6 musthave one-halftheemitterareaof Q 2 and
R6 = 2 R2 = 3 46 k
Toobtain I7 = 20 µA,transistor Q 7 musthave doubletheemitterareaof Q 3 and
R7 = 0 5 R3 = 10 k
Toobtain I8 = 10 µA,transistor Q 8 musthave thesameemitterareaas Q 3 and
R8 = R3 = 20 k
Ex:13.16 RefertoFig.13.23.AtnodeX, I REF = 2 I 1 + 2 β P + 2 I β P
2 I β P + 1 β P + 2
2 I
Thus, I = I REF 2 = 9.5 µA resultingin
IC 1 = IC 2 = IC 3 = IC 4 = 9 5 µA
Ex:13.17 Figure1onnextpageshowsthe determinationoftheloopgainofthefeedback circuitthatstabilizesthebiascurrentsofthefirst stageofthe741opamp.Notethatsince I REF is assumedtobeconstant,wehaveshownits incrementalvalueatnodeXtobezero.Observe thatthiscircuitshowsonlyincremental quantities.Theanalysisshownprovidesthe returnedcurrentsignalas
Ir =− It β P 1 + 2 β P For β P 1,wehave
Ir −β P It andtheloopgain Aβ is
Ex:13.18 I B = 1 2 ( I B 1 + I B 2 ) = 1 2 I β N + 1 + I β N + 1 = I β N + 1 I β N = 9 5 200 = 47 5 nA
I OS = 0 1 × I B = 4 75 nA
Ex:13.19 VC 1 = VCC V EB 8 = 15 0 6 = 14 4 V
Q 1 and Q 2 saturatewhen V ICM exceeds VC 1 by 0.3V.Thus,
V ICM max =+14.7 V
Ex:13.20 r e = VT I E 25 mV 9 5 µA = 2 63 k
gm 1 1 r e = 0 38 mA/V
G m 1 = 1 2 gm 1 = 0 19 mA/V
Rid = (β N + 1) × 4r e = 201 × 4 × 2 63
= 2.1 M
Ex:13.21 r o4 = | V Ap | I = 50 V 9 5 µA = 5 26 M
gm 4 = 0 38 mA/V
r e 2 = 2 63 k
r π 4 = β P gm 4 = 50 0 38 = 131 6 k
Ro4 = r o4 1 + gm 4 (r e 2 r π 4 )
= 5 261 + 0 38(2 63 131 6)
= 10 4 M
10.5M isobtainedbyneglecting r π 4
r o6 = V An I = 125 V 9 5 µA = 13 16 M
gm 6 = 0 38 mA/V
R6 = 1 k
r π 6 = 200 0 38 = 526 3 k
Ro6 = r o6 1 + gm 6 ( R2 r π 6 )
= 13 161 + 0 38(1 526 3)
Exercise13–5
ThisfigurebelongstoExercise13.17.
= 18 2 M
Ro1 = 0 5 × Ro4 = 5 3 M
Ex:13.22 | A v o |= G m 1 Ro1
Using G m 1 givenintheanswertoExercise13.20,
G m 1 = 0 19 mA/V and Ro1 givenintheanswertoExercise13.21,
Ro1 = 5.3 M weobtain
| A v o |= G m 1 Ro1
= 0 19 × 5 3 = 1007 V/V
Ex:13.23
Ro6 = r o6 = | V Ap | IC 6 = 50 V 19 µA = 2 63 M
Ro7 = r o7 [1 + gm 7 ( R4 r π 7 )]
where
r o7 = V An IC 7 = 125 V 19 µA = 6.58 M
gm 7 = IC 7 VT = 19 µA 0 025 V = 0 76 mA/V
R4 = 5 k
r π 7 = β N gm 7 = 200 0 76 = 263 2 k
Ro7 = 6 58[1 + 0 76(5 263 2)] = 31.1 M
Ro = Ro6 Ro7 = 2 63 31 1 = 2 43 M
Ex:13.24 UsingEq.(13.84),weobtain
G m cm = m 2 Ro
= 5.5 × 10 3 2 × 2 43 × 106 = 1 13 × 10 6 mA/V
Figure1
CMRR = G m 1 G m cm = 0 19 1 13 × 10 6 = 1.68 × 105 or 104 5 dB
Withoutcommon-modefeedback,theCMRRis reducedbyafactorequalto β P .Equivalently,
CMRR = 104 5 20 log β P
= 104 5 20 log 50
= 70 5 dB
Ex:13.25 RefertothecircuitinFig.13.29.
