Contentsch.l Introductiorr----ch.2 EnergyBandsand Carrier Concentration -----ch.3 CarrierTransport Phenomena ----ch.4 p-n Junction ch.5 BipolarTransistor andRelated Devicesch.6 MOSFETandRelated Devices--------ch.7 MESFETandRelated Devices------ch.8 MicrowaveDiode,Quantum-Effect andHot-Electron Devices ---------ch.9 Photonic Devices c h . 1 0CrystalGrowthand Epitaxy------c h . r l Film Formation-----ch.t2 Lithography andEtching 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 equidistant nearest neighbors that lie at the cornersof a tetrahedron.Therefore the distance between nearest neighbors in 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 area of d, thereforewe have 2/ &:2/ (5.43,. l0-8)z:618 * 10la / crt atoms Similarlywe havefor (110)plane(Fig. 4a andFig. 6) (2+2xll2+4xll4)/JTo2 :9.6,. 10rs a t o m /sc r 1 . , (Fig. 4aandFig. 6) andfor (111)plane (3x I/2+ 3x r/6) / rlz|mlf : ,ffi" I (9. : 7.83* 10la atoms / crrt. 2. Theheights at X, Y, and Z pointare /0, %,^O %. 3. (a) For the simplecubic,a unit cell contains1/8of a sphere at eachof the eight cornersfor a total of one sphere. Ma><imum fractionof cell filled : no. of sphere x volumeof eachsphere / unit cell volume :1x 4ng/2)3 la3 :52o/o (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 sphere at eachof the six facesfor a total of threespheres.The nearest neighbordistance is l/2(a J; ). Therefore the radiusof eachsphere is 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 sphere at eachof the six facesfor a total of three spheres,and 4 spheres inside the cell. The diagonaldistance (112,0, between 0) and(114, ll4, Il4) shown in Fig. 9a is n:1 2 ;)' .(;)' The radius of the sphere isDl}: - 1Jj 8 Maximum fractionof cell filled )) This is a relatively low percentage compared to otherlatticestructures. : (t + 3+ 4) : 34 'o, :nJT t16 %. ' lyErtjl' 13\8 : la,l : la,l:laol :a 4. la,l 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 reciprocals of theseintercepts we 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 latticeconstant for 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 : 5.33g I cni. + 74.92) I 6.02* 1023 (b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors are formed,because Sn 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 crystalorientation determines the semiconductor's chemicalandelectrical properties, suchas the etch rate, trapdensity,breakage planeetc. (d) Thetemperating of the crusibleandthe pull rate. ErQ): l.l7 4.73x10u T' for Si .'. Es ( 100K) = 1.163 eV, andEs(600K): 1.032 eV E,(D= ,n -t'o-l!j_"t!1,!' r.5 ror GaAs ,.Er( 100 K) : 1.501 eV,and Es(600K) : 1.277 eY . 9. The density of holes in the valenceband is given by integratingthe product N(E)tl-F(DldE from top of the valeri c:e band(En takento be E : 0) to the boffom of the valenceband Rottoml (T + 204) 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 (1) Then from AppendixH and, Eqs. I and2we obtain p : 4Df2mp / h2f3D EtD expI-@r - E) / kT ldE I:""^ - @, Eq.3 becomes Letxt+ E lkT, andletEbooo*: xtDe*dx I wherethe integralon the right is of the standard form and equats G tZ. :2l2Dmo - p kT / h213D exp [-(Ep/ kI)j By refeningto the top of the valence bandas ETinstead of E:0 we have p : 4\-2mo / rtflz (kTlttz exp [-(Ep/ kl)i - exp | - F(E) F (n, - r)lwl e) (3) 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 : 1.03 : 9.4 " 10-3 mo. Similarly, we havefor GaAs ffip:3.9x 10-3 k1 g: 0.43 mo. 1 1 . UsingEq. 19 Ei =(8, +til' . (%)^ (N, I w,) = (Ec* Ey)l 2 + (*T I 4) ln 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 10-5 - 0.5g3ev. Z= 0.5g+ 2.54x 10-3 lr*,1*,)(o%f (1) At 300K : 0.56+ 0.009 : 0.569 + (3.29x 10-sX300; Ei: (1.12/2) eV. At 373K - 0.545 x l0-sx3731 + (3.2g Ei: (1.0912) + 0.012:0.557ey. Because the secondterm on the right-handside of the Eq.l is much smaller comparedto the first term, over the abovetemperafure range,it is reasonable to assume thatEi is in the centerof the forbiddengap. "-@-rrYw6B T2.KE : I::@_EC f: JE -Er"-@-tP)/*r6P l'=rr-rr, 1.5x0.5"G 0.sJ; =? ;or. :9.109 x 10-26 x105 1 3 .(a)p: ftw:9.109 x 10-3r kg-mA h - 6'626x10-1 :4 7.27 x r0-e 1 : m:72.7 A 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 [ 1 5 .From E" - Er: kT lnlNc / (No - N,q)] which can be rewritten as No - N.e: l/c exp IaE, - Er) I kf I Then No-N.a:2.86 * 10re exp(-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 compensated semiconductor can be fabricatedto provide a specific Fermi energylevel. 16. From Fig.28a we candraw the following energy-band diagrams: AT 77K Ec(0.ssevi' EF(0.s3) E;(o) Ev(-0.59) AT 3OOK Ec(0.56 ev) 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: Ep- Ev: kTln(N/ND):0.0259ln(2.66 x 10re / l0r5;:0.26 eY (b) The boron atoms compensate the 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: txiexP I - r"' '\ |' \. kT )' we have : 0.42eY Ep - Ei: kT ln (no I n) : 0.0259 ln (1017 / 9.65* 10e) The resultingflat banddiagramis : 1 . 1e 2V l Ec EF A.rr2eY -EF. l crn3. The numberof electrons that are ionizedis given by 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 : 1015 crn3 forNo: 1017 crn3 forNo: 10lecrn3 10tu l+ e-{Eo-Er)/kr t-l 19. Assuming complete ionization,the Fermi level measuredfrom the intrinsic Fermilevelis 0.35eV for 10rscm-3,0.45 eV for 1017 crn3,and0.54eV for 10le n = Npfl - F(En)l: Np / fl + "-(ro-rr)r*r] Usingthe Fermi levelsgivenabove,we obtainthe numberof ionizeddonorsas Therefore,the assumption of completeionizationis valid onty for the caseof 10ls crn3. 20. No*_ . l0tu l 1.t45 : _ 1016 1+e-0.13s : 5 . 3 3 t 1 0 l sc r n 3 Theneutral donor: 10165.33,.l0lscrr3 : 4.67x 1015 crn3 N; - 4'76 - 0.g76 -The ratioof s.33 N; CHAPTER 3 Si, /4,: 1450, I . (a)For intrinsic l+ :505, andn: p : lti:9.65x lOe We have p - 3.31x 10' C)-cm qntth + {lpltp en,(lt, + pr) (b) Similarly for GaAs, lh: We have p - n: p : ni:2.25x106 9200, Lb : 320,and = 2.92x lOt O-crn qnltn + llPIIp en,(ltn+ po) 2 . For laffice scatterinE,l-h n 73/2 T : 200K, Lh: r3oox #:2388 2 cm2lv-s l3oox T : 4oo K, 1^4,: % 300-rt l l l l-t" : 844 "m'lv-s. 3. Since - = - + - P l-t' 1 -1 = - + -1 p 250 500 4. (a) p:5xl01s 14:4lo : 167 cr# N-s. Fr cm-3 r s1 . 8 6 x 1 0 4 c d 3 , n : n / / p : ( 9 . 6 5 xl 0 e 1 2 l 5 x 1 0 : cm2lv-s, Lh:1300 cm2lv-s p: qpon + qlrpp r I qppp :3 C)-cm (b) p :Nt-No : 2 x 1 0 1 6 - 1 . 5 x 1 0 1 6 : 5 x 1 0 l" t*', n: cm2A/-s, L 8 6 x 1 0 ac m - 3 :290 : !-b: Ib (M + No) I+ (3.5xl0tu) fh: th(Nt + Np): 1000 cm2ny'-s p = qLthn + q[Lpp | qppp :4.3 C)-cm -Nn+ N,q(Gallium):5x10ls (c) p:N,q @oron) cd3, n: L86xl0a cm-3 : 150 cm2A/-s, I+: I$ (M r Np* Ne): 14 Q.05x10tt) l-h: th (Ne * Np* N,q): 520 crr?A/-s p:8.3 C)-cm. 5. AssumeNo- N1>>n;,the conductivity is 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 choose No : 2xl0r7,thenNr : ND*+ Nd-: 3xlQtt, so that Lh x 510 cm2lV-s whichgiveso: 8.16. Furthertrial and effor yields Nn=3.5x1017 crrt.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 ) b+l p * _ q i o ( b nl,4 b + ^ , l b n ,_ P, QFpni(D + l) 7C 2JE i . At the limit when d >> s, CF: lnz :4.53. Thenfrom Eq. 16 V p=ixWxCF - l0 x l0-3 x 5 0 x 1 0 - 0 x 4 . 5 3 - 0 . 2 2C 6) - c m FromFig. 6, CF:4.2 (d/s: 10);using thea/d: 1 curvewe obtain V=p. I/(W.CF)_ x 10-' 0.226 - 10.78 mV. x4.2 50x10-o 8. Hall coefficient, Ru= V,A IB,W - 426.7 cnf tC x ( 3 0 x 1 0 -x nl 0 o ) x 0 . 0 5 2.5xl0-3 x1.6xl0-3 10x10-3 Sincethe signof R'7is positive,the carriersareholes.From Eq.22 l l 1 ' eR, x426.7 l,.6xl0-'n Assuming Ne x p, fromFig. 7 we obtainp: 1.1f)-cm The mobllity pg is givenby Eq. 15b p, : - = 1 I Wp = 380 cm2lv-s. '6 1 . 6 x 1 0 - 'x n1 . 4 6 x 1 0 x 1.1 1 , hence R o. qnph + wl_rp . nl+ + pLrp 9. Since R n pand p- From Einsteinrelation D n lt Hllro=DnlDo-59 R , _ 0.5Rr N olt^ Nolt,+Neqp We haveN.e:50 Nn . 10. The electric potential @is related to electron potential energy by the charge (- q) I Q:*=(Er_ q ni) The electric field for the one-dimensionalsituation is defined as e(x) : -!!-:l dx dEi qdx ft: ni explT): Hence (n- - n \ No(x) Ep- E;:kTh( l *r@) . ) E &) - -( tt\ ll. t dNo(x) dx \q )No(*) (a) FromEq.31, Jn:0 and D, r (x) - H d/d* n q " - - kT No!:a)9- - *kT o N o " n ' q :259 V/cm. (b) E (r): 0.0259 (100) 12. At thermalandelectricequilibria, J , = qgn(x)e + qD, 4:!') -, dx E(x)= --Dn Dn I dn(x) dx N, -No D, Lt, N o * ( N r - N ' ) ( * l N, -ff. F, n(x) L) L p, L N o+ ( N , - N o ) t l0 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 fl:ftno + Ln:No + Ln:10Is +10tt -19ts crn3 p- n? \+4p= e . 6 5x l o n) ' + l o r ,= l o , , c m - , 10" t 4 . \a)Tp = I oor,rry = 5 x l 0 - t 5x 1 0 7x 2 x l 0 t s = 10-t s D oTo 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'=pno p^(o) +roG rlt -,u p * T o S , , ) I \ _ (9.g5 1 l9_')' *t0,,(t _ + 10-, 2 xl}tu (. = lOe cm-'. 10-8x 20 x20 +10-8 3x10-a 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," * QFo p,") - q(l+ + Fp)r oG . ll - - n Dr # t a )J r . a i n - - 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): 1.6x 10-1ex 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 n - o,"l+-4+) : - QDp q[.p,(o) Jr(xo) H,.