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>
(b) For a face-centered cubic, a unit cell contains1/8 of a sphereat eachof the eight cornersfor a total of one sphere. The fcc also containshalf a sphereat eachof the six facesfor a total of threespheres.The nearestneighbordistance is l/2(a J; ). Thereforethe radiusof eachsphereis l/4 1aJz ). -
Maximum fractionof cell filled - (1 + 3) {4[ [(a/2)/ 4It I 3] / a3 :74o/o.
(c) For a diamondlaffice,a unit cell contains1/8 of a sphereat eachof the eight cornersfor a total of one sphere,I/2 of a sphereat eachof the six facesfor a total of three spheres,and 4 spheresinside the cell. The diagonaldistance
between(112,0,0) and(114,ll4, Il4) shownin Fig. 9a is
n:1 2
;)'
.(;)'
The radiusof the sphereisDl}: -
1Jj 8 Maximum fractionof cell filled
: (t + 3+ 4) 'o, :nJT t16: 34%. ' lyErtjl' 13\8
)) This is a relativelylow percentage compared to otherlatticestructures.
:a 4. la,l: la,l: la,l:laol 4*4+4+4:o 4 . ( 4 * 4 + 4* 4 ) : 4 . o: o -4 *4.4 +4. L: o la,l'*4 --d2+ d2coflrz + dcoiln I dcoflr+ ! dz+3 d2coil! 0
- coil:+ [: cos-r +
[ 109.4/!
5. Taking the reciprocalsof theseinterceptswe get ll2, ll3 and l/4. The smallest three integershaving the sameratio are 6, 4, and 3. The plane is referredto as (643)plane. 6. (a) The latticeconstantfor GaAs is 5.65 A, and the atomicweightsof Ga and As are 69.72 and 7492 glmole, respectively. There are four gallium atoms and four arsenicatomsper unit cell, therefore 4/a3: 4/ (5.65x 10-8)3: 4.22x lTn Ga or As atoms/cr*, Density: (no.of atoms/crrfx atomicweight)/ Avogadroconstant : 2.22* 1022(69.72 + 74.92)I 6.02* 1023: 5.33g I cni. (b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors are formed,becauseSn has four valenceelectronswhile Ga has onlv three. The resultingsemiconductor is n-type. 7. (a) The meltingtemperature for Si is l4l2 oC,andfor SiOz is 1600oC. Therefore, SiOz has higher melting temperature.It is more diflicult to breakthe Si-O bondthanthe Si-Si bond. (b) The seedcrystal is usedto initiated the growth of the ingot with the correct crystalorientation. (c) The crystalorientationdeterminesthe semiconductor's chemicalandelectrical
properties,suchas the etch rate, trapdensity,breakageplaneetc. (d) Thetemperatingof the crusibleandthe pull rate. ErQ): l.l7
4.73x10uT'
for Si
.'. Es ( 100K) = 1.163eV, andEs(600K): 1.032eV
E,(D=r.5,n -t'o-l!j_"t!1,!' rorGaAs (T + 204)
,.Er( 100K) : 1.501eV,andEs(600K) : 1.277eY . 9.
The density of holes in the valenceband is given by integratingthe product N(E)tl-F(DldE from top of the valeric:e band(En takento be E : 0) to the boffom of the valenceband Rottoml
p: ytu'n^N(qtl _ F(DldE Ee)*t where I -F(E): I - {t r[t * "(t]: If Er- E >> 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,, va = l_4,E For electricfield of 104V/cm. FoE= 1350x 104= 1.35x 107cm/s= y,r,. The value is comparableto the thermal velocity, the linear relationshipbetween drift velocity and the electricfield is not valid.
CHAPTER 4
l. The impurityprofile is,
(Np-N,a)(cm")
3 xl O r a
x (pm)
c:101ecm-a
The overall space charge neutra
of the semiconductor requires that the total negative
space charge per unit area in the p-side must equal the total positive space charge per unit area in the n-side, thus we can obtain the depletion layer width in the n-side region:
0 . 8 x 8 x 1 0 '- o
W nx 3 x 1 0 ' o
Hence,the n-sidedepletionlayer width is: pm W, =1.067 The total depletionlayer width is 1.867pm. We usethe Poisson'sequationfor calculationof the electric field n(x). In the n-sideregion, L=3-*r+r(x^)=LNox+K d
x
t
"
"
E
s
E ( x , = 1 . $ 6 7F m ) = O + K = - + N , €.,
x l . 0 6 7 x1 0 - a
" ' E ( t , \ = L * 3 x l o r a ( x _ l ' 0 6 7x l o r ) E
E^o, =E ( x, = 0 )= -4. 86 x 103V/cm
In thep-side region,the electricalfield is:
= e ( xP" )' = = n r a x 2+ K ' 2€,
9d x= L *t r^
E( x p = _ 0 . 8 1 m: 0) + K ' = - t *
o * ( 0 . S "t 0 - o ) '
- ( o . s , ,'r o - ' ) ' l . . ,G \ P"' ) : a * o , l* ' ' L
2t,
)
E,o, =E (ro =0) = -4.86 x 103V/cm
The built-in potentialis:
_ l;", (')4 n_s v,,= - I _:,, &W = - l'0,,,(r)d*lo_,,0" ide=0.s2v .
L
From Vo,= -
J
r (*V, , the potentialdistributioncanbe obtained
With zero potential in the neutralp-region as a reference,the potential in the p-side depletionregion is
=- ft *o*l'- (o.r v,(,)=- li e)a* x10')'f* =-ftLir - (o.s * r0-o)' " -l{orx100 = -7.5e6x 10"x
[1"'
- (0.r,ro-')'* -?r(0.t,. to-')']
With the condition Vp(0):V"(0),the potentialin the n-regionis
=-t":' ro''[]"'-t.o67xro-ax.UF,.ro-') v^&) = -4.56xr0' "()*
-1.067x lo-ox
T,.
,o-')
[[!!-
)D--lr-
l[_
l l . - , . !_-.[.-_
m_
l!_.-
!m
n!
n_ ,
l[ _tr-
[!-
, fl_
l!
-____t ,_ [[^
l!-^-[-
m_
l![[-l--
!!t
[[[.
.,
__fll[l-_J_
t----l[-
t![.
f_
___ r ["Jr][+n_Jo_!{
i.
The intrinsic carriersdensity in Si at different temperaturescan be obtained by using Fig22 in Chapter2:
Temperature(K)
lntrinsic carrierdensity(n,)
250
1.50+108
300
9.65+10'
350
2.oo+10"
400
8.50]l10''
450
9.00+10'3
500
2.20+10t4
The Vu canbe obtainedby using Eq.12, and the resultsare listed in the following table.
--lmIml
J!
__m
![-
m nmt m-
__tr
. n[[![_
__tr
l![[[-
m_
ml MD
mlm!-
Iu-
,n,n[!*
m,n
!!-J
Thus,the built-in potentialis decreased as the temperatureis increased. The depletionlayer width and the maximumfield at 300 K are
IT: 4No _mu
2 xll.9 x 8.85x 10-'ox0.717 =0.9715lnt 1.6x10-texl0t5
p f t r _ 1 . 6x 1 0 - ' ex 1 0 ' 5x 9 . 7 1 5x 1 0 - 5 =qN =1.476x 104V/cm. q 1 1 . 9x 8 . 8 5x l O - ' a
' ' x 3 0 [' 9 ] ' t , = l z r n ( + .. E . . y ^ * + ) l " ' = 4 x r 0 s f 2 x t . 6 xl0 )l'" mu L e
[Nr*Nr)) ND > 1 . 7 5 51 x 0 t u=
[ l t . e x 8 . 8 5 x l 0 - [' 1 o 0 ' ' +N " ) ]
r+#
We can selectn-typedopingconcentrationof Nr: 1.755x10t6crnt for the iunction.
5. From Eq. 12 and Eq. 35, we can obtain the l/C2 versus Zrelationship for doping concentration of l0't, 10t6,or lOtt crn3, respectively. ForNr:16r5 "*-:,
+= C,'
'V:, _') -
Qd"Nu ForNo:19r6"*-:,
c,'
ed"Nu
z x ( o . s t-tv ) 1 . 6 x 1 0 - ' nx l l . 9 x 8 . 8 5 x 1 0 - ' nx l 0 '
- = 1 . 1 8 71x0 ,( 6o . t z-tv )
l . 6 x 1 0 - r ex 1 1 . 9 x g . g 5 xl 0 - , 0 ; 1 0 "
= 1 ' 1 8 7 xt o " ( o ' s l o - r )
For Nlr:1017 crn3,
I _ zVu,-v) _
zx(o.gsa-v)
C,'4d,Nul.6xl0-l9'ttffi=L.|87xto'o(o.eso-r,) Whenthe reversedbias is applied,we summarizeatableof I /Ct, vs V for variousNp values as following,
[_
[[_-*JI
l!.-._lI
Lt__-[[ x.
fr! ][ ,.
nl![[!_
3r
l !.
._n rr__Jm
L-
, - fl-l
l!
'-l flt
f[n -
n n m
. f-l
[-JD
TIT]
4U/4U
n . n n n
iL__.__.![
.-n [ "
__[-_.n n-
__-!-_ul
[. nD -n.^---_l
mm
_nn. n[[_
_J____-[!
n.
!*
D[!!D[-
.-n --!
.--8.-.-D [-
J_n._nl
._-[!__^-l!
-_rJJil
][l
Jln__-lil
n
n
n
1!J1_!U
._n
-i--m --[
!
[![I
.-n
[-__n[
Hence,we obtain a seriesof curvesof I/A versus Zas following,
.|II-
!
t r s L tk<{mth4}g
The slopesof the curves is positive proportionalto the values of the doping concentration.
The interceptionsgive the built-in potentialof thep-n junctions.
