硅超大规模集成电路工艺技术-理论、实践与模型课后习题答案

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1.2.Assumingdopantatomsareuniformlydistributedinasiliconcrystal,howfarapartare

theseatomswhenthedopingconcentrationisa).1015cm'3,b).1018cm'3,c).5xl020cm'3.

Answer:

Theaveragedistancebetweenthedopantatomswouldjustbeoneoverthecuberootofthe

dopantconcentration:

,/3

X=NA"

a,x=(lxIO15cm_3J=lxl0-5cm=0.1pm=lOOnm

功x=^xlOl8cm_3J=1x106cm=0.01pm=lOnm

0x=^xlO2Ocm_3J=1.3x10-7cm=0.0013pm=1.3nm

1.3.Considerapieceofpuresilicon100pmlongwithacross-sectionalareaof1^ini2.How

muchcurrentwouldflowthroughthis“resistor“atroomtemperatureinresponsetoan

appliedvoltageof1volt?

Answer:

Ifthesiliconispure,thenthecarrierconcentrationwillbesimplyn；.Atroomtemperature,rii

〜1.45x101()cm\Lndcranappliedfield,thecurrentwillbeduetodriftandhcncc,

I=In+Ip=qAn《n+%》

=^1.6x10-19coul0-8cm2.45x1010carrierscm_3j^000cm2volt-1sec-1

=4.64xlO_12ampsor4.64pA

1.10.Astate-of-the-artNMOStransistormighthaveadrninjunctionareaof0.5x0.5pm.

Calculatethejunctioncapacitanceassociatedwiththisjunctionatanappliedreverse

biasof2volts.Assumethedrainregionisveryheavilydopedandthesubstratedoping

is1xIO16cm-3.

Answer:

ThecapacitanceofthejunctionisgivenbyEqn.1.25.

&二is.二NAN)[i.

-

Axd72(NA+N"@土V)_

Thejunctionbuilt-involtageisgivenbyEqn.1.24.NDisnotspecifiedexceptthatitis

verylarge,sowetakeittobe1020cm'3(roughlysolidsolubility).TheexactchoiceforND

doesn'tmakemuchdifferenceintheanswer.

cm-3¥o16cm-3

kT,NDNA1…s

4>i=_In=(.0259volts)ln

qIn?卜45x1()1°cm)

SinceND»NAinthisstructure,thecapacitanceexpressionsimplifiesto

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港＞［传卜高］

「，.6x10~19coulY1.7)flO16cm~3¥^.86x10-14Fern_1)1°

=IE-4Z|=1.68xlO-8Fcm-o2

(2)(2.934volts)

Giventheareaofthejunction(0.25x10scm2,thejunctioncapacitanceisthus4.2x10'17

Farads.

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3.2.Aboron-dopedcrystalpulledbytheCzochralskitechniqueisrequiredtohavea

resistivityof10Qcmwhenhalfthecrystalisgrown.Assumingthata100gmpure

siliconchargeisused,howmuch0.01Qcmborondopedsiliconmustbeaddedtothe

melt?Forthiscrystal,plotresistivityasafunctionofthefractionofthemeltsolidified.

21

Assumeko=0.8andtheholemobilitypp=550cmvolt"sec".

Answer:

1

p=

Usingthemobilityvaluegiven,andwehave:

10QcmnNA=1.14x10,5cm*3and0.01Qcm=NA=1.14x1018cm'3

FromEqn.3.38,0s=。0卜。(1-f)andwcwantCs=1.14x1015cm'?whenf=

0.5.Thus,solvingfbrCotheinitialdopingconcentrationinthemelt,wehave:

1.14x1A2，入15-3

CQ=­(1—0.5)=1.24x10'cm

__In__#ofimpurities_(DjpingX^lof0.01Qcm)

°VounitvolofmdtVol100gmSi

枭'00则）=。［09典

/.Wfetaddedof0.01QanSi

Theresistivityasafimetionofdistanceisplottedbelowandisgivenby

P(x”号/嗫*="505(1产

FractionSolidified-f

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3.3.ACzochralskicrystalispulledfromameltcontaining10,scm3boronand2x10"cm*3

phosphorus.InitiallythecrystalwillbePtypebutasitispulled,moreandmore

phosphoruswillbuildupintheliquidbecauseofsegregation.Atsomepointthecrystal

willbecomeNtype.Assumingko=0.32forphosphorusand0.8forboron,calculate

thedistancealongthepulledcrystalatwhichthetransitionfromPtoNtypetakesplace.