(a)Currentgainfrom v IP tooutput
= (β1 + 1)(β2 + 1)β P
β1 β2 β P = β N β 2 P
Currentgainfrom v IN tooutput = (β3 + 1)β N
β3 β N = β 2 N
(b)For i L =+10 mA,
currentneededat v IP input
= 10 β N β 2 P = 10 40 × 102 = 2 5 µA
For i L =−10 mA,
currentneededat v IN input = 10 β 2 N = 10 402 = 6 25 µA
Ex:13.26 I Q = 0 4 mA, I = 10 µA, I SN I S 10 = 10, I S 7 I S 11 = 2
UsingEq.(13.93),weobtain 0 4 × 103 = 2 I 2 REF 10 × 10 × 2
where IREF isin µA.Thus, IREF = 10 µA
For i L =−10 mA,then i P = 10 + i N
UsingEq.(13.94),weget
i N (10 + i N ) i N + 10 + i N = 0 2 ⇒ i 2 N 9 6i N + 2 = 0 ⇒ i N = 0 2 mA i P = 10 2 mA
Chapter14
SolutionstoExerciseswithintheChapter
Ex:14.1 A =−20log| T | [dB]
| T | = 10.990.90.80.70.50.10
A 00.1123620 ∞
Ex:14.2 A max =−20log0 9 0 9dB
A min = 20log 1 0 01 = 40dB
Ex:14.3 (a)
T (s ) = K
s 3 + b2 s 2 + b1 s + b0
Thisisalow-passfilterwith3transmissionzeros at s =∞
(b)
T (s ) = Ks
s 3 + b2 s 2 + b1 s + b0
Thereisonetransmissionzeroat s = 0,andtwo transmissionzerosat s =∞ ⇒ Bandpassfilter.
(c)
T (s ) = Ks 2
s 3 + b2 s 2 + b1 s + b0
Therearetwotransmissionzerosat s = 0,and onetransmissionzeroat s =∞ ⇒ Bandpass.
(d)
T (s ) = Ks 3
s 3 + b2 s 2 + b1 s + b0
Threetransmissionzerosat s = 0 ⇒ Highpass.
Ex:14.4
T (s ) = k (s + j 2)(s j 2)
s + 1 2 + j 3 2 s + 1 2 j 3 2
s 2 + 4
= k
s 2 + s + 1 4 + 3 4
= k s 2 + 4
s 2 + s + 1
T (0) = k 4 1 = 1
k = 1 4
∴ T (s ) = 1 4 s 2 + 4
s 2 + s + 1
Ex:14.5
T (s ) = a3 s s 2 + 4
(s + 0 1 + j 0 8)(s + 0 1 j 0 8) × 1
(s + 0 1 + j 1 2)(s + 0 1 j 1 2)
= a3 s s 2 + 4
s 2 + 0 2s + 0 65 s 2 + 0 2s + 1 45
Ex:14.6
T (s ) =− R2 R1 1 1 + s ω0
where R2 / R1 = 10 and ω0 = 105 rad/s.The | T | willbereducedby3dBfromthevalueatdc,at
ω = ω0 = 105 rad/s.Thus,
ω P = 105 rad/s
The | T | at ω = ωs = 3ω0 is | T ( j ωs ) |= 10 1 + 3ω0 ω0 2
Thus,relativeto | T (0) |,thetransmissionwillbe lowerby 20 log 1 + 32 ,whichis10dB.Thus,
A (ωs ) = 10 dB
Ex:14.7 R2 R1 Vi Vo C 104 = 1 CR 1 , R1 = 10k
C = 0 01 µF = 10nF
H.F.gain= R2 R1 =−10
R2 = 100k
Ex:14.8 T (s ) = ω 2 0 s 2 + s ω0 Q + ω 2 0
Formaximallyflatresponse, Q = 1/√2,thus
T (s ) = ω 2 0
s 2 + s √2 ω0 + ω 2 0
T ( j ω) = ω 2 0 (ω 2 0 ω 2 ) + j √2 ωω0
| T ( j ω) |= ω 2 0 (ω 2 0 ω 2 )2 + 2ω 2 ω 2 0
= ω 2 0 ω 4 0 + ω 4 = 1 1 + ω ω0 4 At ω = ω0 ,
| T ( j ω0 ) |= 1 √1 + 1 = 1 √2 whichis3dBbelowthevalueat ω = 0 (0dB). Q E D
Ex:14.9 T (s ) = sK ω0 Q s 2 + s ω0 Q + ω 2 0