6 . The builrin potentialis ( l 0 ' o x 1 0 2 0x I 1 . 9x 2kT,(a'e"kr\ ? 8 . 8 5 x1 0 - ' ax 0 . 0 2 5e ) V"r ,': = - l n l -) --i; l= i*0.0259x hl
3q
\ \ q ' n )i 3
8 x 1.6x l0-'nr (l.os* to' I
t
= 0.5686V From Eq. 38, thejunction capacitancecanbe obtained
" 8.85xlo"o) ' (.,=T=lq;4j - - 4 - l q o t , ', l ' " =L = [[ . 6 x1 0 - ' x' l o' ox ( t t.e-vR) rz(o.s6s6
]"'
At reversebiasof 4V, thejunction capacitanceis 6.866xl0-e F/crf .
7 . FromEq. 35, we canobtain
+=ry#+N,=&*r| 2x4
' . ' v R > > 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 ><10-
No=5.2x1015cm-'
"
I (q.os,. ton)' I r--qrt x l e \ o o 2 s-el l ND
t
L
I l J
(4vrtr -l)=q
=no:'*
J^(-*r)
Ln
[;"t"1't*t-'],
= 2 5 = 1 . 6 x 1 0 - t tx ::>
N t = 5.278x 1016 cm 3 We can selecta p-n diodewith the conditionsof Nn: 5.279xl016crn3 and Np : 5.4x10lscrn3.
12. Assume[s 4p {,r:
1O6s,Dn:21 cr*/sec,andDo: 10 c#/sec
(a) The saturationcurrentcalculation. FromEq. 55aand to =,[Drrr, we canobtain
_ *p o 1 '- _ Q D o P n o* Q D , n p o= o" n | .:N( LDEI ? Le Ln
t NA
F;)
!^)
= r.6xlo-'xe(e.os',0')'[*.,8. +.8] ' [ l o ' ' l l t o - u t o ' u\ l t o - ") = 6.87xl}-t'Ncmt And from the cross-sectional areaA: 1.2x10-5 cm2,we obtain I , = A x J " = 1 . 2 x1 0 - 5x 6 . 8 7x l 0 - t ' = 8 . 2 4 4 x l 0 - t 7 A . (b) The total currentdensityis ( q v
\
\
/
J =J"l"' -tl Thus (
o
r
\
I o r n = 8 . 2 4 4x l 0 - r ? [ e o o r t-,r | | = t . 244x l0-t7 x 5.47x 10" = 4.51x l0-5 A
( u ,
\
4 1 0 - r ? [ e o .-ot " n I _ o . r ,= 8 . 2 4 x r=f
.244x 10-"A.
( q v
\
13. From J = J " l r n - l l \
/
we can obtain
v ="[f+]* r = v :0.025e* to-' ,=)* ,l - 0.78 ,[l, v. l 0.02se l0-'' L\/"/
I
L\t.z++" J I
14. FromEq. 59, andassumeDo: l0 cnflsec,we canobtain
J"- ='1a f i " : * Q ' ' w r, No
t,
* ton)' = r.6x lo.n /-lL (q'os * l Jt o * l o '5
l . 6 x l O - t ex 9 . 6 5 x 1 0 e
l0-u
x lo'5= 0.834V vo,=0.0259,r,-lo'' ( 9 . 6 5x l 0 ' ) '
Thus J n : 5 . 2 6 x1 0 - 'l+ 1 .8 7 2 x1 0 -'J0 .s3 4+ V ^
2 x l l . 9 x 8 . 8 5x 1 0 - r ax ( V o+, V ^ 1.6x10-tex10'5
(
._-[*-__-m l^
Il._
--l--lm [^
[l.-
._ff_-_m [^
m_
J.
"nm !^
m mI
_lm__nI [^
DI
-n__1._nt [^
![^
-J--
.![ !^
m_
_ff__!m
n ^ !m
__n-._.[_-!l [^
__n
m! !^
ll-fi_
ro0ri
m -
m
Jm^
-
E
_!00^
J Jm^
o:B:_.
olo0t0
WhenNo:1017 cn 3,we obtain x l0't= 0 . g 5 3v v ^ ,= 0 . 0 2 5 g , - 1 0 ' ' (9.65x l o')'
J n = 5.26x10-''+1.872* to-'.ft.lso+ tu* llJr]- trri.a.O
u-L
uL R
1 5 . FromEq. 39,
Q o = 4 f , b " - p ^ " \d *
ll"llrtl
u1_
= n[l,o*(ftvrn -t) e-G-',\/'04* The hole diffusion lengthis largerthanthe lengthof neutralregion.
Qo=ef;{0,-p,,)d* =a
ln *Q{
rtcr- l) e-G-.,),t" 4* ';-t
( qv
-t-t \f =epnoct)1"'-ljle Lp-e "
)
)
:r.6xr0-'e x"i#tl(-5xro .,["
-'1" * -" ;)
=8.784x 10-3Clcm'.
16. From Fig. 26, the critical field at breakdownfor a Si one-sidedabrupt junction is about2.8+105V/cm.Thenfrom Eq. 85,we obtain t:'"' zr(breakdownuottugq=E{ = (t, )' 2 2 q .-\, _ 1 1 . 9 x 8 . 8 5 x 1 0x-(' 2o . 8 x 1-0z'-),('1 0 ''/ 2xl.6xro-,,
= 2 s sv 2xll.9x 8.85x l0-'a x 258 1.6x10-rexl0t5
:1.843x 10-3cm= 18.43fem
when the n-region is reducedto 5pm, the punch-throughwill take place first. From Eq. 87, we can obtain Zr' _ shadedar_eain F_is. 2ginsert=(!_)(r_y_)
vB
\w^ )\" w^)
G.w'Iz
(w\( ( s \/ w\ s \ V u ' = V;u l l l2 - - l = 2 5 8 x- 1 l l z - - - : - -l = 1 2 vl It/^ 18.43l\ t8.43 \w ^ )\
)
\
)
Compared to Fig. 29, thecalculated resultis the sameasthe valueunderthe
conditions ofW:5 pm andNa: 1015cnL3. 17. we can usefollowing equationsto determinethe parametersof the diode. J ,' = e' \, p" -, n , ' , u , 0 , * Q w t l ,e c v t 2 k-=r'"\Et t, LNeos -v t k r N, 2T,
vr=t"! " 2 =t,!"' 2 q (*). =
os* lot)' ,kt AJ, . ,=1n" l TE:-eqv o N o , r , + A+ x |Axr.6x . 6 x ll0-re 0 _ ", " Ex i , ;b? e a o 2 5 9 _ 2 . 2 x 1 o - 3
=aH*#5 v,:E+=ff{* ")-r =130
(ru,)-'
Let E.:4x105 Y/cm,wecanobtainNo:4.05x101scrn3. The mobility of minoritycarrierholeis about500 at No:4.05x l0r5 .' Dp:0.0259x500:12.95cnfls Thus,the cross-sectional areaA is 8.6x10-5cm2. 18. As the temperatureincreases,the total reversecurrent also increases.That is. the total electroncurrentincreases. The impact ionizationtakesplacewhen the electrongains enoughenergy from the electricalfield to createan electronhole pair. When the temperatureincreases,total numberof electronincreases resulting in easy to lose their energy by collision with other electron before breakingthe latticebonds.This needhigherbreakdownvoltage. 19. (a) The i.layer is easy to deplete,and assumethe field in the depletion region is
constant.From Eq. 84, we canobtain.
,
"w .( r \u .( e \o t/ IJ o l 0 ' l = - | d*=l=1001.-| x l 0 - ' = l l E " , n i , o t : 4 x 1 0 5x ( 1 0 ) l o= 5 . 8 7x 1 0 5V / c m \4x10'/ [axl05/ 4 V B = 5 . 8 7x 1 0 ' x 1 0 - 3= 5 8 7 V (b) From Fig.26, the critical field is 5 x 105V/cm.
Z"(breakdown uottu):{
=''i;' (wr)'
x 8.85 x 10-ta - 12.4 ab x 105IQ * t',uf' 2 xl.6xl0 = 42.8V.
I - tnl8
=1022cm-o 20, e-on'u 2 xl}-o
y, =Y - +""' l2e"1"'1o1'"' r 3 E
^
L q l
o - 4Er"' |z * n,gl,s,Ls-.:to' 1"' * 1ror,1-,,, 3 1 . 6x l 0 - ' ' L
,j
= 4.84 xl}-,E"3 /t
The breakdown voltage can be determined by a selected t".
21. To calculatethe resultswith appliedvoltageof V = 0.5V , we can use a similar calculation in Example 10 with (1.6-0.5) replacing 1.6 for the voltage. The obtainedelectrostatic potentials are l.lV
and 3.4x l0-o V,
respectively.The depletion widths are
-s -8 3.821xl 0 cm and 1.274 x 10 cm, respectively. Also, bysubstituting V =-5 V to Eqs.90 and 91, the electrostatic potentialsare 6.6V and 20.3x l0-4 V , and the depletionwidths ue 9.359x l0-5 cm and 3.12x l0-8 cm, respectively. The total depletionwidth will be reducedwhen the heterojunctionis forward-biasedfrom the thermalequilibrium condition.On the other hand,when the heterojunctionis reverse-
biased,the total depletionwidth will be increased.
22. Es(0.3)= 1.424+ 1.247x 0.3 = 1.789eV A-E _ ( E o r _ E r r ) / - _Esz r/ _'c / b i_ q _ ( E r r _ E r r ) /q q q
= t.tle* o.^-gh oI \l=orr' -4r 1,.l:,, = r.273 y q 5x10" e 5xl0 r ^,, tV 2N /E$.v^, | l" ^' , - , - , - , LqNo(e,Nr+erNn))
r I 2x12.4xLI.46x8.85x l}-ta x1.273 t 1 . 6x t o - ' e x 5 x 1 o r 5( t z . + + 1 1 . 4 6 )
:4.1x10-5cm. Since Nrr, = N ,txz
.-.xr = xz
. ' .W : 2 x , : 9 . 2 x 1 0 - 5c m = 0 . 8 2 p t m .
1'
CHAPTER 5
l. (a) The common-baseand common-emittercurrentgainsis given by do: 1&r = 0.997x 0.998= 0.995 o
ao l-d^ = 1 9 9.
0.995 l-0.995
(6) Since 1a =0 and lgr=l}xl0-eA,then
lruo is toxlO-e A.Theemittercurrentis
Icno =(t+ p)truo = (t +tr).to x lo-e =2x10{A.
