半导体器件物理与工艺 (施敏 著) 苏州大学出版社 课后答案 130页

'SolutionsManualtoAccompanySEMICONDUCTORDEVICESPhysicsandTechnologynd2EditionS.M.SZEUMCChairProfessorNationalChiaoTungUniversityNationalNanoDeviceLaboratoriesHsinchu,TaiwanJohnWileyandSons,IncNewYork.Chicester/Weinheim/Brisband/Singapore/Toronto课后答案网www.hackshp.cn1 ContentsCh.1Introduction---------------------------------------------------------------------0Ch.2EnergyBandsandCarrierConcentration--------------------------------------1Ch.3CarrierTransportPhenomena--------------------------------------------------7Ch.4p-nJunction--------------------------------------------------------------------16Ch.5BipolarTransistorandRelatedDevices----------------------------------------32Ch.6MOSFETandRelatedDevices-------------------------------------------------48Ch.7MESFETandRelatedDevices-------------------------------------------------60Ch.8MicrowaveDiode,Quantum-EffectandHot-ElectronDevices---------------68Ch.9PhotonicDevices-------------------------------------------------------------73Ch.10CrystalGrowthandEpitaxy---------------------------------------------------83Ch.11FilmFormation----------------------------------------------------------------92Ch.12LithographyandEtching------------------------------------------------------99Ch.13ImpurityDoping---------------------------------------------------------------105Ch.14IntegratedDevices-------------------------------------------------------------113课后答案网www.hackshp.cn0 CHAPTER21.(a)FromFig.11a,theatomatthecenterofthecubeissurroundbyfourequidistantnearestneighborsthatlieatthecornersofatetrahedron.Thereforethedistancebetweennearestneighborsinsilicon(a=5.43Å)is221/21/2[(a/2)+(2a/2)]=3a/4=2.35Å.(b)Forthe(100)plane,therearetwoatoms(onecentralatomand4corneratomseachcontributing1/4ofanatomforatotaloftwoatomsasshowninFig.4a)2foranareaofa,thereforewehave2-821422/a=2/(5.43×10)=6.78×10atoms/cmSimilarlywehavefor(110)plane(Fig.4aandFig.6)2152(2+2×1/2+4×1/4)/2a=9.6×10atoms/cm,andfor(111)plane(Fig.4aandFig.6)æ3ö2142(3×1/2+3×1/6)/1/2(2a)(ç÷a)==7.83×10atoms/cm.è2øæç3ö÷2aç2÷èø2.TheheightsatX,Y,andZpointare31and3.4,4,43.(a)Forthesimplecubic,aunitcellcontains1/8ofasphereateachoftheeightcornersforatotalofonesphere.课后答案网4Maximumfractionofcellfilled=no.ofsphere×volumeofeachsphere/unitcellvolume33=1×4ð(a/2)/www.hackshp.cna=52%(b)Foraface-centeredcubic,aunitcellcontains1/8ofasphereateachoftheeightcornersforatotalofonesphere.Thefccalsocontainshalfasphereateachofthesixfacesforatotalofthreespheres.Thenearestneighbordistanceis1/2(a2).Thereforetheradiusofeachsphereis1/4(a2).4Maximumfractionofcellfilled33=(1+3){4ð[(a/2)/4]/3}/a=74%.(c)Foradiamondlattice,aunitcellcontains1/8ofasphereateachoftheeightcornersforatotalofonesphere,1/2ofasphereateachofthesixfacesforatotalofthreespheres,and4spheresinsidethecell.Thediagonaldistance1 between(1/2,0,0)and(1/4,1/4,1/4)showninFig.9ais2221æaöæaöæaöaD=ç÷+ç÷+ç÷=32è2øè2øè2ø4aTheradiusofthesphereisD/2=384Maximumfractionofcellfilled3é4pæaöù3=(1+3+4)êç3÷ú/a=ð3/16=34%.ë3è8øûThisisarelativelylowpercentagecomparedtootherlatticestructures.4.d=d=d=d=d1234d+d+d+d=01234d•(d+d+d+d)=d•0=01123412d+d•d+d•d+d•d=011213142222224d+dcosè12+dcosè13+dcosè14=d+3dcosè=0-14cosè=3-1-10è=cos()=109.47.35.Takingthereciprocalsoftheseinterceptsweget1/2,1/3and1/4.Thesmallestthreeintegershavingthesameratioare6,4,and3.Theplaneisreferredtoas(643)plane.6.(a)ThelatticeconstantforGaAsis5.65Å,andtheatomicweightsofGaandAsare69.72and74.92g/mole,respectively.Therearefourgalliumatomsand课后答案网fourarsenicatomsperunitcell,therefore3-832224/a=4/(5.65×10)=2.22×10GaorAsatoms/cm,3Density=(no.www.hackshp.cnofatoms/cm×atomicweight)/Avogadroconstant22233=2.22×10(69.72+74.92)/6.02×10=5.33g/cm.(b)IfGaAsisdopedwithSnandSnatomsdisplaceGaatoms,donorsareformed,becauseSnhasfourvalenceelectronswhileGahasonlythree.Theresultingsemiconductorisn-type.7.(a)ThemeltingtemperatureforSiis1412ºC,andforSiO2is1600ºC.Therefore,SiO2hashighermeltingtemperature.ItismoredifficulttobreaktheSi-ObondthantheSi-Sibond.(b)Theseedcrystalisusedtoinitiatedthegrowthoftheingotwiththecorrectcrystalorientation.(c)Thecrystalorientationdeterminesthesemiconductor’schemicalandelectrical2 properties,suchastheetchrate,trapdensity,breakageplaneetc.(d)Thetemperatingofthecrusibleandthepullrate.-424.73x10T8.Eg(T)=1.17–forSi(T+636)Eg(100K)=1.163eV,andEg(600K)=1.032eV-425.405x10TEg(T)=1.519–forGaAs(T+204)Eg(100K)=1.501eV,andEg(600K)=1.277eV.9.ThedensityofholesinthevalencebandisgivenbyintegratingtheproductN(E)[1-F(E)]dEfromtopofthevalenceband(EtakentobeE=0)totheVbottomofthevalencebandEbottom:Ebottomp=òN(E)[1–F(E)]dE(1)0where1–F(E)={[(E-EF)/kT]}(E-EF)/kT-11-1/1+e=[1+e]IfEF–E>>kTthen1–F(E)~exp[-(E-E)kT](2)FThenfromAppendixHand,Eqs.1and2weobtain23/2Ebottom1/2p=4ð[2mp/h]òEexp[-(EF–E)/kT]dE(3)0LetxaE/kT,andletEbottom=-¥,Eq.3becomes23/23/2-¥1/2xp=4ð(2mp/h)(kT)exp[-(EF/kT)]òxedx0wheretheintegralontherightisofthestandardformandequalsp/2.23/24p=2[2课后答案网ðmpkT/h]exp[-(EF/kT)]ByreferringtothetopofthevalencebandasEVinsteadofE=0wehave,23/2p=2(2ðmpkT/h)exp[-(EF–EV)/kT]orp=NVexp[-(EF–EV)/kT]23whereNV=2(2ðmpkTwww.hackshp.cn/h).10.FromEq.1823/2NV=2(2ðmpkT/h)TheeffectivemassofholesinSiis2/32mp=(NV/2)(h/2ðkT)2æ.266´1019´106m-3ö3(-34)26.625´10=ç÷ç÷-23è2ø2p(1.38´10)(300)-31=9.4×10kg=1.03m0.Similarly,wehaveforGaAs-31mp=3.9×10kg=0.43m0.11.UsingEq.193 Ei=(EC+EV)2+(kT2)ln(NVNC)2é3ù=(EC+EV)/2+(3kT/4)lnêë(mpmn)(6)úû(1)At77K-23-19Ei=(1.16/2)+(3×1.38×10T)/(4×1.6×10)ln(1.0/0.62)-5-3=0.58+3.29×10T=0.58+2.54×10=0.583eV.At300K-5Ei=(1.12/2)+(3.29×10)(300)=0.56+0.009=0.569eV.At373K-5Ei=(1.09/2)+(3.29×10)(373)=0.545+0.012=0.557eV.Becausethesecondtermontheright-handsideoftheEq.1ismuchsmallercomparedtothefirstterm,overtheabovetemperaturerange,itisreasonabletoassumethatEiisinthecenteroftheforbiddengap.Etop()(E-E)E-Ee-E-EF/kTdEòCCEC12.KE=Etop-(E-E)/kTxº(E-EC)E-EeFdEòCEC¥3æ5öòx2e-xdxGç÷0è2ø1.5´0.5´p=kT=kT=kT¥1x2e-xdxæ3ö0.5pòGç÷0è2ø3=kT.2-315-2613.(a)p=mv=9.109×10×10=9.109×10kg–m/s-34h6.626´10-9l===7.27×10m=72.7Å-26p9.109´10课后答案网m01(b)l=l=×72.7=1154Å.nm0.063p15-314.FromFig.22whenni=10cm,thecorrespondingtemperatureis1000/T=1.8.SothatT=1000/1.8=555Kor282www.hackshp.cn.15.FromEc–EF=kTln[NC/(ND–NA)]whichcanberewrittenasND–NA=NCexp[–(EC–EF)/kT]1916-3ThenND–NA=2.86×10exp(–0.20/0.0259)=1.26×10cm1616-3orND=1.26×10+NA=2.26×10cmAcompensatedsemiconductorcanbefabricatedtoprovideaspecificFermienergylevel.16.FromFig.28awecandrawthefollowingenergy-banddiagrams:4 17.(a)TheionizationenergyforboroninSiis0.045eV.At300K,allboron15-3impuritiesareionized.Thuspp=NA=10cm292154-3np=ni/nA=(9.65×10)/10=9.3×10cm.TheFermilevelmeasuredfromthetopofthevalencebandisgivenby:1915EF–EV=kT课后答案网ln(NV/ND)=0.0259ln(2.66×10/10)=0.26eV(b)Theboronatomscompensatethearsenicatoms;wehave161615-3pp=NA–ND=3×10–2.9×10=10cmSinceppisthesameasgivenin(a),thevaluesfornpandEFarethesameasin(a).However,themobilitiesandresistivitiesforthesetwosamplesaredifferent.www.hackshp.cn18.SinceND>>ni,wecanapproximaten0=NDand219172-3p0=ni/n0=9.3×10/10=9.3×10cmæEF-EiöFromn0=niexpç÷,èkTøwehave179EF–Ei=kTln(n0/ni)=0.0259ln(10/9.65×10)=0.42eVTheresultingflatbanddiagramis:5 19.Assumingcompleteionization,theFermilevelmeasuredfromtheintrinsic15-317-319Fermilevelis0.35eVfor10cm,0.45eVfor10cm,and0.54eVfor10-3cm.Thenumberofelectronsthatareionizedisgivenbyn-(ED-EF)/kT@ND[1–F(ED)]=ND/[1+e]UsingtheFermilevelsgivenabove,weobtainthenumberofionizeddonorsas15-315-3n=10cmforND=10cm17-317-3n=0.93×10课后答案网cmforND=10cm19-319-3n=0.27×10cmforND=10cmTherefore,theassumptionofcompleteionizationisvalidonlyforthecaseof15-310cm.1616+10www.hackshp.cn1020.ND==1+e-(ED-EF)/kT1+e-0.135161015-3==5.33×10cm11+1.1451615-315-3Theneutraldonor=10–5.33×10cm=4.67×10cmOND4.764Theratioof==0.876.+N5.33D6 CHAPTER391.