where K isthecenter-frequencygain.For ω0 = 105 rad/sand3-dBbandwidth = 103 rad/s, wehave
3-dB BW = ω0 Q
103 = 105 Q
⇒ Q = 100
Also,foracenter-frequencygainof10,wehave
K = 10
Thus,
T (s ) = 104 s s 2 + 103 s + 1010
Ex:14.10
T (s ) = k s 2 + ω 2 n s 2 + s ω0 Q + ω 2 0 For ωn = 1 2 rad/s, ω0 = 1 rad/s, Q = 5,anddc gain= k ω 2 n ω 2 0 = 1,i.e., k = 0 694,
T (s ) = 0 694 s 2 + 1 44
s 2 + 0 2 s + 1
T (∞) = 0 694
Ex:14.11 | T ( j ω) |= 1/ 1 + 2 ( ω ω P )2 N ω P = ω3dB , = 1, and N = 5, | T ( j ω) |= 1/ 1 + ( ω ω3dB )10
= 1/ 1 + ( f f 3dB )10
| T ( j ωs ) |= 1/ 1 + ( 30 10 )10
A (ωs ) =−20 log | T ( j ωs ) | = 47 7 dB
Ex:14.12 = 10 A max 10 1 = 10 1 10 1 = 0 5088
| T ( j ω ) | = 1 1 + 2 ω ω p 2 N A (ω s ) =−20log| T ( j ω s )| = 10log 1 + 2 ω s ω p 2 N
Thus, 10log 1 + 0 50882 × 1 52 N ≥ 30
N = 10: LHS = 29 35dB
N = 11: LHS = 32 87dB
∴ Use N = 11andobtain
A min = 32 87dB
Ex:14.13 Therealpoleisat s = –1
Thecomplexconjugatepolesareat
s =− cos60◦ ± j sin60◦
=−0.5 ± j 3 2
Notethat ω0 = 1 rad/sand Q = 1. j ( )1/3 1 s 30 60 1 e
T (s ) = 1 (s + 1) s + 0 5 + j 3 2 s + 0 5 j 3 2
= 1 (s + 1) s 2 + s + 1
DCgain = 1
Ex:14.14
= 10 A max 10 1 = 10 0 5 10 1 = 0 3493
A (ω s ) = 10log 1 + 2 cosh2 N cosh 1 ω s ω p
= 10log 1 + 0 34932 cosh2 7cosh 1 2
= 64 9dB
For A max = 1dB, = 100 1 1 = 0 5088
A (ω s ) = 10log 1 + 0 50882 cosh2 7cosh 1 2
= 68 2dB
Thisisanincreaseof3.3dB
Ex:14.15 = 10 1 10 1 = 0 5088
(a)FortheChebyshevfilter:
A (ω s )
= 10log 1 + 0 50882 cosh2 N cosh 1 1 5
≥ 50dB
N = 7.4 ∴ choose N = 8
Excessattenuation =
10log 1 + 0 50882 cosh2 8cosh 1 1 5 50
= 55 50 = 5dB
(b)ForaButterworthfilter
= 0 5088
A (ω s ) = 10log 1 + 2 ω s ω p 2 N
= 10log 1 + 0 50882 (1 5)2 N ≥ 50
N = 15 9 ∴ choose N = 16
Excessattenuation =
10log 1 + 0 50882 (1 5)32 50 = 0 5dB
Ex:14.16 Maximallyflat ⇒ Q = 1 √2
ω 0 = 2π × 100 × 103
Arbitrarilyselecting R = 1k ,weget
Q = ω 0 CR ⇒ C = 1 √2 × 2π 105 × 103 = 1125pF L R C Vi Vo Also Q = R ω 0 L ∴ L = R ω 0 Q = 103 2π 105 × 1 √2 = 2.25mH
Ex:14.17 RefertoFig.14.19(c).
ω0 = 1/√ LC = 105 rad/s
3-dB BW = ω0 Q = 1 CR = 103 rad/s
Selecting R = 100 k ,
C = 1 103 R = 1 103 × 105 = 10 8 F = 10 nF
L = 1 ω 2 0 C = 1 1010 × 10 8 = 0 01 H = 10 mH
Ex:14.18 RefertoFig.14.20(a).Selecting,
R1 = R2 = R3 = R5 = R and
C 4 = C gives
Z in = sCR 2 Thus, L = CR 2
Toobtain L = 100 mH,weselect C = 1 nF. Thus,
100 × 10 3 = 1 × 10 9 R 2
⇒ R = 104 = 10 k
Thevaluesof C and R arepracticallyconvenient.
