Solution Manual for Semiconductor Devices--Physics and Technology [Sze, S. M] Solution

Contents ch.l Introductiorr----ch.2 EnergyBandsand Carrier Concentration -----ch.3 CarrierTransportPhenomena ch.4 p-n Junction----ch.5 BipolarTransistorandRelatedDevicesch.6 MOSFETandRelatedDevices--------ch.7 MESFETandRelatedDevices------ch.8 MicrowaveDiode,Quantum-EffectandHot-ElectronDevices ---------ch.9 PhotonicDevices c h . 1 0CrystalGrowthand Epitaxy-----c h . r l Film Formation-------ch.t2 LithographyandEtching ch.13 ImpurityDoping-ch.r4 Integrated Devices---------

--- 0 I ---- 7 --- 16 --------- 32 ---------- 48 -- 60 ----- 68 ----------- 73

92 ---- 99 - 105 -- l 13

CHAPTER 2

1. (a) From Fig. llq

the atom at the center of the cube is surroundby four

equidistantnearestneighborsthat lie at the cornersof a tetrahedron.Therefore the distancebetweennearestneighborsin silicon(a: 5.43A) is l/2 [(a/2)' * (Jzo /2127t/': J-zo/4 : 235 A. (b) For the (100) plane,thereare two atoms(one centralatom and4 corneratoms eachcontributing ll4 of an atom for a total of two atomsas shown in Fig. 4a) for an areaof d, thereforewe have 2/ &:2/ (5.43,. l0-8)z:618 * 10laatoms/ crt Similarlywe havefor (110)plane(Fig. 4a andFig. 6) ( 2 + 2 x l l 2 + 4 x l l 4 ) / J T o 2 : 9 . 6 , . 1 0 r sa t o m /sc r 1 . , andfor (111)plane(Fig. 4aandFig. 6)

(3x I/2+3x r/6)/ rlz|mlf

: ,ffi" I

: 7.83* 10laatoms / crrt.

(9.

2. Theheightsat X, Y, andZ pointare /0, %,^O %. 3. (a) For the simplecubic,a unit cell contains1/8of a sphereat eachof the eight cornersfor a total of one sphere. -

Ma>> kT then

[t *"(E-Eilrt'rlt

| - F(E)- expF (n, - r)lwl

Then from AppendixH and, Eqs. I and2we obtain p : 4Df2mp/ h2f3D EtD expI-@r - E) / kT ldE I:""^ Letxt+ E lkT, andletEbooo*:- @, Eq.3 becomes p : 4\-2mo / rtflz (kTlttz exp [-(Ep/ kl)i

xtDe*dx I wherethe integralon the right is of the standardform and equats G tZ. - p :2l2Dmo kT / h213D exp [-(Ep/ kI)j By refeningto the top of the valencebandas ETinsteadof E:0

or where

p:2Qo';f:"i1ff;Trrlf;, u,/krl Nv:2 (Nmo kT / rtf .

10. FromEq. 18

Nv :2QDmokT I h2f D The effectivemassof holesin Si is mp-- (Nvt 21ztt(rt tzDkT) 1.38x 10-23 3oo)

t kg : 1.03mo. : 9.4 " 10-3 Similarly, we havefor GaAs f f i p : 3 . 9 x 1 0 - 3k1g : 0 . 4 3m o . 1 1 . UsingEq. 19

(1)

we have

e) (3)

Ei =(8, +til'

. (%)^ (N,I w,) = (Ec* Ey)l 2 + (*T I 4) ln

lr*,1*,)(o%f

(1)

At77 K E,: (1.t6/2)+ (3 x 1.3gx t}-,tT) / (4 x 1.6x 10-,r)ln(l.0/0.62) :0.58 + 3.29x 10-5Z= 0.5g+ 2.54x 10-3- 0.5g3ev.

At 300K : 0.56+ 0.009: 0.569eV. Ei: (1.12/2)+ (3.29x 10-sX300; At 373K - 0.545+ 0.012:0.557ey. + (3.2gx l0-sx3731 Ei: (1.0912) Becausethe secondterm on the right-handside of the Eq.l is much smaller comparedto the first term, over the abovetemperafurerange,it is reasonableto assumethatEi is in the centerof the forbiddengap.

T2.KE :

I::@_EC JE

f:

"-@-rrYw6B

-Er"-@-tP)/*r6P

l'=rr-rr,

1.5x0.5"G

0.sJ;

=? ;or. 1 3 .(a)p: ftw:9.109 x 10-3rx105:9.109 x 10-26 kg-mA h - 6 ' 6 2 6 x 1 0 - 1: 4 7 -. 2 7 x r 0 - em : 7 2 . 7 A 1 : p x 9.109 l0-'" m ^ ^ I ( b ) 1 "" 'L : *x 7 2 . 7 : I 1 5 4A . mp 0.063 14. From Fig.22when nr: S o t h a tI :

l0t5 cd3, the correspondingtemperatureis 1000 / T:

l.B.

1 0 0 0 / 1 . 8 : 5 5 5K o r 2 8 2 [

E" - Er: kT lnlNc / (No - N,q)] which can be rewritten as No - N.e: l/c exp IaE, - Er) I kf I

1 5 .From

Then

No-N.a:2.86 * 10reexp(-0.20 /0.0259): 1.26x 1016 crn3

or N o : 1 . 2 6 x 1 0 1 6N + . t : 2 . 2 6* 1 0 1 6 c r n 3 A compensatedsemiconductorcan be fabricatedto provide a specific Fermi energylevel. 16. From Fig.28a we candraw the following energy-band diagrams:

Ec(0.ssevi'

AT 77K

EF(0.s3)

E;(o) Ev(-0.59)

AT 3OOK

Ec(0.56ev) EF(0.38) E i (o) ri

Ev(-0.56)

AT 6OOK

E9(0.50eV)

0.s0)

17. (a) The ionization energy for boron in Si is 0.045 eV. At 300 K, all boron impurities are ionized. Thus pp: N.a: l0ls crn3 np: t?i2 / n.a: (g.6i " K;9f / l}ts :9.3 * lOacrn3. The Fermi level measured from the top of the valence band is given by: E p - E v : k T l n ( N / N D ) : 0 . 0 2 5 9 l n ( 2 . 6 6 x 1 0 r e/ l 0 r 5 ; : 0 . 2 6 e Y (b) The boron atoms compensatethe arsenic atoms; we have pp: Ne_Nn:3 x 10162 _ . 9x l 0 1 6 : l O l s c r n 3 Sincepo is the same as given in (a), the values for no and Ep are the same as in (a). However, the mobilities and resistivities for these two samples are different. 18. Since Np >> ni, wa can approximate e : Nt and po: n?I no:9.3 xl}te I l0l7 : 9.3 x ld crn3 ( "t p From fio: txiexPI

- r"'' \

|'

\. kT )'

we have Ep - Ei: kT ln (noI n) : 0.0259ln (1017/ 9.65* 10e): 0.42eY The resultingflat banddiagramis :

Ec

l

EF

A.rr2eY

1 . 1 e2 V

-E-

l

F.

t-l

19. Assuming complete ionization,the Fermi level measuredfrom the intrinsic Fermilevelis 0.35eV for 10rscm-3,0.45 eV for 1017crn3,and0.54eV for 10le crn3. The numberof electronsthat are ionizedis given by n = Npfl - F(En)l: Np / fl + "-(ro-rr)r*r] Usingthe Fermi levelsgivenabove,we obtainthe numberof ionizeddonorsas n : ! 0 t 5c r n 3 n : 0 . 9 3* 1 0 1 7 crn3 n : 0 . 2 7 * l O l ec r n 3

for Na : 1015crn3 forNo: 1017crn3 forNo: 10lecrn3

Therefore,the assumptionof completeionizationis valid onty for the caseof 10lscrn3. 10tu

20. No*-

l+ e-{Eo-Er)/kr

l0tu

_ .

l 1.t45

: _ 1016 1+e-0.13s

: 5 . 3 3 t 1 0 l sc r n 3

Theneutraldonor: 1016-5.33,.l0lscrr3 : 4.67x 1015 crn3 N; - 4'76-The ratioof 0.g76 s.33 N;

CHAPTER3

I . (a)For intrinsicSi, /4,: 1450,l+ :505, andn: p : lti:9.65x lOe We have p -

9200,Lb : 320,andn: p : ni:2.25x106

(b) Similarlyfor GaAs,lh: We have p -

- 3.31x10' C)-cm

en,(lt, + pr)

qntth + {lpltp

en,(ltn+ po)

qnltn + llPIIp

= 2.92x lOt O-crn

2 . For laffice scatterinE,l-h n 73/2 T : 200K, Lh:

r3oox #:2388

T : 4ooK, 1^4,:l3oox% 300-rt l

l

l

P

l-t'

l-t"