-w) = - QD.H,=, - ql.p Jo(x p^,1?4. ,(o)'o ,inhfw I lL, ) 18. The portion of injectioncurrentthat reaches the oppositesurfaceby diffirsion I S T2 given by J o(W) JoQ) cosh(W lLe) a$= L o= ^ t ' r % = ' & t s o * t o * - 5 x l o - 2c m ;. do - cosh(10-2 /5x 1O-' ) - 0.98 currentcanreachthe opposite Therefore,9SYo of the injected surface. the recombination state, rateatthe surface and in the bulk is equal 19. In steady APr,ou,u LPn,"urf^u minorirv sothattheexcess " *i:;:*..J#'ff : lora. ;,,surrace :g 10-o atthesurface :1013 cm-3 ratecan be determined from the steady-state conditionsin the The generation bulk loto : G: lom crn3s-l l0-6 From Eq. 62, we can write D" ^ a ' L + ! G- & : o Axt To conditions are4(. = - ): l01acrn3and4(* - 0): 1013 crn3 The boundary Hence where 4(i: t to lola( 1- o.ge-' ) : 31.6 pm. Lp:.fio- to-u 20. The potentialbarrierheight Qa= Q^- X: 4.2- 4.0:0.2 volts. occupyingthe energylevel betweenE andE+dE is 2 t . The numberof electrons dn: N(DF(DdE whereN(E^) is the density-of-state function,andF(E) is Fermi-Diracdistribution function. Sinceonly electrons with an energygreaterthan E, + eQ^ and having normal to the surfacecan escape the solid, the thermionic a velocity component currentdensityis +Az.T)% - | = r'v,E%e-G-rr)ln 4g J -JQt, Jrr*qq^ ht where v, is the component of velocity normal to the surfaceof the metal. Since relationship the energy-momentum E-2m n P2 I . ) ) ?r 2m l3 m By changing the momentum 47iP2 dP - dp,dp ,dp " f - Differentiation leads to dE - PdP component to rectangular coordinates, Hence =Ht', J mht lE Z? , S f [- p ,u-'ol p]-z^t1) * p2,+ lzmkr dp ,dp ,dp , p,o dp,f- ,-r)/z^kr dp, o,oo,ll- ,-01r,^0, ^t'r'-,;t,'^rl')u,, "ll 2 wherep',o= Zm(E, + qQ). t Since L e-o"dx -{ ll \a) yield (2dmkT)v'. , the lasttwo integrals - u The first integralis evaluated by setting oi:'9' 2mkT Therefore we 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 integralbecomes mkT 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 x l 0 ox g x t o - ' o ] ' ] :3.r9x l0*. r _{t* [20 4x2x(20-2) L ) 23. Equation79 is the tunnelingprobability px 10n x 10-'o)f _ sinh( 9.99 r'0-,0) - {, * [6x ]-' 0.403 4x2.2x(6-2.2) ) L (q.qq * rinh :. ro' *lo-'l' ]-' = 7.8 Tr0-\' _ {r *' [o x r0-,. -z.z) L^ 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,dp rdp" [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' *' Srnce e-"^dx =[ Z1 J__ \" ) r :ll 2 yield(2dmk|)v,. , thelasttwo integrals pi;29' 2mkT :, . The first integralis evaluated by setting Therefore we 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 integral becomes *0, fr^,o e-"du: mlsT s-tQ./kr - uh). kr)' = A'7, "*"( Hence - 1 =4tq*k' -72"-t0^lw h3 ll. P_tl----------'i-_ 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 (1.054 x 10-'o)t x 3 x 10-'o]' [20x sint(2.17x 100 , - -lr\ r =r.,n,,0* r 2 4 x2 x(20 ) | )-' 13. Equation 79 is thetunneling probability B= x10' x10-'o)I sintr( 9.99 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' =7.8x l0-e. l'' 4x2.2"(a-z.z J) -' 24. FromFig. 22 Ass:103V/s ua= l.3xl06 cm/s(Si) and uax 8.7x106 cm/s(GaAs) t x 77 ps(Si) andt x 11.5 ps (GaAs) AsE :5x104V/s vax 107 cm/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 10n m/s= 9.5x 10u cmls For electric field of 100 vlcm, drift velocity = 1350 va = l_4,E x 100= 1.35 x 105 cm/s<< v,, For electricfield of 104V/cm. = 1.35 = y,r, x 104 x 107 cm/s . FoE= 1350 The value is comparable to 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 the n-sidedepletionlayer width is: Hence, pm W, =1.067 The total depletionlayer width is 1.867pm. of the electric field n(x). equationfor calculation We usethe Poisson's In the n-sideregion, L=3-*r+r(x^)=LNox+K d x t " " E s E( x , = 1 . $ 6 7 Fm)=O+K =-+N, €., E x l . 0 6 7 x1 0 - a xlor) " ' E( t , \ = L * 3 x l o r a ( x _ l ' 0 6 7 E^o, =E ( x, = 0 )= -4. 86 x 103V/cm In thep-side region,the electricalfield is: 9 =L* ^ dx tr ==n rax2 = e ( xP +K' ") ' 2€, o*(0.S" t0-o)' ) + K' =-t* :0 E( x p = _ 0 . 8 1 m L o . s , ,'r o - ' ) ' l * o , l* ' - ( . . ,G \ P" ' ' ):a 2t, ) E,o, =E (ro =0) = -4.86 x 103V/cm The built-in potentialis: _ l;", =0.s2 (r)d*l v,,= - I _:,, &W = - l'0,,, v. _s ide o_,,0" (')4 n 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 0 =- ft *o*l'- (o.r v,(,)=- li e)a* x10' * r0-o)' " -l{orx10 )'f* =-ftLir - (o.s = -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-') [[!!l[_ )D--lr- l l . - , . !_-.[.-_ m_ !m [![[^ l!_.- n! l[ _trl! n_ , , fl_ -____t ,_ l!-^-[l![[-l-[[[. __fll[l-_J_ ., m_ !!t t----l[- t![. ___ r ["Jr][+n_Jo_!{ f_ i. The intrinsic carriersdensity in Si at different temperatures can be obtained by using Fig22 in Chapter2: (K) Temperature lntrinsic carrierdensity(n,) 250 300 350 400 450 500 1.50+108 9.65+10' 2.oo+10" 8.50]l10'' 9.00+10'3 2.20+10t4 The Vu canbe obtained by using Eq.12, and the resultsare listed in the following table. J! --lmIml ![- __m __tr __tr ml MD m nmt m. n[[![_ m,n l![[[- m_ Iu!!-J mlm!,n,n[!* Thus,the built-in potentialis decreased as the temperature is increased. The depletionlayer width and the maximumfield at 300 K are IT: 4No _mu 2 xll.9 x 8.85 x 10-'ox0.717 =0.9715 lnt 1.6x10-texl0t5 x 10-5 p f t r _ 1 . 6x 1 0 - ' ex 1 0 ' 5x 9 . 7 1 5 =qN =1.476 x 104V/cm. q x 8.85 x lO-'a 11.9 ' ' 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 [Nr*Nr)) = ND > 1 . 7 5 51 x0tu L e [ l t . e x 8 . 8 5 x l 0 - [' 1 o 0 ' ' +N " ) ] r+# We can selectn-typedopingconcentration of Nr: 1.755x10t6 crnt 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 "*-:, Qd"Nu ForNo:19r6"*-:, += C,' 'V:, _') - z x ( o . s t-tv ) 1 . 6 x1 0 - ' n xll.9x8.85x 10-'n x l0' -=1.187 (6 o . t z-tv ) 1x 0, c,' ed"Nu x11.9xg.g5x l.6x 10-re l0-,0;10" = 1 ' 1 8 7 xt o " ( o ' s l o - r ) For Nlr:1017 crn3, -v) _ I _ zVu, zx(o.gsa-v) C,'4d,Nul.6xl0-l9'ttffi=L.|87xto'o(o.eso-r,) Whenthe reversed bias is applied,we summarizeatableof I /Ct, vs V for variousNp values as following, [_ l!.-._lI [[_-*JI l! , - fl-l Lt__-[[ x. L- fr! ][ ,. 3r nl![[!_ 4U/4U ._n rr__Jm . f-l '-l flt l !. f[n - n n m n . n n n D[!!D[- [-JD [ " TIT] iL__.__.![ .-n __[-_.n nn. .--8.-.-D [._-[!__^-l! ][l [![I __-!-_ul .-n _J____-[! --! [. nD -n.^---_l !* _nn. n[[_ n n n J_n._nl -_rJJil mm ! ._n --[ 1!J1_!U -i--m Jln__-lil .-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 interceptions give the built-in potentialof thep-n junctions. 6 . The builrin potentialis ( l0'o x1020 2kT,(a'e"kr\ ? x I 1 . 9x 8 . 8 5 x1 0 - ' a x 0.025 e) V"r , -) ': = - l n l - - i ; l= i*0.0259x hl 3q \ \ q ' n )i 3 t x l0-'nr (l.os 8 x 1.6 * to' I = 0.5686 V From Eq. 38, thejunction capacitance canbe obtained ' l o' ox ( t t.e" 8.85x lo"o) ' ( - - 4 - l q o t , ', l ' " =L .,=T=lq;4j = [[ . 6 x1 0 - ' x -vR) rz(o.s6s6 ]"' At reverse biasof 4V, thejunction capacitance is 6.866xl0-e F/crf . 7 . FromEq. 35, we canobtain +=ry#+N,=&*r| '.'vR>>vu,+ No =2(v^) r ' QE" r 2x4 1 . 6x l 0 - t e x 1 1 . 9 x 8 . 8 5 x10-'a x (0.85 x 10-')2 +No =3.43x10"cm-' We can selectthe n-type doping concentrationof 3.43x l015crn3 8. FromEq.56, ,-, G = -rr L :f l [ _ l =l " ocnD,rN, l ln, o,e*n[-;-J] rf Lo,"*ol- )+ .r-l5vln7ylnl5 x l 0 - t 5x 1 0 7x l 0 r t 10-t5 I .lxe.65x10e:3.8exr0'6 and 2 xIl.9 x 8.85 x 10-'o x(0.717 + 0.5) =12.66 x l0-5 cm=1.266 tm 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 ^=il'' no ND we canobtainthe hole concentration at the edgeof the space chargeregion, n., ,^ =#"ii;i' lvD ( or / ) -\r / nc \ -g{#lel;'-'J ,cm-,. =2.42x t0,, 1 0 .J = J o ( r , ) + J , ( - * o )= J , G " ' r ' _ t ) ) v "I - e o . o 25e _l J" v 3 0 . 9 5 e o o 2 s_ el =V = 0.017 V. l l . The parameters are ni: 9.65x10e cm-3 Dn: 21 cm2lsec Do:|0 cm2lseceo:T.o:5xI0-7 sec FromEq. 52 andBq.54 J, (r,\ ' t 7 =1.6x10-tx =n':oLp (suvr* -,)=nF*, z !i-"leltrr J-l I ND L ] f (qv"\ 1 l0 5 ><10- " I (q.os,. ton )' x lIer--qrt ll \ o o 2 s-e ND L t I l J No=5.2x1015cm-' J^(-*r) = =no:'* Ln (4vrtr -l)=q [;"t"1't*t-'], 2 5 = 1 . 6 x1 0 - t t x ::> N t = 5.278 x 1016 cm 3 We can selecta p-n diodewith the conditionsof Nn: 5.279xl016crn3 and Np : 5.4x10lscrn3. 12. Assume[ s 4p {,r: 1O6 s,Dn:21 cr*/sec,and Do: 10 c#/sec (a) The saturation currentcalculation. FromEq. 55aand to =,[Drrr, we canobtain t NA (L _* 1 '- _ Q D o P n o * Q D , n p o= o "n | .:N DE I? po Le Ln !^) F;) = r.6 xlo-'x (e.os',0')'[*.,8. e +.8] ' \ l t o - ") [lo'' llto-u to'u = 6.87xl}-t'Ncmt And from the cross-sectional areaA: 1.2x10-5 cm2, we obtain I , = A x J " = 1 . 2 x1 0 - 5 x 6.87 xl0-t' = 8.244xl0-t7A. (b) The total currentdensityis J =J"l"' -tl \ Thus ( o r \ ( q v \ / x l 0 - r ? [ e o o r t-,r | | = t . 244x l0-t7 x 5.47x 10" = 4.51x l0-5 A Iorn =8.244 t"n 4 1 0 - r ? [ e o .-o I _ o . r ,= 8 . 2 4 x r=f ( u , \ .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, v. l )* ,l - 0.78 0.02se l0-'' L\/"/ I L\t.z++" J I 14. FromEq. 59, andassume Do: l0 cnflsec,we canobtain J"- ='1 afi": *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 2xll.9 x8.85 x10-ra x ( V o+ , V^ 1.6x10-tex10'5 l0-u -lo'' x lo'5- = 0.834 vo,=0.0259,r, V x l0')' (9.65 Thus Jn : 5.26x 1 0 - 'l+ 1 .8 7 2 x 1 0 -'J0 .s3 4 + V^ ( Il._ [l.- ._-[*-__-m l^ --l--lm [^ ._ff_-_m [^ m_ J. "nm !^ m mI DI ![^ _lm__nI [^ -n__1._nt [^ -J-- .![ !^ m_ !m _ff__!m n ^ __n-._.[_-!l [^ __n m! !^ ll-fi_ m - ro0ri m Jm^ E _!00^ J Jm^ o:B:_. olo0t0 WhenNo:1017 cn 3,we obtain x l0't=0.g53 v ^ ,= 0 . 0 2 5 g , - 1 0 ' ' v x l o')' (9.65 J n = 5.26x 10-'' +1.872* to-'.ft.lso + tu* llJr]- trri.a.O u-L R<dIIl uL ll"llrtl u1_ 1 5 . FromEq. 39, Q o = 4 f , b " - p ^ " \d * = 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 ) -t-t \f -ljle Lp-e " =epnoct)1"' ( qv ) :r.6xr0-'e x"i#tl(-5xro .,[" =8.784x 10-3 Clcm'. -'1" * -" ;) 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(breakdown uottugq=E{ = (t, )' 2 2 q , .-\, x-(' 2 o . 8 x 1-0 _ 11.9x8.85x10 z'-)('1 0 ' ' / 2xl.6xro-,, =2ss v x l0-'a x 258 2xll.9x 8.85 1.6x10-rexl0t5 :1.843x 10-3 cm= 18.43fem when the n-region is reducedto 5pm, the punch-through will take place first. From Eq. 87, we can obtain Zr' _ shadedar_eain F_is. 2ginsert =(!_)(r_y_) vB G.w'Iz \w^ )\" w^) (w\( ( s \/ w\ s \ V u ' = V;u l l l2 - - l = 2 5 8 x 1 l l z : - -l = 1 2 v l It/^ 18.43l\ t8.43 \w ^ )\ ) \ ) Compared to Fig. 29, thecalculated resultis the same asthe valueunderthe conditions ofW:5 pm andNa: 1015cnL3. 