2. For an idealtransistor, d'o=f=Q'!))
P ' : & : s e s' Iruo is known and equals to t0 x t0-5 A . Therefore, Iceo = (*
gotruo
: ( l + 9 9 9 ) . 1 0x 1 0 - 6 =10rnA.
junction is forward biased.From Chapter3 we obtain 3. (a) The emitter-base
sl to'''z'-t-9'' o.nru 't! 'l= o.orrnrf =kr 1n(N vo, u. l= q
\ ni- )
Thedepletion-layer widthin thebaseis
t p.65xlo"f J
N n=, gotAa"ptetionlayerwidthof theemitter- basejunction) 4 ' =( ,, I (N, +No ) -
_r,) =^lz,(yolf =_+lr,,u, q 1l [lro )\',n-',o 1 * ro_,, , \. -0's) _ = /z.r.os f:::g:)f ', ,-'' rorj(.;,. roq;Ar.,l(0'es6 t; I =5.364x 10-6 cm =5.364xl0-2trrm. Similarly we obtain for the base-collector function
-l ro''' ro: =o.rn, v^, u. " =o.o25e"[?' " lo'Jr LP.os l
l
4.254x706 cm:4.254x10-2 um . Thereforethe neutralbasewidth is = 0.904Um. Vl =Wn-Wr-W, =l-5.364x10-2-4.254x10'2 (b) UsingEq.l3a .
|
^\2
x l0-'f = l-i' ,rrolor - (9'65 p,(0) = pnoe4vralkr ND2x10t1
=2.543x l0r cm-3 eosloo2ss
4. [n the emitterregion
D, =52 cmls L, =62,10-t = 0 . 7 2 1 x 1 0 ' c m g s- l o -')' =rg .625 nE o=(q . 5x l 0 ' 8
In the baseregion
D o = 4 0 c m l sL o = f f i
= J + o ' 1 0 -=72 x 1 0 -c3m
' ro-nY ^ - n,'-=(q'os = -, =465.613 = +o).1 .
P ro =
Vtr-
"10--
In the collectorregion Dc =ll5 cm/s Z. =fis
3 cm to* =10.724x.10
,rn ' =futrt=9.3r2x lo3 lo'u
The currentcomponents aregivenby Eqs.20, 2I, 22, and23:. l . 6 x l 0 - ' e. 0 . 2 x I 0 - 2. 4 0 . 4 6 5 . 6 1 3 e o s l o o z=s1e. 5 9 6 x l o r A 0.904x 10-o Iro 7 I u o= 1 . 5 9 6 xl 0 - ' A Iuo
Iun _ l . 6 x l 0 , n
. 0 . Z x l 0 - . 2. 5 2 . 1 9 . 6 2 5 ( r o r t r o r r nl_) = t . 0 4 l x I 0 _ i A v 0.721x10-3
. I l 5 . 9 . 31 2x l 0 3 1 . 6x 1 0 - ' '. 0 . 2 x 1 0 - 2 = 3 . 1 9 6 x1 O - t A o 10.724x10-' Iuu = I r o - I r o = 0 1,,
5. (a) The emitter,collector,and basecurrentsaregiven by I, = I to * Irn = l'606x10-5 A 5 I c = I c p * I c n = 1 . 5 9 6x l 0 A I u = I r n * I u , - I c ^ = l ' 0 4 1 x 1 0 - 7A '
(b) We can obtain the emitter efficiency and the basetransport factor: Y'
1,, "' I"
*r -
Iro Itu
1.596x lo-5 = =0.9938 1.606xl0-' 1.596x l0-5 l.596xlo-5
Hence, the common-base and common-emitter current gains are d,=fUr=0.9938 F o= : a o
l-do
:160.3
(c) To improve y,the emitterhasto be dopedmuch heavierthan the base. To improve a, we can makethe basewidth narrower.
6. We cansketchp^(x)l p,(0) curvesby usinga computerprogram:
o c ox vc
0.4
0.4
0.6
D I S T A N C Ex
In the figure, we can see when WlLo.O.L (WlLe=0.05 in this case),the minority carrier distributionapproaches a straightline and can be simplified to Eq. 15.
7 . UsingEq.l4, 1r, is givenby
I,o:A(-ro,H,^) - t.ornEl
=u(nilfo
'ulkr
' a V
e' \e
no
Le
-
I
lL, )
-r)
tt l l
t-rrf = nn.'=O"f L o, l
tl
'alkt
e7 4' Y t lkr _
ll
L
\
.l
rfr ?)l
=uDrf "o'th'hlL,W I
l
'*g)
n rno
Similarly, we can obtain lro:
,r,=u(-no,H,=,) t
, ( w- * \
=^-rr,rln ^.r"'^'*- ttl,*r4) Lp
Izo
)
-a"".n[l] Lp
lL, )
'I'no
,t"hf4l
l_
| =qoTI*'^',,[---L1:{All \1, )
=qAT;E 8.
\1, )
L,*[fJ] L'ry';]i r,
*,-,)."".{fJ] lQ,,
The total excessminority carrierchargecanbe expressedby
g,:nn([r,e)- p,"fdx ,wl
-
-1
= q A l I p , o " o " u , o , 1| l--; l a x Jo W ) " t w
L-""
:qApnoeorol* O-+)l
2prh
otr14tr.. -"evrolkI
2 qAWp"(0) _ 4 L
From Fig. 6, the triangularareain the baseregioni, wpo(O). By multiplying this z
valueby q andthecross-sectional areaA, we can obtainthe sameexpressionas en. In Problem3, 1 . 6x l 0 - t e . 0 . 2 x 1 0 - 2 . 0 . 9 M x l 0 { . 2 . 5 4 3 x l } t l 2 =3.678x10-t5C.
9.
lnEq.27, I c = or, (f avulw -l)+ a,
_ qAD,p,(0) W qAep,(o) =2Do w.2 2 2D
=
*?Q'
'
Therefore,the collectorcurrentis directly proportionalto the minority carrier chargestoredin the base.
10. The basetransportfactoris
-. I l 'l I"^ d ,' Z J =
Ib
-
sllillt
-r)* l{""*'*."./zll \1, ))
|
lL' )
t ""'nI HV L
For t4fLo<<1, cosh(Vf L)=1. Thus, dr=
,,,,,i',hfzl ""rh(L\ \1, )
\1, )
:.""nIlL) \1, )
t-t(w)'
2lL' )
z) =r-0t'lzro
11. The common-emitter currentgain is givenby R: 'o-
do l-oo
Wr l-wr'
Since7 = 1, 8..=
a' l-d,
_ r-frr,'lz4') ,,
t-[-Vy'lzr,')] =Qto'fw')-t . rf WfLp <<1,thenpo=zLo'fW' .
12.
L o = r l D o t , = ^ / 1 0 0 . 3x 1 0 - 7= 5 . 4 7 7x 1 0 - 3c m= 5 4 . 7 7p m Therefore, the common-emitter current gain is
"to.:\'
P' = 2 LP, 2tl W , = z ( s ! . l l
Q"lo-')' = 1500.
13.
In the emitterregion, Fpr =54.3+
l + 0 . 3 7 4x l 0 - r t . 3 x l 0 r 8
=87.6
-87.6= 2.26cmls D t = 0.0259 In the base region, pn =88+
1252 I + 0 . 6 9 8x l 0 - r t . 2 x l 0 t 6
= 1186.63
=30.73cm/s . Dp =0.0259.1186.63
In the collectorregion, Lr.r-= 54.3+
407 .= ,, = 453.82 l+0.374x10-".5x10''
. 453.82 = I l.75cm/s . Dc = 0.0259
14.
In the emifterregion, L, = ^tDp, =^lZ.Z6g.l0*=1.506 x l0 3 cm I
fri
_ _= = rn L,(,, NF
(9.65r'^nY
3r.lir
= 3r.04cm-j
In the baseregion, L, = J3u734.10u= 5.544x10-3 cm
' tonY = 6.os n ^- = Y:::::LJ4656.13 cm-3 2x l0'o ln the collectorregion, L. --"111.754-10{= 3.428x10-3 cm
/ ^\, -19'65 xlol- f = 18624.5cm'3 5x l 0 ' '
,.^
The emiffer current components are given by
, r, :Weo.6ro
o2se = 526.g3x lo-6 A
0 . 5 xl 0 +
..?.?69.3r.04 4o.ozse_r)= x l0_eA . t.OOe \ Qo
'rEP-_1.6xrc-t:..1_0: r.506 x lo-3
Hence,the emittercunent is It = I tna Itu = 526.839x lOj A .
And the collectorcurrentcomponentsare given by ,r^ _l.6xl}'"
.10-o-30.734.4656.13 =526.g3x10_6A eoqo.o2ss 0 . 5x l 0 {
. 1 0 4. 1 1 . 7 5 4 . 1 9 6 2 4 . 5 - 1.6x101e = 1 . 0 2 2 x 1 0 rA n . I'cP - _ 3.428xro-3 Therefore, the collector current is obtained by Ic = Ico '1 Ico :5-268x l0- A .
15. The emitterefficiencycanbe obtainedby I y' - tn - 526'83xlo{- = 0.99998. IE 526.839x l0The base transport factor is
*'fl
_
I"\rt
- 5526.83 2 6 ^ 8x 3l0* *lo-u-''
h Therefore,the common-base currentgain is obtainedby do -- Tdr = I x 0.99998= 0.99998 .
The valueis very closeto unigz. The common-emittercurrentsain is
B^=
oo - 0'99998 = 5oooo. l-a" l-0.99998
16. (a) Thetotalnumberof impuritiesin theneutralbaseregionis wnol(l-r-*/,) % = fr Nnoe-,/tdx= = 2 x l 0 r 8 . 3 x l 0 - 5( - e - ' " ' o ' / ' . ' o - ' ) = 5 . 5 8 3x l 0 t 3 c m - 2
(b) Averageimpurity concentrationis 9c _5.583x10'3 W 8x10-5 =6.979x 1017 cm-3
L7.