3. Since - = - + -

1 -1= - + -1 p 250 500 4. (a) p:5xl01s 14:4lo

p:

Fr: 167 cr# N-s.

cm2lv-s, Lh:1300 cm2lv-s

r

I qppp

:3 C)-cm

: 2 x 1 0 1 6 - 1 . 5 x 1 0 1 6 : 5 x 1 0 l "t * ' , n :

:290 : !-b: Ib (M + No) I+ (3.5xl0tu) fh: th(Nt + Np):

p

: 844"m'lv-s.

c d 3 , n : n / / p : ( 9 . 6 5 xl 0 e 1 2 l 5 x 1 0 :r s 1 . 8 6 x 1 0 4c m - 3

qpon + qlrpp

(b) p :Nt-No

2

cm2lv-s

= qLthn + q[Lpp

1000 cm2ny'-s

| qppp

:4.3 C)-cm

cm2A/-s,

L 8 6 x 1 0 ac m - 3

(c) p:N,q @oron)-Nn+ N,q(Gallium):5x10ls cd3, n: L86xl0a cm-3 I+: I$ (M r Np* Ne): 14 Q.05x10tt): 150 cm2A/-s, l-h: th (Ne * Np* N,q): 520 crr?A/-s p:8.3 C)-cm. 5. AssumeNo- N1>>n;,the conductivityis givenby ox qn[h: elh(No - Nd) We havethat 16 : (1.6xtOae)1^6Qtp10tt) Since mobility is a function of the ionized impurity concentration,we can use Fig. 3 along with trial and error to determine7^6, and No. For example,if we chooseNo : 2xl0r7,thenNr : ND*+ Nd-: 3xlQtt, so that Lh x 510 cm2lV-s whichgiveso: 8.16. Furthertrial and effor yields Nn=3.5x1017crrt.3 and lh x 400 cm2lV-s

which gives 6x 16 (O-cm)-t 6. o- q(lt n + Fop) = ello(bn + ni t n1 Fromthe conditiondddn: 0. we obtain ft:ni I {b

Therefore

p * _ q i o ( b nl,4 b + ^ , l b n ,_) b + l

2JE

P, QFpni(D + l) 7C

i . At the limit when d >> s, CF:

V p=ixWxCF

-

lnz

:4.53. Thenfrom Eq. 16

l0 x l0-3 x 5 0 x 1 0 - 0 x 4 . 5 3 - 0 . 2 2C6) - c m

FromFig. 6, CF:4.2 (d/s: 10);using thea/d: 1 curvewe obtain V=p. I/(W.CF)_

0.226x 10-' - 10.78 mV. 5 0 x 1 0 - ox 4 . 2

8. Hall coefficient, Ru=

V,A IB,W

x1.6xl0-3 10x10-3

- 426.7 cnf tC 2 . 5 x l 0 - 3x ( 3 0 x 1 0 -xnl 0 o ) x 0 . 0 5

Sincethe signof R'7is positive,the carriersareholes.From Eq.22 l

'

l

eR,

1

l,.6xl0-'nx426.7

AssumingNe x p, fromFig. 7 we obtainp: 1.1f)-cm The mobllity pg is givenby Eq. 15b 1 I

p, : - =

Wp

= 380 cm2lv-s. 1 . 6 x 1 0 - 'xn1 . 4 6 x 1 0 x' 61 . 1

9. SinceR n pand p-

1 , hence R o. qnph + wl_rp . nl+ + pLrp

From Einsteinrelation D n lt Hllro=DnlDo-59

R , _

N olt^

0.5Rr Nolt,+Neqp

We haveN.e:50 Nn . 10. The electric potential @is related to electron potential energy by the charge (- q) I

ni)

Q:*=(Er_ q

The electric field for the one-dimensionalsituation is defined as

e(x) : -!!-:l dx

dEi qdx

(n- - n \

ft: ni explT):

No(x)

Hence

Ep- E;:kTh( l

*r@) . )

t

E &) - -( tt\

dNo(x) dx

\q )No(*) ll.

(a) FromEq.31, Jn:0 and D, r (x) - H

d/d* n

" - - kT No!:a)9- - *kT o

q

N

o

"

n

'

q

(b) E (r): 0.0259(100):259 V/cm.

12. At thermalandelectricequilibria, J , = qgn(x)e + qD, 4:!') -, dx Dn

E(x)=

I

F, n(x)

--Dn

dn(x) dx N, -No

N, -ff.

D, Lt, N o * ( N r - N ' ) ( * l

p, L N o+ ( N , - N o ) t l0

L)

L

t , - J[ ' -o

D" p

N, -No LNo + (N, - N, )r

--o-nNL p"

N,

1 3 . N t = L p - T o G , = 1 0 x 1 0 - 6x l 0 t u = 1 0 t t c r n 3 + Ln:10Is +10tt -19ts

fl:ftno + Ln:No

p-

n?

e . 6 5x l o n) ' + l o r ,= l o , , c m - , 10"

\+4p=

t 4 . \a)Tp =

crn3

I

=

oor,rry

5 x l 0 - t 5x 1 0 7x 2 x l 0 t s

D oTo

= 10-t s

9 x 1 0 - 8- 3 x 1 0 { c m

Sr, : v,r,o,N,,, - 107x 2x 10-t6x 10to- 20 cm/s (b) The hole concentration at the surface is given by F,q.67

"11'p^(o)=pno +roG rlt -,u p * T o S , , ) I \

_ (9.g5 1 l9_')'+ 10-,*t0,,(t _ (.

2 xl}tu

10-8x 20

3 x 1 0 - a+ 1 0 - 8x 2 0

= lOe cm-'.

1 5 . 6= QnLIt,* wqy Before illumination fln = tlno, Pn = Pno

After illumination fln=frno*Lrt-ftro*TrG,

pn=pno*Lp-pno*trG

Ao - tqtt,(nno * Ln) + qlrp(p," + 4p)] - (qlt n," * QFop,") - q(l+ + Fp)r oG . ll

t a )J r . a i n- - n Dr # - - l.6x 10-lexIZx . - lx l0rsexp(-x/12) L2xl0-" : 1.6exp(-x/12) Alcrrl (b)

/",drift = J,o,o,-Jp,ain

: 4.8 - 1.6exp(-x/12)Alcni. (c) 't Jn.a,in - qnl+E

-'- 4.8- l.6exp(-xll2): 1.6x10-1ex 1016x 1000x8 E

:3-exp(-x/t2)

V/cm.

1 7 .F o r E : 0 w e h a v e

a__Pn_Pno +D,I* 0t

tp

o Ax'

=g

at steadystate,the boundaryconditionsarep" @ - 0) : p" (0) andpn (, : lV) : Pno.

Therefore

'*F) p ,(x )= p no + [1t,(0)- p^.

'*[fj

- o,"l+-4+) Jr(x-o): - QDp q[.p,(o) H,.n

Jo(x-w) = - QD.H,=, - ql.p p^,1?4. ,(o)'o

,inhfw I lL, )

18. The portion of injectioncurrentthat reachesthe oppositesurfaceby diffirsion I S T2

given by

a$=

J o(W) JoQ)

c o s h ( Wl L e )

L o= ^ t ' r % = ' & t s o * t o * - 5 x l o - 2c m ;. do -

cosh(10-2 /5x 1O-' )

- 0.98

Therefore,9SYo of the injectedcurrentcanreachthe oppositesurface. 19. In steadystate,the recombinationrateatthe surfaceand in the bulk is equal APr,ou,u

-

LPn,"urf^u

minorirv" *i:;:*..J#'ff sothattheexcess

atthesurface

: lora. ;,,surrace :g

:1013 cm-3

10-o

conditionsin the The generationratecan be determinedfrom the steady-state bulk loto : G: lom crn3s-l l0-6

From Eq. 62, we can write

D" ^ a ' L +! G- & : o To

Axt

The boundaryconditionsare4(. = - ): l01acrn3and4(* - 0): 1013crn3 t to lola( 1- o.ge-' )

Hence

4(i:

where

Lp:.fio- to-u: 31.6pm.

20. The potentialbarrierheight Qa= Q^- X: 4.2- 4.0:0.2 volts. 2 t . The numberof electronsoccupyingthe energylevel betweenE andE+dE is dn: N(DF(DdE whereN(E^)is the density-of-state function,andF(E) is Fermi-Diracdistribution function. Sinceonly electronswith an energygreaterthan E, + eQ^ and having a velocity componentnormal to the surfacecan escapethe solid, the thermionic currentdensityis +Az.T)% - | = r'v,E%e-G-rr)ln 4g J -JQt, Jrr*qq^ ht where v, is the componentof velocity normal to the surfaceof the metal. Since relationship the energy-momentum n

P2

E-2m

I

.