17. we can usefollowing equations to determine the parameters of the diode. ,p J, =r '"\E tt ,L Neo svtkr '\ ", n , ' , u , 0 , * Q w t l ,e c v t 2 k ' =e N, 2T, vr=t"! " 2 =t,!"' 2 q (*). = os* lot)' ,kt + Axr.6x AJ, . ,= 1n" l TE:-eqv oNo,r,+A x | . 6 x ll0-re 0_" , "E x i , ;b ?eao259_2.2x1o-3 :E+ = =aH*#5 v, =130 ff{* ")-r (ru,)-' Let E.:4x105 Y/cm,wecanobtainNo:4.05x101s crn3. The mobility of minoritycarrierholeis about500 at No:4.05x l0r5 .' Dp:0.0259x500:12.95 cnfls Thus,the cross-sectional areaA is 8.6x10-5 cm2. 18. As the temperature increases, 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 temperature increases, total numberof electronincreases resulting in easy to lose their energy by collision with other electron before breakingthe latticebonds.This needhigherbreakdown voltage. 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/ l0'l=-x ( 1 0 ) l o= 5 . 8 7x 1 0 5V / c m I | d*=l=1001.-| xl0-' =llE",ni,ot:4x 105 Jo \4x10'/ [axl05/ = 587V 4 V B = 5 . 8 7x 1 0 ' x 1 0 - 3 (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 "1"'1o1'"' y, =Y - +""' l2e r 3 E ^ L q l o - 4E r"' |z * n,g l,s,Ls-.:to' 1"' * 1ror,1-,,, 3 x l0-'' 1.6 L ,j = 4.84 xl}-,E"3 /t The breakdown voltage can be determined by a selected t". 21. To calculate the 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 1.27 4 x 10 cm, respectively. Also, bysubstituting V =-5 V to Eqs.90 and 91, the electrostatic potentials are 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 reduced when the heterojunction is forward-biased from 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 Esz A-E - _ r/ _ ( E o r _ E r r ) / q _ ( E r r _ E r r ) /q _'c / b i_ q q = t.tle * o.^-gh o -4r 1,.l:,, = r.273 y I \l=orr' q 5x10" e 5xl0 r ^,, tV 2N /E$.v^, | l" ^' , - , - , - , LqNo(e,Nr+erNn)) :4.1x10-5cm. Since Nrr, = N ,txz .-.xr = xz r x8.85x l}-ta x1.273 I 2x12.4xLI.46 x t o - ' ex 5 x 1 o r 5 (tz.++11.46) t 1.6 1' . ' .W : 2 x , : 9 . 2 x 1 0 - 5 cm =0.82 ptm. CHAPTER 5 l. (a) The common-base and common-emitter currentgainsis given by = 0.995 x 0.998 do: 1&r = 0.997 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 + 999).10 x 10-6 =10rnA. junction is forward biased. 3. (a) The emitter-base From Chapter3 we obtain sl to'''z'-t-9'' o.nru 't! 'l= o.orrnrf =kr 1n(N vo, u. l= q \ ni- ) xlo"f J t p.65 Thedepletion-layer widthin thebase is N n=, gotAa"ptetion- base junction) layer widthof theemitter 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 =5.364 x 10-6 xl0-2trrm cm =5.364 . Similarly we obtain for the base-collector function - -l ro''' ro: =o.rn, v^, u. " =o.o25e"[?' " lo'Jr LP.os l l (b) UsingEq.l3a 4.254x706 cm:4.254x10-2 um . Therefore the neutralbasewidth is -4.254x10'2 = 0.904 Vl =Wn-Wr-W, =l-5.364x10-2 Um. x l0-'f = l-i' ,rrolor - (9'65 p,(0) = pnoe4vralkr ND2x10t1 . | ^\2 =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 Do=40cmls Lo=ffi P ro = Vtr"10-- =J+o'107 = 2 x 1 0 -c 3 m ' ro-nY ^ - n,'-=(q'os = -, =465.613 = +o).1 . In the collectorregion Dc =ll5 cm/s Z. =fis lo'u 3 cm to* =10.724x.10 ,rn ' =futrt=9.3r2x lo3 The currentcomponents aregivenby Eqs.20, 2I, 22, and23:. .40.465.613 l . 6 x l 0 - ' e. 0 . 2 x I 0 - 2 = eoslooz s1 e. 5 9 6 x l o r A x 10-o 0.904 Iro 7 I u o= 1 . 5 9 6 x l0-'A Iuo Iun _ l . 6 x l 0 , n 1,, .0.Zxl0-.2 .52.19.625 ( r o r t r o r r nl_ ) = t . 0 4 l x I 0 _ iA v 0.721x10-3 .I l 5 . 9 . 3 x 1 0 - ' '. 0 . 2 x 1 0 - 2 1.6 1 2x l 0 3 = 3 . 1 9 6 x1 O - t A o 10.724x10-' Iuu = I r o - I r o = 0 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" Iro Itu x lo-5 1.596 = =0.9938 xl0-' 1.606 x l0-5 1.596 l.596xlo-5 *r - Hence, the common-base and common-emitter current gains are d,=fUr=0.9938 F o= : a o :160.3 l-do (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) curves by usinga computer program: o c ox vc 0.4 0.4 0.6 DISTANCE x 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,^) =u(nilf o no - t.ornEl ' a V 'ulkr e' \e - Le n rno I '*g) lL, ) -r) 'alkt t-rrf = nn.'=O"f L o, l 74 e ' Y t lkr _ =uDrf "o 'th 'hl W L, I \ rfr ?)l t , ( w- * \ Izo ) 'I'no .l L ll tl t l t l l Similarly, we can obtain lro: ,r,=u(-no,H,=,) | =qoTI*'^',,[---L1:{All \1, ) \1, ) *- ttl ^' = ^. rr,rln r"' ^,*r4) Lp -a"".n[l] Lp lL, ) ,t"hf4l l_ = qAT;E 8. *,-,). lQ,, "".{fJ] L,*[fJ] L'ry';]i r, The total excess minority carrierchargecanbe expressed by g,:nn([r,e)- p,"fdx = q A l I p , o " o " u , o , 1| l--; l a x Jo L-"" W ) " t w ,wl -1 :qApnoeorol* O-+)l otr14tr.. -"evrolkI 2prh 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 sameexpression as en. In Problem 3, 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 basetransport factoris -. I l 'l I"^ d ,' Z J = sllillt - | Ib t ""'nI HV L dr= lL' ) -r)* l{""*'*."./zll \1, )) For t4fLo<<1, cosh(Vf L)=1. Thus, ,,,,,i',hfzl ""rh(L\ \1, ) \1, ) :.""nIlL) \1, ) t-t(w)' 2lL' ) z) =r-0t'lzro 11. The common-emitter currentgain is givenby R: 'odo l-oo Wr l-wr' Since7 = 1, 8..= a' l-d, t-[-Vy'lzr,')] =Qto'fw')-t . po=zLo'f rf WfLp <<1,then W' . _ r-frr,'lz4') ,, 12. x 1 0 - 7= 5 . 4 7 7 Lo=rlDot, =^/100.3 x10-3 c m= 5 4 . 7 7 pm Therefore, the common-emitter current gain is ,2 lW,=z(s!.ll P' = 2 L Pt = 1500. "to.:\' Q"lo-')' 13. In the emitterregion, Fpr =54.3+ =87.6 x l 0 - r t . 3x l 0 r 8 l+0.374 -87.6 = 2.26cml D t = 0.0259 s In the base region, pn =88+ 1252 x l0-rt .2xl0t6 I + 0.698 = 1186.63 =30.73 Dp =0.0259.1186.63 cm/s . 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, =1.506 L, = ^tDp, =^lZ.Z6g.l0* x l0 3 cm I n L,(,, _ _= = r NF fri (9.65r'^nY 3r.lir = 3r.04 cm-j In the baseregion, = 5.544x10-3 L, = J3u734.10u cm ' ton 6.os Y = 4656.13 cm-3 n ^- = Y:::::LJ2x l0'o ln the collectorregion, = 3.428x10-3 L. --"111.754-10{ cm ,.^ / ^\, -19'65 xlol- f = 18624.5 cm'3 5x l 0 ' ' The emiffer current components are given by , r, :Weo.6ro 0.5x l0+ o2se = 526.g3x lo-6 A _1.6xrc-t:..1_0: 'rEPr.506 x lo-3 Hence,the emittercunent is ..?.?69.3r.04 4o.ozse_r)= x l0_e t.OOe A . \ Qo It = I tna Itu = 526.839 x lOj A . And the collectorcurrentcomponents are given by ,r^ _l.6xl}'" .10-o -30.734.4656.13 =526.g3x eoqo.o2ss 10_6 A xl0 { 0.5 . 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 obtained by lo{- I tn - 526'83x y = 0.99998 . ' IE 526.839x l0The base transport factor is I"\rt *'fl _ x l0* 526.83 -5 26^8 3*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-emitter currentsain is B^= oo - 0'99998 = 5oooo . l-a" l-0.99998 16. (a) Thetotalnumber of impurities in theneutral base regionis wnol(lr-*/,) % = 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 concentration is 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 = ---------==l--r w2 , (srlo-'I 2Ln' zQ1u " to- ' f = 0.999588 I - ,, Dt Qo D, N ELE = 0.99287 I .5.583x10'3 l+ .10-4 7.77 lOte , 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 voltage dropof kr f q , ,, = o.ol3e A. #= Therefore. I c = F o I a = 1 3 1 . 6 . 0 . 0 1 3= 91 . 8 3 A . 19. FromFig. 100and Eq. 35,we obtain (r,A) 1" 0 5 l0 l5 20 25 1.(mA) 0.20 0.95 2.00 3.10 4.00 4.70 e,=# 150 210 220 180 140 At low 1, , because of generation-recombination current, Fo Bo is not a constant. increases with increasing1, . At high I B, vEBincreases with l, , this in turn causes a reduction of lzr. since Vru+Vur=Ytc=sV. The reduction of Vu, causes a wideningof the neutralbaseregion,thereforepo decreases. The following chart shows Bo as function of I B. lt is obvious that Fo is not a constant. / / o ,1 \ \ \ t / / / \ . 5 1 0 1 5 ls (trA) 2 0 2 5 3 0 20. Comparing the equations with Eq. 32 gives Iro=art, d rI ro = ar, a^Ioo=a, artd I no = dzz Hence, d -' -: * azt or- tZ.%."* LE De p,o crr. - - n- : er, ,r J, _W . - . _ Lc I Dc flco Dp pno 21. In the collectorregion, .3 Lc : ^[Dc%= Jr.tou =1.414 x lo cm =ft,2 = 1 . g 6x 3l 0 ac m - i. ftco "sr c t l f s x l 0 r 5 lN, =(e.O FromProblem20, we have I " 14/ D, . l+-.-.: LE D, npn P,O 9.31 ,*0.5x104.1. lo-3 x lo-2 lo 9.31 = 0.99995 d R Dc 1 2 1.863x lOa , ' -, W ftco + ,, ,- t - *0.5 * " x- l0{ 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.49 x 10-to A 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 ro\ea,ett*t - 1)+ u^l ^o I .. ,.\ = 1 . 7 1 5 x 1 0A r I c = dr I rope'utr' - l)+ I *o L . , . \ = 1 . 7 1 5 x 1A 0{ . Note that thesecurrentsare almostthe same(no basecurrent) for WfLo <kl . 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 boundary conditionat x'= 0 , Cre-oltt :C, =nc(o) n.(o) = nroQ*'"ln -|]1. The solutionis w - tp-'l t" ul n, &) = nro (eav" The collectorcurrentcan be expressed as ,,= n(o,*|.=.J.,n o,#1. n =,) [t*r- t- r^(2"t-. "f) =qAL#(*il m (f+vcnltrar, (dr* lkr - l) - ou 1) . rr - r) Qtrc,r 23. Using E,q.44,the basetransittime is given by =$ji# tu=w'/2D, We can obtainthe cutoff frequency: f, =lf2m u:l'27 Grrz =t.25xro-ro s. From Eq. 41, the common-base cutoff frequency is given by: .f" =.f,I ao=# =1.27 SGHz . The common-emitter cutoff frequency is =2.55MHz . x10e ,fp =(t- a)f" =(t - o.f8)" 1.275 Note that "fp can f, . be expressed by f B = (t - uo)f" =(t - aoYoo * .f, =* Po 24. Neglect thetimedelays of emitter andcollector, thebase transit time is givenby t, "= - J 24, = 3 1 . 8 31 x0 - " s . 2nx5 xl}" FromEq. 44,l/'canbeexpressed by w=^lr4r, . Therefore, w=ffi =2.52x 10-5cm = 0.2521"rn . The neutralbasewidth shouldbe 0.252 pm. 25. :9.8Yoxl .I2 = 110 npV. L,E , D,-.wl "il \ ^, ., B,(t,oo"? : ,) ... (ttomev ll0mev) :0.2g. ( tn-\ 7F-q-exp[ 373r mk ) 26 ffi="*(!o#"J="*L 0.02s9 where Eru(x) -1.424 E rr@) =1.424+ 1.247 x, x < 0.45 = 1 . 9+ 0 . 