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Solution Manual For Electronics, 2nd Edition - Document preview page 1

Solution Manual For Electronics, 2nd Edition - Page 1

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Solution Manual For Electronics, 2nd Edition

With Solution Manual For Electronics, 2nd Edition, you can tackle your textbook assignments with confidence, knowing you have the correct answers.

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Page 1 of 16
Solution Manual For Electronics, 2nd Edition - Page 1 preview imageContentsChapterl....i.ieieeeeeeeenenenanceeanenoaanonassasasssaaanalChapter2....ceceeeneeenneneneasoseseseassansasaaannaess38Chapter3.....i.ieieieececieteeieenereacessccsnacaccassessssl00Chapter4........icececiieiieeneececcccsoncnnncnaseesss.154Chapter5...cuiiiienenreeeeanenecassssssanaasseanasassasa203CHAPLETBueeeeeeeeeeeeeecaseesneeeannecenanennnennsasa244CHAPLET7reneenveeeennnnnnnnnenaeeeeeeeennnnnnnnnnssa284ChapterB8......cieieeeeieteeneencecersecsssonenancesssa328Chapter9.....ieriiieeeeereeenenneasccecesnenennaassssa382Chapter10..tuuceenueeaneeaneeaneenanrenneeenneeenns.ddlChapter1l......ceeeereecneeeaccccnascconncnneonenesess.d467Chapter12.........iiiteeneenenecnencssseascanscnnsacassabll
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Solution Manual For Electronics, 2nd Edition - Page 2 preview imageDownloadedfromStudyXY.com®+StudyXYSdYe.o>\|iF’prE\3SStudyAnythingThisContentHasbeenPostedOnStudyXY.comassupplementarylearningmaterial.StudyXYdoesnotendroseanyuniversity,collegeorpublisher.Allmaterialspostedareundertheliabilityofthecontributors.wv8)www.studyxy.com
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Solution Manual For Electronics, 2nd Edition - Page 3 preview imageChapter1(a)Foranoninvertingamplifiera,=+50andwehave:Vo(t)=Avi(t)=50x0.1sin(2000mt)=5sin(20007nt)(b)ForaninvertingamplifierA,=-50andwehave:vo(t)=Avy(E)=-50x0.1sin(2000mt)=-5sin(2000mt)ReSn+1,SooniVsOv.R:RQ0xY2“bn<>750R;R.R;1Li4A=o—————xXAX=——=—»=300A.=A=-=10vs~Rg+Ryvo©RFRjivRp,vBL_Oo____6A,TyTAxgsg-375G=A;=3.75x10iLoFormaximumpowertransfertotheload,wemusthaveRp,=R,=25Q.ThenasinExercise1.2wefindA,=250,A;=2x10%,andG=5x10°.Rs,RonRos++++YU,R,@{TAuveeKisAvosVisUy1+StudyXxy
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Solution Manual For Electronics, 2nd Edition - Page 4 preview imageR,R,i2i3A=A=A==A=5357vovolRp+Riovo2Roa+Ris'vo3R;=R;;=1000@R,=R_,=300QR,R;i2i1A=AosA5A=4348vovo3Ros+Risvo2Roy+RiqvolRy=Ry;=3000QR,=R,,=1000P=(1.54)x(15V)=22.5WPy=P+P;=P=20.5WPsm=5°x100%=11.1%SsRo=loonoRe{7R.=30054knA,=v_/v.=2000J1500vof’"iRy+RgAggp=20loglAl=63.5dBR,c=(a)%t=1.5x10"VvRLGup=10logG=71.8AB2+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 5 preview imagePB,=10log|-B|=1010g|5107>|=23.0aBwdBWTwI=:P5x10”P4Bm=10logTow=1oleg("257=10log(5)=6.99dBmv20log|_%|=23»v_=1023/20_14.03vTvxAiJi+-R;P+Open7;Ra<8PyVs circait-orload.v200i.R[=]10a=0C2=-__"10°0_R,=1000@R_=20Qvovi1;Ry1oAzAose—>—>+R;4)viSoo7,—SonISRmscVv.5001,11R;=500QR=5003+StudyXxy
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Solution Manual For Electronics, 2nd Edition - Page 6 preview imageA;4++2&HoRoUs4idDeciM0.05;loa_vG__V.R__ooc_"msc'io_—Rnoc=71;v./R;CnscRoRy=500kAiiT(a)WehaveR_<<R,andR,>>R_.ThisisanearlyidealSsiLovoltageamplifier.(b)WehaveR_>>R.andR,<<R_.ThisisanearlyidealSsiLoYcurrentamplifier.(c)WehaveR_<<R;andR,<<R_.ThisisanearlyidealSsiL024transconductanceamplifier.(d)WehaveRy>>R;andRp>>Rg.Thisisanearlyidealtransresistanceamplifier.(e)WehaveR_=R,andR.