(a)ForintrinsicSi,mn=1450,mp=505,andn=p=ni=9.65´10115Wehaver===.331´10W-cmqnm+qpmqn(m+m)npinp6(b)SimilarlyforGaAs,mn=9200,mp=320,andn=p=ni=2.25´10118Wehaver===.292´10W-cm.qnm+qpmqn(m+m)npinp-3/22.Forlatticescattering,mnµT-3/22002T=200K,mn=1300´=2388cm/V-s-3/2300-3/24002T=400K,mn=1300´=844cm/V-s.-3/23001113.Since=+mmm121112=+课后答案网m=167cm/V-s.m2505004.(a)p=5´1015cm-3,n=ni2/p=(9.65´109)2/5´1015=1.86´104cm-322mp=410cm/V-s,www.hackshp.cnmn=1300cm/V-s11r=»=3W-cmqmn+qmpqmpnpp161615-34-3(b)p=NA–ND=2´10–1.5´10=5´10cm,n=1.86´10cm162mp=mp(NA+ND)=mp(3.5´10)=290cm/V-s,2mn=mn(NA+ND)=1000cm/V-s11r=»=4.3W-cmqmn+qmpqmpnpp7 15-34-3(c)p=NA(Boron)–ND+NA(Gallium)=5´10cm,n=1.86´10cm172mp=mp(NA+ND+NA)=mp(2.05´10)=150cm/V-s,2mn=mn(NA+ND+NA)=520cm/V-sr=8.3W-cm.5.AssumeND-NA>>ni,theconductivityisgivenbys»qnmn=qmn(ND-NA)Wehavethat-191716=(1.6´10)mn(ND-10)Sincemobilityisafunctionoftheionizedimpurityconcentration,wecanuseFig.3alongwithtrialanderrortodeterminemnandND.Forexample,ifwe17+-172chooseND=2´10,thenNI=ND+NA=3´10,sothatmn»510cm/V-swhichgivess=8.16.Furthertrialanderroryields17-3ND»3.5´10cmand课后答案网2mn»400cm/V-swhichgiveswww.hackshp.cn-1s»16(W-cm).26.s=q(mn+mp)=qm(bn+n/n)nppiFromtheconditionds/dn=0,weobtainn=n/biTherefore8 1rmqìp(bni/b+bni)b+1==.ri12bqmn(b+1)pip7.Atthelimitwhend>>s,CF==4.53.ThenfromEq.16ln2-3V10´10-4r=´W´CF=´50´10´.453=.0226W-cm-3I1´10FromFig.6,CF=4.2(d/s=10);usingthea/d=1curveweobtain-3.0226´10V=r×I/(W×CF)==10.78mV.-450´10´4.28.Hallcoefficient,-3-3VHA10´10´6.1´103R===4267.cm/CH-3-94IBW5.2´10´(30´10´10)´.005zSincethesignofRHispositive,thecarriersareholes.FromEq.221116-3p===.146´10cm-19qR6.1´10´4267.HAssumingNA»p,fromFig.7weobtain课后答案网r=1.1W-cmThemobilitympisgivenbyEq.15b112m===380cm/V-s.pwww.hackshp.cn-1916qpr6.1´10´.146´10´1.1119.SinceRµrandr=,henceRµqnm+qpmnm+pmnpnpFromEinsteinrelationDµmm/m=D/D=50npnp9 1RNm1Dn=0.5R11Nm+NmDnApWehaveNA=50ND.10.Theelectricpotentialfisrelatedtoelectronpotentialenergybythecharge(-q)1f=+(EF-Ei)qTheelectricfieldfortheone-dimensionalsituationisdefinedasdf1dEiE(x)=-=dxqdxæEF-Eiön=niexpç÷=ND(x)èkTøHenceæN(x)öçD÷EF-Ei=kTlnçn÷èiøækTö1dN(x)(x)=-ç÷DEç÷.qN(x)dx课后答案网èøD11.(a)FromEq.31,Jn=0anddn-axDndxkTN0(-a)ekTE(x)=-=-=+awww.hackshp.cn-axmnqNeqn04(b)E(x)=0.0259(10)=259V/cm.12.Atthermalandelectricequilibria,dn(x)J=qmn(x)E+qD=0nnndxDn1dn(x)Dn1NL-N0E(x)=-=-mn(x)dxmN+(N-N)(xL)Lnn0L0DN-NnL0=-mLN+(N-N)xn0L010 LDN-NDNnL0nLV=ò-=-ln.0mLN+(N-N)xmN0L0n0-61611-313.Dn=Dp=tG=10´10´10=10cmpLn=n+Dn=N+Dn=1015+1011»1015cm-3noD292ni(.965´10)1111-3p=+Dp=+10»10cm.15N10D11-814.(a)t»==10sp-15715snN5´10´10´2´10ptht-8-4L=Dt=9´10=3´10cmppp7-1610S=nsN=10´2´10´10=20cm/slrthssts(b)TheholeconcentrationatthesurfaceisgivenbyEq.67ætSöçplr÷p(0)=p+tG1-nnopLçL+tS÷èpplrø92-8(9.65´10)-817æ10´20ö=+10´10ç1-÷课后答案网16ç-4-8÷2´10è3´10+10´20ø9-3»10cm.15.s=qnm+qpmnpwww.hackshp.cnBeforeilluminationn=n,p=pnnonnoAfterilluminationn=n+Dn=n+tG,nnonopp=p+Dp=p+tGnnonopDs=[qm(n+Dn)+qm(p+Dp)]-(qmn+qmp)nnopnonnopno=q(m+m)tG.npp11 dp16.(a)J=-qDp,diffpdx-19115=-1.6´10´12´´10exp(-x/12)-412´102=1.6exp(-x/12)A/cm(b)J=J-Jn,drifttotalp,diff2=4.8-1.6exp(-x/12)A/cm(c)QJ=qnmEn,driftn-1916\4.8-1.6exp(-x/12)=1.6´10´10´1000´EE=3-exp(-x/12)V/cm.17.ForE=0wehave2¶pp-p¶pnnon=-+D=0p2¶tt¶xpatsteadystate,theboundaryconditionsarepn(x=0)=pn(0)andpn(x=W)=pno.课后答案网ThereforeéæW-xöùêsinhç÷úwww.hackshp.cnêçèLp÷øúpn(x)=pno+[pn(0)-pno]êúêæWöúsinhç÷êçL÷úëèpøû¶pn[]DpæçWö÷Jp(x=0)=-qDp=qpn(0)-pnocothç÷¶xLLx=0pèpø¶pn[]Dp1Jp(x=W)=-qDp=qpn(0)-pno.¶xx=WLpæWösinhç÷çL÷èpø18.Theportionofinjectioncurrentthatreachestheoppositesurfacebydiffusionis12 givenbyJp(W)1a==0J(0)cosh(W/L)pp-6-2LºDt=50´50´10=5´10cmppp1a==.0980-2-2cosh(10/5´10)Therefore,98%oftheinjectedcurrentcanreachtheoppositesurface.19.Insteadystate,therecombinationrateatthesurfaceandinthebulkisequalDpn,bulkDpn,surface=tp,bulktp,surfacesothattheexcessminoritycarrierconcentrationatthesurface-7141013-3Dpn,surface=10×=10cm-610Thegenerationratecanbedeterminedfromthesteady-stateconditionsinthebulk141020-3-1G==10cms-610FromEq.62,wecanwrite2¶DpDpD+G-=0p2t¶xp14-313-3TheboundaryconditionsareDp(x=¥)=10cmandDp(x=0)=10cm课后答案网14-x/LpHenceDp(x)=10(1-0.9e)-6whereLp=10×10=31.6mm.20.Thepotentialbarrierheightf=f-c=4.2-4.0=0.2volts.Bm21.Thenumberofelectronsoccupyingtheenergylevelbetweenwww.hackshp.cnEandE+dEisdn=N(E)F(E)dEwhereN(E)isthedensity-of-statefunction,andF(E)isFermi-Diracdistributionfunction.SinceonlyelectronswithanenergygreaterthanE+qfandhavingFmavelocitycomponentnormaltothesurfacecanescapethesolid,thethermioniccurrentdensityis3¥4p(2m)21J=qv=vE2e-(E-EF)kTdEòxòE+qf3xFmhwherevisthecomponentofvelocitynormaltothesurfaceofthemetal.Sincextheenergy-momentumrelationship2P1222E==(p+p+p)xyz2m2m13 PdPDifferentiationleadstodE=mBychangingthemomentumcomponenttorectangularcoordinates,24pPdP=dpdpdpxyz2q¥¥¥-(p2x+p2y+p2z-2mEf)/2mkTJ=pedpdpdp3òòpp=-¥òp=-¥xxyzmhx0yHence2q¥-(p2-2mE)2mkT¥-p22mkT¥-p22mkT=exfpdpeydpezdp3òxxòyòzmhpx0-¥-¥2wherep=2m(E+qf).x0Fm12¥-ax2æpö12Sinceòedx=ç÷,thelasttwointegralsyield(2ðmkT).-¥èaø2p-2mExFThefirstintegralisevaluatedbysetting=u.2mkTpdpxxThereforewehavedu=mkTThelowerlimitofthefirstintegralcanbewrittenas2m(E+qf)-2mEqfFmFm=2mkTkT¥-u-qfkTsothatthefirstintegralbecomesmkTòedu=mkTemqf/ktm24pqmk2-qfmkT*2æ-qfmöHenceJ=Te=ATexpç÷.3ç÷hèkTø22.Equation79isthetunnelingprobability-31-192mn(qV0-E)2(.911´10)(20-2)(6.1´10)10-1b===.217´10m2-342h(.1054´10)课后答案网2-1ì[0-10]üï20´sinh(.217´10´3´10ï-6T=í1+ý=.319´10.ïî4´2´(20-2)ïþ23.Equation79isthetunnelingprobabilitywww.hackshp.cn-31-192mn(qV0-E)2(.911´10)(6-2.2)(6.1´10)9-1b===.999´10m2-342h(.1054´10)-12ìï[6´sinh(.999´109´10-10)]üï-10T(10)=í1+ý=.0403ïî4´2.2´(6-2.2)ïþ-12ìï[6´sinh(.999´109´10-9)]üï-9-9T(10)=í1+ý=8.7´10.ïî4´2.2´(6-2.2)ïþ14 24.FromFig.223AsE=10V/s66nd»1.3´10cm/s(Si)andnd»8.7´10cm/s(GaAs)t»77ps(Si)andt»11.5ps(GaAs)4AsE=5´10V/s76nd»10cm/s(Si)andnd»8.2´10cm/s(GaAs)t»10ps(Si)andt»12.2ps(GaAs).2E2kTth25.Thermalvelocityv==thmm00-232´1.38´10´300=-311.9´1046=9.5´10m/s=5.9´10cm/sForelectricfieldof100v/cm,driftvelocity5v=mE=1350´100=.135´10cm/s<>VÞN@C=´(0.85´10)RbiDqej-19-14s1.6´10´11.9´8.85´1015-3ÞNd=3.43´10cmWecanselectthen-typedopingconcentrationof3.43课后答案网×1015cm-3.8.FromEq.56,www.hackshp.cnéùêússuNG=-U=êpnthtúniêæEt-EiöæEi-Etöúêsnexpç÷+spexpç÷úëèkTøèkTøûéùê-15-15715úê10´10´10´10ú916=´9.65´10=3.89´10ê-15æ0.02ö-15æ-0.02öúê10expç÷+10expç÷úëè0.0259øè0.0259øûand23 ()-142esVbi+V2´119.´.885´10´(.0717+5.0)-5W===12.66´10cm=.1266mm-1915qNA6.1´10´10Thus1916-5-72J=qGW=6.1´10´.389´10´12.66´10=.7879´10A/cm.gen2ni9.FromEq.49,andp=noNDWecanobtaintheholeconcentrationattheedgeofthespacechargeregion,20.892æ0.8öni()(.965´10)ç0.0259÷17-3p=e0.0259=eèø=.242´10cm.n16N10DqV/kT10.J=J(x)+J(-x)=J(e-1)pnnpsVJÞ=e0.0259-1JsVÞ.095=e0.0259-1ÞV=0.017V.11.Theparametersare课后答案网9-32ni=9.65´10cmDn=21cm/sec2-7Dp=10cm/sectp0=tn0=5´10secFromEq.52andEq.54www.hackshp.cn2éæqVaöùqDppnoqVkTDpnç÷()(/)ièkTøJx=e-1=q´´êe-1úpnLptpoNDêëúûÞ92éæ0.7öù-1910(.965´10)ç÷7=6.1´10´´´eè0.0259ø-1êú-75´10NDêëúûÞ15-3N=2.5´10cmD24 2éæqVaöùqDnnpoDnç÷()(qV/kT)nièkTøJ-x=e-1=q´´êe-1únpLntnoNAêëúûÞ92éæ0.7öù-1921(.965´10)ç÷25=6.1´10´´´eè0.0259ø-1êú-75´10NAêëúûÞ16-3N=.5278´10cmA16-3Wecanselectap-ndiodewiththeconditionsofNA=5.278×10cmandND=15-35.4×10cm.-62212.Assumeôg=ôp=ôn=10s,Dn=21cm/sec,andDp=10cm/sec(a)Thesaturationcurrentcalculation.FromEq.55aandL=Dt,wecanobtainpppqDppn0qDnnp02æç1Dp1Dnö÷J=+=qn+siç÷LLNtNtpnèDp0An0ø-19(9)2æç110121ö÷=6.1´10´.965´10+ç101810-6101610-6÷èø-122=6.87´10A/cm课后答案网-52Andfromthecross-sectionalareaA=1.2´10cm,weobtain-5-12-17I=A´J=2.1´10´.687´10=.8244´10A.sswww.hackshp.cn(b)ThetotalcurrentdensityisqVæöJ=Jçekt-1÷sç÷èøThus0.7æö-170.0259-1711-5I=.8244´10çe-1÷=.8244´10´.547´10=.451´10A0.7Vç÷èø-0.7æö-17ç0.0259÷-17I=.8244´10e-1=.8244´10A.-0.7Vç÷èø25 qVæö13.FromJ=Jçekt-1÷sç÷èøwecanobtainéùé-3ùVæJöæ10ö=lnêçç÷÷+1úÞV=.00259´lnêçç-17÷÷+1ú=.078V..00259ëèJsøûëè.8244´10øû214.FromEq.59,andassumeDp=10cm/sec,wecanobtain2DpniqniWJ@q+RtNtpDg(.9)2.