For N, = 6.979x10tt"*-t, D, =7.77cm3/s,and -3 L, = ,tD^r, = tll.ll . l0{ = 2.787x 10 cm
_l , -I -Ar =
I -
w2 ,
---------==l--r
2Ln'
,, Dt Qo D, N ELE
(srlo-'I
zQ1u " to- ' f
= 0.999588
= 0.99287 I .5.583x10'3 l+ 7 . 7 7 l O t e. 1 0 - 4,
Therefore, ao=wr=0.99246 B o= : u : 1 3 1 . 6
.
l-C/o
18.
The mobility of an averageimpurity concentrationof e .glg x l0r7 cm-3 is about 300 cm'/Vs. The averagebaseresistivity pu is given by
Therefore" Ru--5x to1(po . uI w) = 5x l0' -(o.ozes ltx 10r) = 1.869 Fora voltagedropof kr f q ,
,, = o.ol3e A. #= Therefore. I c = F o I a = 1 3 1 . 6 . 0 . 0 1 3=9 1 . 8 3A .
19. FromFig. 100and Eq. 35,we obtain
1"(r,A)
1.(mA)
e,=#
0
0.20
5
0.95
150
l0
2.00
210
l5
3.10
220
20
4.00
180
25
4.70
140
current, Fo Bo is not a constant.At low 1, , becauseof generation-recombination increaseswith increasing1, . At high I B, vEBincreases with l, , this in turn causes a reductionof lzr. since Vru+Vur=Ytc=sV. The reductionof Vu, causesa wideningof the neutralbaseregion,thereforepo decreases. The following chart shows Bo as function of I B. lt is obvious that Fo is not a constant.
/
,1
\
/
/
\
/
o
\
/
5
t
\
1
0
.
1
5
2
0
2
5
3
ls (trA)
20.
Comparingthe equationswith Eq. 32 gives Iro=art, d rI ro = ar,
a^Ioo=a, artd I no = dzz
Hence,
*d -' -:
azt
or- tZ.%."* LE De
crr. - - n- : er,
,r J, _W . - . _ Lc
21.
p,o
I Dc flco Dp
pno
In the collectorregion, .3 Lc : ^[Dc%= Jr.tou =1.414 x lo cm f t c o= f t , 2l N , = ( e . O"sr c t l f s x l 0 r 5= 1 . g 6x3l 0 ac m - i. FromProblem20,we have
0
I "
14/ D, . l+-.-.: LE D,
npn
9.31 ,*0.5x104.1. lo-3 lo 9.31x lo-2
P,O
= 0.99995 d
,' -, W
R Dc
-
ftco
1
t - *0.5 * " x- l0{
+ ,, ,-
2
1.863x lOa
m 93r.ro-
= 0.876
rro=ctr,=n4+.ry:) . ( r c . 9 . 3 1 x 1 0 21 . 9 . 3 1 ) 1.6x10-te.5xl0-a.l . *-l t0* ) | 0.5x l0* : I . 4 9 x 1 0 - t oA
rno=ozz=r4+.T) ') 2'l'863x 104 = 1.6xl0-re.5 xl0+ .(!2'tt"to' lAA"ll-t ) \ 0.5"10-t = 1 . 7x 1 0 - r 4 A . The emitter and collector currents are I
..
,.-
\
I, = I ro\ea,ett*t - 1)+ u^l ^o
= 1 . 7 1 5 x 1 0A r L .
, . -
\
I c = dr I rope'utr' - l)+ I *o
= 1 . 7 1 5 x 1A 0 {. Note that thesecurrentsare almostthe same(no basecurrent) for WfLo
22.
Refening Eq. 11,the field-freesteady-state continuity equationin the collectorregion is
_ o. D-ld'",y)l_n,(,')-,0" "L t, dtz )
The solutionis given by ( L, = ^lD{, "r(r\
)
= gr"''/r' *Cre-''lLc
Applying the boundaryconditionat x'="o yields
.
Cre4r,+Cre-41.=0 . Hence Cr = 0 . In addition,for the boundaryconditionat x'= 0 , Cre-oltt :C,
=nc(o)
n.(o) = nroQ*'"ln -|]1. The solutionis ulw - tp-'l t" n, &) = nro (eav" The collectorcurrentcan be expressedas
,,=n(-no,*|.=.J.,no,#1. =,) [rr Qtrc,r- r)
=qAL#(*il mt*r- t- r^(2"t-. "f) (f+vcnltrar, (dr* lkr - l) - ou 1) .
23. Using E,q.44,the basetransittime is given by
=$ji# tu=w'/2D,
=t.25xro-ro s.
We can obtainthe cutoff frequency: f, =lf2m u:l'27 Grrz From Eq. 41, the common-base cutoff frequencyis given by:
.f" =.f,I ao=#
=1.27 SGHz .
The common-emittercutoff frequencyis ,fp =(t- a)f" =(t - o.f8)" 1.275x10e=2.55MHz . Note
that
"fp
can
f B = (t - uo)f" =(t - aoYoo * .f, =*
Po
f, .
be
expressed
by
24. Neglectthetimedelaysof emitterandcollector,thebasetransittime is givenby t ,"= - J 24,
2nx5 xl}"
= 3 1 . 8 31x0 - " s .
FromEq. 44,l/'canbeexpressed by
w=^lr4r, . Therefore,
w=ffi =2.52x 10-5cm
= 0.2521"rn . The neutralbasewidth shouldbe 0.252 pm. 25.
L,E, :9.8Yoxl .I2 = 110 npV.
( tn-\
D,-.wl^"il \ , ., : ,) . . .B , ( t , o o " ?
(ttomev ll0mev) :0.2g.
7F-q-exp[
373r
mk )
26 ffi="*(!o#"J="*L 0.02s9
Eru(x) -1.424
where
E rr@) =1.424+ 1.247 x, x < 0.45 = 1 . 9+ 0 . 1 2 5 x+ 0 . 1 4 3 x 0 , . 4 5< x < l . The plot of B,(HBT)/F,
(Bni
is shownin the following graph.
t.e.o.tesx*d.r/t3*
x
Note that B'(IST) 27.
increasesexponentiallywhen x increases.
The impurity concenhationof the nl region is t0racm-3. The avalanche breakdownvoltage(for w rW^) is largerthan 1500V (ry^>l00pm). For a reverseblock vottageof 120V, we can choosea width suchthat punch-through occurs,i.e., v ,
D T
- qNoW' 28, -
-
-
Thus,
*-(#l=r.nuxro-3cm. Whenswitchingoccurs, a,r+a,r=l
That is,
o,=osp'(*) = t - 0 . 4= 0.6 . t J l url | t r I \"0,/
25x70-a = 0.6 0.5 39.6x l0-u =1.51
Therefore. J = 4.5J0= 2.25x l0-5 A/cm2
Area:
-/ ' = "l x" "l 0 - l - = 4 4 . 4 c m t J 2.25xlo-5
28. In the nl -p2 - n2 trarsistor,the basedrive currentrequiredto maintaincurrent conduction is
1,'= (l - u")I* . In addition,the basedrive currentavailableto the nl - p2 - n2 transistorwith a reversegate currert is I, = qI n - I r. Therefore,when use a reversegate current,the condition to obtain tum-off of the thyristor is given by Ir 1I, or qIn-Is
,
-
t s
q+4-1 , o q
/ - t A
'
Note that if we define the ratio of llto 1" as tum-off gain, then the maximum tum-off gatnh* is
% F * = oc,+
, a" _l
.
CHAPTER 6
1.
Ec Ep Ei
\ ...-.................-. - - -
METAL
OXIDE
n-TYPE SEMICONDUCTOR
Ev
2. Ec Ei
Ee Ec
Er Ev
Ei
Ev
3. n- POLYSILICON OXIDE
p-TYPE SEMICONDUCTOR
Ec Ei
Er Ev
-l a",l v
n- POLYSILICON OXIDE
p-TYPE SEMICONDUCTOR
4.
(a)
E (x)
(b)
v6)
5. W, =2
J "n * 8 8 5 x rxoo' o 2\ 6 39 .h +6 xf5l I 0 ' ,
_ - ' .| ,J1
lixtorv"*r;" : 1 . 5x l 0 - sc m : 0 . 1 5I m .
6.
tot
C-,"
d +(e,, /e")W^ W^ =o.l5p m FromProb.5
i'C^in
7.
-ra 3.9x 8.85x l0 = 6.03x l0{ F/cm2. 3'9 *1.5x10-5 8 x l 0 - 7* I1.9
:
Vn =t!n!o:o.o26h q
ni
,"=4atintrinsic e"
to"
= =0.42y
9.65x l0'
v"=vn = 0 . 7 4 x1 0 - 5 c m
_ l . 6 x l 0 - r ex l 0 r 7x 0 . 7 4 x 1 0 - 5 l ' 1 1x l o ' v / c m 1 1 . 9 x8 . 8 5 " 1 0 F = . 1 " ' t' o : 3 . 3 8 xl05 V/cm E o : E r € " : 1 . 1 1x l 0 5 x EJ
3.9
€o*
V: Vo+yr,: Eod + y " :(r rt x lo5x 5 x to-?)+0.+2. :0.59 V.
8.
At the onsetof stronginversion,V ":2V n )
thus, v6=$*r, co
V6:V7
u
F r o mPr o b5. ,W* :0 .t5 fg m,y u :0 .0261n[- t,!to' lr ]:O.O u [9.65x 10'J L o :--------:10*
: 3.45x I 0-7F lcm2
. r / - 1 . 6x 1 0 - t ex 5 x 1 0 1 x6 l . 5 x l 0 - 5
. . v c -
3.45x l0-?
+{.8:0.35+0.8:1.15V.
=ryfft)* s. Q.,=: [:,- yq(rl,, >0,
rc',x ( 1 0 - u) ' l
:8x10-eC/cr* A,V.. =
10.