<<R_.ThisisnotclosetoanySs1Loidealamplifier.Exercise1.14Aun=Voen/Vien=0-1/1=0.1A.naB=20logla|=-20dB1A]3=d_50x10°_CMRR35=20logz——=20log—pG—5—=114dBcm4+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 7 preview imageExercise1.15(@)Vig=Vip"Vip,TLV-1=Viem=2(Via*Vip)SOVA.+A___tM2Vo=AVipTAVi;=—3A/2+A_/2_i!2/%1Aq=Vo/Vig=TTFTA)(6)Vig=Vip"VipSOV=1-Viem=7(Vip*Vip)1VYVo=AyVipTAVipTACAyAen=Vo/ViemTBp72;(¢)1A]A.+A_aq12_201_CMRR=2010g—=20log3G.A20log(100-101)|=40dBcm12Problem1.1Someexamplesofelectronicsystemsareelectronicbrakes,printers,cashregisters,microwaveovens,CDplayers,airportlandingsystems,electronicdoorlocks,andsoforth.Problem1.2Electronicsystemblocksincludeamplifiers,filters,signalsources,wave-shapingcircuits,digitallogicfunctions,digital.memories,powersupplies,andconverters.Problem1.3Someelectronicsystemsprocessinformationinelectronicformandsomepower(hopefullyaslittleaspossible)isconsumed.Inpowerelectronics,thepowerdeliveredtoaloadisthemainconcern.5+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 8 preview imageConversionofanalogsignalstodigitalformisatwo-stepprocess.First,thesignalissampledatperiodicpointsintime.Second,eachsampleisapproximatelyrepresentedbyacodeword.(11Samples110\{lolAndlogI100JouSigned/1ojoloo|Ooo_OfA>ofiloGlogGooool©ojocool0©0e00Jol0(io!Digitalins)wePaleProblem1.5Providedthatitisnottoolargeinamplitude,noisecanbecompletelyremovedfromadigitalsignal.Noisetendstoaccumulateinanalogsignals.Digitalcircuitstendtobeeasierthananalogcircuitstoimplementwithintegratedtechniques.Thusextremelycomplexdigitalsystemsarefeasiblewhileequallycomplexanalogsystemsarenot.Digitalsystemsaremoreadaptablethananalogsystemstoavarietyofuses.6+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 9 preview imageProblem1.6Numberofbitspersecond=16x44.1x10°=705.6kbit/s(formonaural)(1.411Mbits/sareusedforstereo.)Numberofamplitudezones=216=65,536-5-(5)_A=~€58536152.6uvMinimumsamplingrate=2fy=200sample/s_10mv_.._10_N=0.01mV—1000whichrequiresk10atleast(21024)Numberofbitspersecond=200x10=2kbit/sProblem1.8SeeFigure1.6inthebook.Becausewearelimitedtoresistors,theonlyoptionisaresistivevoltagedividerasshownbelow.©-0.34=RIIR—Ry,!I2+R+=Load=“y—R,—ViWedenotethenominalvaluesoftheresistorsasRyandR,.Thehighestloadvoltage(atmost6V)occurswhenI=0,whentheresistorinparallelwiththeloadhasitshighestvalue(whichis1.05R,),andwhentheresistorinserieswiththesourcehasitslowestvalue(whichis0.95R,;).Toachievethedesiredno-loadvoltageweneed7+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 10 preview image1.05R,|PRE0.95R,+1.05R,SolvingforRy/wehaveRy=0.6786Ry(1)Thesmallestloadvoltage(atleast4V)occurswithI,=0.3andresistancevaluesof0.95R,and1.05R,.Forthesevalues,theThéveninvoltageis0.95R.Vv.=14m2th1.05R,+0.95R12andtheloadvoltageisVy=4=VenTRenn0.95R.0.95(1.05)R.R4=14—2_0.3—12(2)1.05R;+0.95R,1.05Ry+0.95R,UsingEquation(1)tosubstituteforR,inEquation(2)andsolvingweobtain:R,=11.06Q1ThenfromEquation(1)weobtain:R,=7.507QThuswecouldusethestandardnominalvaluesofRy=11QandRy=7.5Q.Systemengineersdesigntheblockdiagramsofsystemsincludingspecificationsforeachblock.Circuitdesignersdesignthecircuitsforeachblock.Processengineersdesignthefabricationprocesses.Semiconductorphysicistsresearchfundamentalprocessesusedinelectronicdevices.8+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 11 preview