´-19´.´92´11.9´8.85´10-14´(V+V)-1910965´10161096510biR=1.6´10+-615-6-19151010101.6´10´10191510´10V=.00259ln=.0834Vbi2(9).965´10Thus-11-7J=.526´10+.1872´10.0834+VRR课后答案网www.hackshp.cn26 VRJs01.713E-070.11.755E-070.21.941E-070.32,129E-070.42.316E-070.52.503E-070.62.691E-070.72.878E-070.83.065E-070.93.252E-0713.439E-07课后答案网www.hackshp.cn27 CurrentDensity3.E-073.E-072.E-072.E-07CurrentDensity1.E-075.E-080.E+0000.20.40.60.811.2AppliedVoltage17-3WhenND=10cm,weobtain191710´10V=.00259ln=.0953Vbi2(9).965´10-13-8J=.526´10+.1872´10.0956+VRRCurrentDensity3.00E-082.50E-08课后答案网2.00E-081.50E-08CurrentDensity1.00E-08www.hackshp.cn5.00E-090.00E+0000.20.40.60.811.2AppliedVoltage15.FromEq.39,¥Q=qò(p-p)dxpnnoxn28 ¥qV/kT-(x-xn)/Lp=qp(e-1)edxònoxnTheholediffusionlengthislargerthanthelengthofneutralregion.x"òn()Q=qp-pdxpnnoxnx"n(qV/kT)-(x-xn)/Lp=qpe-1edxònoxnx"-xx-xqVæ-nn-nnöæö=qp(-L)çekT-1÷çeLp-eLp÷nopç÷çç÷÷èøèø92110-19(.965´10)-4æöæ--ö=6.1´10´(-5´10)çe0.0259-1÷çe5-e5÷16ç÷ç÷10èøèø-32=8.784´10C/cm.16.FromFig.26,thecriticalfieldatbreakdownforaSione-sidedabrupt5junctionisabout2.8×10V/cm.ThenfromEq.85,weobtain2EcWesEc-1V(breakdownvoltage)==(N)BB22q2-145119.´.885´10´(8.2´10)(15)-1=10-192课后答案网´1.6´10=258V()-142esVbi-V2´11.9´8.85´10´258-3W=www.hackshp.cn@=1.843´10cm=18.43£gm-1915qNB1.6´10´10Whenthen-regionisreducedto5mm,thepunch-throughwilltakeplacefirst.FromEq.87,wecanobtainVB"shadedareainFig.29insertæWöæWö==ç÷ç2-÷VB(EcWm)/2çèW÷øçèW÷ømmæWöæWöæ5öæ5öVB"=VBçç÷÷çç2-÷÷=258´ç÷ç2-÷=121VèWmøèWmøè18.43øè18.43øComparedtoFig.29,thecalculatedresultisthesameasthevalueunderthe29 15-3conditionsofW=5mmandNB=10cm.17.Wecanusefollowingequationstodeterminetheparametersofthediode.22DpniqV/kTqWniqV/2kTDpniqV/kTJ=qe+e@qeFtN2ttNpDrpD2EcWesEc()-1V==NBD22qÞ2(9)20.7DpniqV/kT-19Dp9.65´100.0259-3AJF=AqeÞA´1.6´10´e=2.2´10tN-7NpD10D2-142EcWesEc()-111.9´8.85´10Ec()-1V==NÞ130=NBD-19D22q2´1.6´10515-3LetEc=4´10V/cm,wecanobtainND=4.05´10cm.15Themobilityofminoritycarrierholeisabout500atND=4.05´10课后答案网2Dp=0.0259´500=12.95cm/s-52Thus,thecross-sectionalareaAis8.6´10cm.www.hackshp.cn18.Asthetemperatureincreases,thetotalreversecurrentalsoincreases.Thatis,thetotalelectroncurrentincreases.Theimpactionizationtakesplacewhentheelectrongainsenoughenergyfromtheelectricalfieldtocreateanelectron-holepair.Whenthetemperatureincreases,totalnumberofelectronincreasesresultingineasytolosetheirenergybycollisionwithotherelectronbeforebreakingthelatticebonds.Thisneedhigherbreakdownvoltage.19.(a)Thei-layeriseasytodeplete,andassumethefieldinthedepletionregionis30 constant.FromEq.84,wecanobtain.66W4æEö4æEö-3516510ç÷dx=1Þ10ç÷´10=1ÞE=4´10´(10)=.587´10V/cmò0è4´105øè4´105øcritical5-34V=.587´10´10=587VB5(b)FromFig.26,thecriticalfieldis5×10V/cm.2EcWesEc-1V(breakdownvoltage)==(N)BB22q-1452124.´.885´10´(5´10)16-1=(2´10)-192´1.6´10=428.V.182´1022-420.a==10cm-42´103/21/22EW4Ecé2esù()-1/2V==aBêú33ëqû3/21/24E-14cé2´11.9´8.85´10ù(22)-1/2=êú´10-193ë1.6´10û-83/2=4.84´10EcThebreakdownvoltagecanbedeterminedbyaselected课后答案网Ec.21.TocalculatetheresultswithappliedvoltageofV=0.5V,wecanuseasimilarcalculationinExample10with(1.6-0.5)replacing1.6forthevoltage.Theobtainedelectrostaticwww.hackshp.cn-4potentialsare1.1Vand3.4´10V,respectively.Thedepletionwidthsare-5-83.821´10cmand1.274´10cm,respectively.Also,bysubstitutingV=-5VtoEqs.90and91,theelectrostaticpotentialsare6.6V-4-5-8and20.3´10V,andthedepletionwidthsare9.359´10cmand3.12´10cm,respectively.Thetotaldepletionwidthwillbereducedwhentheheterojunctionisforward-biasedfromthethermalequilibriumcondition.Ontheotherhand,whentheheterojunctionisreverse-31 biased,thetotaldepletionwidthwillbeincreased.22.Eg(0.3)=1.424+1.247×0.3=1.789eVEg2DECV=--(E-E)/q-(E-E)/qbiF2V2C1F1qq1718kT4.7´10kT7´10=1.789–0.21–ln–ln=1.273V1515q5´10q5´1011é2Neevù2é-14ù2A12bi2´12.4´11.46´8.85´10´1.273x1=ê()ú=ê-1915úëqNDe1ND+e2NAûë1.6´10´5´10(12.4+11.46)û-5=4.1´10cm.SinceNDx1=NAx2x1=x2-5W=2x=8.2´10cm=0.82mm.1课后答案网www.hackshp.cn32 CHAPTER51.(a)Thecommon-baseandcommon-emittercurrentgainsisgivenbya=ga=0.997´0.998=0.9950Ta00.995b0==1-a1-0.9950=199.-9-9(b)SinceIB=0andICp=10´10A,thenICBOis10´10A.TheemittercurrentisI=(1+b)ICEO0CBO()-9=1+199×10´10-6=2´10A.2.Foranidealtransistor,a=g=0.9990a0b==999.01-a0-6Iisknownandequalsto10´10A.Therefore,CBOI=(1+b)ICEO0CBO-6课后答案网=(1+999)×10´10=10mA.3.(a)Theemitter-basejunctionisforwardbiased.FromChapterwww.hackshp.cn3weobtainkTæNNöé5´1018×2´1017ùV=lnçAD÷=0.0259lnêú=0.956V.biqçn2÷92èiøêë(9.65´10)úûThedepletion-layerwidthinthebaseis33 æNöW=çA÷(Totaldepletion-layerwidthoftheemitter-basejunction)1ç÷èNA+NDø2esæNAöæ1ö()=ç÷ç÷V-VqçN÷çN+N÷bièDøèADø-12æ18ö2×1.05´105´10æ1ö=ç÷ç÷(0.956-0.5)-19ç17÷18171.6´10è2´10øè5´10+2´10ø-6-2=5.364´10cm=5.364´10mm.Similarlyweobtainforthebase-collectorfunctioné2´1017×1016ùVbi=0.0259lnêú=0.795V.92êë(9.65´10)úûand-12æ16ö2×1.05´1010æ1öW2=-19çç17÷÷ç1617÷(0.795+5)1.6´10è2´10øè10+2´10ø-6-2=4.254´10cm=4.254´10mm.Thereforetheneutralbasewidthis-2-2W=W-W-W=1-5.364´10-4.254´10=0.904mm.B12(b)UsingEq.13a2课后答案网n2(9.65´109)p(0)=peqVEBkT=ieqVEBkT=e0.50.0259=2.543´1011cm-3.nno17ND2´104.Intheemitterregionwww.hackshp.cn-8-3D=52cmsL=52×10=.0721´10cmEE(9)2.965´10n==18.625.EO185´10Inthebaseregion34 -7-3D=40cmsL=Dt=40×10=2´10cmpppp2(9)2n9.65´10p=i==465.613.no17ND2´10Inthecollectorregion-6-3D=115cmsL=115×10=10.724´10cmCC(9)29.65´103n==9.312´10.CO1610ThecurrentcomponentsaregivenbyEqs.20,21,22,and23:-19-26.1´10×2.0´10×40×465.6130.50.0259-5I=e=1.596´10AEp-4.0904´10-5I@I=1.596´10ACpEp-19-26.1´10×2.0´10×52×18.625(0.50.0259)-7I=e-1=1.041´10AEn-3.0721´10-19-236.1´10×2.0´10×115×.9312´10-14I==3.196´10ACn-310.724´10I=I-I=.0BBEpCp5.(a)Theemitter,collector,andbasecurrentsaregivenby-5I=I+I=1.606´10A课后答案网EEpEn-5IC=ICp+ICn=1.596´10AI=I+I-I=1.041´10-7A.BEnBBCn(b)Wecanobtaintheemitterefficiencyandthebasetransportfactor:www.hackshp.cnI1.596´10-5Epg===0.9938-5IE1.606´10I-5Cp1.596´10a===1.T-5IEp1.596´10Hence,thecommon-baseandcommon-emittercurrentgainsarea0=gaT=0.9938a0b==160.3.01-a035 (c)Toimproveg,theemitterhastobedopedmuchheavierthanthebase.Toimprovea,wecanmakethebasewidthnarrower.T6.Wecansketchp(x)p(0)curvesbyusingacomputerprogram:nn1.00.80.6W/L<0.1p(0)n1(x)/pn0.42p50.210200.00.00.20.40.60.81.0DISTANCExInthefigure,wecanseewhen课后答案网WLp<0.1(WLp=0.05inthiscase),theminoritycarrierdistributionapproachesastraightlineandcanbesimplifiedtoEq.15.www.hackshp.cn7.UsingEq.14,IEpisgivenby36 ædpnöI=Aç-qD÷Epçpdx÷èx=0øìé1æW-xöùé1æxöùüïê-coshç÷úê-coshç÷úïïqVkTêLpçèLp÷øúêLpçèLp÷øúï=A(-qD)p(eEB-1)+ppínoêúnoêúýïêæWöúêæWöúïsinhç÷sinhç÷ïêçL÷úêçL÷úïîëèpøûëèpøûþx=0ìéæWöùéùüïêcoshç÷úêúïDppnoïqVkTêçèLp÷øúê1úï=qA(eEB-1)+íêúêúýLpïêæWöúêæWöúïsinhç÷sinhç÷ïêçL÷úêçL÷úïîëèpøûëèpøûþéùêúDppnoæWöêqVkT1ú=qAcothç÷(eEB-1)+.LçL÷êæWöúpèpøêcoshç÷úêçL÷úëèpøûSimilarly,wecanobtainICp:ædpnöI=Aç-qD÷Cpçpdx÷èx=Wøìé1æW-xöùé1æxöùüïê-coshç÷úê-coshç÷úïïqVkTêLpçèLp÷øúêLpçèLp÷øúï=A(-qD)p(eEB-1)+ppínoêúnoêúýïêæçWö÷úêæçWö÷úïsinhsinhïêçL÷úêçL÷úïî课后答案网ëèpøûëèpøûþx=WìéùéæWöùüïêúêcoshç÷úïDppnoïê1úêçèLp÷øúï=qA(eqVEBkT-1)+íêúêúýLpïêæçWö÷úêæçWö÷úïsinhsinhïwww.hackshp.cnêçL÷úêçL÷úïîëèpøûëèpøûþDppno1éqVkTæWöù=qAê(eEB-1)+coshç÷ú.LpæWöêçèLp÷øúsinhç÷ëûçL÷èpø8.Thetotalexcessminoritycarrierchargecanbeexpressedby37 WQ=qA[p(x)-p]dxBònno0Wéxù=qApeqVEBkT(1-)dxò0ênoWúëûW2qVEBkTx=qApnoe(x-)2W0qAWpeqVEBkTno=2qAWpn(0)=.2Wpn(0)FromFig.6,thetriangularareainthebaseregionis.Bymultiplyingthis2valuebyqandthecross-sectionalareaA,wecanobtainthesameexpressionasQ.BInProblem3,-19-2-4111.6´10×0.2´10×0.904´10×2.543´10Q=B2-15=3.678´10C.9.InEq.27,I=a(eqVEBkT-)+1aC2122qADp(0)课后答案网pn@W2DqAQp(0)pn=2W22Dwww.hackshp.cnp=Q.2BWTherefore,thecollectorcurrentisdirectlyproportionaltotheminoritycarrierchargestoredinthebase.10.Thebasetransportfactoris38 1éæWöùê(eqVEBkT-1)+coshç÷úæWöêçL÷úsinhç÷ëèpøûIçL÷CpèpøaT@=.IEpéùêúæçWö÷ê(qVEBkT)1úcothç÷êe-1+úèLpøêæçWö÷úcoshêçL÷úëèpøûForWLp<<1,cosh(WLp)@1.Thus,1a=TæWöæWösinhç÷×cothç÷çL÷çL÷èpøèpøæWö=sechç÷çL÷èpø21æWö=1-ç÷2çL÷èpø(22)=1-W2Lp.11.Thecommon-emittercurrentgainisgivenbya0gaTbº=.0课后答案网1-a01-gaTSinceg@1,aTb0@www.hackshp.cn1-aT(22)1-W2Lp=[(22)]1-1-W2Lp(22)=2LW-1.p22IfWLp<<1,thenb0@2LpW.