1 2 5 x +0.143x0 , .45 < 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. increases exponentially when x increases. The impurity concenhationof the nl region is t0racm-3 . The avalanche (for w rW^) is largerthan 1500V (ry^>l00pm). For a breakdown voltage 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 x l0-u 0.5 39.6 =1.51 Therefore. = 2.25 J = 4.5J0 x l0-5 A/cm2 /' lx "" l0-l = " - =44.4cmt J 2.25xlo-5 Area: 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<G-ur)I*. Using Kirchhoffs law, we have I*=Iu-Is. Thus, the condition for hrm-on of the thyristor is t s , - / - t A q+4-1 , o q ' 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 \ ...-.................-. - - - Ev METAL OXIDE n-TYPE SEMICONDUCTOR 2. Ec Ei Ee Ec Ei Er Ev 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 _ 1 - ' .| ,J o' o 2 6 J "n * 8 8 5 x rxo 3h +f lI0', 5 \ 9 . 6x lixtorv"*r;" : 1 . 5x l 0 - sc m : 0 . 1 5 I m. 6. C-," tot d +(e,, /e")W^ W^ =o.l5p m FromProb.5 : i'C^in -ra 3.9x 8.85x l0 = 6.03x l0{ F/cm2. 3'9 *1.5x10-5 8 x l 0 - 7* I1.9 7. Vn =t!n!o:o.o26h q ni to" 9.65x l0' = =0.42y ,"=4atintrinsic e" v"=vn = 0 . 7 4 x1 0 - 5 c m x l 0 r 7x 0 . 7 4 x 1 0 - 5 _ l.6xl0-re l ' 1 1x l o ' v / c m 11.9x 8.85"10F= . 1 "'t 'o :3.38 xl05 V/cm E o : E r € " : 1 . 1 1x l 0 5 x EJ €o* 3.9 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 ) V6:V7 thus, v6=$*r, co u From Pr o b5 . ,W* :0 .t5 fg m,y u :0 .0261n[- t,!to' lr ]: O.O u x 10'J [9.65 L o :--------:10* . . v c - : 3.45x I 0-7F lcm2 6 l.5x l0-5 . r / - 1 . 6x 1 0 - t ex 5 x 1 0 1 x x l0-? 3.45 +{.8:0.35+0.8:1.15V. =: =ryfft)* s. Q., >0, [:,- yq(rl,, :8x10-eC/cr* A,V.. = QO' x (10-u rc', )'l - co 8xlo-e - = 2 . 3 z x l o - 'v . xl0-' 3.45 10. Q* =i lo rp",Q)dy po,=e x5x10tt{x), ( where 6 ( _ r= ) 1l @ ' l r a 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-' :+ I 1. Q",:; I 10-" rlo{ x1.6 x t O - 'x nl " S " 1 0 " 1 1 0 * ; , . 3 y ( q x 5 x L o 2x 3 y)dy l, :2.67x10'8Clcn? = 7 ' 7 4 x 1 0 - 'V ' :' LV" -Qo' -2'67x10-B x 10-' C 3.45 =% 12. LVr, fD C = +"4x C d NM N, is thearea density of'e*. ^,where ) 0 ' 3 - x J ' + } 1 0? = 6 . 4 7 x 1 0 'c , rn2. 1.6x10 - lt{,-!'vou*c"q 13. Since Z, << (Vo -Vr) , the first termin Eq. 33 canbe approximated as lt,C"(vo -ht/ u )Vo. I Performing Taylor'sexpansion on the 2nd termin Eq. 33, we obtain (vo+2tyu ) 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 7 -z n = T' l't ' '{'" t ' l - -l*" .E 'n\'*'1I'^ Lo L rl =(|)p,c"(vG -vr)vD where v,=ayu.JO'tTy 14. Whenthe drain and gateareconnected together,Vo =Vo andthe MOSFET is operated in saturation(Vo > Vo"*) . I o canbe obtainedby substituting V, =V^* in Eq.33 = n, "{W 2v,u)v^", I ol,o*o 1 {ry)[,""^ VD"o,= ^l2e"Own(Vo,* *2Vr) +Attu-Vc =0 . L, For Vo,o, <<2eu, the above equatioh reduces to -rfv)'''l +2vru)''' where Vo,o, is givenby Eq. 38. Insertingthe condition Q,(y = L)=0 into Eq. 27 yields -J'4@J vo=vo " c o =v, . +2tyu Thereforea linearextrapolation from the low currentregionto I o =0 will yield the thresholdvoltagevalue. | ). kT. .N, " --l= = o.o26rn o.oo u I s to'u V =-ln(-) q n i /-------:-;Y:''lt'QNn C 1 9 . 6x 5l 0 - ' J x l0-7 3.45 :s-0.8 +(0.27)' Irl1+ ':3.42V I ,l^ (o.zt),' -- 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 :2.55x10-2 A. 16. The deviceis operated in linearregion,since V o = 0 . 1 Y < ( V o- V r ) = 0 . 5 V a l t T -vr) =f Therefore, r, : F,C "(vo frl n=*^, :5 *500x3.45x1 ' x0 0.5 0.25 : 1 .7 2 x103 S. 17. oln I = z c,vo g. = frlvpuo,"t Tlt 5 x500x3.45x10-'x0.l 0.25 :3.45x10-4 S. - 1 8 . V, =Vru + Ay u V,o=6 -:!-= lol? ,Vn -= 0 . 0 2 6 t { - - : i - - ) x10"9.65 E" o" -7 -un- 1.6x1O-x te 5x1Oto c" 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 x l 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., can alsobeobtained fromFig.8 to be- 0.9S V). t9. 0.7= 0 . 3 3 + a'B 3 . 4 5x l 0 - ? 0.37x 3.45x 10-? =8xlOttcrn2 FB _ 1 . 6x 1 0 - ' e nF -0.14 20. f., = -ti*U/a =-0.56 +0.42= V v,=e., ? "W - 2.[€JN 2rru (- x 8.85x10*'axl.6x 10-'t xl0t7 x0.42 - 0.02 - 0.84 = -0.14 - 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 bandgap in degenerately dopedSi is aroundleV dueto bandgapnarrowingeffect.Therefore, Q ^ "= - 0 . 1 4 + 1 = 0 . 8 6V - 0 . 4 9= - 0 . 4 9 V . . ' . V ,= 0 . 8 6 - 0 . 0 2 - 0 . 8 4 z). to'' -]:0.0, u v. :o.o26r[ [ 9 . 6 5x l 0 ' / Vr=Q^" ?+2ryu+'ry xl0tt = - 0 . 9 g- l ' 6 x 1 0 - t e *0.*O co = 4.14* l 5 ' 2x l 0 - 8 co ,v co - 3 . 9x 8 . 8 5 x l0-'o d - x 10-'3 3 . 4d5 x 10-'' 3.45 .'.d > 4.57x l0-5cm : 0.457prn v r> 2 0 =d . l l t l t 9 , ' > 2 0 . 1 4 24. Vr =0.5Y at Io = 0.lfg{ = --9.1 iJ- olr.=o = -12 .'.I olro* = I x 10-''A. bg 1rl vo=o 25. -7 -logI ^vr=a*utp;-lv;) ' c o V n = o ' 0 2 6 n' ( 9 6t5 =. - o ' '^ -. ) ' =0.42Y x 10" 3'9x 8'85x-l0-ro a" 6.9x ro-?Frcrfi 5x 1 0 ' ' AY, = 0.I V if we want to reduce Ip at V6: 0 by one orderof magnitude, sincethe subthreshold swins is 100mV/decade. '' x4.9x8.85xr0 r/2xl.6xl0 o1 7 xr'0 0.,_ ( r o r a . r r __ J o 8 4 ) 6 . 9x 1 0 - ' i.Vu,=0.83V. 26. Scaling factor r :10 = Switching energy *,a . A)v' C'-t* -tC d A A,{ V'=L If .'. scalingfactorfor switchingenergy: A reductionof one thousand times. n.--.---.-=----:.=- 1 1 t< t( i 1 1 1000 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 rectangula r region) _ [no. / r ' c o nW^,L + L', _QN ---:Co 2L' 28. Pros: QNnW, Co qN nW,r, ( f W r ,rT--r C,L rj [1 ,') !. I l . Higher operation speed. 2 . High device density Cons: l . More complicated fabricationflow. 2 . High manufacturing cost. 29. The maximum width of the surface depletion region for bulk MOS n l , m _ - L"1 l - l e , k T l n ( N n / n , ) Y Q,Nn =-1/ r. 1.6;m = 4.9xl0* cm = 49 rnn F o r F D - S O Id , ,i 1W^ =49rwr. =v-^ 3 0 . v+2tu*4Nnd"' t t D rD C. -+ -+^(+)= -f -o026tnt##) = = -,o, = v,u u e^" =z*!!n(L) 2uru q n Co= =0.92Y i ".V, = 3 . 9 x8 . 8 5 x 10-'a = 8.63 x10-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+ 8 . 6 3 x1 0 - i = -0.1+0.28 = 0.18 V. At, .7 3 1 . l\l/ ' qNnLd = ri C o x5 xl0tt x5x 10-7 1 . 6 x1 0 - t n 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 =3.45x 10_rF c: A'"/a , Forthetrench capacitor A:4x 7 prfr+I pm2: 29 pn? C :29 x3.45 x10-1s F : 100 x l0-1s F. 33. ''t a I = C d V = 5 x 1 0 - t* -=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. 34. ' V r= 7 V 35. V, =Vru +2r,ltu Co=3'9 x 8.854x l0-'o =3.45x 10-' Flcm2 l0-' =Q^, vou ( r '-o ' i ) V" =0.0261n1 = l =0 .4 2 Y xl0'/ (9.65 - tO = ,- ; - vE u -o =-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 3.45x 10-8 co = - 2 . 3 2V -2.32 Vr = 4 . 3 4 =2.02Y. CHAPTER 7 1. FromEq.l, the theoretical banierheightis :0.54 eV Qa,: Q. - X= 4.55- 4.01 We can calculateZ, as yr5= v, " =!! q ND x 1o'' g.g259 *2'86 , = 0.r88v 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 _t ) -f N" ^= | l=4.7x10'ucm-' qe" Ld(l /C') I dV ) -l lk- = vs.6259 ^( +: * to".l= o.* u " =t!n q ND \4.7 x l0'" / From Fig.6,the intercept of the GaAs contactis the built-in potentialV6i, which is equalta 0.7 V. Then,the barrierheightis *Vn =0.76V Q u ,= V o , (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 Thebarrier height fromcapacitance is 0.04V or 5Yo larger. (c) For V: -l Y =2.22x 10-t cm=0.222 tm 4N'w =1.43x105 v/cm es C =L=5.22x10-8F/cm2 W 3. The barrierheightis - 4.01= = Q^- x= 4.65 0.64 V Qu, *vr(2.8orlo'n v, l=0.,r, u " =L! yr{s-=6.6259 q ND [ 3xlo'u ) The built-in potentialis = 0.463V V o= , Qr,-V,=0.64-0.177 The depletion width is -V) 2e,(Vo, 4No The maximumelectricfield is x 1 0 ' ' )x ( 3x 1 0 1 u x)( 1 . 4 2 x t0r) n y4t =QN = (1.6 (r= o)l l..l=1. q x 10-'n 11 . 9x ( 8 . 8 5 ) =6.54 x 100V/cm. 2 x l l . 9 x 8 . 8 5x l 0 - ' o 1.6x10-tex3x10t6 = 0.142hn 4. The unit of C needs to be changed from pF to F/c#, so -- t.74xl01s -2.r2xl}ts 4 1crfinf 1rc2 Therefore, we obtainthe built-inpotential at IlC2 :0 ,^ "' l'57xlotj= 0.74y 2.12x10" From the given relationship betweenC 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'' vu " ={1n &- =0.025e*[z.so*]=0.,u, q ND I S.0xl0'" J We can obtain the barrier height =Vo, 10 . 9 0 1 V . * V n = 0 . 7 4 + 0 . 1 6= Qun 5. The built-in potentialis vt,=Qan_T" *O " ro"] - o.o25gr[z'se =0.8 - 0.195 = 0.8 = 0.605 V Then, the work function is Q^ =Qr^ * X :0.8 + 4.01 = 4.81V. 6. The saturation cunent densityis I t.s><lo" ) J. = ArT2 "*o(- qQu") \rr ) \0.025e / ', = I lox (3oo), >( "*ol, ,-,9 ) :3.81 x l0*1 Alcm2 The injectedhole currentdensityis ) 'r , ^ . . Q D o n , ' - 1 . 6 x 1 0 - 'x , n1 2 x ( 9 . 6 5 x 1 0 '= J l.lex l0-"Ncm' *= i:t= ffi Hole cunent J ,"1eqn'k' -11 Electron current J"(etvlkr -l) = J r o_ t . l g x l o - r r= 3 x l o _ 5 . J" 3.81x 10-' 7. The differencebetween the conduction bandand the Fermi level is given by V^=0.0259 t"l :* l:0.04V. \ 1 x 1 0 ") The built-in potentialbarrieris then Vt, =0.9- 0.04= 0.86 V For a depletionmodeoperation,VTis negative. Therefore, From Eq.38a V r = 0 . 8 6 - V ,< 0 qa'N^ l.6xlO-'na'xlO't ( dt 'to" \ 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 _ e7 x l 0 t 6 x ( 3 x 1 0 - s ) 2 Q N o a ' _ 1 . 6 x 1 O - tx _ A A., 2e. 2x12.4 x 8.85 x l0-'o ' o " 6 m = 5 x 10-o x 4500x12.4x8.85 x 10-'a Es2 0.3x10-o x1.5x10-a :1.28x10-3 S : 1 . 2 8m S . il*o.t 9 . (a) The built-in voltageis v, =0.9 - 0 . 0 2 5 h1 v o= , Q u-n Io 6 1 , t 0l':' 0 . 8 V [ 10" / At zerobias, the width of the depletionlayer is 2x1.09x10-" x0.86 [.6x10-tnx10t7 :1.07 x 10-c sm : 0 . 1 0 7p m SinceI/ is smallerthan0.2 pm, it is a depletion-mode device. (b) The pinch-off voltageis x 1 0 t t ( 2 lx0 r ) l . 6 xl 0 - ' n :2.92y v'" _ e N o a ' _ 2e" 2 x 1 2 . 4 x 8 . 8 5l x 0-'o and the threshold voltageis - 2.92: -2.06V. Vr : Vri- Vp : 0.86 10. From Eq.3lb, the pinch-offvoltage is ,, _QNoa' _l.6xlo , - - - 7 - - ' nx l o ' t ( 2 x l o 2t, 2 x l 2 - 4x 8 . 