)

)

?r

2m

l3

Differentiation leads to dE -

PdP

m By changing the momentum 47iP2dP - dp,dp ,dp " Z? , S

f -

Hence =Ht', J mht

f

[-

component

p ,u-'ol

to

rectangular

p]-z^t1)lzmkr * p2,+ dp ,dp ,dp ,

dp,f- ,-r)/z^kr dp, o,oo,ll- ,-01r,^0, ^t'r'-,;t,'^rl')u,,

p,o

wherep',o= Zm(E, + qQ). t

Since

lE

L

"ll

e-o"dx -{ ll

2

, the lasttwo integralsyield (2dmkT)v'.

\a)

- u The first integralis evaluatedby setting oi:'9' 2mkT Thereforewe have du - P'dP' mkT The lower limit of the first integralcan be written as 2m(E, +qQ)-2mE, _qQ* 2mkT kT so thatthe first integralbecomesmkT

fr^,0,

e-" du - mkT e-qL-lkr

- A*7, "*(-fg^\ Hence, -4tqmkz 72o-a0^lk, h ' \ k T ) 22. Equation79 is the tunnelingprobability

x s i n h2( . 1 7 xl 0 ox g x t o - ' o ] ' ] :3.r9x l0*. r _{t* [20 4x2x(20-2) L ) 23. Equation79 is the tunnelingprobability

psinh( 9.99x 10nx 10-'o)f _ r'0-,0) - {, * [6x ]-' 0.403 4x2.2x(6-2.2) ) L * rinh(q.qq :. ro' *lo-'l' ]-' = 7.8x r0-,. Tr0-\' _ {r *' [o -z.z)

L^

coordinates,

4 x2.2"(e

l

t4

PdP m By changing the momentum component to 2 47d,dp : dp,dp,dp " Differentiationleadsto dE =

rectangular coordinates,

t = +[, dp,dprdp" [l =*o,r-rr2'+pi+p2,-znr1)''^o' [=Hence = "to'-'^Et)l2^trp,dp,!-_ "-oll'^o'dp, dp, f* "^o1/'^0, ,r2o ff,, rvherep',o= 2m(E, + eQ). r

:ll 2

r' *' Srnce e-"^dx =[ Z1 J__ \" )

yield(2dmk|)v,. , thelasttwo integrals

The first integralis evaluatedby setting

pi;29' 2mkT

:,

.

Thereforewe have 4y - P'dP' mkT The lower limit of the first integralcanbe written as 2 m ( E ,+ q Q ) - 2 m E o =qQ. 2mkT kT so thatthe first integralbecomes*0,

fr^,o

Hence - 1 =4tq*k' -72"-t0^lw= A'7, "*"( h3 ll.

e-"du: mlsTs-tQ./kr - uh). kr)'

Equation79 is the tunnelingprobability - 2 ) ( 1 . 6 x1 0 - ' e ) o _ l2*,(qVo E) _ 2(9.lIx t0_3'x20 = 2 . 1 7x 1 O t o m - t

P_tl----------'i-_

(1.054 x 10-'o)t

[20x sint(2.17x 100x 3 x 10-'o]' = r . , n , , 0 * ,r - -lr\ r 4 x 2 x ( 2 0- 2 ) | )-' 13. Equation79 is thetunnelingprobability

B= sintr( 9.99x10' x10-'o)I r'0-,0) = {r* [6x ] -2.2) x 2 . 2 x ( 6 4 |. )

= o.oo,

( q . q qrIo ' * l o - ' l ' r ' 0 - n l' = [ * [ o * r i n h I ' = 7 . 8 xl 0 - e .

l''

4 x 2 . 2 " ( a - z . zJ )- '

24. FromFig. 22 Ass:103V/s ua= l.3xl06 cm/s(Si) and uax 8.7x106cm/s(GaAs) t x 77 ps(Si) andt x 11.5ps (GaAs) AsE :5x104V/s vax 107cm/s(Si) and,uax 8.2x106cm/s(GaAs) t x l0 ps (Si) andt = l2.2ps (GaAs). 25. Thermal velocitv 2 x|.38x 10-2'x 300 = 9.5x 10nm/s= 9.5x 10ucmls For electric field of 100 vlcm, drift velocity = 1350x 100= 1.35x 105cm/s > v u , +N o = 2 ( v ^ )r ' QE"

r

1 . 6x l 0 - t e x 1 1 . 9 x 8 . 8 5 x10-'a

x (0.85x 10-')2

+No =3.43x10"cm-' We can selectthe n-type doping concentrationof 3.43x l015crn3

8. FromEq.56,

,-, G = -rr L : fl

" ocnD,rN,

l

o,e*n[-;-J]l n , rf Lo,"*ol- )+

[ _ l =l

.r-l5vln7ylnl5 x l 0 - t 5x 1 0 7x l 0 r t 10-t5

I

.lxe.65x10e:3.8exr0'6 and

2 x I l . 9 x 8 . 8 5x 1 0 - ' ox ( 0 . 7 1 7+ 0 . 5 ) =12.66x l0-5 cm=1.266tm 1.6x10-'e xl0tt

Thus J g " , = q G W r= 1 . 6 x l O t nx 3 . 8 9 x 1 0 ' ux 1 2 . 6 6 x l 0 - t = 7 . 8 7 9 x l 0 - t A / c m ' .

9 . F r o m E q . 4 9 , a n dp -no^ = i l ' ' ND

we canobtainthe hole concentrationat the edgeof the spacechargeregion, (

or

n., ,^ =#"ii;i' lvD

/ )

-\r

/

nc \

-g{#lel;'-'J

=2.42x t0,,,cm-,.

1 0 .J = J o ( r , ) + J , ( - * o )= J , G " ' r ' _ t ) " I - e o . ov2 5 e _l

)

J" v 3 0 . 9 5 e o o 2 s_el

= V = 0 . 0 1 7V . l l . The parametersare

ni: 9.65x10ecm-3 Dn: 21 cm2lsec Do:|0 cm2lseceo:T.o:5xI0-7sec FromEq. 52 andBq.54

-,)=nF*, z

J ,' ( r , \

=n':oLp

f (qv"\

1

!i-"leltrr J-l I ND L ]

(suvr*

7 = 1 . 6 x 1 0 - t xt

l0 5 > 2 0 . 1 4

3 . 4 5x 1 0 - ' ' .'.d > 4.57x l0-5cm : 0.457prn

24. Vr =0.5Y at Io = 0.lfg{

9.1= ---

iJ-

-7 -logI

olr.=o = -12 .'.I olro* = I x 10-''A. bg 1rl vo=o 25.

^vr=a*utp;-lv;) ' c o V n = o ' 0 2 6 n' (9=. -6t5- o ' '^-. )' = 0 . 4 2 Y x 10" 3'9x 8'85x-l0-ro a" 6.9x ro-?Frcrfi 5x 1 0 ' ' AY, = 0.I V if we want to reduceIp at V6: 0 by one orderof magnitude, sincethe subthresholdswins is 100mV/decade.

r / 2 x l . 6 x l 0' ' x 4 . 9 x 8 . 8 5 x r 0x-r'0o1 7 0.,_ ( r o r a . r r __ J o 8 4 ) 6 . 9x 1 0 - '

i.Vu,=0.83V. 26. Scalingfactor r :10 Switchingenergy=

*,a

. A)v'

C'-t* -tC d A A,{

V'=L If

.'. scalingfactorfor switchingenergy:

1 1 t< t( i

n.--.---.-=----:.=-

1

1 1000

A reductionof one thousandtimes.

27. From Fig. 24 wehave (r, + L)2 : (r, +[ry^\' -W.' .'.A2+ 2Lr, -2lf/.r . = 0 L=-rj+\l;;zw* L'= L-2A,

L + - 2 1 - 2 ^ = r - A= t - 2 ( 2

L

2

L

L

L

I

From Eq. 17we have (spacechargein tre toapezaid al regio! - spacechargein tre rectangular region) _ [no. / r '

c

nW^,L + L', _QN ---:Co 2L'

o

QNnW, Co

qN nW,r, ( f W r ,rT--r C,L rj [1

28. Pros: l . Higher operationspeed.

2 . High devicedensity Cons: l . More complicatedfabricationflow.

2 . High manufacturingcost. 29. The maximum width of the surface depletion region for bulk MOS ,nml

_ - L"1 l - l e , k T l n ( N n / n , ) Y

=-1/ r.