-7-312.L=Dt=100×3´10=5.477´10cm=54.77mmpppTherefore,thecommon-emittercurrentgainis39 (-4)222254.77´10b@2LW=0p2(-4)2´10=1500.13.Intheemitterregion,407m=54.3+=87.6pE-17181+0.374´10×3´10D=0.0259×87.6=2.26cms.EInthebaseregion,1252m=88+=1186.63n-17161+0.698´10×2´10Dp=0.0259×1186.63=30.73cms.Inthecollectorregion,407m=54.3+=453.82pC-17151+0.374´10×5´10D=0.0259×453.82=11.75cms.C课后答案网14.Intheemitterregion,-6-3L=Dt=2.269×10=1.506´10cmEEEwww.hackshp.cn22(9)ni9.65´10-3p===31.04cm.EO18NE3´10Inthebaseregion,-6-3L=30.734×10=5.544´10cmn(9)29.65´10-3n==4656.13cm.po162´10Inthecollectorregion,-6-3L=11.754×10=3.428´10cmC40 (9)29.65´10-3p==18624.5cm.CO155´10Theemittercurrentcomponentsaregivenby-19-41.6´10×10×30.734×4656.130.60.0259-6I=e=526.83´10AEn-40.5´10-19-41.6´10×10×2.269×31.04(0.60.0259)-9I=e-1=8.609´10A.Ep-31.506´10Hence,theemittercurrentis-6IE=IEn+IEp=526.839´10A.Andthecollectorcurrentcomponentsaregivenby-19-41.6´10×10×30.734×4656.130.60.0259-6I=e=526.83´10ACn-40.5´10-19-41.6´10×10×11.754×18624.5-14I==1.022´10A.Cp-33.428´10Therefore,thecollectorcurrentisobtainedby-4IC=ICn+ICp=5.268´10A.课后答案网15.Theemitterefficiencycanbeobtainedby-6I526.83´10g=En==0.99998.-6www.hackshp.cnIE526.839´10Thebasetransportfactoris-6I526.83´10a=Cn==1.T-6IEn526.83´10Therefore,thecommon-basecurrentgainisobtainedbya=ga=1´0.99998=0.99998.0TThevalueisveryclosetounity.Thecommon-emittercurrentgainis41 a00.99998b==@50000.01-a01-0.9999816.(a)Thetotalnumberofimpuritiesintheneutralbaseregionis0-xl(-Wl)QG=òNAOedx=NAOl1-eW185810-5310-513-2-(-´´)=2´10×3´101-e=5.583´10cm.(b)Averageimpurityconcentrationis13QG5.583´10=-5W8´1017-3=6.979´10cm.17-3317.ForN=6.979´10cm,D=7.77cms,andAn-6-3L=Dt=7.77×10=2.787´10cmnnn22(-5)W8´10aT@1-2=1-2=0.9995882L2(2.787´10-3)n11g===0.99287.DQ15.583´10131课后答案网+EG1+×DnNELE7.771019×10-4Therefore,a=ga=0.99246www.hackshp.cn0Ta0b==131.6.01-a017-318.Themobilityofanaverageimpurityconcentrationof6.979´10cmisabout2300cmVs.TheaveragebaseresistivityrisgivenbyB1r==0.0299W-cm.B()qmnQGW42 Therefore,-3()-3(-5)R=5´10rW=5´10×0.02998´10=1.869W.BBForavoltagedropofkTq,kTI==0.0139A.BqRBTherefore,I=bI=131.6×0.0139=1.83A.C0B19.FromFig.10bandEq.35,weobtainIB(mA)IC(mA)DICb0=DIB00.20---50.95150102.00210153.1022020课后答案网4.00180254.70140b0isnotaconstant.Atlowwww.hackshp.cnIB,becauseofgeneration-recombinationcurrent,b0increaseswithincreasingI.AthighI,VincreaseswithI,thisinturncausesBBEBBareductionofVsinceV+V=V=5V.ThereductionofVcausesaBCEBBCECBCwideningoftheneutralbaseregion,thereforebdecreases.0ThefollowingchartshowsbasfunctionofI.Itisobviousthatbisnota0B0constant.43 2502000b150100051015202530I(mA)B20.ComparingtheequationswithEq.32givesI=a,aI=aFO11RRO12aI=a,andI=a.FFO21RO22Hence,a211a==Fa111+W×DE×nEO课后答案网LEDppnoa121a==.Ra221+W×DC×nCOwww.hackshp.cnLCDppno21.Inthecollectorregion,-6-3L=Dt=2×10=1.414´10cmCCC22(9)154-3n=nN=9.65´105´10=1.863´10cm.COiCFromProblem20,wehave44 11a==FWDn-4EEO0.5´1019.311+××1+××LDp10-3109.31´10-2Epno=0.9999511a==RWDn0.5´10-421.863´104CCO1+××1+××LDp1.414´10-3109.31´10-2Cpno=0.876æDpDnöçpnoEEO÷IFO=a11=qAç+÷èWLEøæ2×ö-19-410×9.31´1019.31=1.6´10×5´10×ç+÷ç-4-4÷è0.5´1010ø-14=1.49´10AæDpDnöçpnoCCO÷IRO=a22=qAç+÷èWLCøæ24ö-19-410×9.31´102×1.863´10=1.6´10×5´10×ç+÷ç-4-3÷è0.5´101.414´10ø-14=1.7´10A.TheemitterandcollectorcurrentsareI=I(eqVEBkT-1)+aIEFORRO-4=1.715´10A课后答案网IC=aFIFO(eqVEBkT-1)+IRO-4=1.715´10A.Notethatthesecurrentsarealmostthesame(nobasecurrent)forWLp<<1.www.hackshp.cn22.ReferringEq.11,thefield-freesteady-statecontinuityequationinthecollectorregioniséd2n(x")ùn(x")-nCCpoDCê2ú-=0.ëdx"ûtCThesolutionisgivenby(L=Dt)CCCn(x")=Cex"LC+Ce-x"LC.C12Applyingtheboundaryconditionatx"=¥yields45 Ce¥LC+Ce-¥LC=0.12HenceC=0.Inaddition,fortheboundaryconditionatx"=0,1Ce-0LC=C=n(0)22Cn(0).=n(eqVCBkT-1).CCOThesolutionisn(x)=n(eqVCBkT-1)e-x"LC.CCOThecollectorcurrentcanbeexpressedasædpnöædnCöI=Aç-qD÷+Aç-qD÷Cçpdx÷çCdx"÷èx=Wøèx"=0øDppno(qVEBkT)æçDppnoDCnCOö÷(qVCBkT)=qAe-1-qA+e-1WçèWL÷øC=a(eqVEBkT-1)-a(eqVCBkT-1).212223.UsingEq.44,thebasetransittimeisgivenby(-4)220.5´10-10t=W2D==1.25´10s.Bp2´10Wecanobtainthecutoff课后答案网frequency:f@12pt=1.27GHz.TBFromEq.41,thecommon-basecutofffrequencyisgivenby:www.hackshp.cn1.27´109f@fa==1.275GHz.aT00.998Thecommon-emittercutofffrequencyis()()9fb=1-a0fa=1-0.998´1.275´10=2.55MHz.Notethatfbcanbeexpressedby1f=(1-a)f=(1-a)a´f=f.b0a00TTb046 24.Neglectthetimedelaysofemitterandcollector,thebasetransittimeisgivenby11-12t===31.83´10s.B92pf2p´5´10TFromEq.44,WcanbeexpressedbyW=2Dt.pBTherefore,-12W=2´10´31.83´10-5=2.52´10cm=0.252mm.Theneutralbasewidthshouldbe0.252µm.25.DEg=9.8%×1.12@110meV.æDEgöb~expç÷oçkT÷èøbo(100°C)æ110meV110meVö=expç-÷=0.29.bo(0°C)è373k273kø课后答案网b0(HBT)æEgE-EgBöéEgE(x)-.1424ù26.=expçç÷÷=expêúb(BJT)kT.002590èøëûwherewww.hackshp.cnE(x)=.1424+.1247x,x£.045gE=9.1+.0125x+.0143x.0,45W)islargerthan1500V(W>100mm).Forammreverseblockvoltageof120V,wecanchooseawidthsuchthatpunch-throughoccurs,i.e.,2qNWV=D.PT2esThus,课后答案网æç2esVPTö÷-3W==3.96´10cm.çqN÷èDøWhenswitchingoccurs,www.hackshp.cna+a@1.12Thatis,LpæJöa=0.5lnç÷1ç÷WèJ0ø=1-0.4=0.6-4æJö0.625´10lnç÷=ç÷-4èJ0ø0.539.6´10=1.51.48 Therefore,-52J@4.5J=2.25´10Acm0-3Is1´102Area===44.4cm.J-52.25´1028.Inthen1-p2-n2transistor,thebasedrivecurrentrequiredtomaintaincurrentconductionisI=(1-a)I.12KInaddition,thebasedrivecurrentavailabletothen1-p2-n2transistorwithareversegatecurrentisI=aI-I.21AgTherefore,whenuseareversegatecurrent,theconditiontoobtainturn-offofthethyristorisgivenbyII.gAa2NotethatifwedefinetheratioofIAtoIgasturn-offgain,thenthemaximumturn-offgainbismaxa2bmax=课后答案网.a+a-112www.hackshp.cn49 CHAPTER61.VG=VTECEFEiEVyS=2yB课后答案网METALOXIDEn-TYPESEMICONDUCTORwww.hackshp.cn50 2.ECEiEFEFECEVEiEV3.+nPOLYSILICONOXIDEp-TYPESEMICONDUCTORECEi课后答案网EFVG=VFB=ffmsEVwww.hackshp.cnfms+nPOLYSILICONOXIDEp-TYPESEMICONDUCTOR51 4.rS(x)QPqND-d0Wx(a)QME(x)-d0W(b)x课后答案网y(x)www.hackshp.cn-d0W(c)xysVVo52 æNAöeskTlnçç÷÷nèiø5.Wm=22qNAæ16ö-145´1011.9´8.85´10´0.026lnç÷ç9÷è9.65´10ø=2-19161.6´10´5´10-5=1.5×10cm=0.15mm.eox6.C=mind+(e/e)WoxsmW=0.15mmFromProb.5m-143.9´8.85´10-82Cmin==6.03´10F/cm.-73.9-58´10+´1.5´1011.917kTNA107.y=ln=0.026ln=0.42VB9qni9.65´10qNWAE=atintrinsicy=yssBes-142esys2´119.´.885´10´.042-5W===.074´10cm-1917qN6.1´10´10A-1917-51.6´10´10´0.74´105E==1.11´10V/cms-14课后答案网11.9´8.85´10Ees511.95o=Es=1.11×10×=3.38×10V/cme3.9ox5-7V=Vo+y=Eod+y=(3.38´10´5´10)+0.42.ss=0.59V.www.hackshp.cn8.Attheonsetofstronginversion,y=2yÞVG=VTsBqNWAmthus,VG=+2yBCoæ16ö5´10FromProb.5,Wm=0.15£gm,y=0.026lnç÷=0.4VBç9.65´109÷èø-143.9´8.85´10-72Co==3.45×10F/cm-610-1916-51.6´10´5´10´1.5´10V=+0.8=0.35+0.8=1.15V.G-73.45´1053 -6-19110176.1´10117-629.Q=yq(10)dy=[´10´(10)]otò0-6d102-92=8´10C/cm-9Qot8´10-2DV===.232´10V.FB-7C.345´10o1d10.Q=yr(y)dyotòotd0-7ì¥,y=5´1011rot=q´5´10d(x),whered(x)=í-7î,0y¹5´10111-7-19Q=´5´10´5´10´6.1´10ot-610-82=4´10C/cm-8Qot4´10DV===.012VFB-7C.345´10o1-19123-63=´6.1´10´´5´10(10).-6103110-62311.Q=y(q´5´10´y)dyotòd0-82=2.67´10C/cm-8Qot.267´10-2DV===.774´10V.FB-7C.345´10o课后答案网12.DVFB=Qot=q´d´Nm,whereNmistheareadensityofQm.CCdoowww.hackshp.cn-7DVFB´Co3.0´.345´1011-2ÞN===.647´10cm.m-19q6.1´1013.SinceV<<(V-V),thefirstterminEq.33canbeapproximatedasDGTZmC(V-2y)V.noGBDLndPerformingTaylor’sexpansiononthe2terminEq.33,weobtain3/23/23/231/23/231/2(V+2y)-(2y)@(2y)+(2y)V-(2y)=(2y)VDBBBBDBBD22Equation33cannowbere-writtenas54 Zé22esqNA3ùI@mCê(V-2y)V-´2yVúDnoGBDBDLêë3Co2úûZìïé2esqNA(2yB)ùüï@mnCoíVG-ê2yB+úýVDLïîêëCoúûïþZ@()mC(V-V)VnoGTDL2eqN(2y)sABwhereV=2y+.TBCo14.Whenthedrainandgateareconnectedtogether,V=VandtheMOSFETisGDoperatedinsaturation(V>V).IcanbeobtainedbysubstitutingDDsatDV=VinEq.33DDsatZìïæVDsatö2æ2esqNAö3/23/2üïIDV=V=mnCoíç-2yB÷VDsat-ç÷[(VDsat+2yB)-2(yB)]ýDGL23çC÷ïîèøèoøïþwhereVisgivenbyEq.38.InsertingtheconditionQ(y=L)=0intoEq.Dsatn27yields1V=2eqN(V+2y)+2y-V=0DsatsADsatBBGCoForV<<2j,theaboveequationreducestoDsatB2eqN(2y)sABV=V=+2y=V.GDBTCoThereforealinearextrapolationfromthelowcurrentregiontoI=0willyieldDthethresholdvoltagevalue.