Answer:

WecancalculatethepointatwhichthecrystalbecomesNtypefromEqn.3.38asfollows:

,4-068

Cs(Phos)=Coko(l-f)%T=^X10)0.32X1-f)-

k-1152

Cs(Boron)=Coko(l-f)°=^O^0.8X1-f)°-

Atthepointwherethecross-overoccurstoNtype,thesetwoconcentrationswillbeequal.

Solvingfbr「wefind

f=0.995

Thusonlythelast05%ofthecrystalisNtype.

3.6.Supposeyourcompanywasinthebusinessofproducingsiliconwafersforthe

semiconductorindustrybytheCZgrowthprocess.Supposeyouhadtoproducethe

maximumnumberofwafersperboulethatmetafairlytightresistivityspecification,

a).WouldyouprefertogrowNtypeorPtypecrystals?Why?

b).WhatdopantwouldyouuseingrowingN-typecrystals?Whatdopantwouldyou

useingrowingPtypecrystals?Explain

Answer:

a).Boronhasthesegregationcoefficientclosesttounityofallthedopants.Thusitproducesthemostuniform

dopingalongthelengthofaCZcrystal.ThusPtypewouldbethenaturalchoice.

b).ForPtype,theobvious(andonlyrealchoice)isboronasexplainedinparta).ForN

typecrystalsFig.3-18showsthateitherPorAswouldbeareasonablechoicesincetheir

segregationcoefficientsarequitecloseandarebetterthanSb.Table3-2indicatesthatP

mightbeslightlypreferredoverAsbecauseitskovalueisslightlycloserto1.

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4.1.AnICmanufacturingplantproduces1000wafersperweek.Assumethat

eachwafercontains100die,eachofwhichcanbesoldfor$50ifitworks.The

yieldonthesechipsiscurrentlyrunningat50%.Iftheyieldcanbeincreased,

theincrementalincomeisalmostpureprofitbecauseall100chipsoneachwafer

aremanufacturedwhethertheyworkornot.Howmuchwouldtheyieldhaveto

beincreasedtoproduceanannualprofitincreaseof$10,000,000?

Answer:

At1000wafersperweek,theplantproduces52,000wafersperyear.Ifeach

waferhas50gooddieeachofwhichsellsfor$50,theplantgrossincomeis

simply

Income=(52,000)(50)(S50)=$130,000,000peryear.

lbincreasethisincomeby$I(),000,000requiresthattheyieldincreaseby

A,%

4.3.AsMOSdevicesarcscaledtosmallerdimensions,gateoxidesmustbe

reducedinthickness.

a.Asthegateoxidethicknessdecreases,doMOSdevicesbecomemoreorless

sensitivetosodiumcontamination?Explain.

b.Asthegateoxidethicknessdecreases,whatmustbedonetothesubstratedoping(or

alternativelythechannelVTHimplant,tomaintainthesameVTH?Explain.

Answer:

a).Fromthetext,Na-contaminationcausesthresholdvoltageinstabilitiesinMOSdevices.

AlsofromEqn.4.1,thethresholdvoltageisgivenby

vvm'gqNA%)qQv

VV

TH=FB+2(|)r++

Asthegateoxidethicknessdecreases,Coxincreases,sothesameamountofmobilecharge

QMwillhavelesseffectonVTHasoxidesgetthinner.ThereforeMOSdevicesarcless

sensitivetosodiumcontamination.

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b).UsingthesameexpressionfbrVTHasinparta),weobservethatastheoxidethickness

decreases,(Coxincreases),tomaintainthesameVTH,NAwillhavetoincrease.NAwill

actuallyhavetoincreasebythesquarerootoftheoxidethicknessdecreasetokeepVTH

constant.

4.4.AnewcleaningprocedurehasbeenproposedwhichisbasedonH2OsaturatedwithO2

asanoxidant.ThishasbeensuggestedasareplacementfortheH2O2oxidizingsolution

usedintheRCAclean.SupposeaSiwafer,contaminatedwithtraceamountsofAu,Fe

andCuiscleanedinthenewH2O/O2solution.Willthiscleanthewafereffectively?