Q,Nn

1.6;m

= 4.9xl0* cm = 49 rnn F o r F D - S O Id, , i 1 W ^ = 4 9 r w r .

,') !.

I

3 0 . v-t =v-^ +2tu*4Nnd"' t D rD C.

=e^" =-+ -+^(+)=-f -o026tnt##) =-,o,u v,u =z*!!n(L) 2uru q

Co=

n

=0.92Y

i

3 . 9 x8 . 8 5 x1 0 - ' a = 8.63x10-t F/cm' 4xl}-' x 5 x 1 0 t tx 3 x l 0 - u l . 6 x1 0 - t e -1.02+0.92+

".V, =

8 . 6 3 x1 0 - i

= -0.1+0.28 = 0 . 1 8V . 3 1 . l\l/A t ,.7'

qNnLd =

C

ri

o

1 . 6 x1 0 - t nx 5 x l 0 t t x 5 x 1 0 - 7 8 . 6 3x 1 0 - ? = 46mV Thus, the range of 27"is from (0.18-0.046): 0.134 V to (0.18+0.046) : 0.226 V .

32. The planar capacitor ^ . F , z_ ----a.lo* = ( 1 x 1 0 r ) ' 3 . 9 x 8 . 8 6 x1 0 - 1 4= 3 . 4 5 x 1 0 _ r F, c: A'"/a

Forthetrenchcapacitor A:4x 7 prfr+I pm2: 29 pn? C :29 x3.45 x10-1s F : 100 x l0-1sF.

33.

34.

''t I = C d V = 5 x 1 0 - t*a -=3.1x10-"A d 4x10-' 7 f" =;!-tpci(V, -Vr) 4x l0-5:A(-5-Vr) I'10-5:A(-5+2) LY =7 -(-2)=9Y.

35. V, =Vru +2r,ltu

' V r= 7 V

Co=3'9 x

8 . 8 5 4 xl 0 - ' o =3.45x 10-' Flcm2 l0-'

( r'-o ' i ) V" =0.0261n1 = l =0 .4 2 Y ( 9 . 6 5x l 0 ' / - tO = ,- ; - vEu

-o

=Q^, vou

=-0.56 -0.42 = -{.98 V

2 V -= 4 . 9 8 + 0 . 4 2 + 3.45x 10-8

=-0.98+0.42+4.9 = 4 . 3 4Y LV-- =

Q,

5x10" xl.6xlO-'e

co

3.45x 10-8

= - 2 . 3 2V Vr = 4 . 3 4- 2 . 3 2 =2.02Y.

CHAPTER 7

1. FromEq.l, the theoreticalbanierheightis Qa,: Q. - X= 4.55- 4.01:0.54 eV We can calculateZ, as x 1o'' g.g259 *2'86 , = 0.r88v

v," =!! yr5= q

ND

2xl0r6

Therefore,the built-in potentialis V o , = Q r-,V n = 0 . 5 4 - 0 . 1 8 = 8 0.352V.

2. (a) FromEq.l1 de I C,) _ (6.2- L6):I0', = _2.3x10,0(cm,/F),/V -2-0 dv

-l

_t )- f N "^ = | l=4.7x10'ucm-' qe" Ld(l /C') I dV )

lk- = vs.6259 ^( +: * to".l=o.* u " =t!n q

ND

\4.7 x l0'" /

From Fig.6,the interceptof the GaAs contactis the built-in potentialV6i, which is equalta 0.7 V. Then,the barrierheightis Q u ,= V o ,* V n = 0 . 7 6 V (b) J" =5x10'7Ncmt Ar = 8 NK2-cr* for n- type GaAs J,

= A.T'e-qQBt1

tkr

r (1ggJ' ={'{I\, =0.025erf eu. I I =o.rr.,, q /" 5xl0-' |

J

Thebarrierheightfromcapacitance is 0.04V or 5Yolarger.

(c) For V: -l Y

=2.22x 10-t cm=0.222tm 4N'w =1.43x105 v/cm es

C =L=5.22x10-8F/cm2 W 3. The barrierheightis

Qu,= Q^- x= 4.65- 4.01=0.64V *vr(2.8orlo'n v," =L! yr{s-=6.6259 l=0.,r, u q

ND

[ 3xlo'u )

The built-in potentialis V o= , Q r , - V , = 0 . 6 4 - 0 . 1 7 7= 0 . 4 6 3 V The depletionwidth is 2e,(Vo,-V) 4No

2 x l l . 9 x 8 . 8 5x l 0 - ' o

= 0.142hn

1.6x10-tex3x10t6

The maximumelectricfield is

n y4t = ( 1 . 6x 1 0 ' ' )x ( 3x 1 0 1 ux)( 1 . 4 2 xt 0 r ) l..l=1.(r= o)l=QN q 11 . 9x ( 8 . 8 5x 1 0 - ' n )

4. The unit of C needsto be changedfrom pF to F/c#, so 1rc2-- t.74xl01s-2.r2xl}ts 4 1crfinf Therefore,we obtainthe built-inpotentialat IlC2 :0 , ^" ' -

l'57xlotj= 0.74y 2.12x10"

=6.54 x 100V/cm.

From the given relationshipbetweenC and Vo,we obtain

d(J-) \ c ' ) : -Z.l2xl}ts 1cn?ny2V dn FromEq.t 1 Nr=

_t

2f

-1

I

_

|

q e , l a g t c ' )d/ vl 2 _

[

r

'

)

ffil.rlr.ro"j = 5 . 6 x1 0 " c m - '

r0'' v" ={1n &- =0.025e*[z.so*]=0.,u,u q

ND

I S . 0 x l 0 ' "J

We can obtain the barrier height Q u n= V o ,* V n = 0 . 7 4 + 0 . 1 6=10 . 9 0 1 V . 5. The built-in potentialis

vt,=Qan_T" *O " ro"] =0.8- o.o25gr[z'se = 0.8- 0.195 = 0.605V

I t . s > < l o ")

Then, the work function is Q^ =Qr^ * X :0.8 + 4.01 = 4.81V. 6. The saturationcunent densityis

J. = ArT2 "*o(-

qQu")

\rr )

= I lox (3oo), >("*ol,,-,9', )

\0.025e /

:3.81 x l0*1 Alcm2

The injectedhole currentdensityis , ^ . . l0-"Ncm' J,* -=Q D o n , ' - 1 . 6 x 1 0 - 'xn1 2 x ( 9 . 6 5 x 1 0 '=) 'rl.lex i:t= ffi Hole cunent

J ,"1eqn'k' -11

Electroncurrent

J"(etvlkr -l) = J r o_ t . l g x l o - r r= 3 x l o _ 5 . J" 3 . 8 1 x1 0 - '

7. The differencebetweenthe conductionbandand the Fermi level is given by ( dt 'to" \

V^=0.0259 t"l :* l:0.04V. \ 1 x 1 0 ") The built-in potentialbarrieris then Vt, =0.9- 0.04= 0.86V For a depletionmodeoperation,VTis negative.Therefore,From Eq.38a V r = 0 . 8 6 - V ,< 0 qa'N^

l.6xlO-'na'xlO't

rr=-zE:=m>0'86

1 . 6x 1 0 - 2 ----------------a" >0.86 2 . 1 9x l 0 - ' ' a > 1.08x 10-5cfir= 0. 108 i m.

8. From 8q.33we obtain

VP _

' o " 6 m

=

Q N o a ' _ 1 . 6 x 1 O - t xe 7 x l 0 t 6 x ( 3 x 1 0 - s ) 2 _ A A . ,

2e.

2 x 1 2 . 4x 8 . 8 5x l 0 - ' o

5 x 10-ox 4500x12.4x8.85 x 10-'a Es2

il*o.t

0.3x10-o x1.5x10-a : 1 . 2 8 x 1 0 - 3S : 1 . 2 8m S .