课后答案网16kTNAé5´10ù15.y=ln()=0.026lnêú=0.40V-9qnië9.65´10ûeqNwww.hackshp.cn-14-19-16sA11.9´8.85´10´1.6´10´5´10Kº=-7Co3.45´10=0.272æç2VGö÷V@V-2y+K1-1+DsatGBçK2÷èø210=5-8.0+(.027)[1-1+]2(.027)=3.42V-7ZmnCo10´800´.345´102I=(V-V)=(5-7.0)DsatGT2L2´1-2=2.55´10A.55 16.Thedeviceisoperatedinlinearregion,sinceV=1.0V<(V-V)=5.0VDGT¶IDZTherefore,g==mC(V-V)dVG=const.noGT¶VLD5-7=´500´.345´10´5.00.25-3=1.72´10S.¶IDZ17.g==mCVm¶VVD=const.LnoDG5-7=´500´.345´10´1.00.25-4=3.45´10S.2eqNy17sAB1018.V=V+2y+,y=0.026ln()TFBBB9Co9.65´10QE6.1´10-19´5´1010fgV=f-=--y-FBmsB-7C2.345´10o=-.056-.042-.002=-1V-1417-192119.´.885´10´10´.042´6.1´10V=-1+.084+T-73.45´10=-1+.084+.049=0.33V(fcanalsobeobtainedfromFig.8tobe–0.98V).ms课后答案网qFB19.7.0=.033+-73.45´10-7.037´.345´1011-2F==8´10cm.B-191.6´10www.hackshp.cnEg20.f=-+y=-.056+.042=-.014VmsB2Qf2esqNDyBV=f--2y-TmsBCCoo-14-19172119.´.885´10´6.1´10´10´.042=-.014-.002-.084--73.45´10=-.149V.56 qFB21.-7.0=-.149+-73.45´10-7.079´.345´1012-2F==7.1´10cm.B-191.6´1022.ThebandgapindegeneratelydopedSiisaround1eVduetobandgap-narrowingeffect.Therefore,f=-.014+1=.086VmsV=.086-.002-.084-.049=-.049V.T课后答案网www.hackshp.cn57 æ17ö1023.y=0.026lnç÷=0.42VBç9.65´109÷èøqQ2eqNyfsABVT=fms-+2yB+CCoo-1911-1417-191.6´10´10211.9´8.85´10´10´0.42´1.6´10=-0.98-+0.84+CCoo-815.2´10=-0.14+Co-14-139.3´.885´10.345´10C==odd-8d´152.´10V>20Þ>20.14T-133.45´10-5d>.457´10cm=0.457mm.24.V=5.0VatI=1.0£gATd-1ælogI-logIöSubthresholdswing=çDVG=VTDVG=0÷çV-0÷èTø5.01.0=-7-logIDVG=0-12logI=-12I=1´10A.DV=0DV=0GG25.2qeNsA(课后答案网)DV=2y+V-2yTBBSBCo1710y=.0026ln()=.042VB99.65www.hackshp.cn´10-149.3´.885´10-72C=9.6´10F/cmo-75´10DV=1.0VifwewanttoreduceIDatVG=0byoneorderofmagnitude,Tsincethesubthresholdswingis100mV/decade.-19-14172´6.1´10´9.4´.885´10´10()1.0=.084+V-.084-7BS9.6´10V=.083V.BS26.Scalingfactork=1012Switchingenergy=(C×A)V258 eoxC¢==kCdAA¢=2kVV¢=k1111scalingfactorforswitchingenergy:k××==223kkk1000Areductionofonethousandtimes.27.FromFig.24wehave222(rj+D)=(rj+Wm)-Wm2D+2Dr-2Wr=0jmjD=-r+r+2WrjjmjL"=L-2DL+L"2L-2DDrjæç2Wmö÷==1-=1-1+-12L2LLLçèrj÷øFromEq.17wehave(spacechargeinthetrapezaidalregion-spacechargeintherectangularregion)DV=TCoqNAWmL+L"qNAWmqNAWmrjæç2Wö÷=()-=-1+-1.C2LCCLçr÷oooèjø28.Pros:课后答案网1.Higheroperationspeed.2.HighdevicedensityCons:1.Morecomplicatedfabricationflow.2.Highmanufacturingcost.www.hackshp.cn29.ThemaximumwidthofthesurfacedepletionregionforbulkMOSekTln(N/n)sAiW=2m2qNA-14179119.´.885´10´.0026´ln(5´10/.965´10)=2-19176.1´10´5´10-6=9.4´10cm=49nmForFD-SOI,d£W=49nm.sim59 qNdAsi30.V=V+2y+TFBBCo17EkTN.112æ5´10ögAV=f=--ln()=--.0026lnç÷=-.102VFBms2qn2ç9÷iè.965´10økTNA2yB=2´ln()=.092Vqni-149.3´.885´10-72C==.863´10F/cmo-74´10-1917-66.1´10´5´10´3´10V=-.102+.092+T-7.863´10=-1.0+.028=.018V.qNDdAsi31.DV=TCo-1917-76.1´10´5´10´5´10=-7.863´10=46mVThus,therangeofVTisfrom(0.18-0.046)=0.134Vto(0.18+0.046)=0.226V.32.Theplanarcapacitor-42-14C=Aeox=(1´10)9.3´.886´10=.345´10-15Fd1´10-6Forthetrenchcapacitor课后答案网222A=4×7µm+1µm=29µmC=29×3.45×10-15F=100×10-15F.dV-145.2-1133.I=C=5´10www.hackshp.cn´=1.3´10A-3d4´10Z34.g=mC(V-V)opioTL-54×10=A(-5-V)V=7VTT-51×10=A(-5+2)DV=7-(-2)=9V.eqNysAB35.V=V+2y+TFBBC060 -14.8854´10-82C=9.3´=.345´10F/cm0-51017æ10öy=.0026lnç÷=.042VBç9÷è.965´10øQEfgV=f-=--y-0FBmsB22=-0.56-0.42=-0.98V-1417-192119.´.885´10´10´.042´6.1´10V=-.098+.042+T-8.345´10=-0.98+0.42+4.9=4.34VQ11-19f5´10´6.1´10DV=-=-FB-8C.345´100=-2.32VV=.434-.232T=2.02V.课后答案网www.hackshp.cn61 CHAPTER71.FromEq.1,thetheoreticalbarrierheightisf=f-c=.455-.401=.054eVBnmWecancalculateVnas19kTNC2.86´10V=ln=0.0259ln()=0.188Vn16qND2´10Therefore,thebuilt-inpotentialisV=f-V=.054-.0188=.0352V.biBnn2.(a)FromEq.11214d(1/C)(2.6-6.4)´101422==-3.2´10(cm/F)/VdV-2-02é-1ù16-3ND=ê2ú=7.4´10cmqesëd(1/C)/dVûæ17ökTNC4.7´10V=ln=0.0259lnç÷=0.06VnqNç4.7´1016÷D课后答案网èøFromFig.6,theinterceptoftheGaAscontactisthebuilt-inpotentialVbi,whichisequalto0.7V.Then,thebarrierheightiswww.hackshp.cnf=V+V=.076VBnbin172(b)J=5´10A/cms*22A=8A/K-cmforn–typeGaAsJ=A*T2e-qfBn/kTs*22kTATé8´(300)ùfBn=ln()=.00259lnê-7ú=.072eVqJsë5´10ûThebarrierheightfromcapacitanceis0.04Vor5%larger.62 (c)ForV=–1V-122es(Vbi+VR)2´.109´10(7.0+1)W==-1916qND6.1´10´7.4´10-5=.222´10cm=0.222mmqNWD5E==.143´10V/cmmeses-82C==.522´10F/cm.W3.Thebarrierheightisf=f-c=.465-.401=.064VBnmæ19ökTNC2.86´10V=ln=0.0259´lnç÷=0.177VnqNç3´1016÷DèøThebuilt-inpotentialisV=f-V=.064-.0177=.0463VbiBnnThedepletionwidthis-142es(Vbi-V)2´119.´.885´10W=课后答案网=´.0463=.0142ìm-1916qN6.1´10´3´10DThemaximumelectricfieldis1916-5www.hackshp.cnqND(6.1´10)´(3´10)´(.142´10)4E=E(x=0)=W==.654´10V/cm.m-14e119.´(.885´10)s24.TheunitofCneedstobechangedfrommFtoF/cm,so21515221/C=1.74´10–2.12´10Va(cm/F)2Therefore,weobtainthebuilt-inpotentialat1/C=015.157´10V==.074Vbi152.12´1063 FromthegivenrelationshipbetweenCandVa,weobtainæ1ödç÷2èCø1522=-.212´10(cm/F)/VdVaFromEq.112é-1ùN=Dê2úqesëd(1/C)/dVû2æ1ö=ç÷-19-14156.1´10´119.´.885´10è.212´10ø15-3=6.5´10cmæ19ökTNC2.86´10V=ln=0.0259lnç÷=0.161VnqNç5.6´1016÷DèøWecanobtainthebarrierheightf=V+V=.074+.0161=.0901V.Bnbin5.Thebuilt-inpotentialis课后答案网kTNCV=f-lnbiBnqNDæ19öç2.86´10÷=0.8-0.0259lnç16÷è1.5´10ø=0.8-0www.hackshp.cn.195=0.605VThen,theworkfunctionisf=f+cmBn=8.0+.401=.481V.6.Thesaturationcurrentdensityis64 *2æ-qfBnöJs=ATexpç÷èkTø2æ-0.8ö=110´(300)´expç÷è0.0259ø-72=3.81´10A/cmTheinjectedholecurrentdensityisqDn2-1992pi6.1´10´12´(.965´10)-112J===.119´10A/cmpo-316LN1´10´5.1´10pDqV/kTHolecurrentJpo(e-1)=qV/kTElectroncurrentJ(e-1)sJ.118´10-11po-5===3´10.-3J.381´10s7.ThedifferencebetweentheconductionbandandtheFermilevelisgivenbyæ17ö4.7´10V=0.0259lnç÷=0.04V.nç1´1017÷èøThebuilt-inpotentialbarrieristhenV=9.0-.004=.086VbiForadepletionmodeoperation,VTisnegative.Therefore,FromEq.38aV=.086-V<0TP课后答案网qa2N6.1´10-19a2´1017DV==>.086P-142e2´124.´.885´10swww.hackshp.cn-26.1´102a>.086-12.219´10-5a>1.08´10cm=0.108ìm.65 8.FromEq.33weobtainIVPPg=Vm2D2VV+VpGbiZmeVnsP=VDaLVbi2-1916-52qNa6.1´10´7´10´(3´10)DV===.462VP-142e2´124.´.885´10s-4-145´10´4500´124.´.885´10.462g=´1m-4-43.0´10´5.1´10.084-3=1.28´10S=1.28mS.9.(a)Thebuilt-involtageis17æ7.4´10öV=f-V=9.0-.0025lnç÷=0.86VbiBnnç1017÷èøAtzerobias,thewidthofthedepletionlayeris-122esVbi2´.109´10´.086W==课后答案网-1917qN6.1´10´10D-5=1.07´10cm=0.107µmwww.hackshp.cnSinceWissmallerthan0.2µm,itisadepletion-modedevice.(b)Thepinch-offvoltageis2-1917-5qNa6.1´10´10(2´10)DV===2.92VP-142e2´124.´.885´10sandthethresholdvoltageisVT=Vbi–VP=0.86–2.92=–2.06V.10.FromEq.31b,thepinch-offvoltageis66 2-1917-5qNa6.1´10´10(2´10)DV===0.364VP-142e2´124.´.885´10sThethresholdvoltageisVT=Vbi–Vp=0.8–0.364=0.436VandthesaturationcurrentisgivenbyEq.39Zesmn2I=(V-V)DsatGT2aL-4-1450´10´124.´.885´10´45002-4=(0-.0436)=7.4´10A.-4-42´(5.0´10)´(1´10)17-19æ7.4´10ö6.1´10´ND211.0.85–0.0259lnç÷-a=0ç÷-14N2´124.´.885´10èDø16-3ForND=4.7×10cm11723æ7.4´10ö(7.3´10)a=ç.085-.00259ln÷ç÷èNDøND-5=1.52×10cm=0.152µmForN课后答案网17-3D=4.7×10cm-5a=0.496×10cm=0.0496µm.www.hackshp.cn12.FromEq.48thepinch-offvoltageisDECVP=fBn--VTq=0.89–0.23–(–0.5)=0.62Vandthen,2-19182qNd6.1´10´3´10´dD11V===0.62VP-142e2´123.´.885´10s67 -6d1=1.68×10cmd1=16.8nmTherefore,thisthicknessofthedopedAlGaAslayeris16.8nm.13.Thepinch-offvoltageis2-1918-7qNd6.1´10´10´(50´10)D1V===1.84VP-42e2´123.´.885´10sThethresholdvoltageisDECV=f--VTBnPq=0.89–0.23–1.84=–1.18V12-2Whenns=1.25×10cm,weobtain-14123.´.885´10[]12n=´0-(-.118)=.125´10s-19-76.1´10´(50+d+8)´100andthen课后答案网d0+58.5=64.3d0=5.8nmThethicknessofthewww.hackshp.cnundopedspaceris5.8nm.14.Thepinch-offvoltageis2-1917-72qNd6.1´10´5´10´(50´10)D1V===1.84VP-142e2´123.