Whyorwhynot?Explain.

Answer:

Asdescribedinthetext,cleaningmetalionsoffofsiliconwafersinvolvesthefollowing

chemistry:

M<->Mz++ze-

Thecleaningsolutionmustbechosensothatthereactionisdriventotherightbecausethis

putsthemetalionsinsolutionwheretheycanberinsedoff.Sincedrivingthereactiontothe

rightcorrespondstooxidation,vveneedanoxidizingsolationtocleanthewafer.

H2O/O2iscertainlyanoxidizingsolution.Butwhetheritcleanseffectivelyornotdependson

thestandardoxidationpotentialofthevariouspossiblereactions.FromTable4-3inthetext,

wehave:

Oxidant/StandardOxidationOxidation-ReductionReaction

ReductantPotential(volts)

SiO/Si0.84

2+

Si+2H20oSi02+4H+4e-

0.17

Fe3+/FeFe―Fe3++3e-

-0.34

Cu^+/CuCu—Cu2++2e-

-1.23

+

°2/H2O2H20co2+4H+2e-

-1.42

AU3+/AUAucAu"+3c-

Thestrongerreactions(dominating)areatthebottom.

ThustheH2O/O2reactionwillcleanFeandCu,butitwillnotcleanAuoffthewafer.

4.5.Explainwhyitisimportantthatthegenerationlifetimemeasurementillustratedin

Figure4-19isdoneinthedark.

Answer:

Themeasurementdependsonmeasuringcarriersgeneratedthermallyinthesiliconsubstrate

(oratthesurface).Iflightisshiningonthesample,thenabsorbedphotonscanalsogenerate

therequiredcarriers.Asaresult,theextractedgenerationlifetimewiththelightonwould

reallybemeasuringtheintensityoftheincidentlightandnotabasicpropertyofthesilicon

material.

实用文档

5.1.Calculateandplotversusexposurewavelengththetheoreticalresolutionanddepthof

focusforaprojectionexposuresystemwithaNAof0.6(aboutthebestthatcanbedone

today).Assumeki=0.6andkz=0.5(bothtypicalva加es).Considerwavelengths

between100nmand1000nm(DUVandvisiblelight).).Indicatethecommonexposure

wavelengthsbeingusedorconsideredtodayonyourplot(g-line,i-line,KrFandArF).

WillanArFsourcebeadequateforthe0.13gmand0.1pmtechnologygenerations

accordingtothesesimplecalculations?

Answer:

Therelevantequationsaresimply

九九XX

R=ki=0.6—andDOF=±kT=±0.5T

1NA0.622(NA)2(0.6)2

Theseequationsarcplottedbelow.NotethattheArF(193nm)willnotreach0.13

pmor0.1|imresolutionaccordingtothesesimplecalculations.Infact,withmore

sophisticatedtechniquessuchasphaseshiftmasks,offaxisilluminationetc.,ArF

isexpectedtoreach0.13pmandperhapsthe0.1pmgenerations.

5.3.AnX-rayexposuresystemusesphotonswithanenergyof1keV.Ifthe

separationbetweenthemaskandwaferis20estimatethediffraction

limitedresolutionthatisachievablebythissystem.

Answer:

Theequivalentwavelengthof1keVx-raysisgivenby

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g..xloTcVsccXxloMcmsccT)

E=hvJhc

X=

E-103eV

=1.24x10-7cm=1.24nm

X-raysystemsoperateintheproximityprintingmode,sothatthetheoreticalresolutionis

givenbyEqn.5.12:

=7^g=J1.24xl0-3Hmj20|im)=0.15pm

Resolution

5.8.Asdescribedinthischapter,therearenoclearchoicesforlithographysystemsbeyond

opticalprojectiontoolsbasedon193-nmArFeximerlasers.Onepossibilityisanoptical

projectionsystemusinga157-nmFzexcimerlaser.

a.Assuminganumericalapertureof0.8andk(=0.75,whatistheexpectedresolution

ofsuchasystemusingafirstorderestimateofresolution?

b.Actualprojectionsforsuchsystemssuggestthattheymightbecapableofresolving

featuressuitableforthe20090.07pmgeneration.Suggestthreeapproachestoactually

achievingthisresolutionwiththesesystems.