9 . (a) The built-in voltageis

v o=, Q u-nv , = 0 . 9- 0 . 0 2 5 h1I o1 , t 0l':' 0 . 8 V 6 [

10"

/

At zerobias,the width of the depletionlayer is 2 x 1 . 0 9 x 1 0 - "x 0 . 8 6 [.6x10-tnx10t7

: 1 . 0 7 x 1 0 - cs m : 0 . 1 0 7p m SinceI/ is smallerthan0.2 pm, it is a depletion-modedevice. (b) The pinch-off voltageis l . 6 x l 0 - ' nx 1 0 t t ( 2 lx0 r ) : 2 . 9 2 y v'" _ e N o a ' _ 2e" 2 x 1 2 . 4 x 8 . 8 5l x0 - ' o and the thresholdvoltageis Vr : Vri- Vp : 0.86- 2.92: -2.06V. 10. From Eq.3lb, the pinch-offvoltageis

,, _QNoa' _l.6xlo , - - - 7 - -

2t,

' nx l o ' t ( 2 x l o

l): o.ruo u

2 x l 2 - 4x 8 . 8 5 xl 0 - ' o

The thresholdvoltageis Vr : Vni- Vo: 0.8- 0.364: 0.436Y and the saturationcurrentis given by Eq. 39

:t44v" I r"o,

-Vr)'

2 a L \ s

50x 1:0r;+2'1:j':851t0: 4s00 : 4.7x10*A. (0 0.436), " ) x ( l x l 0: -io ) 2x(0.5x10

ll'

) 0 ' 8 5 - o ' o z s g h ( 4 ' 7 " 1l0- -" l u x l ' - r e x N ^ a- : 0 . N , 2 * 1 2 . 4 , 8 J 5 l 0 " \ ) For No : 4.7 x 1016crn3 (

a: - | 0.85-0.0259^4'7

t

xl}"

_ , t ,

)"

ND )

(3'7xr}')

^lN"

: 1 . 5 2 x l 0 - sc m : 0 . 1 5 2p m ForNl : 4.7 x 1017 crn3 a : 0 . 4 9 6 " 1 0 - sc m : 0 . 0 4 9 6p m .

12. From Eq.48the pinch-off voltageis v , = Q. n n- - 'M - v , q :0.89 -0.23 - (-{.5) :0.62V andthen, " " y' . = -e:N n d , ' 1 . 6 x l 0 - t xe 3 x 1 0 t t' a ' : 0 . 6 2 V 2e. 2 x 1 2 . 3x 8 . 8 5x 1 0 - ' o

d r : 1 . 6 8 ' 1 0 6c m dt: 16.8nnr Therefore,this thicknessof the dopedAlGaAs layer is 16.8nm.

13. The pinch-offvoltageis l . 6 x l O - r nx l 0 ' sx ( 5 0 xl 0 - 7 ) : l . g 4v v , - _Q N o d l 2e" 2 x 1 2 . 3x 8 . 8 5x l 0 - o The thresholdvoltageis AE, _ r./ r, V,=Qu,-" q : 0 . 8 9 - 0 . 2 3- 1 . 8 4 : - 1 . 1 8V Whennr:1.25x 1012 crn2,weobtain

'" :

1 2 . 3x 8 . 8 5x l O - t o

ffi

ri - (-l'18)] =r'25xrot2 x[o

and then do+58.5: 64.3 d o : 5 . 8n m The thicknessof the undopedspaceris 5.8 nm.

14. The pinch-offvoltageis rl r '

-

q' NUr d |l

l . 6\ x l- 0 -- ' e -x 5' x l -0 x' 7 ( 5 o0 x' 1 0| ' ) \' 1

_

28"

2 x 1 2 . 3 x 8 . 8x51 0 - ' o

The barrierheightis *AF c * V , = - 1 . 3+ 0 . 2 5 +1 . 8 4= 0 . 7 9 V Q u n= V , + q

.

'

The 2DEG concentrationis 12.3x8.85x10-'o

* [o l . 6 x l 0 r nx ( 5 0 + l 0 + 8 ) x l 0 - ?

frr=

15.

- 1-t.:yf = |.zgx t o, cm-2

The pinch-off voltage is

, , - Q N o d l- r < - l . 6 x l o - t nx l x l o t 8 x d f ' 2e 2 x 1 2 . 3 x 8 . 8x51 0 * The thicknessof the dopedAlGaAs is

dr=

1 . 5x 2 x 1 2 . 3x 8 . 8 5x I n - r +

ffi:4.45x10-8cm:44.5nm

*_=S_ _ Vp =0.9 _ 0.23_ 1.5= _0.93V. vr = QB, _ AF

t6.

The pinch-off voltage is aN^ v' ^ 2- ' e" d :

1 . 6 x 1 0 -x' t3 x 1 O t t

(rs* to' )': z.tv

2x12.3x8.85x10-to the threshold voltage is V, = Qr, -

Atr *c -V, q

: 0 .8 9-0 .2 4-2 .7 : -2.05Y Therefore, thetwo-dimensional electrongasis flr=

12.3x8.85x10*'o

* [o- 1-z.os)]= 3.zx1o" crn2

l.6x 10-'nx (35 + 8) x 10-7

CHAPTER8

l.

Zo:

r

L = C: . 2 t o= 2 * 1 0 - 1 2x 7 5 2= 1 1 . 2 5n H .

2.

:{ro)'.0;GI 3 x 108m/s r:== x { l 1 6= 1 . 6 1 6G t l z . "a

3.

: 1.48V Vo,= (E, / q) +V^ *Vo : 1.42+0.03+0.03

'=^W

1 /q I N , N , )

={

t-r-''

\lo"xro''/'

= 1.89x 10-6cm = 18.9nm = 6 . 1 3x l o ' F / c m z. c =k-r'l6xlo-t2 W 1 . 8 9x 1 0 - '

4r="(t)"*? t).,,*o(#) From Fig. 4, We note that the largestnegativedifferentialresistanceoccursbetween Vp0.5 pm 2L t 3596: 735 particles/r#with particlesizes> 1.0 pm 100 35OO: 157particles/# with particlesizes> 2.0 1tm Therefore,(a) 3500-735 : 2765particles/nfbetween0.5 and I pm (b\ 735-157:578 particles/m3 befween1 and2 pm (c) particles/# 157 above2 pm. **

r00

:.

y = fr,r-o,n A :50 mm2:0.5cm2 l r _ e 4 ( 0 . r x 0 . s*)" 4 ( o 2 s x o . s )* " - t { t x o . s ; = g - r . z = 3 0 . l y o

I

.

The availableexposureenergyin an hour is 0.3mW2/cmzx 3600s:1080 mJlcr* For positiveresist,the throughputis _1 0 8 0= 7 wafers/ty 140 For negativeresist,the throughputis loSo 120 wafervtrr. 9

1. (a) The resolutionof a projectionsystemis givenby L u ' u^ 9:193fgt= o.r7g pm t ^ = K t f r f . =9.6,1 0.65

o,le,3f DoF=k,-L= u* '(NA)' o.sl(0.65)'l: o.r28 L

(b)

I

We can increaseNA to improve the resolution.We can adopt resolutionenhancement techniques(RET) such as optical proximity correction(OPC) and phase-shiftingMasks (PSM). We can also developnew resiststhat provide lower h and higher k2 for better resolutionand depthof focus.

(c) PSM techniquechangeskr to improveresolution. j.

(a) Using resistswith high y value can result in a more vertical profile but tkoughput decreases. (b) Conventionalresistscan not be usedin deepUV lithographyprocessbecausetheseresists have high absorptionand require high dose to be exposedin deep [fV.

This raisesthe

concernof damageto stepperlens,lower exposurespeedand reducedthroughput.

6. (a)

A shapedbeam system enablesthe size and shapeof the beam to be varied, thereby minimizing the number of flashesrequired for exposing a given area to be patterned. Therefore,a shapedbeamcan savetime and increasethroughputcomparedto a Gaussian beam.

(b)

We can makealignmentmarkson wafersusing e-beamand etchthe exposedmarks.We can then usethem to do alignmentwith e-beamradiationand obtainthe signal from thesemarks for wafer alignment. X-ray lithographyis a proximity printing lithography.lts accuracyrequirementis very high, thereforealignmentis difficult.

(c)

X-ray lithographyusing synchrotronradiationhasa high exposureflux so X-ray has better throughputthan e-beam.

7. (a) To avoid the mask damageproblem associatedwith shadowprinting, projection printing

exposuretools have been developedto project an image from the mask. With a 1:l projectionprinting systemis much more difficult to producedefect-freemasksthan it is with a 5:l reductionstep-and-repeat system. (b) It is not possible. The main reasonis that X-rays cannot be focusedby an optical lens. When it is through the reticle. So we can not build a step-and-scanX-ray lithography system.

8.

As shown in the figure, the profile for eachcaseis a segmentof a circle with origin at the initial mask-film edge. As overetchingproceedsthe radius of curvatureincreasesso that the profile tendsto a verticalline.

9.