Ex:14.19
f 0 = 10kHz, f 3dB = 500Hz
Q = f f 3dB = 104 500 = 20
UsingEqs.(14.53)and(14.54)andselecting, C = 1 2nF,
weget,
R = 1 ω 0 C
= 1 2π104 × 1 2 × 10 9 = 13 26k
R6 = QR = 20 × 13 26 = 265k
Now,
K = center-frequencygain = 10
Thus, 1 + r 2 / r 1 = 10
Selecting r 1 = 10k ,weobtain r 2 = 90k .
Ex:14.20 RefertotheKHNcircuitinFig.14.24.
Choosing C = lnF,weobtain
R = 1 ω 0 C = 1 2π 104 × 10 9 = 15.9k
UsingEq.(14.65)andselecting R1 = 10k , weget
R f = R1 = 10k
UsingEq.(14.66)andsetting R2 = 10k , we obtain
R3 = R2 (2 Q 1) = 10 (2 × 2 1) = 30k
High-frequencygain = K = 2 1 Q = 1.5V/V
Thetransferfunctiontotheoutputofthefirst integratoris
Vbp
Vi =− 1 sCR × Vhp Vi = sK / (CR ) s 2 + s ω 0 Q + ω 2 0
Thusthecenter-frequencygainisgivenby
K
CR Q ω 0 = KQ = 1 5 × 2 = 3V/V
Ex:14.21 RefertoFig.14.25(b)
CR = 1 ω 0 ⇒ C = 1 2π 104 × 104 = 1 59nF
Rd = QR = 20 × 10 = 200k
Centerfrequencygain = KQ = 1
∴ K = 1 Q = 1 20
R g = R / K = 20 R = 200k
Ex:14.22 RefertoFig.14.27(a)andEqns. (14.80)-(14.85).Selecting
C 1 = C 2 = C = 1 nF, and,
R3 = R then,
R = 10 4 10 9 = 105 = 100 k
R3 = R = 100 k
R4 = R 100 = 1 k
Thecenter-frequencygainisgivenbyEq. (14.85),
K =−2 Q 2 =−2 × 25 =−50 V/V
Ex:14.23 RefertoFig.14.29andEqns. (14.94)-(14.101).Selecting
R1 = R2 = R
C 4 = C then
C 3 = C 4 Q 2 = C 4 × 1 2 = C 2
Thus,ifweselect C = 1 nFthen
C 4 = 1 nF
C 3 = 0 5 nF Now,
CR = 2 Q ω0 = 2 × 1/√2 105 = 1 414 × 10 5 s
Thus, R = 1 414 × 10 5 1 × 10 9 = 1 414 × 104 = 14 14 k
⇒ R1 = R2 = 14 14 k
Finally,fromthetransferfunctioninEq.(14.94)
DCgain = 1 V/V
Ex:14.24 FromEq.(14.108),
C 3 = C 4 = ω 0 Tc C = 2π 104 × 1 200 × 103 × 20 = 6 283pF
FromEq.(14.110),
C 5 = C 4 Q = 6 283 20 = 0 314pF
FromEq.(14.111),
R4 = R 4 Q 2 = R 4 × 25 = R 100 and CR = 2 Q ω0 = 2 × 5 105 = 10 4 s
Centre-frequencygain= C 6 C 5 = 1
C 6 = C 5 = 0 314pF
Chapter15
SolutionstoExerciseswithintheChapter
Ex:15.1 Polefrequency f 0 = 1kHz
Center-frequencygain = 1 Amplifiergain
= 1 2 V/V
Ex:15.2 RefertoFig.15.2(a).
ω0 = 1 √ LC = 1 √0 1 × 10 3 × 10 × 10 9 = 106 rad/s
Theloopgainat ω0 = (1 + r 2 r 1 ) × 1 = 1 01
Thus,theconditionofoscillationissatisfied.
Ex:15.3 ω0 = 1 √ LC (a)
L −→ 1 01 L
ω0 + ω0 = 1 √1 01 LC = 0 995 √ LC
Thus,
ω0
ω0 × 100 =−0 5%
(b)
C −→ 1 01C
ω0 + ω0 = 1 √ L × 1 01C = 0.995 √ LC
ω0
ω0 × 100 =−0 5% (c)
R −→ 1 01 R
Since ω0 is not afunctionof R ,changing R has noeffecton ω0 ,thatis, ω0 = 0.