Answer

a).Thesimpleformulafbrresolutionis

_,九一-0.157.m八I-

R=ki=0.75=0.147|im

1NA0.8

b).Thecalculatedresolutioninpartaisafactoroftwolargerthanrequiredfbrthe0.07pm

generation.Thereforesome“tricks”willhavetobeusedtoactuallyachievesuchresolution.

Therearcanumberofpossibilities:

I.Useofphase-shiftmasks.Thistechnique,discussedinthischapter,hasthepotentialfbr

significantresolutionimprovements.Itworksbydesigningamoresophisticatedmask.

Simplemasksaredigital-blackorwhite.Phaseshiftingaddsasecondmaterialtothemask

features,usuallyattheedgeswhichshiftstheopticalphaseandsharpensuptheaerialimage.

Sophisticatedcomputerprogramsarcrequiredtodesignsuchmasks.

2.Useofopticalproximitycorrectioninthemaskdesign.Thisisanotherapproachto

designingabettermaskandasdiscussedinclass,canalsoimproveresolutionsignificantly.

Theapproachinvolvesaddingextrafeaturestothemask,usuallyatcomerswherefeatures

aresharp,tocompensatefbrthehighfrequencyinformationlosttodifiractioneffects.

3.Off-axisilluminaticn.Thisallowstheopticalsystemtocapturesomeofthehigherorder

diffractedlightandhencecanimproveresolution.

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5.9.Currentopticalprojectionlithographytoolsproducediffractionlimitedaerialimages.A

typicalaerialimageproducedbysuchasystemisshowninthesimulationbelowwherea

squareandrectangularmaskregionsproducetheimageshown.(Themaskfeaturesarethe

blackoutlines,thecalculatedaerialimageisthegrayscaleinsidetheblackrectangles.)The

majorfeatureoftheaerialimageisitsroundedcornerscomparedtothesharpsquare

cornersofthedesiredpattern.Explainphysicallywhythesefeatureslookthewaytheydo,

usingdiffractiontheoryandthephysicalpropertiesofmodernprojectionoptical

lithographytools.

Answer:

Modemopticalprojectionlithographysystemsarelimitedintheresolutiontheycanachieve

bydiffractioneffects.Thefinitesizeofthefocusinglensmeansthatthehighorder

diffractioncomponentsareTost”andarethereforenotavailabletohelpinprintingareplica

ofthemaskimage.Butthehighfrequencyspatialcoinponentsareexactlythecompcnents

thatcontaininformationabout“sharp”features,i.c.cornersetc.Thustheprojectedaerial

imagelosesthisinfbmiationandcomersbecomerounded.Theonlywaystoimprovethe

imagearebyusingshorterwavelengthlight,orahigherNAlens.

5.10.Futureopticallithographysystemswilllikelyuseshorterexposurewavelengthsto

achievehigherresolutionandtheywillalsolikelyuseplanarizationtechniquesto

provide"flat"substratesonwhichtoexposetheresistlayers.Explainwhy"flat”

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substrateswillbemoreimportantinthefuturethantheyhavebeeninthepast.

Answer:

Asthewavelengthoftheexposuresystemdecreases,thedepthoffbcusoftheexposure

systemalsodecreases.Thusitwillbenecessarytomakesurethattheresistinwhichthe

imageistobeexposed,isflatanddoesnotrequiremuchdepthoffbcus.Planarization

techniqueswillberequiredtoaccomplishthis.ThiscouldmeanCMPtoplanarizethe

substratebeforetheresistisapplied,oritcouldmeanusingaspunonresistwhich

planarizesthesubstrateandwhichisthencoveredwithathin,unifonnimagingresistlayer.

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6.4.ConstructaHFCVplotforaP-typesiliconsample,analogoustoFig.6-9.

Explainyourplotbasedonthebehaviorofholesandelectronsinthe

semiconductorinasimilarmannertothediscussioninthetextforFig.6.9.