´.885´10sThebarrierheightisDECf=V++V=-3.1+.025+.184=.079VBnTPq68 The2DEGconcentrationis-14123.´.885´10[]12-2n=´0-(-3.1)=.129´10cm.s-19-76.1´10´(50+10+8)´1015.Thepinch-offvoltageis2-19182qNd6.1´10´1´10´dD11V==5.1=P-42e2´123.´.885´10sThethicknessofthedopedAlGaAsis-145.1´2´123.´.885´10-8d==4.45×10cm=44.5nm1-19186.1´10´10DECV=f--V=0.8-0.23-1.5=-0.93V.TBnPq16.Thepinch-offvoltageisqND2V=dP12es-1918=6.1´10课后答案网´3´10(35´10-7)2=2.7V-142´12.3´8.85´10thethresholdvoltageisDEwww.hackshp.cnCV=f--VTBnpq=0.89–0.24–2.7=–2.05VTherefore,thetwo-dimensionalelectrongasis-14123.´.885´10[]12-2n=´0-(-.205)=2.3´10cm.s-19-76.1´10´(35+8)´1069 CHAPTER8L1.Z0=C2-122L=C×Z=2´10´75=11.25nH.0222cæmöænöæpö2.fr=ç÷+ç÷+ç÷2èaøèbøèdøc()2()2=10+0+4283´10m/s=´116=.1616GHz.23.V@(E/q)+V+V=1.42+0.03+0.03=1.48Vbignp2eæN+NöSAD()W=ç÷V+Vç÷biqèNANDø课后答案网-1219192´.116´10æ10+10ö=ç÷(.148-.025)-19ç1919÷6.1´10è10´10ø-6=.189´10cm=18.9nm-12eS.116´10-72C==www.hackshp.cn=.613´10F/cm.-6W1.89´10æVöæVöæqVö4.I=Ipçç÷÷expçç1-÷÷+I0expç÷èVPøèVPøèkTøFromFig.4,WenotethatthelargestnegativedifferentialresistanceoccursbetweenVp>t,t>>,4B2then,x=Btsimilarly,2Awhent>>t,t>>,4BBthen,x=(t+t)A-32-24.At980(=1253K)and1atm,B=8.5×10µm/hr,B/A=4×10µm/hr(fromFigs.6163and7).SinceAa2D/k,B/A=kC0/C1,C0=5.2×10molecules/cmandC1=22-32.2×10cm,thediffusioncoefficientisgivenbyAkAæBC1öBæC1öD==ç×÷=ç÷22çAC÷2çC÷è0øè0ø-3228.5´102.2´102=mm/hr2165.2´1032=1.79´10mm/hr-92=4.79课后答案网´10cm/s.5.(a)ForSiNxHySi=1www.hackshp.cn=1.2Nx4x=0.83100yatomic%H==201+0.83+y4y=0.46TheempiricalformulaisSiN0.83H0.46.28-33.3×1.211(b)ñ=5×10e=2×10Ù-cmAstheSi/Nratioincreases,theresistivitydecreasesexponentially.95 6.SetTa2O5thickness=3t,e1=25SiO2thickness=t,e2=3.9Si3N4thickness=t,e3=7.6,area=AtheneeA10C=Ta2O53t1ttt=++CeeAeeAeeAONO203020eeeA230C=ONO()e+2et23CTa2O5e1(e2+2e3)25(9.3+2´6.7)===.537.C3ee3´9.3´6.7ONO237.SetBSTthickness=3t,e1=500,area=A1SiO2thickness=t,e2=3.9,area=A2Si3N4thickness=t,e3=7.6,area=A2theneeAeeeA1012302=3t(e+2e)t23A1=课后答案网.00093.A28.LetTa2O5thickness=3t,www.hackshp.cne1=25SiO2thickness=t,e2=3.9Si3N4thickness=t,e3=7.6area=AtheneeAeeA1020=3td3et2d==.0468t.e196 9.Thedepositionratecanbeexpressedasr=r0exp(-Ea/kT)whereEa=0.6eVforsilane-oxygenreaction.ThereforeforT1=698Kr(T)éæ11öù2=2=expê6.0çç-÷÷úr(T1)ëèkT1kT2øû6.0éæ300300öùln2=êçç-÷÷ú.00259ëè698T2øû4T2=1030K=757.10.Wecanuseenergy-enhancedCVDmethodssuchasusingafocusedenergysourceorUVlamp.AnothermethodistouseborondopedP-glasswhichwillreflowattemperatureslessthan900.11.Moderatelylowtemperaturesareusuallyusedforpolysilicondeposition,andsilanedecompositionoccursatlowertemperaturesthanthatforchloridereactions.Inaddition,silaneisusedforbettercoverageoveramorphousmaterialssuchSiO2.12.Therearetworeasons.Oneistominimizethethermalbudgetofthewafer,reducingdopantdiffusionandmaterialdegradation.Inaddition,fewergas课后答案网phasereactionsoccuratlowertemperatures,resultinginsmootherandbetteradheringfilms.Anotherreasonisthatthepolysiliconwillhavesmallgrains.Thefinergrainsareeasiertomaskandetchtogivesmoothanduniformedges.www.hackshp.cnHowever,fortemperatureslessthan575ºCthedepositionrateistoolow.13.Theflat-bandvoltageshiftisQotDVFB=0.5V~C0-14eox9.3´.885´10-8-2C===9.6´10F/cm.0-8d500´104Numberoffixedoxidechargeis97 -85.0C05.0´9.6´1011-2==1.2´10cm-19q6.1´10Toremovethesecharges,a450heattreatmentinhydrogenforabout30minutesisrequired.14.20/0.25=80sqs.Therefore,theresistanceofthemetallineis5´50=400W.15.ForTiSi230´2.37=71.1nmForCoSi230´3.56=106.8nm.16.ForTiSi2:Advantage:lowresistivityItcanreducenative-oxidelayersTiSi2onthegateelectrodeismoreresistanttohigh-field-inducedhot-electrondegradation.Disadvantage:bridgingeffectoccurs.课后答案网LargerSiconsumptionduringformationofTiSi2LessthermalstabilityForCoSi2:Advantage:www.hackshp.cnlowresistivityHightemperaturestabilityNobridgingeffectAselectivechemicaletchexitsLowshearforcesDisadvantage:notagoodcandidateforpolycides98 17.(a)L-613R=r=.267´10´=2.3´10W-4-4A0.28´10´0.3´10-14-44-6eAeTL3.9´8.85´10´0.3´10´1´10´10-13C====2.9´10F-4dS0.36´105-15RC=3.2´10´9.2´10=.093nsL-613(b)R=r=7.1´10´=2´10W-4-4A0.28´10´0.3´10-14-4eAeTL8.2´.885´10´3.0´10´1-13C====1.2´10F-4dS0.36´103-13RC=2´10´1.2´10=.042ns.042(c)WecandecreasetheRCdelayby55%.Ratio==0.45.093L-61318.(a)R=r=.267´10´=2.3´10W44A0.28´10´0.3´10-14-4eAeTL9.3´.885´10´3.0´10´1´3-13C====7.8´10F4dS0.36´103-13RC=3.2×10×8.7×10=2.8ns..(b)R=rLA=7.1´10课后答案网-6´0.28´1041´0.3´104=2´103W-14-4eAeTL8.2´.885´10´3.0´10´1´3-13C====3.6´10F4dSwww.hackshp.cn0.36´103-13RC=2´10´7.8´10=5.2ns3-13RC=2.3´10´7.8´10=5.2ns.19.(a)Thealuminumrunnercanbeconsideredastwosegmentsconnectedinseries:20%(or0.4mm)ofthelengthishalfthickness(0.5µm)andtheremaining1.6mmisfullthickness(1µm).Thetotalresistanceisél1l2ù-6é.016.004ùR=rê+ú=3´10ê-4-4+-4-4úëA1A2ûë10´1010´(5.0´10)û=72Ù.99 ThelimitingcurrentIisgivenbythemaximumallowedcurrentdensitytimescross-sectionalareaofthethinnerconductorsections:52-4-4-3I=5×10A/cm×(10×0.5×10)=2.5×10A=2.5mA.Thevoltagedropacrossthewholeconductoristhen-3V=RI=72W´5.2´10A=0.18V.20.0.5mm0.540nmmmCu=Al60nmh:height,W:width,t:thickness,assumethattheresistivitiesofthecladdinglayerandTiNaremuchlargerthanrandrAlCullR=r´=7.2AlAlh´W(5.0-1.0)´5.0llR=r´=7.1CuCuh´W(5.0-2t)´(5.0-2t)WhenR=RAlCu7.27.1Then=24.0´5.0(课后答案网5.0-2t)Þt=0.073mm=73nm.www.hackshp.cn100 CHAPTER1231.WithreferencetoFig.2forclass100cleanroomwehaveatotalof3500particles/mwithparticlesizes³0.5µm212´3500=735particles/mwithparticlesizes³1.0µm1005.42´3500=157particles/mwithparticlesizes³2.0µm1003Therefore,(a)3500-735=2765particles/mbetween0.5and1µm3(b)735-157=578particles/mbetween1and2µm3(c)157particles/mabove2µm.9-D1A2.Y=Pen=122A=50mm=0.5cm-4(0.1´0.5)-4(0.25´0.5)-1(1´0.5)-1.2Y=e´e´e=e=301.%.3.Theavailableexposureenergyinanhouris2220.3mW/cm×3600s=1080mJ/cmForpositiveresist,thethroughputis1080=7wafers/hr140Fornegativeresist,thethroughputis1080=120wafers/hr.9课后答案网4.(a)Theresolutionofaprojectionsystemisgivenbyl0.193£gml=k=0.6´=0.178µmm1NA0.65léwww.hackshp.cn.0193mmùDOF=k22=5.0ê2ú=0.228µm(NA)ë(.065)û(b)WecanincreaseNAtoimprovetheresolution.Wecanadoptresolutionenhancementtechniques(RET)suchasopticalproximitycorrection(OPC)andphase-shiftingMasks(PSM).Wecanalsodevelopnewresiststhatprovidelowerk1andhigherk2forbetterresolutionanddepthoffocus.(c)PSMtechniquechangesk1toimproveresolution.5.(a)Usingresistswithhighgvaluecanresultinamoreverticalprofilebutthroughputdecreases.(b)ConventionalresistscannotbeusedindeepUVlithographyprocessbecausetheseresistshavehighabsorptionandrequirehighdosetobeexposedindeepUV.Thisraisesthe101 concernofdamagetostepperlens,lowerexposurespeedandreducedthroughput.6.(a)Ashapedbeamsystemenablesthesizeandshapeofthebeamtobevaried,therebyminimizingthenumberofflashesrequiredforexposingagivenareatobepatterned.Therefore,ashapedbeamcansavetimeandincreasethroughputcomparedtoaGaussianbeam.(b)Wecanmakealignmentmarksonwafersusinge-beamandetchtheexposedmarks.Wecanthenusethemtodoalignmentwithe-beamradiationandobtainthesignalfromthesemarksforwaferalignment.X-raylithographyisaproximityprintinglithography.Itsaccuracyrequirementisveryhigh,thereforealignmentisdifficult.(c)X-raylithographyusingsynchrotronradiationhasahighexposurefluxsoX-rayhasbetterthroughputthane-beam.7.(a)Toavoidthemaskdamageproblemassociatedwithshadowprinting,projectionprintingexposuretoolshavebeendevelopedtoprojectanimagefromthemask.Witha1:1projectionprintingsystemismuchmoredifficulttoproducedefect-freemasksthanitiswitha5:1reductionstep-and-repeatsystem.(b)Itisnotpossible.ThemainreasonisthatX-rayscannotbefocusedbyanopticallens.Whenitisthroughthereticle.Sowecannotbuildastep-and-scanX-raylithographysystem.课后答案网8.Asshowninthefigure,theprofileforeachcaseisasegmentofacirclewithoriginattheinitialmask-filmedge.Asoveretchingproceedstheradiusofcurvatureincreasessothattheprofiletendstoaverticalline.www.hackshp.cn9.