QO'
co
-
8 x l o - e- = 2 . 3 z x l o - ' v. xl0-' 3.45
l r a
Q* =i lo rp",Q)dy po,=e x5x10tt{x),
( w h e r e6 ( _ r =) 1l @ '
10,
/ = 5xl0-7 y * 5xl0-7
I .'.O^," 5 x 1 0 " x 5 x 1 0 - ' x. l6x 1 O r n 10' :4x10-8C/cnf 4x10-8= 0 ' 1 2V :'LV'o-Qo'C, 3.45x10-'
:+
x 1 . 6x t O - 'xnl " S " 1 0 " 1 1 0 * ; , .
10-"
I 1. Q",:;
I
3
rlo{
y ( q x 5 x L o 2 x3 y ) d y l, :2.67x10'8Clcn?
: ' L V " - Q o ' - 2 ' 6 7 x 1 0 - B= 7 ' 7 4 x 1 0 - 'V ' C 3 . 4 5x 1 0 - '
=% 12. LVr, fD C
= +"4x C
d
- lt{,-!'vou*c"q
NM N, is theareadensityof'e*. ^,where ) 0 ' 3 - x J ' + } 1 0? = 6 . 4 7 x 1 0 ' c, r n 2 . 1.6x10
13. Since Z, << (Vo -Vr) , the first termin Eq. 33 canbe approximated as 7
lt,C"(vo -ht/ u )Vo. I PerformingTaylor'sexpansionon the 2ndtermin Eq. 33, we obtain ( v o + 2 t y u) t , , - ( 2 V u ) t , , = ( 2 V u ) t , , + ) { z w u ) r , ' v o - ( 2 V u ) r , , = } f r w , l , r v o Equation33 can now be re-writtenas
- zT 'l ' tn' '{'" = t ' l - -l*" .E 'n\'*'1I'^ Lo L rl =(|)p,c"(vG -vr)vD
where v,=ayu.JO'tTy 14. Whenthe drain and gateareconnectedtogether,Vo =Vo andthe MOSFET is operatedin saturation(Vo > Vo"*) . I o canbe obtainedby substituting V, =V^* in Eq.33
= n, "{W 2v,u)v^", I ol,o*o 1 {ry)[,""^
-rfv)'''l +2vru)'''
where Vo,o,is givenby Eq. 38. Insertingthe condition Q,(y = L)=0 into Eq. 27 yields VD"o,= ^l2e"Own(Vo,* *2Vr) +Attu-Vc =0 .L , For Vo,o,<<2eu, the aboveequatiohreducesto
-J'4@J vo=vo " c o
=v, . +2tyu
Thereforea linearextrapolationfrom the low currentregionto I o =0 will yield the thresholdvoltagevalue.
| ).
kT. .N, " = o.o26rn u I s to'u--l=o.oo V =-ln(-) q
n
i
/-------:-;Y:''lt'QNn C
1 9 . 6x5l 0 - ' J
3.45x l0-7
:s-0.8 +(0.27)' Ir-l 1 +' -
--
I ,l^ (o.zt),'
:3.42V 0 --? -vr) _ 1 0 x 8 0 0 x 3 . 4 5 x 1 (s =zt!:l' vo r D"o, 0.7)' \ U 2 L
2xl :2.55x10-2A.
16. The deviceis operatedin linearregion,since V o = 0 . 1 Y < ( V o- V r ) = 0 . 5 V a l t T =f Therefore, r, : F,C"(vo- v r ) frl n=*^,
:5 * 5 0 0 x 3 . 4 5 x 1x 00 -. 5' 0.25 : 1 .7 2 x10S.3
17.
oln I = z c,vo g. = frlvpuo,"t Tlt 5 x500x3.45x10-'x0.l 0.25 :3.45x10-4 S. -
lol?
1 8 . V, =Vru + Ay u V,o=6
, V n- = 0 . 0 2 6 t { - - : i - - ) 9 . 6 5x 1 0 " -
o " - 7E"- u n -
-:!-=
c"
1 . 6 x 1 O - xt e5 x 1 O t o
3.45 x10-7 = - 0 . 5 6- 0 . 4 2 - 0 . 0 2 : - l V
"'Vr = -1+ 0.84
2 . l l l . 9x 8 . 8 5 xl 0 - ' ox 1 0 " x 0 . 4 2 x 1 . 6 x 1 0 - ' n 3 . 4 5x l 0 - ?
: - 1 +0 .8 4+0 .4 9 : 0 . 3 3V (Q., canalsobeobtained fromFig.8 to be- 0.9SV).
t9.
nF
a'B 3 . 4 5x l 0 - ? 0.37x 3.45x 10-? =8xlOttcrn2 FB _ 1 . 6x 1 0 - ' e
0.7= 0 . 3 3 +
20. f., = -ti*U/a =-0.56+0.42=-0.14V
v,=e.,
?
"W - 2.[€JN 2rru (-
x 8.85x10*'axl.6x 10-'t xl0t7 x0.42 = -0.14- 0.02- 0.84- h?;,:.lIl.9
: - 1 . 4 9V
3.45 xl0-7
21.
-0J : -1.4g*
nFu = 3.45xl0-'
, ," - o ' 7 9 x 3 ' 4 5 x l o r = l . 7 x l o ' ' c r n 2 . l . 6 xl 0 - ' ' 22.
The bandgapin degeneratelydopedSi is aroundleV dueto bandgapnarrowingeffect.Therefore, Q ^ "= - 0 . 1 4 + 1 = 0 . 8 6V . ' . V , = 0 . 8 6 - 0 . 0 2 - 0 . 8 4- 0 . 4 9 = - 0 . 4 9 V .
z).
to'' -]:0.0, u v. :o.o26r[ [ 9 . 6 5x l 0 ' /
Vr=Q^" ?+2ryu+'ry = - 0 . 9 g- l ' 6 x 1 0 - t ex l 0 t t * 0 . * O co
co
l 5 ' 2x l 0 - 8
= 4.14*
-
,v
co
3 . 9x 8 . 8 5x l 0 - ' o d
-
3 . 4d5x 1 0 - ' 3
v r> 2 0 =d . l l t l t 9 , ' > 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 Vp
0. By neglectingthe
#=(+-T)"*l t)
-+)=, #=(+.#).*[' . ' . V- 2 V p= 2 x 0 . 1 = 0 . 2 V
-0 = 0367 #|"'.=ifr#) %#]"4' -27.2,,. ^=(#1,,")-'=
5(a) *":l ( n a*=+t:&d-=#* |
_
,\2
(l2xl0-,1 2 ( 5 x l } - a ) 1 . 0 5x 1 0 - ' ' x l 0 '
= 1 3 7{ l
(b) The breakdownvoltagefor Np: 101scrn3and W: 12 pm is 250V (Referto Chapter4). The voltagedue to {sc is 1 R r .= ( t O '* Sx t 0 ' ) x 1 3 7= 6 8 . 5V The totalappliedvoltageis then250+ 68.5:318.5 V. 6. (a) The dc input power is l00V(10-'a1: tOW.For 25Yoefficiency,the power dissipatedas heatis I0W(l-25%):7.5 W. L T - - 7 . 5 Wx ( 1 0 ' C / W =) 7 5 ' C ' = 4 . 5V ( b ) A V B= ( 6 0 m V / " C ) x 7 5C The breakdown voltageat roomtemperature is (100-4.5):95.5V.
7. (a) For a uniform breakdownin the avalancheregion,the maximumelectricfield is E^:4.4 xld V/cm. The total voltageat breakdownacrossthe diodeis
4 =E,xu.[--f), w-xn)
= 4Axtd(0.+xt on)*(+.+. rO-
1.6xl0-'ex 1.5x1d2
1.09x10-"
3-0.a)x 10*
)r
=17.6+57.2=748V (b) The averagefield in the drift regionis 572 =Z.2xl05V/cm ( 3 0 . 4 ) xl 0 - " This field is high enoughto maintainvelocity saturationin the drift region. u' lot. (c)f =, ,= , , =l9GHz. 2ltr-xn) 2(3-0.4)10-4
8. (a) In thep layer s r ( x ) :r ^ - q y ' t
o
E2(x)+.^-W
b< x< Il:
E"
q
t21tm
E2(-r)shouldbe largerthan 105V/cm for velocity saturation " ' E ^- Q N ' b ' 1 g ' ts
o r E m > 1 0 5+
qN'b
q
= 1 0 5+ 4 . 6 6 x 1 0 - " N 1 .
This equationcoupledthe plot of E, versusN in Chapter3 gives Nt :7 t 1015 cm3for E^: 4.2 x lOsV/cm . . . vo *= ( e .
-er)b x3xl0{ +E2w:(4.2-l)x105 +' -l ' "s x g x l 0 " 2 z
:138V (b) Transittimet :
W-b _(12-3)x10-: Qxlg-rs: 90 ps . ;=ffi
9. (a) For transit-timemode,we requiren6L > l0rz cni2. no xl}tz / L =lotz /1x lor - lorucm-' (b) r: L/v: I}a / 107- l0-rrs: 10ps (c) The thresholdfield for InP is 10.5kV/cm; the correspondingappliedvoltageis
v =(to's:to')(,',0-')=o.52sv [
2
)
"
The current is I = JA= (qnry, )a =(t.A" I 0-tex 4600xl016x 5.25x 10,)x 10{ : 3.g6A The power dissipatedin the deviceis then P=IV=2.02W.
10. (a) Refeningto Chapter2, we have
Ncu:r(ry\, =*,,(;), = 4 . 7 x 1 g , r (t ' z * o ) ' = o . r x 1 0 r 7x 7 l = 3 . 3 3 xl 0 , , c m - . \0I7m0 ) (b) For T":300 K NCU
_ 0'3
lev ) - t, lkq): ztxexn[ffiff (- 1r.e7) Ncz-exp(-aE )=rtxexp o u n , . - AE t L , r , \ - ? r . , ^ - . ^ (
:4.4x104 (c)ForL:1500K
-0'3iev
Ncu
o-^(-,rt]It k L.r\ = 7| . ^-,J exP(-Ar: "*pl T) --r., .rr/.r'oxlpoo
)
l :oo))=
7| " "*p (- n94)
= 6.5 Therefore,at T" : 300 K most electronsare in the lower valley. However,at T" : 1500 K , 87Yo,i.e.,6.5/(6.5+D,of the electrons arein the uppervalley. I l.