8 5 x l0-'o l): o.ruo u The threshold voltageis Vr : Vni- Vo: 0.8- 0.364: 0.436Y and the saturation currentis given by Eq. 39 :t44v" I r"o, -Vr)' 2 a L \ s 50x 1:0r;+2'1:j':851t0: 4s00 : 4.7x10* (0 0.436), A. : -io ) ")x(lxl0 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 1016 crn3 ( a: - | 0.85-0.0259^4'7 xl}" _ , t , t ND ) )" (3'7xr}') ^lN" :1.52xl0-s pm cm:0.152 ForNl : 4.7 x 1017 crn3 a:0.496 " 10-s pm. cm:0.0496 12. From Eq.48the pinch-off voltageis . - - 'M - v , v, = Q nn q :0.89 -0.23 - (-{.5) :0.62V andthen, e 3 x 1 0 t t" ' a"' : 0 . 6 2 V y' . = -e:N n d , ' 1 . 6 x l 0 - t x 2e. x 8.85 2x12.3 x 10-'o dr:1.68' 106 cm dt: 16.8nnr Therefore, this thickness of the dopedAlGaAs layer is 16.8nm. 13. The pinch-offvoltage is l.6x lO-rn x l 0 ' sx ( 5 0 xl 0 - 7 ) : l . g 4v v , - _Q N o d l 2e" 2x12.3 x 8.85 x l0-o The threshold voltageis AE, _ r./ r, V,=Qu,-" q : 0 . 8 9 - 0 . 2 3- 1 . 8 4 : -1.18 V When nr:1.25x 1012 crn2,we obtain '" : 1 2 . 3x 8 . 8 5x l O - t o ffi ri - (-l'18)] =r'25 x[o xrot2 and then do+58.5: 64.3 d o : 5 . 8n m The thickness of the undoped spacer is 5.8 nm. 14. The pinch-offvoltage is r l r ' q NUr d |l ' 28" l.6 x l- 0 -- ' e -x 5' x l -0 x '7 ( 5 o0 x' 1 0 ' ) \' \ 1 | 2 x 1 2 . 3 x 8 . 8x 51 0 - ' o - _ . ' The barrierheightis AF c =V, + * = 0.79V * V , = - 1 . 3+ 0 . 2 5 + 1.84 Qun q The 2DEG concentration is frr= * [o l . 6 x l 0 r nx ( 5 0 + l 0 + 8 ) x l 0 - ? 12.3x8.85x10-'o - 1-t.:yf = |.zgx t o, cm-2 15. The pinch-off voltage is xlxlot8xdf , , - Q N o d l- r < - l . 6 x l o - t n ' 2e x5 10* 2x12.3x8.8 The thickness of the dopedAlGaAs is 1 . 5x 2 x 1 2 . 3 x 8 . 8 5x I n - r + dr= ffi:4.45x10-8cm:44.5nm AF *_=S_ _ Vp =0.9 _ 0.23_ 1.5= _0.93 vr = QB, _ V. t6. The pinch-off voltage is aN^ v' ^ 2 - ' e" d : 1 . 6 x 1 0 -x ' t3 x 1 O t t 2x12.3x8.85x10-to the threshold voltage is V, = Qr, Atr *c -V, q (rs * to' )': z.tv : 0 .8 9 -0 .2 4-2 .7 : -2.05Y Therefore, gasis thetwo-dimensional electron flr= 12.3x8.85x10*'o l.6x 10-'nx (35 + 8) x 10-7 " crn2 * [o- 1-z.os)]= 3.zx 1o CHAPTER 8 l. Zo: r 3 x 108 m/s r:== x { l 1 6= 1 . 6 1 6 Gtlz. " a L=C : . 2t o= 2 * 1 0 - 1 2x 7 5 2= 1 1 . 2 5n H . 2. :{ro)'.0;GI 3. : 1.48V Vo, = (E, / q) +V^ *Vo : 1.42+0.03+0.03 '=^W ={ t-r-'' 1 /q I N , N , ) \lo"xro''/' = 1.89 = 18.9nm x 10-6cm = 6 . 1 3x l o ' F / c m z c =k-r'l6xlo-t2 . W 1.89 x 10-' 4r="(t)"*? t).,,*o(#) From Fig. 4, We note that the largest negativedifferentialresistance occursbetween Vp<V<Vq The conesponding voltagecan be obtainedfrom the condition &ttdV: second term in Eq. 5, we obtain 0. By neglectingthe -+)=, #=(+.#).*[' =2x0.1=0.2V . ' . V- 2 V p #=(+-T)"*l t) -0 = 0367 #|"'.=ifr#) %#]"4' -27.2,,. ^=(#1,,")-'= 5(a) *":l ( n a*=+t:&d-=#* _ (l2xl0-,1 2(5 xl}-a ) 1.05 x 10-'' x l0' | ,\2 = 1 3 7{ l (b) The breakdown voltagefor Np: 101s crn3and W: 12 pm is 250V (Referto Chapter 4). The voltagedue to {sc is = 6 8 . 5V 1 R r .= ( t O '* Sx t 0 ' ) x 1 3 7 The totalapplied voltage is then250+ 68.5:318.5 V. 6. (a) The dc input power is l00V(10-'a1: tOW.For 25Yo efficiency,the power dissipated as heatis I0W(l-25%):7.5 W. LT --7.5W x ( 1 0 ' C / W= ) 75'C ' = 4 . 5V ( b ) A V B= ( 6 0 m V / " C ) x 7 5C The breakdown voltage at roomtemperature is (100-4.5):95.5V. 7. (a) For a uniform breakdown in the avalanche region,the maximumelectricfield is E^:4.4 xld V/cm. The total voltage at breakdown across the diodeis -f), w-xn) 4 =E,xu.[- = 4Axt xt on)*(+.+. rOd(0.+ xl0-'ex 1.5 1.6 x1d2 1.09x10-" )r 3-0.a)x 10* =17.6+57.2=748V (b) The average field in the drift regionis 572 -" =Z.2xl05V/cm (3 0.4)x l0 This field is high enoughto maintainvelocity saturation in 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 E" o<x<b:31tm E2(x)+.^-W q b< x< Il: 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 equation coupledthe plot of E, versus N in Chapter3 gives Nt :7 t 1015 cm3for E^: 4.2 x lOsV/cm ...v o *= ( e . -er)b x3xl0{ +E2w:(4.2-l)x105 + l '" s xgxl0 " '2 z :138V (b) Transittime t: W-b _(12-3)x10-: : 90 ps . Qxlg-rs ;=ffi 9. (a) For transit-timemode,we requiren6L > l0rz cni2. no xl}tz / L =lotz /1x lor - lorucm-' - l0-rrs: 10ps (b) r: L/v: I}a / 107 (c) The threshold field for InP is 10.5kV/cm; the corresponding appliedvoltageis v =(to's:to' o.52sv )(,',0-')= [ 2 ) " The current is I = JA= (qnry, )a =(t.A" I 0-te x 4600x x 5.25 l016 x 10,)x 10{ : 3.g6A The power dissipated in 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 7 x 7l=3.33x l0,,cm-. \0I7m0 ) (b) For T":300 K lev ) - t, lkq): ztxexn[ffiff (- 1r.e7) Ncz-exp(-aE )=rtxexp o u n , . - AE t L , r , \ - ? r . , ^ - . ^ ( NCU _ 0'3 :4.4x104 (c)ForL:1500K o-^(-,rt] tk L.r\ = 7| . ^-,J exP(-Ar: I "*pl T) --r., .rr/.r'oxlpoo Ncu -0'3iev ) l :oo))= 7| " "*p (- n94) = 6.5 Therefore,at T" : 300 K most electrons are 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 deepquantum well is En= h" n' :+ l*:1#e'x',)l L .'. LEt = 3 meV, 8 m't LE,=4* 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 barrierthickness mustbe reduced to 1.25nm for AlAs. The resonant-tunneling currentis relatedto the integrated flux of electronswhoseenergyis in the rangewherethe transmission coefficientis 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) pm):3xlOa crnr oc(0.6 The net incidentpoweron the sampleis the total incidentpowerminus the reflected power,or 10 mW I 10"[ _ e-3,10't{ =J 5 x t O r , \ W:0.231 urn The portion of eachphoton'senergythat is converted to heatis hv - E, _2.07-1.42 =3l.4yo hv 2.07 The amountof thermalenergydissipated to the latticeper second is 3l.4Yox5:1.57 mW. 2. For l": 0.898pm, the conesponding photonenergy is l'24 = E = 1.38 eV h FromFig. 18,we obtain"r 1arc.fCa0.7As):3.38 -+ sin 0.2958 ) o"= 17"12' . Q = fiz =+= 3.38 z' t z')] - 4n'n'(r c?:o")- + x t x g'ls[t- c9!(l Effi cie ncy (1+3.38f @,*fi,)' = 0 . 0 3 1= 53 . I S Y.o 3. FromEq. 15 _ f3.39-ll=0.294 R=l-l + 1/ \3.39 (a) The mirror loss / \ 2 fill)= L \R/ t, 300x10-" \0.296) h,,f-l-')=40.58cm-,. (b) The threshold currentreductionis - J,o(R,= 0.296,R,= 0.9) J,o(R= 0.296) _L -l o*rr[--r-)l l,f1)l lo*! r 2r \R/J L [R,R,/j o*!"fr') L \Ri t t 40.58) ,hf x l0-- \0.296'0.90/ 2.300 _ 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 - sine" 1x 10 * r = I - exp(- c Li d)= t - ""p(- 8 x r0' . 3.6(1 ). ) 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 aboveequation with respectto), we obtain ndm ^,dn /1e-+m-zL_dL d^ SubstitutingZitll" for m andlettingdmldJ,: - Lm/L2,yield {-}!\*z! 1 =2LE il" \^2) :. a,).= fttm | \n )\dL)) g\l znrl r-(1\( and 6. From Eq.22 r:!I..|'(fl andEq 15 R= (=)' Rr : 0.317,R2: 0.31 I g(a+" )= #a["'.#'(#)] g(zs' )= #*['*. **o"'(#)] 2L' (R.0.99l t00xl0-a (R/ For Rr :0.317 =270 =zt. t rnf ' )= ' hfl) *^Gi "*l ---1-_-'(#), L,-50 pn F o rR 2: 0 . 3 1 1 #'["#-) =-*-'(#) >Li=5043 1t' t 7. FromEq.23a ForR :0.317 ' r =ro. * J,o , h( )l= ,ooo Ncm', ' 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 2xl00xl0-* \0.311x0.99)) h( | )'l = ,uu Alcm-, ' andso I,a =766 x 100 x 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 which can be solved as (2sa) (25b) There are severalvariationsof + in this solutions.Take the solution which is the only g fuo, practical givesl"s :1.3296 or 1.3304 one,i.e.,fuB pm. d1'33 =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 llo I = o.oo9 r (, c)-', If To:50 oC,thetemperature coefficient becomes 1 t-= ! = 50 . 0 02 ( ,\C ) '/. which is largerthanthat for To: 110oC.Therefore the laserwith To: 50 oC is worse for high-temperature operation. 10. (a)Atr: QQt"+ 1:n\ enA -'Ln =electron-holeoairs=, 2.83x l0-' M oltt,+ prb,a =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 .sooo to* rooo " ro-'' rP _=n( ) ' \ 'o.ss. ).I 3"q)\ t0xlO-o ) =2.55x10-u = 2.551tA 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 wavelength 2"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 : 1.6x l0-reJ. Therefore, hv/q:1.24 / ),fum) Thus, / Land R: (rZU / 1.24. 4: (Rx1.24) 13. The electricfield in the p-layeris givenby ot< x < b , r,(x)=u^-T'Nt €" whereE,nis the maximumfield. In the p layer,the field is essentially a constant given by r, (x)=,^ -9!t!ts b <x <w The electricfield requiredto maintainvelocity saturation of holesis - lOs V/cm. Therefore qN'b E- -JI-J" > 16s €r or E. >105 * 4 1 ! ' b= 1 0 ' + 4 . 6 6 x 1 0 - , , N r . q From the plot of the critical field versusdoping,the corresponding Em areobtained: N1 :7xl01s cm-3 E^:4.2x105V/cm The biasingvoltageis given by =":,q*-lo*) +ro, ,, =fui& *E2w (rxro-o) = 138 V. The transittime is , - (w u)_n1 ps. lg* =exro-,,=eo vs 107 14. (a) For a photodiode, only a naffow wavelength rangecentered at the optical signal wavelengthis important;whereas for a solarcell, high spectralresponse over a broad solarwavelength rangeare required. (b) Photodiode are small to minimizejunction capacitance, while solarcells are large-area device. (c) An importantfigure of merit for photodiodes is the quantumeffrciency(numberof pairs generated electron-hole by incidentphoton),whereas the main concernfor solar cells is the power conversion efficiency(powerdeliveredto the load per incidentsolar energy). =AqN,N,(+^E.!^T-)"-",u 15 (a) 1" '[N, Nolto 1C ) = zft.sx xto'nfu..66x to-'n[z.so 10'n )x (= = t * . 8 = 2.43x to2o (s.al * 1g+p-+t z = 2 . 2 8 xl 0 - ' ' A I:1,Q*lr')t, es * t o - 'J I t . z * 1 9 t e ! 1 9 -5 sx r o '1 *, =^ E ) , s - ,-, , " , 1 r , lll=tr-1"(etvtr' -r1 0.3 / 0.4 1 0.5 0.6 0.65 0.7 1520 (mA) 94.5 68.2 -84 2.5x10-1.1x100.55 26.8179 95 95 nt" 'o'.-)= o.u, (b)v^.=u^( !-\= 0.025e*[ u q [1" , \.2.28x10-") -t)-trrt (c) P= I,VQevrw dP dV - t) * t, o = 1"(rev/t' I,etvlkr (l + tt) - t, IL #enrlr, - I, .'