imageProblem1.11Thecomponentsofintegratedcircuitsandtheirinterconnectionsaremanufacturedconcurrentlyonasemiconductorwaferbyasequenceofphotolithographicprocessingsteps.Thecomponentsofadiscretecircuitaremanufacturedseparatelyandtheninterconnected,usuallyonacircuitboard.Oftenoverallcostcanbereducedbyintegratingthesystemontoasfewchipsaspossiblebecausechipcostisnearlyindependentofcomplexity(withincertainbounds).Problem1.12Theareaconsumedbyeachtransistoris(10pm)x(10um)=107%n?.Thechipareais(2cm)x(2cm)=4x107%n?.Thusthenumberoftransistorsthatcanbeplacedonthechipis(4x107%)71070)=4x10°transistors/chipProblem1.13Gainisanegativenumberforaninvertingamplifier,andtheoutputsignalisaninvertedversionoftheinputsignal.Gainisapositivenumberforanoninvertingamplifier,andtheoutputsignalhasthesamepolarityastheinputsignalateachinstantoftime.Problem1.14"Loadingeffects"refertothefactthattheinputvoltageofanamplifierislessthantheinternalsourcevoltagebecauseofthevoltagedropacrosstheinternalsourceimpedance.Alsotheamplifieroutputvoltageislessthantheopen-circuitvoltagegaintimestheinputvoltagebecauseofthevoltagedropacrosstheoutputimpedanceoftheamplifier.9+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 12 preview imageProblem1.15Ry=50RAR.=1a+Vv+R,Ros=(OO)wv>4HJuL100jo)mVamos-havR.oLA==A——=_=50VvAvoRp+R,vsAvoRy+RoRy+RyR.=1_6A;=A,g=1.25x10LG=AA.=6.25x107voiUsingtheunity-gainamplifierwehave:R5=/00knRo=/00n+“0Ovwl50SV-rome5R.Rp—-1—=v2-AS=VoR,TR.R+&_152VrmsandPy=Vo/Ry,=45.9mWis'L<)Withthesourceconnecteddirectlytotheload,wehave:10+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 13 preview imageRy=/00knA°+ORLY°sv’rmsRpVo=Vsmw"2-5mVrmsLSsP=vV?/R=125nWoo’"LThustheoutputpowerismuchlargeriftheunity-gainamplifierisused.P.=V2JR.=0.333pWinin’TinP=vV%/R=3.135W[3]o/Ry,h==12G=Po/Pin=9.376x10Ra++ViRo>Av;V.Ro~iJ]_~Y=loov;_10kaRy410A=90=A=F=100——M——vvoRy+RyrR,+104SolvingwefindthatR,=1.11kQ11+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 14 preview imageProblem1.19Withtheswitchopenwehave:R.R.iLVo=50nv=v,——1aA__>(1)oSR,+10°voRp"+RjWiththeswitchclosedwehave:RLVo=100mV=VoApp(2)LoDividingtherespectivesidesofEquation(1)bythoseofEquation(2),wehave:50mv_Ry100mVR.+10°iSolvingweobtainR;=1MQ.Problem1.20IfwecascadetwoamplifiersAandBtheequivalentcircuitis:AB—_————>R+Roa+RogAlun{7AsiaRigAvonVinTheopen-circuitvoltagegainofthecascadedamplifieris:R.iBA=AATsvoVOA'voBRoa+RigProblem1.21SeethefigureshowninthesolutionforProblem1.20.WhentheamplifiersarecascadedintheorderA-B,wehave:12+StudyXY
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Solution Manual For Electronics, 2nd Edition - Page 15 preview imageRy=Ria=3kQRy=Rop=20QR.iB4A=AESTF5=4.998x10voVOA"voBRoa+RipOntheotherhandfortheB-Acascadewehave:Ry=Rip=1MQRy=Roa=400QR.iA4A=AATp—=4.967x10voVOA"VOBR+RiaKea[M4nRo.+S++{522;Vea=AvoaVis°°Resn?2VvR../n=°B_iBA|e]Via[Aven“RE.+R/n?mx|vosOAiBR..n=aiB[sel[vn)|vos|ACOAiB13+StudyXxy
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Solution Manual For Electronics, 2nd Edition - Page 16 preview image22ala,—0=Ja«|a%Rip="RoaRipdn'VoAVvoBm’rR__+R)2OAiBSolvingfornwehave:Ripn=|=OAProblem1.23Theinternalsourceimpedanceis:R=open-circuitvoltage_20x1073=20kQSsshort-circuitcurrent1076Thedesiredvoltagegainisrequiredtobeatleast:vApe=g2=—X0_~s00Ss20x10Ifwecascadenstages,connectthesourcetotheinput,andconnecttheloadtotheoutput,thevoltagegainisgivenby:n-1PRU.SEN.SxSLanvsRy+RgRy+RyRp+RyvoSubstitutingvaluesandreducingweobtain:A,=0.02381x(0.9091)™71x(10)Bytrialanderrwedeterminethatwemusthaven=5toachieveAinexcessof500.vsProblem1.24Toavoidexcessiveloadingeffectsattheinput,weshouldchoosethefirststagesuchthatitsinputresistanceislargerthanthesourceresistance.ThereforewechoosetypeAastheinputstage.Toavoidexcessiveloadingeffectsattheoutput,14+StudyXY
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Document Details

Subject
Electrical Engineering & Electronics
Textbook
Electronics by Allan R. Hambley

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