(a) 20 sec 0.6 x 20/60:0.2 pm....(100) plane 0.6/16x 20/60: 0.0125pm........(ll0) plane 0.6/100x 20/60: 0.002pm.......(tI 1)plane

wo= wo- .l-zt= 1.5- "li x0.z= 1.22pm ( b ) 40 sec

0.6x 40/60:0.4pm....(100)plane

0.6/16x 40/60: 0.025pm....(l t0) ptane 0.6/100x 40/60: 0.004pm.....(lI l) plane

wu=wo-Jzt = 1.5-Ji xo.a:0.93 pm ( c ) 6 0s e c 0 . 6x l : 0 . 6 p m . . . . ( 1 0 0 ) p l a n e xl: 0.0375 pm....(l l0) plane 0.6/16 x l: 0.006pm.....(lI l) plane 0.6/100 W o : W o - J z t = 1 . 5 -J 2 x 0 . 6 : 0 . 6 5p m . [-singthedatain Prob.9, theetchedpattemprofileson -Siareshownin below. (at 20 sec I :0112 ltm,Wo= Wu= 1.5pm (br -10sec l:0.025 ltm, Wo- l[t = 1.5 pm ( c t 6 0 s e c l : 0 . 0 3 7 5 1 t mW o- W u = 1 . 5p m .

lf u'e protectthe IC chip areas(e.g.with SfNa layer)and etchthe wafer from the top, the ri idth of the bottomsurfaceis r r ' = t T t + J z t = 1 0 0 0* J T * 6 2 5 = 1 8 8 4p m l-he fractionof surfaceareathat is lost is (r'' -t4/:)/w2 x 100%:(1g842-10002) /t8842x100%:71.8% In termsof the wafer area.we havelost 7 1.8% x tdl5 / 2)' :127 cr* Another methodis to definemaskingareason the backsideand etch from the back. The width trf eachsquaremaskcenteredwith respectof IC chip is given by v [ = W r - ^ l - z t = 1 0 0 0J- i x 6 2 5 : 1 1 6 p m [- sing this method,the fractionof the top surfaceareathat is lost can be negligibly small.

I Pa: 7.52mTon

PV: NRT 7.52/760, 10-3: n/V x0.082x 273 :2-7 ,l}ta cm3 n/V: 4.42x l0-7mole/liter:4.42x l0-7x 6.02" 1023/1000 mean-free-path 2 , = 5 x I 0 - 3/ P c m : 5 x l 0 - 3x 1 0 0 07/ . 5 2 : 0 . 6 6 4 9c m : 6 6 4 9 p m l50Pa: ll28 m Torr PV

nRT

ll28l 760 x 10-': n/V x 0.082x 273 :-4 x 1016 n/V : 6.63x 10-smole/liter: 6.63x lQ-sxg.Q/x 1023/1000 crn3 mean-free-path : 0.0044cm : 44 um. 2": 5 xl}-' /Pcm : 5* 10-3x1000/1128 x n, x7/' ,r-tiA' l-1. Si EtchRate(nm/min):2.86 x 10-13

: 2 . 8 6 x l 0 - r 3x l x l S r s x (92t 1 / ,, r # :224.7 nm/min.

*3" Iytsx(2Oq%^"#: 11. SiOzEtchRate(nm/min): Q.Sl{x 10-13 Etchselectivityof SiOz overSi : or etchrate(Sio2)/etchrate(Si)' :

#=

5.6nm/min

0.025

':u-)! - 0.025. ""t-t'u*'ot/n'7x2e8 2.86

15. A three-stepprocessis requiredfor polysilicon gate etching. Step 1 is a nonselectiveetch processthat is usedto removeany native oxide on the polysilicon surface.Step 2 is a high polysilicon etch rateprocesswhich etchespolysiliconwith an anisotropicetch profile. Step3 is a highly selectivepolysiliconto oxide processwhich usuallyhasa low polysilicon etch rate. I 6. If the etch rate can be conholledto within l0 %o,thepolysilicon may be etched 10 o/olonger or for an equivalentthicknessof 40 nm. The selectivityis therefore 40 nm/l nm:40. 17. Assuminga30Yooveretching,and that the selectivityof Al over the photoresistmaintains3. The minimum photoresistthicknessrequiredis

(l+ 30%)x I pm/3:0.433 pm:433.3 nm.

t8.

(I)" =-

qB me

2 n x 2 . 4 5x 1 0 e=

[.6x10-'exB 9.lx 10-31

B: 8.75x 10-2(tesla) :875 (gauss). l 9 Traditional RIE generateslow-densityplasma(10e crn3) with high ion energy.ECR and ICp

generatehigh-densityplasma(10rr to 1012crn3) with low ion energy.Advantagesof ECR and ICP are low etch damage,low microloading,low aspect-ratiodependentetching effect, and simple chemistry. However,ECR and ICP systemsare more complicatedthan traditional RIE systems. I ( J . The corrosionreactionrequiresthe presenceof moistureto proceed. Therefore,the first line of

defensein controlling corrosionis controlling humidity. Low humidity is essentiaf.especially if coppercontainingalloys are being etched. Secondis to removeas much chlorineas possible from the wafersbeforethe wafersare exposedto air. Finally, gasessuchas CF+and SF6can be usedfor fluorine/chlorineexchangereactionsand polymericencapsulation.Thus, Al-Cl bonds are replacedby Al-F bonds.WhereasAl-Cl bonds will react with ambient moistureand start the corrosionprocess, Al-F bondsare very stableand do not react.Furthermore,fluorine will not catzlyzeany corrosionreactions.

CHAPTER 13

l. E"(boron):3.46 eV, D :0.76 cn?/sec From Eq. 6,

- F -3'46 D = D o e x ? ( # ) = o . l e r *. o ( 1o-"cm'ls t g . 6 t 4 x l o _x1223 5 )=o't+zx

L = J D t = 4 . 1 4 2 x l 0 - "x 1 8 0 0= 2 . 7 3 x 1 0 ac m FromEq. 9,

c (x)= c,ertu (fi)= 1.8x I 02oerfc (17{6;)

I f x = 0 , C ( 0 ) = 1 . 8 x 1 0 2a 0 i o m s/ c m 3 ; x : 0 . 0 5 t 1 0 4 , C ( 5 * 1 0 5 ) : 3 . 6 x l 0 l e atoms/cm3;x : 0.075 r I 0-4, C(7.5x 10-6) : g.4 x 1gls atoms/crrf;, :0. I " I 0-a, C(10-s): 1.8 x l0r8 atoms/cm3; x : 0 . 1 5 x 1 0 4 ,C 1 l . 5 x l 0 - 5 ) : l . g x l 0 1 6a t o m s / c # .

The x.,=ZJDI (efc 'gs!--) =o.l5Arn Totalamountof dopantintroduced: Q(0 1

:#C,L 4n

= 5.54xlOtaatoms/c#.

-3.46

2' D=Do*(#)=oz6exn(

8.614xl}-s x1323) = FromEq. 15, Cs = C(Q,t)= + Jtilt

o' ' ux1o- ' ocm ' /s

3 =2.342x 10'natomVcm

" =2.342*to'nerrc[ C(x\ =C-"rf.[a) ) " \2L) \ 2 . 6 7 3 x r 0 -)' Ifx :0, C(0): 2.342x 10leatoms/crrf ;*:0.lxl0-a, C(l0r) : l.4lxl0te atoms/cm3; x : 0.2x104,C12xl0't): 6.7911018 atoms/c nf ; * :0.3 x l6-a,C(3x lo-5;: 2.65x 10t8

atoms/cm'; x : 0 . 4 x 1 0 4 , c 1 4 x 1 0 - s ;9: . 3 7 x 1 0 ta7t o m s / c m 3: ;0x. 5 x 1 0 - 4c, 1 5 x l O - s ) :1 . g 7 x 1 g t z atoms/cm3; x : 0.6x104,c(6x 10-t): 3.51*1016 atoms/crrt; t :0.7x10-a,c(7x l0-5): 7.03xl0r5 atoms/cm3; :5.62x101aatoms/cm3. .r :0.8x104, C(8x10-s) The x, =

= 0.72qm. 4Dtht-CuJ tDr

3 . 1 x 1 0 "= t * t O " " * p [

l 0 -' ) 4 x 2 . 3 x 1 0 ' ')t

t : 1 5 7 3s : 2 6 m i n For the constant-total-dopant diffusion case,Eq. 15 gives C, = -L *Dt t. = 3.4xlo*atomVcm

s = 1x lottm

4. The processis calledthe rampingof a diffusionfumace.For the ramp-downsituation,the furnacetemperatureT is given by T:To-rt where To is the initial temperatureand r is the linearramp rate.The effectiveDt product duringa ramp-downtime of tr is given by

(Dt)"r= l^' ogyat In a typical diffirsionprocess,rampingis carriedout until the diffirsivity is negligibly small.Thusthe upperlimit t1 con be takenas infinity: l l l r t -=-* (l+-+...) T T -rt T' T 0