. eon/o' - 1" (1 + v ) V m = 0.64 V r P^ = I,V^ = I t l v o , L I _t!n(r.sL\_tr1 q \. 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 y_ -_nl -t)t, t r . ( r ", + l. l)= _rl r . ( r-, \ rll | q \1" - ) q 1.1"1 I, = It€-qvlkr = 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 0.13.000.3 0 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 0.5 3 2.80 48 0 . 54 2 . 71 0 . 55 2 . 5 7 46 4l 2.00 € I a 1.50 B 1.00 0.50 0.00 0.5 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 mAx0.375V:35.6 mW, 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). Solar cells needed under l-suncondition ion) x P,,(concennat ion) 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 14.2o/o xl - l8'5% x 10= 1300 cells for 1000 - sun. 1 4 . 2 Yxo l =ll,Y|19 =ns cellsfor loo-sun CHAPTER 1O l . C o : 1 0 1 7c r n 3 /.g(As in Si):0.3 Cs ftoCo(l - 1,11M0)*-t : 0.3x1017(1- il50)0.7 x)-oz: 3x 101641 0.2 /(cm) Cs(cm-3) l0 3.5x 10 0.4 0.6 30 0.9 45 4.28x10 5.68x l0 16 14 o b 10 8 6 2 0 2. (a) The radiusof a silicon atomcan be expressed as f =-g It 8 t; VJ so r =-x5.43=1.175A 8 (b) The numbers of Si atom in its diamondstructure are 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 3. ltu:0.8 for boronin silicon M /N4o:0.5 The density of Si is 2.33g / ctrl. " 22 :2.33 gl cm3 g/ ctlf . The acceptor concentration for p : 0.01Clcm is 9x 1018 crn3. The dopingconcentration Cs is given by C =k^C^(l--' 1&-t M^' Ilf Therefore -o ------------7-- C" 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" boron atoms 2.338 So that 'A#ffi,=o.7sg ro.8g/more 4. (a) The molecular weightof boronis 10.81. The boronconcenhation can be siven as / I" ^=- boron . numberof boronaioms volumeof siliconwa fer x 10-g x . 0 2 x1 0 ' 3 3 / 1 0 . 8 1 96 _ 5.41 1 0 . 0x 23 . 1 4 x 0 . 1 3 = 9.78x l0r8atomVcm (b) The average occupiedvolumeof everyone boronatomsin the wafer is tt =L =---l -cm3 nb x l0'" 9.78 We assume the volumeis a sphere, so the radiusof the sphere ( r ) is the average distance between two boron atoms.Then r =^l-' =2.9xI0-tcm. \4n 5. The cross-sectional areaof the seedis - ' - - )' / o.ss | =0.24cm2 tv 7d ( 2 ) The maximumweight that can be supported by the seedequalsthe productof the critical yield strength and the seed's cross-sectional area: = 4 . 8 x1 0 5 ( 2x 1 0 6 ) x 0.24 g = 4 8 0k g The corresponding weight 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 0.2 0.06 0.08 0.8 1. cslc0 0.05 0.t2 0.23 rjl E I ru. tm_ ! m - m q _ m _ lJ-Sflo"mm 7 . The segregation coefficient of boronin siliconis 0.72.It is smallerthanunity, so the solubility of B in Si undersolid phaseis smallerthanthat of the melt. Therefore, the excess B atomswill be thrown-off into the melt, then the concentration of B in the melt will be increased. The tailend of the crystal is the last to solidify. Therefore,the concentration of 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 temperature is higher in the centerpart than at the perimeter. Therefore, the solubility is higher in the centerpart,causinga higher impurity concentration there. 9 . The segregation coefficientof 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(-0.3 x 1)= 6.52 : 0.38 at x/L:2. atxlL=l 11. For the conventionally-doped silicon,the resistivity variesfrom 120C)-cm to 155f)-cm. The corresponding dopingconcentration variesfrom 2.5xl0t' to 4" 1013 crn3.Therefore the ranseof breakdown voltagesof p* - n junctions is given by t E 2 ' v,=.t(il,) "(3x105)2 - 1 . 0 5 x 1 0 -x ( N r ) - , = 2 . 9 x 1 0 ,/,N B = 7 2 5 0t o l 1 6 0 0 V 2xl.6x 10-'t -7250 = 4350 LVB= 11600 V -. =+30%o | ll7250 \ 2 ) For the neutronirradiatedsilicon,p: 148+ 1.5C)-cm. The doping concentration is 3xl0l3 (tl%). The rangeof breakdown voltageis v B = 1 . 3x l 0 t 1/ N B = 2 . 9 x r 0 t 7/ 3 x 1 0 1(3 tl%s =9570to9762Y. LVB-9762 -9570=192V (n v ^ \ ( tv"\ l-^a ll9570=*tyo. \ 2 ) 12. We have M , _ w e i g ho t f GaAsatTo _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 pressure of As is much higher than that of the Ga. Therefore, the As content will be lost when the temperatureis increased.Thus the composition of liquid GaAsalwaysbecomes gallium rich. = 5 x1022 14. n": Nexp(-n" / kT) er,p(2.3 ey / kz) = 5 x ro" "*p[- {'8--l : 1.23x x 0 at 27o l0-t6cm-3 C = 300K :6.7x10''cm-3 : 6 . 7x l O ' u c m * ' L(r/3oo)j =1173K at 9000C a t 1 2 0 0C 0= 1 4 7 3 K . 15. n, =JNi/ :W exp(-y, /zkT) * " - 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 0 K :2.l4xl0ta at 90fC : 1173 K. 1 6 . 3 7 x 4 : 1 4 8c h i p s In termsof litho-stepper considerations, there are 500 pm spacetolerancebetweenthe mask boundary of two dice. We divide the wafer into four symmetrical parts for convenient dicing, and discardthe perimeterpartsof the wafer. Usually the quality of the perimeter partsis the worst dueto the edgeeffects. Totot Dies, 148 300 nrn I I rt =ji_- ll vf,dv tykr l_ ff f'o' \l781/t 4 ( M \ ' ' ' " ( =ro, where v'"*o[ M ! \ n =Glfr) I M: Molecularmass k Boltzmann constant:1.38x10-23 J/k T: The absolute temperature v: Speed of molecular So that '*= 16. L= 2 2 x 1 . 3 8 x 1 0 -x2 33 00 29xI.67 xl}-" J; = 4.68xlOncm/sec m/sec . " 0.66 P( in Pa) ;. P 19. o'!u = 0'66pa. 4.4xto-3 L 150 For close-packing affange, there are 3 pie shapedsections in the equilateral triangle. Each section correspondsto 1/6 of an atom. Therefore . l number of aloms contained intre tiangle ^. _ = '1Y." JX_ - 6 areaof fire hiansle 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-t Pafor Gaand13pa for AszThe arrivalrateis givenby theproduct of theimpringement rate and NnL2 : Arrivat rate: 2.64xto2o(-LY4) l"lvt )\'t' 1 :2.9x101sGa molecules/c# -s The growth rate is determined by the Ga arrival rate and is given by (2.9xI 0rs)x (6xt0ta) : I 3.5 A./s: 8 I 0 A"imin 2.81 . (b) The pressure at 700"Cfor tin is 2.66xl0-6Pa.The molecular weightis 118.69. Therefore the anival rate is 2.64xrc^( ]j54Y-ll 9 9 7 3. / \n x 1 2 ' ) \ "/l18.6x .s =2.zlxt0,0 molecul art cm, If sn atomsare fully incorporated and activein the Ga sublattice of GaAs, we have an electron concentration of ( z .ztx to - 'o " .4 '-" )(4 2x to ^ "" ) I t-tll | =1.74x10tt cm-'. (z.e"lo',J[ 2 ) 21. The.r valueis about0.25,which is obtained from 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 CHAPTER 11 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,therefore l= 0.31pm. The time required to grow 0.45pmoxide is t = * G ' + A x ) == l r =' 1 0 . 4 + 0.45)=g.72tr:44min. 50 t .31x B' 0.47 2. After a window is openedin the oxide for a second oxidation,the rate constants are B: 0.01 pmz/hr, A: 0.116 pm (B/A:6 'I0'2 pm /hr). If the initial oxide thicknessis 20 nm : 0.02 pm for dry oxidation, the value oflcan be obtained as followed: (0 +[) Q.0D2+ 0.166(0.02):0.0r or J:0.372tu. For an oxidation time of 20 min (:1/3 hr), the oxide thicknessin the window areais :0.01(0.333+ ,t + 0.166x 0.372\: 0.007 or x :0.0350 pm:35 nm (gate oxide). For the field oxide with an original thickness 0.45 pm, the effective[is given by p:11r, + Ax)=-J-10.+S, =27.72ttr. + 0.166 x 0.45) ' B' 0.01 ,2+ 0.166x: 0.01(0.33 3+27.72):0.2g053 or x :0.4530 pm (an increase of 0.003pm only 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 then, x2 :Bt similarly, s) r;tu, when/ then.x: !-ft*c\ A ' At . 48, A' . 48, 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.rsion coefficientis given by o=L = 4 (! . r ' ) = g [ - E - ) 2 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 - = - = r . ! si I ,^ N x x :0.83 o/o atomic H= looY =20 l+0.83+y -v:0.46 Theempirical formula is SiN6s3FI0 a6. 3" '' :2, (b) != 5x 1028e-33 10t' [-cm As the SiA{ ratio increases,the resistivity decreasesexponentially. 6. : 3t, e1:25 Set Th2O5 thickness : t, ez: 3.9 SiOzthickness : 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 -er(er+Zer) 3Er€, _ z s ? . g + z x t .= a5 ) .37. 3 x 3 . 9x 7 . 6 7. Set : 3t, e1: 500,area: 41 BST thickness : t, a2:3.9, area: Az SiOz thickness : t, E3:7.6, area: Az Si:Nathickness then 444, = 3t A2 o' = o.oonr. 8. Let :3t" e1:25 TazOs thickness : t, e2: 3.9 SiOzthickness : 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-oxygen reaction. Thereforefor Tt : 698 K r(r,) t )I =2=€XDlo.{ ' L L_ 4r,) (k4 kr,)) ) ) ,'z: 06 [fry_3oo I] o.o2s9 r, L\6e8 _ Tz:1030 K: 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 temperatures are usually used for polysilicon deposition,and silane decompositionoccurs at lower temperaturesthan that for chloride reactions. In addition, silane is used for better coverageover amorphous materials suchSiO-,. 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. Anotherreason is 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 deposition rateis too low. 13. The flat-bandvoltaseshift is LVFBD[l![[ QO, co c^ " =to'd x lo-'a 3 ' 9 x8 ' 8 5 =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 x 10-'' 1.6 To removethese charges,a 450 heat treatmentin hydrogen for about 30 minutesis required. 14. 20/0.25: 80 sqs. Therefore, the resistance of the metalline is 5x50:400 O. 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-oxide layers TiSiu on the gate electrodeis more resistantto high-fieldinducedhot-electron degradation. Disadvantage: bridging effectoccurs. LargerSi consumption during formationof TiSiu Lessthermalstabilifv F o rC o S b : Advantage: lowresistivity High temperature stability No bridgingeffect A selective chemicaletchexits Low shearforces Disadvantage: not a goodcandidate for polycides 1 7(.a ) O = e + = 2 . 6 7 x t 0 6 x tA eTL =3.2x103Q xa 0.3xl0-a 0.28x10- 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 x 2 . 9 x l } - t s= 0 . 9 3 n s R C= 3 . 2 x 1 0 5 ( b )R =p + = r . 7 x 1 0x -6 -.=+ A =2xI03C) x 10-a x 0 . 