(a)20sec0.6×20/60=0.2µm…..(100)plane0.6/16×20/60=0.0125µm……..(110)plane0.6/100×20/60=0.002µm…….(111)planeW=W-2l=5.1-2´2.0=.122µmb0(b)40sec0.6×40/60=0.4µm….(100)plane102 0.6/16×40/60=0.025µm….(110)plane0.6/100×40/60=0.004µm…..(111)planeW=W-2l=5.1-2´4.0=0.93µmb0(c)60sec0.6×1=0.6µm….(100)plane0.6/16×1=0.0375µm….(110)plane0.6/100×1=0.006µm…..(111)planeW=W-2l=5.1-2´6.0=0.65µm.b010.UsingthedatainProb.9,theetchedpatternprofileson<100>-Siareshowninbelow.(a)20secl=0.012µm,W=W=5.1µm0b(b)40secl=0.025µm,W=W=5.1µm0b(c)60secl=0.0375µmW=W=5.1µm.0b11.IfweprotecttheICchipareas(e.g.withSi3N4layer)andetchthewaferfromthetop,thewidthofthebottomsurfaceisW=W+2l=1000+2´625=1884µm1课后答案网Thefractionofsurfaceareathatislostis222222(W-W)/W×100%=(1884-1000)/1884×100%=71.8%1www.hackshp.cnIntermsofthewaferarea,wehavelost2271.8%×p(15/2)=127cmAnothermethodistodefinemaskingareasonthebacksideandetchfromtheback.ThewidthofeachsquaremaskcenteredwithrespectofICchipisgivenbyW=W-2l=1000-2´625=116µm1Usingthismethod,thefractionofthetopsurfaceareathatislostcanbenegligiblysmall.12.1Pa=7.52mTorr103 PV=nRT-37.52/760×10=n/V×0.082×273-7-72314-3n/V=4.42×10mole/liter=4.42×10×6.02×10/1000=2.7×10cmmean–free–path-3-3l=5´10/Pcm=5×10×1000/7.52=0.6649cm=6649µm150Pa=1128mTorrPV=nRT课后答案网www.hackshp.cn104 -31128/760×10=n/V×0.082×273-5-52316-3n/V=6.63×10mole/liter=6.63×10×6.02×10/1000=4×10cmmean-free-path-3-3l=5´10/Pcm=5×10×1000/1128=0.0044cm=44µm.1-Ea13.SiEtchRate(nm/min)=2.86×10-13×n´T2´eRTF-2.48´103-13151=2.86×10×3×10×(298)2´e1.987´298=224.7nm/min.-3.76´103-1315114.SiO2EtchRate(nm/min)=0.614×10×3×10×(298)2´e1.987´298=5.6nm/min6.5EtchselectivityofSiO2overSi==.0025224.7.0614(-3.76+2.48)Oretchrate(SiO2)/etchrate(Si)=´e1.987´298=.0025.2.8615.Athree–stepprocessisrequiredforpolysilicongateetching.Step1isanonselectiveetchprocessthatisusedtoremoveanynativeoxideonthepolysiliconsurface.Step2isahighpolysiliconetchrateprocesswhichetchespolysiliconwithananisotropicetchprofile.Step3isahighlyselectivepolysilicontooxide课后答案网processwhichusuallyhasalowpolysiliconetchrate.16.Iftheetchratecanbecontrolledtowithin10%,thepolysiliconmaybeetched10%longerorforanequivalentthicknessof40www.hackshp.cnnm.Theselectivityistherefore40nm/1nm=40.17.Assuminga30%overetching,andthattheselectivityofAloverthephotoresistmaintains3.Theminimumphotoresistthicknessrequiredis105 (1+30%)×1µm/3=0.433µm=433.3nm.qB18.w=eme-1996.1´10´B2p´.245´10=-319.1´10-2B=8.75×10(tesla)=875(gauss).9-319.TraditionalRIEgenerateslow-densityplasma(10cm)withhighionenergy.ECRandICP1112-3generatehigh-densityplasma(10to10cm)withlowionenergy.AdvantagesofECRandICParelowetchdamage,lowmicroloading,lowaspect-ratiodependentetchingeffect,andsimplechemistry.However,ECRandICPsystemsaremorecomplicatedthantraditionalRIEsystems.20.Thecorrosionreactionrequiresthepresenceofmoisturetoproceed.Therefore,thefirstlineofdefenseincontrollingcorrosioniscontrollinghumidity.Lowhumidityisessential,.especiallyifcoppercontainingalloysarebeingetched.Secondistoremoveasmuchchlorineaspossiblefromthewafersbeforethewafersareexposedtoair.Finally,gasessuchasCF4andSF6canbeusedforfluorine/chlorineexchangereactionsandpolymericencapsulation.Thus,Al-ClbondsarereplacedbyAl-Fbonds.WhereasAl-Clbondswillreactwithambientmoistureandstartthecorrosionprocess,Al-Fbondsareverystableanddonotreact.Furthermore,fluorinewill课后答案网notcatalyzeanycorrosionreactions.www.hackshp.cn106 CHAPTER1321.Ea(boron)=3.46eV,D0=0.76cm/sec-Eaæ-.346ö-152FromEq.6,D=Dexp()=.076expç÷=.4142´10cm/s0-5kTè.8614´10´1223ø-15-6L=Dt=.4142´10´1800=.273´10cmx20æxöFromEq.9,C(x)=Cserfc()=8.1´10erfcç-6÷2Lè.546´10ø203-4-619Ifx=,0C(0)=8.1´10atoms/cm;x=0.05×10,C(5×10)=3.6×103-4183-4atoms/cm;x=0.075×10,C(7.5×10-6)=9.4×10atoms/cm;x=0.1×10,-5183C(10)=1.8×10atoms/cm;-4-5163x=0.15×10,C(1.5×10)=1.8×10atoms/cm.C-1subThex=2Dt(erfc)=.015mmjCsTotalamountofdopantintroduced=Q(t)=2CsL=.554´10课后答案网14atoms/cm2.pwww.hackshp.cnæ-Eaöæ-.346ö-1422.D=Dexpç÷=.076expç÷=.496´10cm/s0-5èkTøè.8614´10´1323øS193FromEq.15,C=C(,0t)==.2342´10atoms/cmSpDtæxö19æxöC(x)=CSerfcç÷=.2342´10erfcç-5÷è2Løè.2673´10ø193-4-5193Ifx=0,C(0)=2.342×10atoms/cm;x=0.1×10,C(10)=1.41×10atoms/cm;-4-5183-4-518x=0.2×10,C(2×10)=6.79×10atoms/cm;x=0.3×10,C(3×10)=2.65×10107 3atoms/cm;-4-5173-4-517x=0.4×10,C(4×10)=9.37×10atoms/cm;x=0.5×10,C(5×10)=1.87×103atoms/cm;-4-5163-4-515x=0.6×10,C(6×10)=3.51×10atoms/cm;x=0.7×10,C(7×10)=7.03×103atoms/cm;-4-5143x=0.8×10,C(8×10)=5.62×10atoms/cm.SThex=4Dtln=.072mm.jCpDtB-81518æ10ö3.1´10=1´10expç÷ç-13÷è4´3.2´10tøt=1573s=26minSFortheconstant-total-dopantdiffusioncase,Eq.15givesC=SpDt18-13132S=1´10p´3.2´10´1573=4.3´10atoms/cm.4.Theprocessiscalledtherampingofadiffusionfurnace.Fortheramp-downsituation,the课后答案网furnacetemperatureTisgivenbyT=T0-rtwww.hackshp.cnwhereT0istheinitialtemperatureandristhelinearramprate.TheeffectiveDtproductduringaramp-downtimeoft1isgivenbyt1(Dt)=Dt)(dteffò0Inatypicaldiffusionprocess,rampingiscarriedoutuntilthediffusivityisnegligiblysmall.Thustheupperlimitt1canbetakenasinfinity:111rt=»(1++...)TT-rtTT000108 andæ-Eaöé-Eartù-Ea-rEat-rEatD=D0expçkT÷=D0expê(1++...)ú=D0(exp)(exp2...)»D(T0)exp2èøëkT0T0ûkT0kT0kT0whereD(T0)isthediffusioncoefficientatT0.SubstitutingtheaboveequationintotheexpressionfortheeffectiveDtproductgives2¥-rEtkTa0(Dt)»D(T)expdt=D(T)effò00kT20rE0a2Thustheramp-downprocessresultsinaneffectiveadditionaltimeequaltokT0/rEaattheinitialdiffusiontemperatureT0.Forphosphorusdiffusioninsiliconat1000°C,wehavefromFig.4:-142D(T0)=D(1273K)=2×10cm/s1273-773r==.0417K/s20´60Ea=3.66eVTherefore,theeffectivediffusiontimefortheramp-downprocessiskT2-课后答案网232.138´10(1273)0==91s»5.1min.-19rE.0417(.366´6.1´10)awww.hackshp.cn5.Forlow-concentrationdrive-indiffusion,thediffusionisgivenbyGaussiandistribution.ThesurfaceconcentrationisthenSSæEaöC(,0t)==expç÷pDtpD0tè2kTø3/2dCSæEaöæ-töC=expç÷ç÷=-5.0´ç÷dtpD0è2kTøè2øt109 dCdtor=-5.0´Ctwhichmeans1%changeindiffusiontimewillinduce0.5%changeinsurfaceconcentration.dCSæEaöæ-EaöEa=expç÷ç÷=-C22dTpD0tè2kTøè2kTø2kT-19dC-EdT-6.3´6.1´10dTdTaor=´=´=-169.´-23C2kTT2´1.38´10´1273TTwhichmeans1%changeindiffusiontemperaturewillcause16.9%changeinsurfaceconcentration.18-36.At1100°C,ni=6×10cm.Therefore,thedopingprofileforasurfaceconcentration18-3of4×10cmisgivenbythe“intrinsic”diffusionprocess:æxöC(x,t)=Cerfcç÷sç÷è2Dtø18-3-142whereCs=4×10cm,t=3hr=10800s,andD=5x10cm/s.The课后答案网diffusionlengthisthen-5Dt=.232´10cm=.0232mmwww.hackshp.cn18æxöThedistributionofarsenicisC(x)=4´10erfcç÷-5è.464´10øThejunctiondepthcanbeobtainedasfollows1518æxjö10=4´10erfcç÷ç-5÷è.464´10ø-4xj=1.2×10cm=1.2mm.18-318-37.At900°C,ni=2×10cm.Forasurfaceconcentrationof4×10cm,givenby110 the“extrinsic”diffusionprocess--4.05´1.6´10-19Ea18D=DekT´n=458.e1.38´10-23´1173´4´10=.377´10-16cm2/s018n2´10i-16-6x=6.1Dt=6.1.377´10´10800=.323´10cm=323.nm.j8.Intrinsicdiffusionisfordopantconcentrationlowerthantheintrinsiccarrierconcentrationniatthediffusiontemperature.Extrinsicdiffusionisfordopantconcentrationhigherthanni.9.Forimpurityintheoxidationprocessofsilicon,equilibriumconcentrationofimpurityinsiliconsegregationcoefficeint=.equilibriumconcentrationofimpurityinSiO2113´10310.k===.0006.135´1050011.–0.5=–1.1+qFB/Ci-14es9.3´.885´10-7C===.345´10ox-6d10-7.345´1012-2F=´6.0=3.1´10cmB-19课后答案网1.6´1010-512-19t=3.1´10´6.1´102p(10.16)www.hackshp.cntheimplanttimet=6.7s.12.TheiondoseperunitareaisIt10´10-6´5´60Nq-191.6´10122===2.38´10ions/cmAA102p´()2FromEq.25andExample3,thepeakionconcentrationisatx=Rp.Figure.17indicatesthespis20nm.111 Therefore,theionconcentrationis12S.238´1017-3==.474´10cm.-7s2p20´102pp13.FromFig.17,theRp=230nm,andsp=62nm.Thepeakconcentrationis15S2´1020-3==.129´10cm-7s2p62´102ppFromEq.25,2é-(x-R)ù1520jp10=.129´10expêú2êë2spúûxj=0.53mm.