Ex:15.4 RefertoFig.15.5.Notingthatthe sinusoidatthepositiveinputterminaloftheop amphasapeakamplitudeof0.9V,toobtaina 5-Vpeakamplitudeattheoutput,theclosed-loop gainoftheopampmustbe
1 + r 2 r 1 = 5 0 9 = 5 6
⇒ r 2 r 1 = 4 6
Ex:15.5 (a) L (s ) = 1 + R2 R1 Z P Z P + Z S = 1 + R2 R1 1 1 + Z S Y P = 1 + 20 3 10 1 1 + R + 1 sC 1 R + sC
= 3 03
3 + sCR + 1 sCR
where R = 10k and C = 16nF
Thus,
L (s ) = 3 03
3 + s 16 × 10 5 + 1 s × 16 × 10 5
Theclosed-looppolesarefoundbysetting L (s ) = 1;thatis,theyarethevaluesofs, satisfying
3 + s × 16 × 10 5 + 1 s × 16 × 10 5 = 3 03
⇒ s = 105 16 (0 015 ± j )
(b)Thefrequencyofoscillationis 105 /16 rad/s orapproximately1kHz.
(c)UsingEq.(15.11),thepositivepeakamplitude canbefoundas
v o+ = 1 3 × 15 + 1 + 1 3 × 0 7 / 2 3 1 3 × 1 3
= 10 68 V
Sincethelimitercircuitissymmetric,the negativepeakamplitudealsowillbe10.68V. Thus,thepeak-to-peakamplitudeoftheoutput sinusoidwillbe
ˆ v op p = 2 × 10 68 = 21 36 V
Ex:15.6 (a)Foroscillationstostart,weneed fromEq.(15.10), R2 / R1 = 2; thusthe potentiometershouldbesetsothatitsresistance togroundis 20k
(b)FromEq.(15.9), f 0 = 1 2π RC = 1 2π 10 × 103 × 16 × 10 9 = 1kHz
Ex:15.7 Figure1showsthecircuittogetherwith theanalysisdetails.Thecurrent I canbefoundas I = Vo R f 1 + 1 sCR + Vo sCRR f 2 + 1 sCR
Finally, Vx canbefoundfrom Vx =− 1 sCR f Vo 2 + 1 sCR I sC =− 1 sCR f Vo 2 + 1 sCR Vo sCR f 1 + 1 sCR
ThisfigurebelongstoExercise15.7.
Vx Vo = 3 sCR f + 4 s 2 C 2 RR f + 1 s 3 C 3 R 2 R f
Vo Vx =− s 2 C 2 RR f 4 + s 3CR + 1 sCR
For s = j ω ,
Vo Vx = ω 2 C 2 RR f 4 + j 3ω CR 1 ω CR
Ex:15.8 Fromtheloop-gainexpression
Vo Vx = ω 2 C 2 RR f 4 + j 3ω CR 1 ω CR weseethatthephasewillbezeroat
3ω0 CR = 1 ω0 CR
⇒ ω0 = 1 √3CR
Atthisfrequency,wehave
Vo Vi = ω 2 0 C 2 RR f 4 = 1 12 R f R
Thustheminimumvaluethat R f musthavefor oscillationstostartis
R f = 12 R
UsingthevaluesinFig.15.10,weobtain
ω0 = 1 √3 1 16 × 10 9 × 10 × 103 = 3 608 krad/s
f 0 = ω0 2π = 3.608 × 103 2π = 574 3 Hz
R f ≥ 12 × 10 = 120 k
Ex:15.9 ω 0 = 1 CR ⇒ CR = 1 2π 103
For C = 16nF, wehave R = 10k . Thepeak-to-peakamplitudeofthesquarewaveis 1.4V,whichhasafundamental-frequencycomponentof 4 × 1 4 π = 1 8 Vpeak-to-peak.Since theoutputistwiceaslargeasthevoltageacross theresonator,itspeak-to-peakamplitudeis3.6V.
Ex:15.10 RefertoFig.15.14(a).Theadmittance seenbythetransistorbetweendrainandsourceis
Y = sC 1 + 1 sL + 1 sC 2 = s 2 LC 1 + C 1 C 2 + 1 sL + 1 sC 2 = s 2 + 1/ L C 1 C 2 C 1 + C 2 s 1 C 1 + 1 sLC 1 C 2
Y ( j ω) = ω 2 + 1/ L C 1 C 2 C 1 + C 2 j ω C 1 1 1 ω 2 LC 2 fromwhichweseethatat
ω = ω0 = 1/ L C 1 C 2 C 1 + C 2 thenumeratoriszero and Y ( j ω0 ) = 0.Q.E.D.
Ex:15.11 RefertothecircuitinFig.15.15(a).
R = Rcoil r o R L = Q ω0 C 1 100 × 103 100 × 103 = 100 106 × 0 01 × 10 6 105 105
Figure1