3x 1 0 r 0.28 €A {L 2.8x 8.85 x l0-'o x 0.3x 10-a x 1( _= =-: 2 'I x 1 0 - ' 3 F d s x l0-o 0.36 R C = 2 x 1 0 tx 2 . 1 xl 0 - ' 3= 0 . 4 2 n s (c) We candecrease theRC delayby 55%.Ratio: o4) ,r; :0.45. 18. (a) R = 'eA L=2.67x l0* = 3 . 2x l 0 ' C ) 0.28xlOox0.3xl0o C =4={Ld .S 3 . 9 x 8 . 8 51 x0 - ' a x 0 . 3 x1 0 { x l x 3 =g.7xl0-,rF x l0o 0.36 : 2.8ns. x8.7x 10-13 RC: 3.2 x 103 ( b ) R = 'O l0rC) A: = l . 7 x 1 0 - 6x 4 = 2 x 0.28xl0ox0.3xlOo a4 {L d s 2 . 8 x 8 . 8 5 x 1 0x - '0 o. 3 x l 0 { x l x 3 = 6 ' 3x l 0 - t 3F x 10a 0.36 T 1 32 . 5 n s R C = 2 x 1 0 3x 8 . 7 x 1 0 - t= x 8 . 7x l 0 - ' 3= 2 . 5n s . R C = 3 . 2x 1 0 3 19. (a) The aluminumrunnercan be considered as two segments connected in series: (or 0.4 mm) of the length is half thickness(0.5 pm) and the remaining 20%o 1.6mm is full thickness (lpm). Thetotal resistance is o'oo * =' J L u f ' 1 6 , * , lA , . ! " 1 = g * r o -9 Ar) :72U. -l [10-*xl0-*l0-ox(0.5x10-')l The limiting current1 is given by the maximum allowed current density times cross-sectional areaofthe thinnerconductorsections: ./: 5x10s a5 ; :x 1 0 3A : 2 . 5 m A . N c r * " ( 1 0 - a x 0 . 5 x 1 0 -2 The voltagedrop across the 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) h xW x 0.5 ( =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 reference to Fig.2 for class100cleanroom we havea total of 3500 particlesinf with particle sizes>0.5 pm 2L t 3596: 735 particles/r# > 1.0 pm with particlesizes 100 > 2.0 1tm 35OO: 157particles/# with particlesizes (a) 3500-73 Therefore, 5 : 2765particles/nfbetween 0.5 and I pm :578 particles/m3 (b\ 735-157 befween 1 and2 pm (c) particles/# pm. 157 above 2 ** r00 :. y = fr,r-o,n A :50 mm2:0.5cm2 *) " 4 ( o 2 s x o . s )* " - t { t x o . s ; = g - r . z = 3 0 . l y o lr _ e4(0.rx0.s . I The availableexposure energyin an hour is x 3600s:1080 mJlcr* 0.3mW2/cmz For positiveresist,the throughputis 1080 _ = 7 wafers/ty 140 For negativeresist,the throughputis loSo 120 wafervtrr. 9 1. (a) The resolution of a projection system is givenby L 9:193fgt u'u = o.r7g pm ^ t ^ = K t f r f . =9.6,1 0.65 o,le,3f =k,-L= DoF o.r28 u* '(NA)' o.sl (0.65)'l: L I (b) We can increaseNA to improve the resolution.We can adopt resolutionenhancement techniques(RET) such as optical proximity correction(OPC) and phase-shifting Masks (PSM). We can also developnew resiststhat provide lower h and higher k2 for better resolutionand depthof focus. (c) PSM techniquechanges kr 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 lithographyprocess because theseresists have high absorptionand require high dose to be exposedin deep [fV. This raisesthe concernof damage to stepper lens,lower exposure speed and reduced throughput. 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 increase throughputcomparedto a Gaussian beam. (b) 6. (a) We can makealignmentmarkson wafersusing e-beam and etchthe exposed marks.We can then usethem to do alignmentwith e-beam radiationand obtainthe signal from thesemarks for wafer alignment. X-ray lithographyis a proximity printing lithography.lts accuracyrequirement is very high, thereforealignmentis difficult. (c) X-ray lithographyusing synchrotron radiationhasa high exposure flux so X-ray has better throughput than e-beam. 7. (a) To avoid the mask damageproblem associated with 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-scan X-ray lithography system. 8. As shown in the figure, the profile for eachcaseis a segment of a circle with origin at the initial mask-film edge. As overetchingproceeds the radius of curvatureincreases so that the profile tendsto a verticalline. 9. (a) 20 sec 0.6 x 20/60:0.2 pm....(100) plane : 0.0125 x 20/60 0.6/16 pm........(l l0) plane x 20/60: 0.002 pm.......(t 0.6/100 I 1)plane - "li x0.z = 1.5 = 1.22 wo= wo- .l-zt pm ( b ) 40 sec 0.6x 40/60:0.4pm....(100)plane : 0.025 x 40/60 0.6/16 pm....(l t0) ptane x 40/60: 0.004 0.6/100 pm.....(l I l) plane wu=wo-Jzt = 1.5pm Ji xo.a:0.93 ( 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: 0.006 pm.....(l 0.6/100 I l) plane W o : W o - J z t = 1 . 5 -J 2 x 0 . 6 : 0 . 6 5 pm. pattem [-singthedatain Prob. profiles 9, theetched on <100>-Si areshown in below. (at 20 sec I :0112 ltm,Wo= Wu= 1.5pm (br -10 sec 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 bottomsurface is r r ' = t T t+ J z t = 1 0 0 0 *JT*625 = 1884 pm l-he fractionof surface areathat 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 definemaskingareas on the backside and etch from the back. The width trf eachsquare maskcentered with respect of IC chip is given by -ix625:116 pm v [ = W r - ^ l - z t = 1 0 0 0J [- sing this method,the fractionof the top surface areathat is lost can be negligibly small. I Pa: 7.52mTon PV: NRT : n/V x0.082 x 273 7.52/760, 10-3 :2-7 ,l}ta cm3 n/V: 4.42x l0-7mole/liter:4.42x l0-7x 6.02" 1023/1000 mean-free-path / .52:0.6649 2,= 5xI0-3 / P c m : 5 x l 0 - 3x 1 0 0 07 c m : 6 6 4 9p m l50Pa: ll28 m Torr PV nRT ll28l 760 x 10-': n/V x 0.082x 273 : 6.63x lQ-sxg.Q/x :-4 x 1016 n/V : 6.63x 10-s mole/liter 1023/1000 crn3 mean-free-path : 0.0044 x1000/1128 2": 5 xl}-' /Pcm : 5* 10-3 cm : 44 um. x n, x7/' ,r-tiA' l-1. Si EtchRate(nm/min):2.86 x 10-13 :2.86x l0-r3 xlxlSrsx( 92 t 1 / ,, r # :224.7 nm/min. *3" Iytsx(2Oq% 11. SiOz EtchRate(nm/min): Q.Sl{x 10-13 ^"#: Etchselectivity of SiOz overSi : or etchrate(Sio2)/etch rate(Si) ' : 0.025 5.6nm/min #= ':u-)! - 0.025. ""t-t'u*'ot/n'7x2e8 2.86 15. A three-stepprocessis requiredfor polysilicon gate etching. Step 1 is a nonselectiveetch process that is usedto removeany native oxide on the polysilicon surface.Step 2 is a high polysilicon etch rateprocess polysiliconwith an anisotropic which etches etch profile. Step3 is polysiliconto oxide process a highly selective which usuallyhasa low polysilicon etch rate. I 6. If the etch rate can be conholledto within l0 %o, thepolysilicon may be etched 10 o/o longer or for an equivalent thickness of 40 nm. The selectivityis therefore 40 nm/l nm:40. 17. Assuming a30Yooveretching, and that the selectivity of Al over the photoresist maintains3. The minimum photoresist thickness requiredis (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 generates low-densityplasma(10e crn3) with high ion energy.ECR and ICp generate high-densityplasma(10rr to 1012 crn3) with low ion energy.Advantages of ECR and ICP are low etch damage, low microloading,low aspect-ratio dependent etching effect, and simple chemistry. However,ECR and ICP systems are more complicated than traditional RIE systems. I ( J . The corrosionreactionrequiresthe presence of moistureto proceed. Therefore,the first line of defensein controlling corrosionis controlling humidity. Low humidity is essentiaf.especially if coppercontainingalloys are being etched. Second is to removeas much chlorineas possible from the wafersbeforethe wafersare exposed to air. Finally, gases suchas CF+and SF6can be usedfor fluorine/chlorine exchange reactions and polymericencapsulation. Thus, Al-Cl bonds are replacedby Al-F bonds.Whereas Al-Cl bonds will react with ambient moistureand start the corrosionprocess, Al-F bondsare very stableand do not react.Furthermore, fluorine will not catzlyze any corrosionreactions. CHAPTER 13 :3.46 eV, D :0.76 cn?/sec l. E"(boron) From Eq. 6, - F -3'46 D=Doex?(#)= o.ler* 1o-"cm'ls . o( 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 = 2.73x10a cm FromEq. 9, c (x)= c,ertu (fi)= 1.8 x I 02oerfc (17{6;) I f x = 0 , C ( 0 ) = 1 . 8 x 1 0 2a 0i o m s /cm3;x:0.05 t104,C(5* 105):3.6 x l0le 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 : Q(0 Totalamount of dopant introduced :#C,L 4n 1 = 5.54xlOtaatoms/c#. 2' D=Do*(#)=oz6exn( FromEq. 15, Cs = C(Q,t)= + Jtilt -3.46 8.614x l}-s x1323 )= 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 -)' x 10leatoms/crrf Ifx :0, C(0): 2.342 ;*:0.lxl0-a, C(l0r) : l.4lxl0te atoms/cm3; : 6.7911018 x : 0.2x104,C12xl0't) atoms/c nf ; * :0.3 x l6-a,C(3x lo-5;: 2.65 x 10t8 atoms/cm'; :; 0 ,15xlO-s): x : 0 . 4 x 1 0 4 , c 1 4 x 1 0 - s ;9 : .37x10t 7 a toms/cm3 x. 5 x 1 0 - 4c 1.g7x1gtz atoms/cm3; : 3.51*1016 : 7.03x x : 0.6x104, c(6x 10-t) atoms/crrt; t :0.7x10-a, c(7x l0-5) l0r5 atoms/cm3; :5.62x101a .r :0.8x104, C(8x10-s) atoms/cm3. The x, = = 0.72qm. 4Dtht-CuJ tDr 3 . 1 x 1 0 "= t * t O " " * p [ t: 1573 s:26 min l 0 -' ) 4 x 2 . 3 x 1 0 ' ') t For the constant-total-dopant diffusion case,Eq. 15 gives C, = -L *Dt s = 1x lottm t. = 3.4xlo*atomVcm 4. The process is calledthe rampingof a diffusionfumace.For the ramp-downsituation,the furnacetemperatureT is given by T:To-rt where To is the initial temperature and r is the linearramp rate.The effectiveDt product duringa ramp-down time 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 l-r .l 3 " _X e x p - # . . . ) =D\( T D=Do e x p [ " /h " * p l ; " r y # r l = D o 1 + A * ; * . . . ) u / T r u r / l : O o ( " * p \ Io klo kfoLkIo kT^, J where D(To) is the diffusion coefficientat T6.Substituting the aboveequationinto the expression for the effectiveDr productgives =[i otr,\"*p-]#dt (Dt)"n =D(r)+ ' '"''Eo kTo' Thus the ramp-downprocess resultsin an effbctiveadditionaltime equalto Kls2/rgu at the initial diffi.rsion temperature T6. For phosphorus difflrsionin siliconat 1000"C, we havefrom Fig. 4: |{7d : D (1273K) : Zx 10-ra cm2ls -773 1273 r=-=0.417K/s 20x 60 E":3.66 eY Therefore, the effectivediffusiontime for the ramp-down process is oro' ,Eo _. l.3g xlo-r'(1273)' = 9ls = 1.5min . 0.417(3.66 x. 6x 1 0 - ' n ) 1 5. For low-concentration drive-in diffusion,the diffusion is given by Gaussian distribution. The surfaceconcentration is then ,s =+=---exDl C(0.t) 4d)t Jil; .