-14QC0DVT9.3´.885´10´111-214.Doseperunitarea====6.8´10cm-8-19qq250´10´6.1´10FromFig.17andExample3,thepeakconcentrationoccursat140nmfromthesurface.Also,itisat(140-25)=115课后答案网nmfromtheSi-SiO2interface.15.ThetotalimplanteddoseisintegratedfromEq.25Q=¥Sexpéê-(www.hackshp.cnx-Rp)2ùúdx=Sìï1+éê1-erfc(Rp)ùúüï=S[2-erfc(3.2)]=S´.19989Tò2íý0s2pê2sú2ïîês2úïþ22pëpûëpûThetotaldoseinsiliconisasfollows(d=25nm):é-(-)2ùìé-ùü¥SxRpSïRpdïSSQSi=òexpê2údx=í1+ê1-erfc()úý=[2-erfc(.187)]=´.19918ds2pê2sú2ïîês2úïþ22pëpûëpûtheratioofdoseinthesilicon=QSi/QT=99.6%.112 16.Theprojectedrangeis150nm(seeFig.17).Theaveragenuclearenergylossovertherangeis60eV/nm(Fig.16).60×0.25=15eV(energylossofboronionpereachlatticeplane)2-183thedamagevolume=VD=p(2.5nm)(150nm)=3×10cmtotaldamagelayer=150/0.25=600displacedatomforonelayer=15/15=120-3damagedensity=600/VD=2×10cm20222×10/5.02×10=0.4%.17.Thehigherthetemperature,thefasterdefectsannealout.Also,thesolubilityofelectricallyactivedopantatomsincreaseswithtemperature.Q118.DV=1V=tCoxwhereQ1istheadditionalchargeaddedjustbelowtheoxide-semiconductorsurfacebyionimplantation.COXisaparallel-platecapacitanceperunitareaesgivenbyC=oxd(distheoxidethickness,课后答案网esisthepermittivityofthesemiconductor)-141V´9.3´.885´10F/cm-7CQ=DVC==8.63´101tox-620.4´10cmcmwww.hackshp.cn-7.863´10122=5.4×10ions/cm-191.6´10124.5´10132Totalimplantdose==1.2×10ions/cm.45%19.ThediscussionshouldmentionmuchofSection13.6.Diffusionfromasurfacefilmavoidsproblemsofchanneling.Tiltedbeamscannotbeusedbecauseofshadowingproblems.Iflowenergyimplantationisused,perhapswithpreamorphizationbysilicon,thentokeepthejunctionsshallow,RTAisalso113 necessary.20.FromEq.35Sd1æ4.0-6.0ö=erfcç÷=.084S2è2.02øTheeffectivenessofthephotoresistmaskisonly16%.Sd1æ1-6.0ö=erfcç÷=.0023S2è2.02øTheeffectivenessofthephotoresistmaskis97.7%.-u21e-521.T==102puu=.302d=Rp+4.27sp=0.53+4.27×0.093=0.927µm.课后答案网www.hackshp.cn114 CHAPTER141.EachU-shapesection(refertothefigure)hasanareaof2500mm×8mm=2×422410mm.Therefore,thereare(2500)/2×10=312.5U-shapedsection.Eachsectioncontains2longlineswith1248squareseach,4cornersquares,1bottomsquare,and2halfsquaresatthetop.Thereforetheresistanceforeachsectionis1kW/¡(1248×2+4×0.65+2)=2500.6kWThemaximumresistanceisthen8312.5×2500.6=7.81×10W=781MW课后答案网www.hackshp.cn2.ThearearequiredonthechipisC(30´10-7)(5´10-12)d-52A===.435´10cm-14e9.3´.885´10ox32=4.35×10mm=66×66mm115 RefertoFig.4aandusingnegativephotoresistofalllevels+(a)Ionimplantationmask(forpimplantationandgateoxide)(b)Contactwindows(2×10mm)(c)Metallizationmask(usingAltoformohmiccontactinthecontactwindowandformtheMOScapacitor).Becauseoftheregistrationerrors,anadditional2mmisincorporatedinallcriticaldimensions.课后答案网www.hackshp.cn116 课后答案网www.hackshp.cn3.Ifthespacebetweenlinesis2mm,thenthereis4mmforeachturn(i.e.,2×n,for2-62oneturn).Assumetherearenturns,fromEq.6,L»m0nr»1.2×10nr,wherercanbereplacedby2×n.Then,wecanobtainthatnis13.117 4.(a)Metal1,(b)contacthole,(c)Metal2.(a)Metal1,(b)contacthole,(c)Metal2.课后答案网www.hackshp.cn118 5.Thecircuitdiagramanddevicecross-sectionofaclampedtransistorareshownin(a)and(b),respectively.课后答案网6.(a)Theundopedpolysiliconisusedforisolation.(b)Thepolysilicon1isusedasasolid-phasediffusionsourcetoformthewww.hackshp.cnextrinsicbaseregionandthebaseelectrode.(c)Thepolysilicon2isusedasasolid-phasediffusionsourcetoformtheemitterregionandtheemitterelectrode.7.(a)For30keVboron,Rp=100nmandDRp=34nm.AssumingthatRpandDRpforboronarethesameinSiandSiO2thepeakconcentrationisgivenby11S8´1016-3==4.9´10cm-72pDR2p(34´10)p119 Theamountofboronionsinthesiliconis2Q¥Sé(x-Rp)ù=expê-údxqòd2DR22pDRpêëpúûSéæR-döùçp÷=ê2-erfcú2êçè2DRp÷øúëû118´10éæ750öù=ê2-erfcç÷ú2ëè2´340øû11-2=.788´10cmAssumethattheimplantedboronionsformanegativesheetchargeneartheSi-SiO2interface,then-1911æQö6.1´10´(.788´10)DV=qç÷/C==.091VTç÷ox-14(-7)èqø9.3´.885´10/25´10(b)For80keVarsenicimplantation,Rp=49nmandDRp=18nm.Thepeakarsenic16S1021-3concentrationis==.221´10cm.-72pDRp´(18´10)p课后答案网www.hackshp.cn120 课后答案网8.(a)Because(100)-orientedsiliconhaslower(~onetenth)interface-trappedchargeandwww.hackshp.cnalowerfixedoxidecharge.(b)Ifthefieldoxideistoothin,itmaynotprovidealargeenoughthresholdvoltageforadequateisolationbetweenneighboringMOSFETs.(c)Thetypicalsheetresistanceofheavilydopedpolysilicongateis20to30W/¡,whichisadequateforMOSFETswithgatelengthslargerthan3mm.Forshortergates,thesheetresistanceofpolysiliconistoohighandwillcauselargeRCdelays.Wecanuserefractorymetals(e.g.,Mo)orsilicidesasthegatematerialtoreducethesheetresistancetoabout1W/¡.121 (d)Aself-alignedgatecanbeobtainedbyfirstdefiningtheMOSgatestructure,thenusingthegateelectrodeasamaskforthesource/drainimplantation.Theself-alignedgatecanminimizeparasiticcapacitancecausedbythesource/drainregionsextendingunderneaththegateelectrode(duetodiffusionormisalignment).(e)P-glasscanbeusedforinsulationbetweenconductinglayers,fordiffusionandionimplantationmasks,andforpassivationtoprotectdevicesfromimpurities,moisture,andscratches.9.Thelowerinsulatorhasadielectricconstante1/e0=4andathicknessd1=10nmTheupperinsulatorhasadielectricconstante2/e0=10andathicknessd2=100nm.UponapplicationofapositivevoltageVGtotheexternalgate,electricfieldE1andE2areestablishedinthed1andd2respectively.Wehave,fromGauss’law,thate1E1=e2E2+QandVG=E1d1+E2d2whereQisthestoredchargeonthefloatinggate.Fromtheseabovetwoequations,weobtain课后答案网VGQE=+1()()d+de/ee+ed/d12121212ìüïwww.hackshp.cn7ï-7ï10´10Qï5J=sE1=10í+ý=2.0-.226´10Qï10+100æç4ö÷é4+10æ10öùú´.885´10-14ïêç÷ïîè10øëè100øûïþ5(a)IfthestoredchargedoesnotreduceE1byasignificantamount(i.e.,0.2>>2.26×10úQú,wecanwritet-6-8Q=sEdt"»2.0Dt=2.0´(.025´10)=5´10Cò10-8Q5´10DV===.0565VT(-14)(-7)C10´.885´10/100´102122 5-7(b)whent®¥,J®0wehaveQ®2.0/.226´10@8.84×10C.-7Q.884´10ThenDV===.998V.T(-14)-5C10´.885´10/10210.课后答案网www.hackshp.cn123 课后答案网www.hackshp.cn11.TheoxidecapacitanceperunitareaisgivenbyeSiO2-72C==5.3´10F/cmoxdandthemaximumcurrentsuppliedbythedeviceis1W215mm-72I»mC(V-V)=5.3´10(V-V)»5mADSoxGTGT2L25.0mm124 andthemaximumallowablewireresistanceis0.1V/5mA,or20W.Then,thelengthofthewiremustbe-82R´Area20W´10cmL£==.0074cm-8r7.2´10W-cmor740mm.Thisisalongdistancecomparedtomostdevicespacing.Whendrivingsignalsbetweenwidelyspacedlogicblockshowever,minimumfeaturesizedlineswouldnotbeappropriate.12.课后答案网www.hackshp.cn125 课后答案网www.hackshp.cn126 13.Tosolvetheshort-channeleffectofdevices.14.Thedeviceperformancewillbedegradedfromtheboronpenetration.Therearemethodstoreducethiseffect:(1)usingrapidthermalannealingtoreducethetimeathightemperatures,consequentlyreducesthediffusionofboron,(2)usingnitridedoxidetosuppresstheboronpenetration,sinceboroncaneasilycombinewithnitrogenandbecomeslessmobile,(3)makingamulti-layerofpolysilicontotraptheboronatomsattheinterfaceofeachlayer.15.Totalcapacitanceofthestackedgatestructureis:e1e2æe1e2ö725æ725öC=´çç+÷÷=´ç+÷=2.12d1d2èd1d2ø5.010è5.010ø9.3=2.12d9.3d==1.84nm.2.1216.DisadvantagesofLOCOS:(1)hightemperatureandlongoxidationtimecauseVTshift,(2)bird’sbeak,(3)notaplanarsurface,(4)exhibitsoxidethinningeffect.Advantagesofshallowtrenchisolation:(1)planarsurface,(2)nohigh课后答案网temperatureprocessingandlongoxidationtime,(3)nooxidethinningeffect,(4)nobird’sbeak.www.hackshp.cn17.Forisolationbetweenthemetalandthesubstrate.18.GaAslacksofhigh-qualityinsulatingfilm.127 19.(a)(-4)æLöæAöæ-51öé-141´1´10ùRC=çr÷çeox÷=ç10´-8÷ê9.3´.885´10´-4úèAøèdøè1´5.0´10øë5.0´10û(-14)-9=2000´69.03´10=.138´10s=.138ns.(b)ForapolysiliconrunneræLöæAöRC=çRsquare÷çeox÷èWøèdøæ1ö(-14)-7=30ç÷69.03´10=.207´10s-4è10ø=207nsThereforethepolysiliconrunner’sRCtimeconstantis150timeslargerthanthealuminumrunner.20.Whenwecombinethelogiccircuitsandmemoryonthechip,weneedmultiplesupplyvoltages.Forreliabilityissue,differentoxidethicknessesareneededfordifferentsupplyvoltages.21.(a)11+1C=CCtotalTa2O5nitridehenceEOT=75+10=173.Å9.3课后答案网257(b)EOT=16.7Å.www.hackshp.cn128'