The energyE" for infinitely deepquantumwell is En=
h"
n'
8 m't
:+ l*:1#e'x',)l LE,=4*
L .'. LEt = 3 meV,
LE2= 11meV .
12. From Fig. 14 we find that the first excitedenergyis at 280 meV and the width is 0.8 meV. For sameenergybut a width of 8 meV, we usethe samewell thicknessof 6.78 nm for GaAs, but the barrierthicknessmustbe reducedto 1.25nm for AlAs. The resonant-tunneling currentis relatedto the integratedflux of electronswhoseenergyis in the rangewherethe transmissioncoefficientis large. Therefore,the currentis proportionalto the width A,En,andsufficientlythin baniersarerequiredto achievea high currentdensity.
CHAPTER 9
= I .24/ 0.G 2.07eY (F romEq.9) 1. fw (0.6gtllrr) oc(0.6pm):3xlOa crnr The net incidentpoweron the sampleis the total incidentpowerminus the reflected power,or 10 mW I
10"[
,
\
_ e - 3 , 1 0 ' t {=J 5 x t O r
W:0.231 urn The portion of eachphoton'senergythat is convertedto heatis hv - E, _2.07-1.42 =3l.4yo hv 2.07 The amountof thermalenergydissipatedto the latticeper secondis 3 l . 4 Y o x 5 : 1 . 5 7m W .
2. For l": 0.898pm, the conesponding photonenergyis l'24 = E = 1.38 eV h FromFig. 18,we obtain"r 1arc.fCa0.7As):3.38 -+
sinQ = =+= 0.2958 ) o"= 17"12' . fiz 3.38
4n'n'(r- c?:o")- + x t x g'ls[t- c9!(lz' t z')] Efficiency(1+3.38f @,*fi,)' = 0 . 0 3 1=53 . I S Y.o
3. FromEq. 15 /
\ 2
_ f3.39-ll=0.294 R=l-l \ 3 . 3 9+ 1 / (a) The mirror loss
t,
fill)= L
\R/
h,,f-l-')=40.58cm-,.
300x10-" \0.296)
(b) The thresholdcurrentreductionis J,o(R= 0.296)- J,o(R,= 0.296,R,= 0.9)
lo*!r l,f1)l-l o*rr[--r-)l 2r \R/J L
_L
[R,R,/j
o*!"fr') L \Ri
t t 40.58) ,hf 2 . 3 0 0x l 0 - - \ 0 . 2 9 6 ' 0 . 9 0 / _ l 0 + 40.58 =36.6%o.
4. From Eq. 10 s i n @= \ - i r = i r . s i n O . n2 From Eq. l4
r = I - exp(- c Li d)= t - ""p(- 8 x r0' . 3.6(1- sine"). 1x 10 * ) F o r4 : 8 4 o n r : 3 . 5 8n : 0 . 7 9 4 Q : 7 8 on z : 3 . 5 2 r ; : 0 . g g S.
5. From Eq. 16we have mL=2iL. Differentiatingthe aboveequationwith respectto), we obtain ndm ^,dn /1e-+m-zL_dL d^
SubstitutingZitll"
for m andlettingdmldJ,: - Lm/L2,yield
{-}!\*z! 1 =2LE il" \^2)
fttm
:. a,).=
znrlr-(1\(g\l |
\n )\dL))
and
6. From Eq.22
r:!I..|'(fl
andEq 15R=(=)'
Rr : 0.317,R2: 0.31I
g(a+" )=#a["'.#'(#)] g(zs' )=#*['*. **o"'(#)] t rnf '
)=
'
=270 =zt.
hfl)
2L' (R.0.99l t00xl0-a (R/ For Rr :0.317
*^Gi
"*l
---1-_-'(#), L,-50 pn
F o rR 2: 0 . 3 1 1
#'["#-)
7. FromEq.23a
=-*-'(#)
>Li=50431t' t
ForR :0.317
' r J,o=ro. * , h( )l= ,oooNcm', ' ftoo 2 x l 0 0 x [ 0 ' \ 0 . 3 1 7 x 0 . 9 9 ) ) andso I,n : 1000x100 xl0-o x5x 10+ = 5 nrA F o rR : 0 . 3 1 1 64 : 0.I 3 FnI]> Fz: 0.1x0.998/0.7
J,o=7.66,.[roo L
h(
|
)'l = ,uu Alcm-, '
2xl00xl0-* \0.311x0.99))
andso I,a =766 x 100x 10* x 5 x 10-o= 3.83rnA .
8. From the equation,we have for m:0:
4 t 4 n L 4 + 4 n L L= o
(2sa)
which can be solved as
(25b)
There are severalvariationsof + in this solutions.Take the solution which is the only practicalone,i.e.,fuBg fuo,givesl"s :1.3296 or 1.3304pm. d-
1'33 =0.196 rrn 2x3.4
9. The thresholdcurrentin Fig. 26b is given by I*: Io exp(Zl 10). Therefore
,oE ' - = J-dI I,o dT
I = o.oo9 r (, c)-',
llo
If To:50 oC,thetemperature coefficientbecomes
1
t-= ! =50 . 002 ( ,\C ) '/. which is largerthanthat for To: 110oC.Thereforethe laserwith To: 50 oC is worse for high-temperature operation.
10. (a)Atr: QQt"+ 1:n\enA M
-'-Ln =electron-holeoairs=,
.
oltt,+ prb,a
2 . 8 3 xl 0 - '
-
=tu
cill
l.6x l0-''(3600 +t700{10/0.0)(zx tx l0-' ,f 3 xlo 0) ? - 2'83 =l2ous 23.6
(c) nn(t)=^n exp(-tlc)=r0""*n[-#h]
=2.5x10e cm-3.
l l . F r o mE q . 3 3
.o" ro-''.sooo to* rooo rP_=n( ) ' \ 'o.ss. ).I 3"q)\
t0xlO-o
)
= 2 . 5 5 x 1 0 - u= 2 . 5 5 1 t A andfrom Eq.35 3000.6 x l0-'o .5000 (raln= tt-E =--=t ' r 1o; r-,
12. FromEq. 36
=fglP\=of41) ,' :( ( qL\('--l' )\nvl lP.o,)\q) \q) The wavelength2"of light is relatedto its frequencyv by v = c/L wherec is the velocity of light in vacuum.Thereforehv /q: hc I 2q andh: 6.625x10-34 J-s,c : 3.0xl0l0 cmls,q : 1.6x10-re coul.I eV : 1.6xl0-reJ.
Therefore, hv/q:1.24 / ),fum) Thus,4: (Rx1.24) / LandR: (rZU/ 1.24.
13. The electricfield in the p-layeris givenby r , ( x ) = u ^ - T ' N tot< x < b , €" whereE,nis the maximumfield. In the p layer,the field is essentiallya constantgiven by r, (x)=,^ -9!t!-
b
ts
The electricfield requiredto maintainvelocity saturationof holesis - lOs V/cm. Therefore qN'b E- -JI-J" > 16s €r
or E . > 1 0 5* 4 1 ! ' b = 1 0 ' + 4 . 6 6 x 1 0 - , , N r . q
From the plot of the critical field versusdoping,the correspondingEm areobtained: N1 :7xl01s cm-3 E^:4.2x105V/cm The biasingvoltageis given by
=":,q*-lo*) +ro,(rxro-o) ,, =fui& *E2w = 138 V. The transittime is
, - (w u)_n1lg* =exro-,,=eops. vs
107
14. (a) For a photodiode,only a naffow wavelengthrangecenteredat the optical signal wavelengthis important;whereasfor a solarcell, high spectralresponseover a broad solarwavelengthrangeare required. (b) Photodiodeare small to minimizejunction capacitance, while solarcells are large-area device. (c) An importantfigure of merit for photodiodesis the quantumeffrciency(numberof electron-holepairs generatedby incidentphoton),whereasthe main concernfor solar cells is the power conversionefficiency(powerdeliveredto the load per incidentsolar energy).
15 (a)1"=AqN,N,(+^E.!^T-)"-",u '[N, Nolto 1C
)
= zft.sxto-'n[z.soxto'nfu..66x 10'n )x
(= = t * . 8
*,
=^ E ) , s - ,-, , " , 1 r ,
I t . z * 1 9 t e ! 1 9 -5sx r o '1es * t o - 'J
= 2.43x to2o(s.al * 1g+p-+t z = 2 . 2 8 xl 0 - ' ' A I:1,Q*lr')-
t,
lll=tr-1"(etvtr' -r1 0.3
0.4
0.5 0.6 0.65 0.7
2.5x10-1.1x10-0.55 26.8179 /
1
95
95
94.5 68.2 -84
nt" 'o'.-)= o.u,u (b)v^.=u^( !-\= 0.025e*[ q
[1" ,
(c) P= I,VQevrw-t)-trrt
\.2.28x10-")
1520
(mA)
dP
o = 1"(rev/t'- t) * t,
dV
#enrlr,
- I,
I,etvlkr (l + tt) - t, IL
.'. eon/o' -
1" (1 + v ) V
m
= 0.64 V r
I
P^ = I,V^ = I t l v o ,
_t!n(r.sL\_tr1 q
L
\.
kr)
q)
- 0.025e = 9sx 10-3 n(t + z+.t)- 0.}zssl [0.68 =52 mW
1 5 0
16. From Eq.38 and 39 I = I,("0,0, and
-t)-
t,
t r . ( r ", + l. l)= _rl rr l .l ( r-, \|
y_ -_nl
q
I,
\1"
= It€-qvlkr
) -
q
1.1"1
= 2.493 x 1 0 - t 0 A 3. e-o6fao7ss5
I = 3 - 2.493 xlO-ro . evloo2sssand P : I V
V
I
P
0 3.000.00 0 . 1 3 . 0 0 0 . 30 0.2 3.000.60 0.3 3.000.90 0.4 3 . 0 0 20 0.5 2.94 .47 0.512.91 48 0 . 5 22 . 8 6 4 9 0 . 53 2 . 8 0 4 8 0 . 54 2 . 71 0 . 55 2 . 5 7
2.00 a
1.50
€ I
B 1.00 0.50 0.00 0.5
46 4l
V (volts)
0.6 0.00 0.00 .'.Maximumpoweroutput: 1.49W Fill Factor pp:I^V^ = P- = t.4 9 IrV* I,V* 0.6x3 = 0.83.