|1il, ,s ( r-i" \ i '\2kr) { =_L.,*l,&Y-r"' *g ) =_0.5 '\2ftrl( dt 2 ) t dC dt =-0.5xC t which meanslYo change in diffusion time will induce0.5Yochange in surface concentration. ----:'s.vr-[- dc dT ,s (E-\(-E-) ' \zkr )\2kT' ) dT T E_ zkT' dT dT T Jrilo, dc - E a - 3.6x 1.6 x lo-'' nr ,.............._- 1L i.. C 2kT 2xl.38xl0-"x1273 T which meanslo/ochange in diffirsiontemperature will cause16.9%changein surface concentration. 6. At I 100oC, ni : 6x 1018 crn3.Therefore, the dopingprofile for a surface concentration of 4 x 1018 crn3 is givenby the "intrinsic"diffi.rsion process: c(x.t)= c"..r"[4] \2''lDt ) where C, : {x 1018 crn3,t: 3 hr: 10800 s, andD : 5x10-1a cm2ls. The diffusion length is then cm = 0.232 J Dt = 2.32x10-5 tm The distribution of arsenic is C(x) = 4 x 1018 "rf"[---:-) [4.64 xl0-' ) Thejunction depthcan be obtainedas follows 10"=4xl0"".frf t' x 1 0 -'J ( 4 .6 4 -l xi: I.2x 104 cm: 1.2pm. 7. At 900oC,ni:2x 1018 crn3. For a surface concentration of 4x10tt "-', given by the "extrinsic" diffusion process -E, -4.05x1.6x10-le " 2r*rl?l =45.8et 38"10 D = Doe kr x ni "l# =3.77x10-'ucm2ls = 3.23x = 32.3 l0*ucm nm. xt =1.6J Dt = 1.6m 8. lntrinsic diffusion is for dopant concentrationlower than the intrinsic carrier concentration ni at the diffusion temperature. Extrinsic diffi.rsionis for dopant concentration higher than n;. 9. For impurity in the oxidationprocess of silicon, segregatio ncoefficein 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 i4r0.16)' ; t = 1 . 3x 1 0 ' t x l . 6 x l 0 - t n 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 concentration is at x : Ro.Figure.l7 indicates the oo is 20 nm. Therefore, the ion concentration is ^9 oo42n 2.38 x 10'' -3 = -___-___-___-_ :. = 4.74 x l0'tCm . 20x10-'42tr 13. FromFig. 17,the &:230 The peakconcentration is nm,andoo = 62 nm. 2x10'5 : ---------------= l.29x l0'ocm-l oo{2tt 62x10-'42n FromEq. 25, S t.2ex,o" "*[3:i{l | x;:0.53 pm. 260- J : perunitarea 14. Dose xl 9 =CoLVr - 3'9x8'85x 10*ta = 8 ' 6x 1 0 " c m - 2 q q 2 5 0x 1 0 - s x l - 6x 1 0 - ' n From Fig. 17 andExample3, the peakconcentration occursat 140nm from the surface. Also, it is at (140-25): 115nm from the Si-SiOzinterface. 15. The total implanteddoseis integrated from Eq. 25 ur - lo e,= f @- s-"*[-t*I{l*=l{r*fr| 2oo. J z I L Re-,-l}=lo 's -erfc(z. errcl 3)l=r-xt'ee8e 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 projected rangeis 150nm (seeFig. 17). The average nuclearenergylossover the rangeis 60 eV/nm (Fig. 16). 60x 0.25: 15eV (energy lossof boronion per eachlatticeplane) thedamage volume: Vo: n (2.5nm;21tSO nm):3" 10-18 cm3 total damage layer: 150/0.25: 600 displaced atomfor one layer: l5l15 : I : 600/Vo : 2" l02ocm3 damage densitY : 0.4%. zxlo2o / 5.02x102 17. The higher the temperature, the fasterdefectsannealout. Also, the solubility of electricallyactivedopantatomsincreases with temperature. o 18. LV ' ,=lV:=' Cor where p1 is the additionalchargeaddedjust below the oxide-semiconductor surfaceby ion implantation.Cox is a parallel-plate per unit area capacitance givenby C"" =1 a (d is the oxide thickness,€r is thepermittivity of the semiconductor) x 10-'4F/cm c; - 1v x 3'9x 8'85 : g.63x r0-' e, = LV,c". 0 . 4x l 0 - 6 c m cm' 8'63x 1o-' : 5.4x1ol2 ions/c# 1 . 6x 1 0 - t e Totalimplant dose: t'0,\!9" :1.2 x 1013 ions/c#. 45% 19. The discussion shouldmentionmuchof Section13.6.Diffi.rsion from a surface film avoids problems of channeling. Tilted beams cannot be used because of shadowing problems. If low energy implantation is used, perhaps with preamorphization by 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 effectiveness of the photoresist mask is only 16%o. s, = l.rr"ig) .s 2 \0.2J2 ) =0.023 The effectiveness of the photoresist maskis 97.7%. 21. r=]-{u 24n - lo-, u . ' . u= 3 . 0 2 = 0.927 x 0.093 + 4.27 d: &+ 4.27 pm. op: 0.53 CIIAPTER 14 l. section(referto the figure) hasan areaof 2500 pm x 8 pfft:2 x EachU-shape there 10apm2. Therefore, are(250(|)2/2*104:312.5 U-shaped section. Each each,4 cornersquares, 1 bottom sectioncontains2long lineswith 1248squares the resistance for eachsectionis square, and 2 halfsquaresat the top. Therefore +4x0.65+2):2599.6 kO l kO/il (1248x2 The maximumresistance is then *t 1 0 8 3r2.5x2500.6:7.8 C ) : 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 :4.35 " 103 p m z: 6 6 x 6 6 p m Referto Fig.4aand usingnegative photoresist of all levels (a) Ion implantation mask(for p+ implantationand gateoxide) (b)Contact windows(2x10pm) (c) Metallizationmask (using Al to form ohmic contact in the contact window and form the MOS capacitor). Because of the registrationerrors,an additional2 pm is incorporated in all critical dimensions. (o) 56 pm I I {l+2Fm I Zli* (b) 73 If the space betweenlines is 2 p*, then there is 4 pm for eachturn (i.e., 2xn, for * 1.2 x 10-6r?r, one tum). Assumetherearen turns,from 8q.6, [, x 1tan2r wherer can be replaced by 2 x n. Then,we canobtainthat n is 13. 1 4. (a) Metal 1, (b) contacthole,(c) Metal2. (a) Metal l, (b) contacthole, (c) Metal2. I 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 polysiliconis usedfor isolation. 6 . (a) The undoped (b) The polysilicon I is usedas a solid-phase diffusion sourceto form the extrinsicbaseregionand the baseelectrode. (c) The polysilicon2 is usedas a solid-phase diffusion sourceto form the emiffer region and the emitterelectrode. 7. (a) For 30 keV boron,& : 100 nm and A,Ro: 34 nm. Assuming that .Ro and A,Ro for boronarethe samein Si and SiOz the peakconcentration is given by J-zntn, J:19f- tl2nQa x l0-') =9.4xro,u cm-, The amountof boron ions in the silicon is f;=rffiq"*lW s [2 _ e rf,c ==1 l _ +_ (n"-d.tl 2L [^/2m,.,1-1 ^ ( -=:- 7 5 0 ) l I I ll =-t 8xlo"[^ 2 z-Eltvt L \Jz*340)l = 7.88 x 10"cm-' Assumethat the implanted boron ions form a negative sheetchargenearthe SiSiO2interface, then x 10") =-J:6 1 10-''x (7'88 LVr = re)/ ' l q ) Co, (b) For 80 keV arsenic implantation , &:49 3.e*8.8m=o'elv nm andA & : 1Snm. The peakarsenic 10'u - = 2.21x102, concentration i, Jcm-, . 42nA,R" r/ax(l8x l0-') Rp=49oi (As) tu;\ \ \.o"r.*. \ \ \ \ \ I \ /ar\ {ol5 (LOWER ScaLsl 1 BORON Re-toooi (B) I d -tso.r-zsoi o tooo -zm x(i) - FoRcHAls{EL Rectfi 8. (a) Because (100)-oriented silicon 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 adequate isolationbefween neighboringMOSFETs. (c) The typical sheetresistance of heavily dopedpolysilicongate is 20 to 30 Q fl, which is adequate for MOSFETswith gate lengthslargerthan 3 pm. For shortergates,the sheetresistance of polysilicon is too high and will causelarge RC delays. We can use refractory metals (e.g., Mo) or silicides as the gate materialto reducethe sheetresistance to aboutI O /t1. (d) A self-aligned gatecanbe obtainedby first definingthe MOS gate structure,then using the gate electrodeas a mask for the source/drain implantation.The selfaligned gate can minimize parasitic capacitance caused 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 implantation masks,and for passivation to protect devicesfrom impurities, moisfure,and scratches. 9. The lower insulatorhas a dielectricconstantAlq: 4 anda thicknessdr l0 nm The upperinsulatorhasa dielectricconstant A/q : 10 and a thicknessdz: 100 nm. Upon applicationof a positivevoltage Vc to the extemalgate,electric field E 1 and E2 arc established inthe 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 stored charge doesnot reduce E1by a significant amount(i.e.,0.2>> 2.26x10s lQ l, *" can write Lt = o.zx (o.zsxtO-u)= 5 x t0-sC Q = foadf x 0.2 LV ' -:Q =, 5xlo-8 c2 (lo*8.s@=o'565 v = 8.84x10-7 (b) when t -) a,J -+}we havelgl-+ 0.212.26x10s C. Then L,Vr=t= , 8 . 8 4 xl 0 - ? ( t 0 x 8 . 8 5x 1 g - t 4 ) 7 1 6 - 5 =9.98 V. 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^- = €"rn ''7 = 3.5x l0 t F/cm2 d and the maximumcurrentsupplied by the deviceis -v,)' =::+3.5xt0r (vo -v,)'= 5mA r", =!\ rc,,(vo 2 L20.5ttn and the maximum allowablewire resistance is 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. si3N4 x Ill,*f x v - EPITAXY (o) (bl i tap* FIELD OXIDE (d) RESIST (e) 1 3 .To solvethe short-channel effect ofdevices. 14 . The deviceperformance will be degraded from the boron penetration. There are methodsto reducethis effect: (1) using rapid thermal annealingto reducethe time at high temperatures, consequently reduces the diffusion of boron,(2) using nitrided oxide to suppress the boron penetration, sinceboron can easily combine with nitrogenand becomes lessmobile,(3) makinga multi-layerof polysilicon to trap the boron atomsat the interfaceof eachlayer. 1 5 . Total capacitance gatestructure of the stacked is : "z l(z.e-) c_q d, d,l d, \d, ) 3 ' 9= 2 . 1 2 d .'.d = 3'9 =1.84 nm. = -x-/ 7 2s/(7 0 . s l 0 l \ 0 . 5 t o) l-+-l:2.12 25\ 2.12 1 6 . Disadvantages of LOCOS:(1) high temperature and 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 processing temperature and long oxidationtime, (3) no oxide thinning effect, (4) no bird'sbeak. t 7 . For isolationbetween the metalandthe substrate. 1 8 .GaAs lacksof high-qualityinsulatingfilm. 19. (a) : 2000* (el.o: x l0-'o)= 1.38 x 10-'s= 1.38 ns. (b) For a polysilicon runner RC = | R,ouo,"# ll t",- | w)\ d) \ ( ( r.\( l) --l- l(ol.or>< = 391 t0-'o)=2.07 xl0-7 s \10-",/' = 2 0 7n s Therefore the polysiliconrunner'sRC time constant is 150 times largerthan the aluminumrunner. 20. Whenwe combinethe logic circuitsand memoryon the chip, we needmultiple supplyvolrages.For reliability issue,differentoxide thicknesses are needed for different supplyvoltages. 21' (a) * t \, /r^^, = /cr^,o, /r,u,,o" uo%'='%r* 1%=fi 3 A hence (b)Eor : 16.7 A.
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