0

0

and

(-n -E-,. -rEt -rEt rt -r D = D oe x p [ " /hr u/rr\l/= D-o " * pl 1 l + A * ; * . . . ) . ll : O o ( " *3p" _X e x p - # . . . ) =D\( T ; " r y # u / T \ Io klo kfoLkIo kT^, J where D(To) is the diffusion coefficientat T6.Substitutingthe aboveequationinto the expressionfor the effectiveDr productgives

=[i otr,\"*p-]#dt (Dt)"n =D(r)+ ' '"''Eo kTo' Thus the ramp-downprocessresultsin an effbctiveadditionaltime equalto Kls2/rguat the initial diffi.rsiontemperatureT6. For phosphorus difflrsionin siliconat 1000"C,we havefrom Fig. 4: |{7d : D (1273K) : Zx 10-racm2ls 1273-773 r=-=0.417K/s 20x 60 E":3.66 eY Therefore,the effectivediffusiontime for the ramp-downprocessis

oro' ,Eo

_. l.3g xlo-r'(1273)' = 9ls = 1.5min. 0 . 4 1 7 ( 3 . 6 61x. 6x 1 0 - ' n )

5. For low-concentrationdrive-in diffusion,the diffusion is given by Gaussiandistribution. The surfaceconcentrationis then

,s

,s C(0.t)=+=---exDl 4d)t Jil;

( r-i" \ i '\2kr)

{ =_L.,*l,&Y-r"' *g ) =_0.5 '\2ftrl( dt 2 .|1il,

)

t

dC

dt =-0.5xC t

which meanslYo changein diffusion time will induce0.5Yochangein surface concentration.

dc

,s

----:'s.vr-[-

dT

Jrilo, dc

nr

'

,.............._-

C

- E a

2kT

(E-\(-E-)

E_

\zkr )\2kT' )

zkT'

- 3.6x 1.6x lo-''

dT T

dT

1L

i..

2xl.38xl0-"x1273 T

dT T

which meanslo/ochangein diffirsiontemperaturewill cause16.9%changein surfaceconcentration.

6. At I 100oC,ni : 6x 1018crn3.Therefore, the dopingprofile for a surfaceconcentration of 4 x 1018crn3 is givenby the "intrinsic"diffi.rsionprocess:

c(x.t)=c"..r"[4] \2''lDt ) whereC, : {x 1018crn3,t: 3 hr: 10800s, andD : 5x10-1a cm2ls.The diffusion length is then J Dt = 2.32x10-5cm = 0.232tm The distributionof arsenicis C(x) = 4 x 1018 "rf"[---:-) [4.64 xl0-' ) Thejunction depthcan be obtainedas follows

10"=4xl0"".frf

t'

-l

( 4 .6 4x 1 0 -'J

xi: I.2x 104cm: 1.2pm.

7. At 900oC,ni:2x

1018crn3. For a surfaceconcentrationof 4x10tt "-',

given by

the "extrinsic" diffusion process -4.05x1.6x10-le

-E,

" 2r*rl?l =45.8et 38"10 D = Doe kr x ni

=3.77x10-'ucm2ls

"l#

xt =1.6JDt = 1.6m 8.

= 3.23xl0*ucm= 32.3 nm.

lntrinsic diffusion is for dopant concentrationlower than the intrinsic carrier concentrationni at the diffusion temperature.Extrinsic diffi.rsionis for dopant concentrationhigher than n;.

9. For impurity in the oxidationprocessof silicon, segregationcoefficein t =

3x10" 3 10. f=---..-=-:-=0.006. 5x 10'' 500 1 1 .[ 0 . 5 : [ 1 . 1 + q F p / C i " " * --€Z r'

3.9x8.85x10-ra =3.45x10-7

10-u

, ," - 3 ' 4 5 x l o r x o . 6 = 1 . 3x l o , c m - ' I.6 x l0-'" 10-t

; t = 1 . 3x 1 0 ' t x l . 6 x l 0 - t n

i4r0.16)'

the implanttime r :6.7 s. 12. The ion doseper unit areais f t 1 0 x 1 0 {x 5 x 6 0 N = q = l . 6 x l Q - t e = 2 . 3 g x 1 0 1i 2 ons/cm2 A A ,I0.,

n "\T)-

From Eq. 25 andExample3, the peakion concentrationis at x : Ro.Figure.l7 indicatesthe oo is 20 nm.

Therefore,the ion concentrationis ^9 oo42n

2.38x 10'' -3 = -___-___-___-_ :. = 4.74 x l0'tCm . 20x10-'42tr

13. FromFig. 17,the&:230

nm,andoo = 62 nm.

The peakconcentrationis S

2x10'5 : ---------------= l.29x l0'ocm-l oo{2tt 62x10-'42n FromEq. 25,

t.2ex,o" "*[3:i{l |

260-

J

x;:0.53 pm.

14. Doseperunitarea:

9 =CoLVr - 3'9x8'85x 10*taxl = 8 ' 6x 1 0 " c m - 2 q q 2 5 0x 1 0 - sx l - 6x 1 0 - ' n

From Fig. 17 andExample3, the peakconcentrationoccursat 140nm from the surface.Also, it is at (140-25): 115nm from the Si-SiOzinterface. 15. The total implanteddoseis integratedfrom Eq. 25

Re-,-l}=lo 's -erfc(z. errcl 3)l=r-xt'ee8e

s-"*[-t*I{l*=l{r*fr-

ur - lo e,= f @-

|

2oo. J

z

I

L

o,.lz l)

z-

The totaldosein siliconis as follows(d:25 nm):

Q,,=I:;fu"*|#]*=i{'-['_",f"(*i,]}=;'-er|c(|.87)]={x the ratioof dosein the silicon: Qs/Qr:99.6%.

16. The projectedrangeis 150nm (seeFig. 17). The averagenuclearenergylossover the rangeis 60 eV/nm (Fig. 16). 60x 0.25: 15eV (energylossof boronion per eachlatticeplane) thedamage volume: Vo: n (2.5nm;21tSO nm):3" 10-18 cm3 total damagelayer: 150/0.25: 600 displacedatomfor one layer: l5l15 : I damagedensitY: 600/Vo : 2" l02ocm3 zxlo2o/ 5.02x102: 0.4%. 17. The higher the temperature,the fasterdefectsannealout. Also, the solubility of electricallyactivedopantatomsincreaseswith temperature. o 1 8 . L V' , = l V : = '

Cor

where p1 is the additionalchargeaddedjust below the oxide-semiconductor surfaceby ion implantation.Cox is a parallel-platecapacitanceper unit area givenby C"" =1 a

(d is the oxide thickness,€r is thepermittivity of the semiconductor)

1v x 3'9x 8'85x 10-'4F/cm : g.63xr0-' c; e, = LV,c".0 . 4x l 0 - 6 c m

cm'

8'63x 1o-' : 5.4x1ol2ions/c# 1 . 6x 1 0 - t e Totalimplantdose:

t'0,\!9" :1.2 x 1013 ions/c#. 45%

19. The discussionshouldmentionmuchof Section13.6.Diffi.rsionfrom a surfacefilm avoids problems of channeling. Tilted beams cannot be used because of shadowing problems. If low energy implantation is used, perhaps with preamorphizationby silicon, then to keep the junctions shallow, RTA is also

necessary.

20. FromEq.35

o-g)= & = 1"rp"[o 0.84 ^s 2 [0.2J2) The effectivenessof the photoresistmask is only 16%o.

s, = l.rr"ig) .s 2

=0.023

\0.2J2)

The effectivenessof the photoresistmaskis 97.7%.

21. r=]-{u 24n

- lo-, u

. ' . u= 3 . 0 2 = 0.927pm. d: &+ 4.27 op: 0.53+ 4.27x 0.093

CIIAPTER 14

l.

EachU-shapesection(referto the figure) hasan areaof 2500 pm x 8 pfft:2 x thereare(250(|)2/2*104:312.5U-shaped 10apm2.Therefore, section.Each sectioncontains2long lineswith 1248squareseach,4 cornersquares,1 bottom square,and 2 halfsquaresat the top. Thereforethe resistancefor eachsectionis l kO/il (1248x2+4x0.65+2):2599.6 kO The maximumresistanceis then 3 r 2 . 5 x 2 5 0 0 . 6 : 7 . 8*t 1 0 8C ) : 7 8 1M O

2.5 mm 4Pm (PITCH)

2. The arearequiredon the chip is

,

co Eo,

( 3 0 x l o - ' X 5 x l 0 -=, , ) 3.9x 8.85x l0-tu

4.35x 10-5cm2

: 4 . 3 5 " 1 0 3p m z: 6 6 x 6 6 p m

Referto Fig.4aand usingnegativephotoresistof all levels (a) Ion implantationmask(for p+ implantationand gateoxide) (b)Contactwindows(2x10pm) (c) Metallizationmask (using Al to form ohmic contact in the contact window and form the MOS capacitor). Becauseof the registrationerrors,an additional2 pm is incorporatedin all critical dimensions.