TheC-VplotlooksbasicallythesameastheNsubstrateexampleinthetext,thatwe

discussedinclass,exceptthatthehorizontalaxisisflipped.Fornegativeappliedgate

voltages,themajoritycarrierholesinthesubstrateareattractedtothesurface.Thisisthe

accumulationregiona)above.WemeasurejustCoxforthecapacitancesincethereisno

depletioninthesubstrate.For+VG,theholesaredrivenawayfromthesurfacecreating

firstadepletionregionasinb)andfinallyaninversionlayerofelectronsasinc).The

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measuredcapacitancedropsaswemoveintodepletionandfinallyreachesaminimum

valueafteraninversionlayerforms.

TheC-Vcurvesshownarehighfrequencycurves.Asdiscussedinthetext,thecapacitance

remainsatitsminimumvaluefbr+VGvaluesgreaterthanVTHbecausetheinversionlayer

electronscannotbecreatedordestroyedasfastasthesignalischanging.Hencethesmall

ACsignalmust“wiggle”thebottomofthedepletionregiontobalanceAVG.

6.6.InasmallMOSdevice,theremaybeastatisticalvariationinVTduetodifferencesin

QFfromonedevicetoanother.Ina0.13pmtechnologyminimumdevice(gateoxide

area=O.lgmx0.11im)witha2.5nmgateoxide,whatwouldthedifferenceinthreshold

voltagebefordeviceswith0or1fixedchargeinthegateoxide?

Answer:

Theoxidecapacitanceis

C°X-d-2,5x1O-7

=1.38xl0-16

Thechangeinthresholdvoltageisgivenby

AV=-9^£=

T16=1.1mV

Cox1.38x10-

Thisshowsthatasingleelectrontrapinagateoxidewillhaveanegligibleeffectonthe

thresholdvoltageatthistechnologygeneration.

6.12AsiliconwaferiscoveredbyanS1O2film0.3pmthick.

a.Whatisthetimerequiredtoincreasethethicknessby0.5pmbyoxidationinH2O

at1200℃?

b.RepeatforoxidationindnrO2at1200°C.

Answer:

Wewillperformthecalculationfor<111>siliconwafers.For<100>wafers,thelinearrate

constantshouldbedividedby1.68.

a.At1200℃,inH2O

B=3.86xl02e:=0.829pm2/hr

=1.63x10sexp(一音)=15.86pm/hr

A=0.052jam

Theinitialoxide,ifgrownat1200℃wouldhavetakenthislongtogrow

二£2^^»=。.皿「

B0.829

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Thetimerequiredtogrow°・8at1200℃is

xf+AXj（0.8）2+（0.052）（0.8）

T=-1L=""、=0.822hr

B0.829

Thus,thetimerequiredtoaddtoanexisting°,3filmis

0.822—0.127=0.695hror417minutes.

b.At1200℃,indryoxygen

B=7.72x102expI-]=0.048um2/hr

k（1200+273»

旦=6.23x106exp（一言）=0.899pm/hr

A

A=0.053pm

Theinitialoxidewouldhavetaken2.206hourstogrowindryoxygen,itwouldrequire

14.217hourstogrow—m,thuswouldrequireanadditional12hourstoadd

toanexisting°3即film.

6.13.Supposeanoxidationprocessisusedinwhich（100）wafersareoxidizedinO2forthree

hrs.at1100℃,followedbytwohrs.inH2Oat900℃,followedbytwohrsinO2at

1200℃.UseFigs.6-19and6-20inthetexttoestimatetheresultingfinaloxide

thickness.Explainhowyouusethesefigurestocalculatetheresultsofamulti-step

oxidationlikethis.

Answer:

Wecanusethesefigurestoestimatetheoxidethicknessasfollows.First,weuse

Fig.6-19fbrthefirstdryoxidationcycle

Athreehouroxidationat1100℃producesanoxidethicknessofabout0.21|im.Wenext

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useFig.6-20fbrthewetoxidationasshownbelow.Theoxidationis2hrsinH2Oat9()°C.

Wcstartbyfindingthepointonthe900℃curvethatcorrespondsto0.21pmsincethisis

thestartingoxidethickness.ThisispointA.Wethenmovealongthe900℃cunebytwo

hourstopointB.Thiscorrespondstoathicknessofabout0.4pmwhichisthethicknessat

theendofthewetoxidation.