17. FromFig. 40 The outputpowersfor &: Pr(&:0):95
0 and &:
5 C)canbe obtainedfrom the area
m A x 0 . 3 7 5 V : 3 5 . 6m W ,
Pz(&: 5 C)): 50mAx0.l8v:9.0 mW .'.For4,:6
Pr/ Pt : L00%o
F o r & : 5 O P z lP t : 9 / 3 5 . 6 : 2 5 . 3 Y o .
18. Theefficiencies are14.2o/o (1 sun),16.2%(1O-sun), 17.g%(100-sun), and 18.5%
(1000-sun). Solarcellsneeded underl-suncondition
ion) x P,,(concennation) 4(concentrat
rfi - sun)* P,,(l - sun) -16 ' 2 Yo x l o= 1 1 .4 ce l l s fo r1 o -su n xl 14.2Yo
=ll,Y|19 =ns cellsfor loo-sun
14.2o/o xl - l8'5% x 10= 1300 cells for 1000- sun. 1 4 . 2 Yxol
CHAPTER 1O
l . C o : 1 0 1 7c r n 3 /.g(Asin Si):0.3
Cs ftoCo(l - 1,11M0)*-t : 0.3x1017(1x)-oz: 3x 101641- il50)0.7
0.2 /(cm)
l0
Cs(cm-3)
3 . 5 x1 0
0.4
4.28x10
0.6
0.9
30
45
5 . 6 8 xl 0
16 14 o -
10 b
8 6
2 0
2. (a) The radiusof a silicon atomcan be expressedas
It
f =-g
8 t;
VJ so r =-x5.43=1.175A 8
(b) The numbersof Si atom in its diamondstructureare 8. So the densityof silicon atomsis 8 o 3 = 5.0x 1022 n =---==---+ atomVcm o' (5.434)3 (c) The densityof Si is
"
M / 6.02x10 28.09x5x l0 = -ffi O = -ff
22
gl cm3:2.33g/ ctlf.
3. ltu:0.8 for boronin silicon M /N4o:0.5 The densityof Si is 2.33g / ctrl. The acceptorconcentration for p : 0.01Clcm is 9x 1018crn3. The dopingconcentrationCs is given by Ilf
C =k^C^(l--' 1&-t M^'
Therefore C"
-o ------------7--
k"Q-!)a'
9x10" 0 ' 8 ( 1 -0 ' 5 ) - o '
: 9 . 8x l 0 t t c m - ' The amountof boronrequiredfor a 10 kg chargeis 10'000 x 9.8x l0'' = 4.2x|0" boronatoms 2.338 So that
'A#ffi,=o.7sg ro.8g/more
boron .
4. (a) The molecularweightof boronis 10.81. The boronconcenhationcan be siven as numberof boronaioms volumeof siliconwa fer 3 / 1 0 . 8 1 96x . 0 2 x1 0 ' 3 _ 5 . 4 1x 1 0 - g 1 0 . 0x23 . 1 4 x 0 . 1 3 = 9.78x l0r8atomVcm
/ I"^ = -
(b) The averageoccupiedvolumeof everyoneboronatomsin the wafer is
tt =L =---l -cm3 nb
9 . 7 8x l 0 ' "
We assumethe volumeis a sphere,so the radiusof the sphere( r ) is the averagedistancebetweentwo boron atoms.Then
tv
r =^l-' =2.9xI0-tcm. \4n 5. The cross-sectional areaof the seedis - ' - - )' / o.ss | =0.24cm2
7d
( 2 )
The maximumweight that can be supportedby the seedequalsthe productof the critical yield strengthand the seed'scross-sectionalarea: ( 2 x 1 0 6 ) x0 . 2 4= 4 . 8 x1 0 5g = 4 8 0k g The correspondingweight of a 2O0-mm-diameter ingot with length/ is
(2.33!cm' ),{ry)' / : 48oooo s \
a z
l
/
I = 6 5 6 c m= 6 . 5 6 m .
6. We have
c,/co=^(-#)^' Fractional 0 solidified
cslc0
0.2
0.05
0.8 1.
0.06
0.08
0.t2
0.23
rjl E
I
ru. tm_ m - m q _ m _
!
lJ-Sflo"mm
7 . The segregation coefficientof boronin siliconis 0.72.It is smallerthanunity, so the solubility of B in Si undersolid phaseis smallerthanthat of the melt. Therefore,the excessB atomswill be thrown-off into the melt, then the concentrationof B in the melt will be increased.The tailend of the crystal is the last to solidify. Therefore,the concentrationof B in the tail-end of grown crystalwill be higherthan that of seed-end. The reason is that the solubility in the melt is proportional to the temperature,and the temperatureis higher in the centerpart than at the perimeter. Therefore,the solubility is higher in the centerpart,causinga higher impurity concentrationthere. 9 . The segregationcoefficientof Ga in Si is 8 x10-3 FromEq. 18 C" / Co = 1- (1- k)"-o't We have
= 2501n(1.102) =24cm.
10. We havefrom Eq.18 Cs = Co[ -(l - k") exP(krxI L)] Sothe ratio Cs/C0 =[1-(1- k")exp(4rx/Q] : 1- (1- 0.3). exp(-0.3x 1)= 6.52 : 0.38 at x/L:2.
atxlL=l
11. For the conventionally-doped silicon,the resistivityvariesfrom 120C)-cmto 155f)-cm. The corresponding dopingconcentration variesfrom 2.5xl0t' to 4" 1013crn3.Thereforethe ranseof breakdownvoltagesof p* - n junctions is given by t E 2 ' v,=.t(il,)
- 1 . 0 5 x 1 0 -x" ( 3 x 1 0 5 ) (2N r ) - , = 2 . 9 x 1 0 ,/,N B = 7 2 5 0t o l 1 6 0 0 V 2xl.6x 10-'t LVB= 11600-7250 = 4350V
(n v ^ \
-. =+30%o | ll7250 \ 2 ) For the neutronirradiatedsilicon,p: 148+ 1.5C)-cm.The doping concentrationis 3xl0l3 (tl%). The rangeof breakdownvoltageis v B = 1 . 3x l 0 t 1/ N B = 2 . 9 x r 0 t 7/ 3 x 1 0 1(3t l % s =9570to9762Y. LVB-9762 -9570=192V
( tv"\ l-^a ll9570=*tyo. \ 2 ) 12. We have M , _ w e i g hot f G a A s a t T o_ C ^ - C , _ s Mr weight of liquid at To C, -C^ I Therefore,the fractionof liquid remained/can be obtainedas following - I = 3o =0.65. "f = , M , M,+M, s+/ 16+30
13. From the Fig.ll, we find the vapor pressureof As is much higher than that of the Ga. Therefore, the As content will be lost when the temperatureis increased.Thus the compositionof liquid GaAsalwaysbecomesgallium rich.
14. n": Nexp(-n" / kT)= 5 x1022er,p(2.3 ey / kz) = 5 x ro" "*p[- {'8--l
L(r/3oo)j
: 1.23xl0-t6cm-3x 0 at 27oC = 300K
15. n, =JNi/
:6.7x10''cm-3
a t 9 0 0 0 C= 1 1 7 3 K
: 6 . 7x l O ' u c m * '
a t 1 2 0 0C0= 1 4 7 3 K .
exp(-y, /zkT)
:W
* " - r . t e v t 2 k r= 7 . 0 7 x l 1 t o x e - t 0 . 7 / ( r / 3 o o )
: 5 . 2 7x 1 0 - ' t a t 2 7 o C : 3 0 0K :2.l4xl0ta at 90fC : 1173K. 1 6 . 3 7 x 4 : 1 4 8c h i p s In termsof litho-stepperconsiderations, there are 500 pm spacetolerancebetweenthe mask boundary of two dice. We divide the wafer into four symmetrical parts for convenientdicing, and discardthe perimeterpartsof the wafer. Usually the quality of the perimeterpartsis the worst dueto the edgeeffects.
300 nrn
Totot Dies, 148
I
I
rt
ll vf,dv tykr l_
=ji_-
ff f'o'
\l781/t
4 ( M \ ' ' ' " ( =ro, where v'"*o[ M ! \ n =Glfr) I M: Molecularmass k Boltzmannconstant:1.38x10-23 J/k T: The absolutetemperature v: Speedof molecular So that
'*= "
16. L=
0.66 P( in Pa)
2
J;
2 x 1 . 3 8 x 1 0 -x2330 0 29xI.67 xl}-"
m/sec= 4.68xlOncm/sec.
;. P 19.
o'!u = 0'664.4xto-3pa. L 150
For close-packing affange, there are 3 pie shapedsections in the equilateral triangle. Each section correspondsto 1/6 of an atom. Therefore .
number of aloms contained intre tiangle ^. _ = '1Y."
areaof fire hiansle
l
JX_
6
-
1 d , - c l x -J1 2
2
2
=_-
Jia'- -Jlg.as"ro*)'
:5.27 xlOto atomVcmt.
20. (a)Thepressure at970C (:1243K)is 2.9x10-tPafor Gaand13pa for Asz-The arrivalrateis givenby theproductof theimpringement rate andNnL2 : Arrivatrate: 2.64xto2o(-LY4)
l"lvt )\'t' 1
:2.9x101sGa molecules/c#-s The growth rate is determinedby the Ga arrival rate and is given by (2.9xI 0rs)x2.81 (6xt0ta) : I 3.5 A./s: 8 I 0 A"imin. (b) The pressure at 700"Cfor tin is 2.66xl0-6Pa.The molecularweightis 118.69. Thereforethe anival rate is
2.64xrc^( ]j54Y-ll
\ " / l 1 8 . 6 x9 9 7 3. / \n x 1 2 ' )
=2.zlxt0,0molecul art cm,.s
If sn atomsare fully incorporatedand activein the Ga sublatticeof GaAs, we have an
electronconcentration of ( z .ztx -to- 'o)(4" .4 ' - "2 x^to "" ) I t-t-
ll
(z.e"lo',J[ 2
| = 1 . 7 4 x 1 0 t tc m - ' .
