cmos相关资料cmos集成电路作业答案

Chapter1Solution:IftheMoore’slawinmicroprocessorswouldstillhold,thenthetransistoronleadmicroprocessorsdoubleeverytwoyears.Fortheintegrationof486in1990is1MT,thusin2015theintegrationcomplexityofmicroprocessorsis5792.6MT.Leadmicroprocessorsfrequencyalsodoubleeverytwoyears,thusby2015itwillbeupto72407MHz.128GbitDRAMshouldbeavailableonasinglechipatthatpointintime.Solution:Theapplicationclassoftheleading-edgedevicesisbelow:Theleading-edgemicroprocessorisIntelcore2quadQ6600,whichintegrates582millionelectronictubes,itscoreareais143mm2,anditsfrequencyis2.93GHz.Theleading-edgesignalprocessorbasicARMarchitectureARMCortex-A8,whosefrequencyisupto1GHz.Theleading-edgeDRAMisbasicDDR3,itsfrequencyismorethan1GHz.Solution:TheexampleofvarietiesoftheabstractionlevelsSystem:SOC,CPU,MCU,ARMModule:Adder,Multiplier,C-element,PLLGate:and,or,xorCircuit:CMOSinverter,CMOSSchmittTrigger,Flip-FlopcircuitDevice:thepowerdevice,theGeAsdevice,MOSdevice,bipolar

Chapter3Solution:a)b)Wherethatfromtheaboveexpression,wecanobtainc)SPICEasfollow:RESISTORDIODECIRCUIT.OPTIONLISTNODEPOSTV1102.5R1122KR2342KD123DND240DN.MODELDNDIS=1E-14.TEMP27.PROBE.PRINTDCV(1)V(2)V(4)V(0)I(R1).ENDwhenIs=10-14AandT=300K,then(Figure1)Figure1(whenIS=10-14Ad)WhenUsetheaboveexpression,wecancalculateWhenUsetheaboveexpression,wecancalculateFigure2(whenIS=10-16ASolution:a)ifweusemoreaccuratemodel,,b)thediodeisreversebiased.c)theexpressionofwhere,,Thuswecancalculatethatd)where,thuswecangete)wecanseethatThenewvoltagereducesthereversebiasofthePNjunction,hencethewidthofdecrease.Asyoubringtheplatesofacapacitortogether,thecapacitanceincrease.Solution:Forashortchanneldevice,whereTobeginwith,theoperationregimeneedtobedetermined.Formanyofthosedatatobeinsaturation,VTshouldbe:Thatis=>VT>1.4VThisisaquitehighvalueinourprocess.Thuswecanassumethatalldataaretakeninvelocitysaturation,wewillcheckthisassumptionlater.Invelocitysaturation:Using1&2=>Using2&3=>Using2&4InputthedataintoWecancalculatethat①Using2&5InputthedataintoWecancalculatethat②BothofthesevaluessatisfyVT<1.4V,soallthedatainourtableatetakeninthevelocitysaturation.①&②canbeusedalongwith,Wecancalculatethat=0.6VandUsing2alone,WecancalculatethatSolution:a)whenR=10k,=>Assumethedeviceisinsaturation(thatneedtobeverifiedeventually)=>=>saturationverified.Thereforeb)assumethedeviceisinlinear.=>linearverified.c)wouldincrease.Becauseisfixedduetoconstantcurrent.wouldtrytoincreasethecurrentmorethanavailable,thusneedstoreducebyincreasing.Solution:Thisoneshouldimmediatelysignalyoutolookatacurve(s)thatdon’thavebody-effect.Thatmeans.Picktwopoints,eachfromdifferentcurvesthatsatisfythenobody-effectcondition.Makesurethey’reinthesameoperatingregiontoo.ThereasonwhyIchoosepointswiththesamewillbeevidentonceIworkthroughthemath.Asyoucansee,inorderformetoisolate,IneededtomakesureIcancancelasmanyvariablestobeabletosolvethisequation.Wecanusethesamemethodologyasabove.Thistime,wewanttokeepconstant.Itshouldn’tbeasurprise,butthatleavesustokeepalmosteverythingconstantexceptfor.Nowsolveforusingthefollowingequation:Solution:a)Where①Becauseistypicallysmall,canbeneglected.From①,wecangetthatThereforeComparingbythegivenshortchannelequation,revealsthatb),weinputthedataintotheequationof=ThisvaluedependsonW,independentofchannellength.

Chapter4Solution:i=c100fF*50/5ns=100mAR1=R2=R3=R4=R5=R6=R7=L/w=150*5mm/3um=250kC1=C2=C5=(0.3*5*10um)*88aF/um+2*(0.3*5*10um)*54aF/um=213fFC3=C4=C6=C7=213fF+100fF=323fFc)R1C1+R1C2+R1C3+R1C4+(R1+R5)C5+(R1+R5)C6+(R1+R5+R7)C7solutiona)C=(3.6um*10um)*30aF/um+2*(10um)*40aF/um=1.88pFR=0.08/*1cm/3.6um=222tp=0.38RC=0.38*1.88pf*222=160psb)Rt0=5KCi=10fF=5*k*Ci+（5+5/k+0.222）1880fF+5*k*Ci对求导并令其等于零得k3P=CV*Fmax=1.88*10*2.5*2.5*0.25*20*10=59uwSolution:thedelayofmetalwireC=(3.6um*10um)*30aF/um+2*(10um)*40aF/um=1.88pFR=0.08/*1cm/3.6um=222thedelayofmetalwiretp=0.38RC=0.38*1.88pf*222=160psb)Rt0=17.5kCi=10fF=17.5*k*Ci+(17.5+17.5/k+0.222)1880fF对求导并令其等于零得k14Solution:a)criticallengthis:当互联线超过这个临界长度Lcri时，RC延时才占主要地位b)tp=0.69RsCw+0.38RwCw（Rw=rLCw=cL）(RwCw)/2>RsCw时，即L>2Rs/r时，导线电阻引起的延时变成主要延时了。L>=62.5cm

Chapter5

Chapter6

Chapter7Solution:a)该电路是一个脉冲寄存器，它利用三个反相器产生延迟，因此会出现Clk与Clkd同时为正的情况。脉冲的宽度由3个反相器决定。波形图如下图所示：b)此电路的建立时间和维持时间是以短脉冲时钟的上升沿为参照衡量的，上升时间基本上是零，维持时间基本上等于脉冲的宽度，而传播延迟等于数据D经过两个门的延迟，即经过2个反相器的延迟。c)此电路的功耗为：忽略P的第二项，即可得到Solution:a)Whentheinputgeneratethesignal“D”,the“D”signalispassedthough“Data—transitionlook-ahead”and“PulseGenerator”togeneratetheCKIandCKIBsignal.QholdorQ=DisdeterminedbytheCKIandCKJBsignal.WhenCKJ=1,CKJB=0,Q=D.CKI=0,CKIB=1,Qhold