WenowgobacktoFig.6-19forthefinaldryO2cycle.Thisprocessis2hrsat1200℃,We

startbyfindingthepointonthe12()0°Ccurvethatcorrespondstoastarlingoxidethickness

of0.4pm.ThisispointAbelow.Wcthenincrementthetimeby2hrsalongthe1200℃

curve,toarriveatafinaloxidethicknessofabout0.5

6.18.SilicononInsulatororSOIisanewsubstratematerialthatisbeingconsideredfor

futureintegratedcircuits.Thestructure,shownbelow,consistsofathinsinglecrystal

siliconlayeronaninsulating(S1O2)substrate.ThesiliconbelowtheSiOzprovides

mechanicalsupportforthestructure.Oneofthereasonsthistypeofmaterialisbeing

considered,isbecausejunctionscanbediffusedcompletelythroughthethinsilicon

layertotheunderlyingS1O2.Thisreducesjunctioncapacitancesandproducesfaster

circuits.Isolationisalsoeasytoachieveinthismaterial,becausethethinSilayercan

becompletelyoxidized,resultingindevicescompletelysurroundedbySiOi.ALOCOS

processisusedtolocallyoxidizethroughthesiliconasshownontherightbelow.

AssumingtheLOCOSoxidationisdoneinHzOat1000℃,howlongwillittaketo

oxidizethroughthe0.3siliconlayer?CalculateanumericalanswerusingtheDeal

Grovemodel.

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(13gm

(HMI)SilicimSuli*tnit<,(HHI)SilinmSulwlmtr

StartingMuicriulARcrLOCOSholation

Answer:

Tooxidizecompletelythrougha0.3jimsiliconlayer,wewillneedtogrow(2.2)(0.3jim)=

0.66gmofSiO2.At1000℃inH2O,theDealGroverateconstantsaregivenby(Table

6-2):

B=3.86xl02expi-0.78eV>=0.316|im2hr-1

kTJ

©=回心邛(一书)=0.747即.

A1.68RkTJ产

(0.66)20.66

.,.t=211-+=2.25hours

0.3160.747

6.23.AspartofanICprocessflow,aCVDSiOzlayer1.0gmthickisdepositedona<100>

siliconsubstrate.Thisstructureisthenoxidizedat900℃for60minutesinanH2O

ambient.WhatisthefinalSiOzthicknessafterthisoxidation?Calculateananswer,do

notusetheoxidationchartsinthetext.

Answer:

At900℃inH2O,theoxidationrateconstantsarcgivenby:

0.78

B=3.86xl02cxpl-7pm2hr_1=0.17^m2hr-1

(g.62xio「5)1173〃

B1.63x102.05pmhr_1=0.152pmhr'1

A1.688.62x10-511173),

Theinitialoxideonthewaferis1.0pmthick.ThiscorrespondstoaTof

(冉0.17

T=一-=12.46hours

0.17

Thusthefinaloxidethicknessisgivenby

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f_______1

0.17IE13.46J

=---11+-1r=1.064

°o(2X0.152)J(LU)2

I(HO。.

Thusnotmuchadditionaloxidegrows.

Cha—ter7Problems

7.1.Aresistorforananalogintegratedcircuitismadeusingalayerofdepositedpolysilicon

0.5thick,asshownbelow.

Polysilicon

(a)(a)Thedopingthepolysiliconis1'<10cm.Thecarriermcbility

N=100cm\secjslowbecauseofscatteringatgrainboundaries.Ifthe

resistorhasL=100|im,W=10pm,whatisitsresistanceinOhms?

(b)(b)Athermaloxidationisperformedonthepolysiliconfbr2hoursat90C,oCin

.AssumingB/Afbrpolysiliconis2/3thatof<111>silicon,whatisthe

polysiliconthicknessthatremains.

(c)(c)Assumingthatallofthedopantremainsinthepolysilicon(i.e.doesnot

segregateiooxide),whatisihenewvalueoftheresisiorin(a).Assumethemobility

doesnotchange.

Answer:

1__________

Pnqg(lx1016V=6.25Qcm

J(1,6X10-19)100)

p6.25八

pc===125kQ

x；0.5x10

R=-^-ps=1.25MQ(10squares)

(b)Thelinearratecoefficientat900℃is

="fl.63xl08