)
21. The.r valueis about0.25,which is obtainedfrom Fig. 26. 22. The latticeconstants for InAs, GaAs,Si and Ge are6.05,5.65,5.43, and.5.65 A, respectively(AppendixF). Therefore,thef valuefor InAs-GaAssystemis = _0.066 f = (5.65- 6.05)/6.05 And for Ge-Sisystemis = -0.39. f = (s.43- s.65)15.65
CHAPTER11
l. FromEq. l1 (with [:0) x"+Ax: Bt From Figs.6 and 7, we obtain B/A :r.5 pm /hr, 8=0.47 pm2/hr,thereforel= 0.31pm. The time requiredto grow 0.45pmoxide is
t = * G ' + A x ) == l r =' 1 0 . 4 + 50 t . 3 1 x0 . 4 5 ) = g . 7 2 t r : 4 4 m i n . B' 0.47
2. After a window is openedin the oxide for a secondoxidation,the rate constantsare B: 0.01 pmz/hr,A: 0.116pm (B/A:6 'I0'2 pm /hr). If the initial oxide thicknessis 20 nm : 0.02 pm for dry oxidation, the value oflcan be obtainedas followed: Q.0D2+ 0.166(0.02):0.0r(0 +[) or
J:0.372tu. For an oxidation time of 20 min (:1/3 hr), the oxide thicknessin the window areais ,t + 0.166x:0.01(0.333+ 0.372\: 0.007 or x :0.0350 pm:35 nm (gateoxide). For the field oxide with an original thickness0.45 pm, the effective[is given by p:11r, + Ax)=-J-10.+S,+ 0.166 x 0.45)=27.72ttr. ' B' 0.01 ,2+ 0.166x: 0.01(0.33 3+27.72):0.2g053 or x :0.4530 pm (an increase of 0.003pmonly for the field oxide). 3.
*+Ax:B(r+c) G ' * !21 ' - 'A ' 4 - B ( t + r \
s*!t' =tl#+tr+zl] w h e n f2 ) t t t r u
A' . 48,
then,x2 :Bt similarly, w h e n /s ) r ; t u ,
At . 48,
then.x: !-ft*c\ A '
4. At980 1:1253K)and I atm,,B:8.5x10-tlr^'/lo,B/A:
4xl0'2pm lfu (from Figs.6
and7).SinceI r>2D/k, B/A: kColCr,Co:5.2x1016molecules/cm3 and C, : 2.2xI022crn3 , the diffi.rsioncoefficientis given by
o = L2 = 4 (! . r ' ) = g [ - E - ) 2\A Co) 2\c,)
8.5 x 10-3 2.2x1022 =-?;ffi1tm'tttr
=1.79xl03 1tm2/hr =4.79xlO-ncm2 /s.
5. (a)ForSil{"It si I ,^
- = - = r . !
N
x
x :0.83 o/oH= atomic
looY =20 l+0.83+y -v:0.46
Theempiricalformulais SiN6s3FI0 a6. 3" '' :2, (b) != 5x 1028e-33
10t' [-cm
As the SiA{ ratio increases,the resistivity decreasesexponentially.
6.
SetTh2O5thickness: 3t, e1:25 SiOzthickness: t, ez: 3.9 : t, E3:7.6, area: A SLN+thickness then cToror:
qqA 3t
t = _+_+_ Co*o Er€oA ereoA er4A I
co*o=ffi, Ct"ro, Co*o
7.
-er(er+Zer) 3Er€,
_ z s ? . g + z x t .= a5 ) .37. 3 x 3 . 9x 7 . 6
Set BST thickness: 3t, e1: 500,area: 41 : t, a2:3.9, area: Az SiOzthickness Si:Nathickness: t, E3:7.6, area: Az then
444, = 3t
o' = o.oonr.
A2
8.
Let :3t" e1:25 TazOsthickness SiOzthickness: t, e2: 3.9 : t, 4:7.6 Si:Nqthickness area: A then qqA _4€oA
d =3t't =0.46gt. q
9 . The depositionratecan be expressed as r: ro exp(-E^/kT) whereE": 0.6 eV for silane-oxygenreaction. Thereforefor Tt : 698 K
r(r,)=2=€XDlo.{ t )I ' L L_
(k4 kr,))
4r,)
, ' z : 0 6 [ f r y _ 3 o oI ] o.o2s9 r, L\6e8
) )
_ Tz:1030K: 757 1 0 . We can use energy-enhanced CVD methodssuch as using a focused energy sourceor UV lamp. Another methodis to use boron doped P-glass which will reflow at temperatures lessthan 900 .
I 1. Moderatelylow temperaturesare usually used for polysilicon deposition,and silane decompositionoccurs at lower temperaturesthan that for chloride reactions. In addition, silane is used for better coverageover amorphous materialssuchSiO-,.
12. There are two reasons.One is to minimize the thermal budget of the wafer, reducingdopant diffusion and material degradation. In addition, fewer gas phasereactionsoccur at lower temperatures,resulting in smootherand better adheringfilrns. Anotherreasonis that the polysiliconwill havesmall grains.The finer grains are easierto mask and etch to give smooth and uniform edges. However,for temperatures lessthan 575"C the depositionrateis too low.
13. The flat-bandvoltaseshift is
LVFBD[l![[
c ^" = t o ' d
QO,
co
3 ' 9 x 8 ' 8 5x l o - ' a =6.9 xl0* F/cm-rI 500x10-o
- Numberof fixed oxide charseis
0 . 5 C ,_ 0 . 5x 6 . 9x l 0 - 8 = 2 . 1 x l 0 r ,c m - , q 1 . 6x 1 0 - ' ' To removethese charges,a 450 heat treatmentin hydrogen for about 30 minutesis required. 14.
20/0.25: 80 sqs. Therefore,the resistanceof the metalline is 5 x 5 0 : 4 0 0O .
15.
For TiSiz 3 0x 2 . 3 7 : 7 l . l n m For CoSio 30x 3.56: 106.8nm.
16.
For TiSiz: Advantage: lowresistivity lt can reducenative-oxidelayers TiSiu on the gate electrodeis more resistantto high-fieldinducedhot-electrondegradation. Disadvantage: bridging effectoccurs. LargerSi consumptionduring formationof TiSiu Lessthermalstabilifv F o rC o S b : Advantage: lowresistivity High temperaturestability No bridgingeffect A selectivechemicaletchexits Low shearforces Disadvantage:not a goodcandidatefor polycides
1 7(.a ) O = e + = 2 . 6 7 x t 0 6 x
=3.2x103Q
0 . 2 8 x 1 0 -xa0 . 3 x l 0 - a
tA
eTL
d
s
3 . 9x 8 . 8 5x l 0 - r ax 0 . 3x 1 0 + x l x l 0 a x 1 0 - 6 = 2.9 x l0 -r' F0 . 3 6x l 0 - a
R C = 3 . 2 x 1 0 5x 2 . 9 x l } - t s= 0 . 9 3 n s
( b )R =p + = r . 7x 1 0 x- 6- . = + A
=2xI03C)
0 . 2 8x 1 0 - ax 0 . 3x 1 0 r
{L 2.8x 8.85x l0-'o x 0.3x 10-ax 1 ( _= -€A =-: 2 'I x 1 0 - ' 3F d s 0 . 3 6x l 0 - o R C = 2 x 1 0 tx 2 . 1 xl 0 - ' 3= 0 . 4 2 n s (c) We candecrease theRC delayby 55%.Ratio:
18. (a)
R = 'eAL = 2 . 6 7 x l 0 *
C =4={Ld .S
o4)
:0.45.
,r;
= 3 . 2x l 0 ' C ) 0.28xlOox0.3xl0o
3 . 9 x 8 . 8 51x 0 - ' a x 0 . 3 x1 0 { x l x 3 =g.7xl0-,rF 0.36x l0o
: 2.8ns. RC: 3.2 x 103x8.7x 10-13
T
1
( b ) R = 'OA: = l . 7 x 1 0 - 6x 4 = 2 x l0rC) 0.28xl0ox0.3xlOo a4 {L d s
2 . 8 x 8 . 8 5 x 1 0 x- '0o. 3 x l 0 { x l x 3 = 6 ' 3x l 0 - t 3F 0 . 3 6x 1 0 a
R C = 2 x 1 0 3x 8 . 7 x 1 0 - t=32 . 5 n s R C = 3 . 2x 1 0 3x 8 . 7x l 0 - ' 3= 2 . 5n s . 19. (a) The aluminumrunnercan be consideredas two segmentsconnectedin series: 20%o(or 0.4 mm) of the length is half thickness(0.5 pm) and the remaining 1.6mm is full thickness(lpm). Thetotal resistance is
* =' Jl AL, . !A"r1) = g * r o -9u'f1 6 , * , o ' o o
-l
[10-*xl0-*l0-ox(0.5x10-')l
:72U.
The limiting current1 is given by the maximum allowed current density times cross-sectional areaofthe thinnerconductorsections:
. / : 5 x 1 0 sN c r * " ( 1 0 - a x 0 . 5 x 1 0 -2a5; :x 1 0 3A : 2 . 5 m A . The voltagedrop acrossthe whole conductoris then
V : N = 7 2 Q x 2 . 5 xl 0 - ' A : 0 . 1 8 V . 20. 0 . 5p m 4 0n m
6 0n m
h:height, W;width, r: thickness, assumethatthe resistivities of the cladding layerand TiN aremuch largerthan p n, and pru ( I =2.7 R,,=D,,* ( 0 . 5 0 . 1 )x 0 . 5 h xW ( =1.7 R . . .= p , . ., - J hxtr (0.5- 2t) x(0.5 -2t) When Rn,= Rru lnen
=
2.7 1.7 =0.4x 0.5 (0.5-2t)' t : 0 . 0 7 3 p m : 7 3n m .
CHAPTER 12
I . With referenceto Fig.2 for class100cleanroom we havea total of 3500 particlesinf with particlesizes>0.5 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 <100>-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.