(o) 56 pm

I

II {l+2Fm

Zli*

(b)

73

1

If the spacebetweenlines is 2 p*, then there is 4 pm for eachturn (i.e., 2xn, for one tum). Assumetherearen turns,from 8q.6, [, x 1tan2r* 1.2 x 10-6r?r,wherer can be replacedby 2 x n. Then,we canobtainthat n is 13.

4. (a) Metal 1, (b) contacthole,(c) Metal2. (a) Metal l,

(b) contacthole,

I

(c) Metal2.

I

5. The circuit diagramand devicecross-section of a clampedtransistorare shown in (a) and (b), respectively.

(o) COLLECTOR

EMITTER

Si02

(b) P-SUBSTRATE

oHilrc CONTACT

6 . (a) The undopedpolysiliconis usedfor isolation. (b) The polysilicon I is usedas a solid-phasediffusion sourceto form the extrinsicbaseregionand the baseelectrode. (c) The polysilicon2 is usedas a solid-phasediffusion sourceto form the emiffer region and the emitterelectrode. 7. (a) For 30 keV boron,& : 100 nm and A,Ro: 34 nm. Assumingthat .Roand A,Ro for boronarethe samein Si and SiOz the peakconcentrationis given by

J-zntn,

J:19f-

tl2nQa x l0-')

=9.4xro,u cm-,

The amountof boron ions in the silicon is

f;=rffiq"*lW (n"-d.tl

s [2 _ e rf,c ==1 l _ +_

2L

=-t

8xlo"[^

2

ll

[^/2m,.,1-1

z-Eltvt

^ ( -=:- 7 5 0 ) l

L

I I

\Jz*340)l

= 7 . 8 8x 1 0 " c m - ' Assumethat the implantedboron ions form a negativesheetchargenearthe SiSiO2interface,then

10-''x (7'88x 10") LVr = re)/ ' l q ) Co,=-J:6 1

3.e*8.8m=o'elv

(b) For 80 keV arsenicimplantation , &:49

nm andA & : 1Snm. The peakarsenic

10'u - = 2.21x102,cm-, . concentration i, J42nA,R" r/ax(l8x l0-')

Rp=49oi (As)

tu;\ \

\.o"r.*. \ \ \ \ \

BORON

I \

(LOWER ScaLsl 1

/ar\

Re-toooi (B) {ol5

o

I d

-tso.r-zsoi

tooo

-zm

x(i) - FoRcHAls{ELRectfi

8.

(a) Because(100)-orientedsilicon has lower (- one tenth) interface-trapped chargeand a lower fixed oxide charge. (b) If the field oxide is too thin, it may not provide a large enoughthreshold voltagefor adequateisolationbefweenneighboringMOSFETs. (c) The typical sheetresistanceof heavily dopedpolysilicongate is 20 to 30 Q fl, which is adequatefor MOSFETswith gate lengthslargerthan 3 pm. For shortergates,the sheetresistanceof polysilicon is too high and will causelarge RC delays. We can use refractory metals (e.g., Mo) or silicides as the gate materialto reducethe sheetresistanceto aboutI O /t1.

(d) A self-alignedgatecanbe obtainedby first definingthe MOS gate structure,then using the gate electrodeas a mask for the source/drainimplantation.The selfaligned gate can minimize parasitic capacitancecaused by the source/drain regions extending underneath the gate electrode (due to diffusion or misalignment). (e)

P-glasscan be usedfor insulationbetweenconductinglayers,for diffi.rsionand ion implantationmasks,and for passivationto protect devicesfrom impurities, moisfure,and scratches.

9. The lower insulatorhas a dielectricconstantAlq:

4 anda thicknessdr

l0 nm The

upperinsulatorhasa dielectricconstantA/q : 10 and a thicknessdz: 100 nm. Upon applicationof a positivevoltage Vc to the extemalgate,electric field E 1 and E2 arc establishedinthe dt and dz respectively.We have,from Gauss'law, that e1E1: e2E2 +Q andVc: Erdt *Ezdz whereQ is the storedchargeon the floating gate.From theseabovetwo equations,we obtain

"-l

vo Q -' + d, d,(q ltr) g * er(a,tar) a

I

r

-

'

, ^ - ,I

loxlo?

|

| 4 \

O |

/r^ t,l

.'tffiJ],.r.rsx ro-'o f Iro+roolrJ

(a) If the storedchargedoesnot reduceE1by a significantamount(i.e.,0.2>> 2.26x10s

lQ l, *" can write Q = foadf x 0.2Lt = o.zx (o.zsxtO-u)=5 x t0-sC L V' - : Q = ,

5xlo-8

c2 (lo*8.s@=o'565

v

= 8.84x10-7 (b) when t -) a,J -+}we havelgl-+ 0.212.26x10s C.

T h e nL , V r = t = ,

8 . 8 4 xl 0 - ?

=9.98 V.

( t 0 x 8 . 8 5x 1 g - t 4 ) 7 1 6 - 5

10.

+

+

(o) p-TUB

+

+

(b) POLYSILICOT{GATE

+

+

(c) n-TYPE O|FFUSION

+

+

+ {d) p-TypE DtFFUstoN

+

+

trn (e) CONTACT WINDows

trn

+

(f) METALLTZATIOTTI

11. The oxidecapacitance per unit areais givenby C^- =

''7 = 3.5x l0 t F/cm2 d

€"rn

and the maximumcurrentsuppliedby the deviceis

-v,)' =::+3.5xt0r (vo-v,)'= 5mA r", =!\ rc,,(vo 2 L20.5ttn

and the maximum allowablewire resistanceis 0.1 V/5 mA. or 20Q. Then. the lengthof the wire must be -

t

- _

Rx Area

20Ox I0-tcm'

p

2 . 7x l } ' Q - c m

= 0.074cm

or 740 pm. This is a long distancecomparedto most device spacing. When driving signalsbetweenwidely spacedlogic blocks however,minimum feature sizedlineswould not be appropriate.

12.

x

si3N4

x

Ill,*f

v - EPITAXY

(o)

(bl

i tap*

FIELDOXIDE

(d)

RESIST (e)

1 3 .To solvethe short-channeleffect ofdevices. 14 . The deviceperformancewill be degradedfrom the boron penetration. There are methodsto reducethis effect: (1) using rapid thermal annealingto reducethe time at high temperatures, consequentlyreducesthe diffusion of boron,(2) using nitrided oxide to suppressthe boron penetration,sinceboron can easily combine with nitrogenand becomeslessmobile,(3) makinga multi-layerof polysilicon to trap the boron atomsat the interfaceof eachlayer. 1 5 . Total capacitance of the stackedgatestructureis :

"z l(z.e-) c_q d, d,l d, \d,

)

7

= -x-/

2s/(7

25\

l-+-l:2.12

0 . s l 0 l \ 0 . 5 t o)

3 ' 9= 2 . 1 2 d .'.d =

3'9 =1.84 nm.

2.12

1 6 . Disadvantages of LOCOS:(1) high temperatureand long oxidationtime causeV1 shift, (2) bird's beak,(3) not a planarsurface,(4) exhibits oxide thinning effect. Advantages of shallow trench isolation: (l) planar surface, (2) no high temperatureprocessingand long oxidationtime, (3) no oxide thinning effect, (4) no bird'sbeak.

t 7 . For isolationbetweenthe metalandthe substrate. 1 8 .GaAs lacksof high-qualityinsulatingfilm.

19. (a)

: 2000* (el.o: x l0-'o)= 1.38x 10-'s= 1.38ns. (b) For a polysiliconrunner

(

r.\(

l)

RC = | R,ouo,"#ll t",- | w)\ d) \ (

t

\,

= 391--l- l(ol.or>< t0-'o)=2.07 xl0-7 s \10-",/' = 2 0 7n s Thereforethe polysiliconrunner'sRC time constantis 150 times largerthan the aluminumrunner. 20. Whenwe combinethe logic circuitsand memoryon the chip, we needmultiple supplyvolrages.For reliability issue,differentoxide thicknessesare neededfor different supplyvoltages. 21' (a)

/r^^,

=

/cr^,o,

*

/r,u,,o"

uo%'='%r* 1%=fi 3 A hence (b)Eor : 16.7 A.