Solution3 - 3.15 Consider a layout of an nMDS transistor...

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Unformatted text preview: 3.15 Consider a layout of an nMDS transistor shown in Flgflfi. The process parameters are: ' ND: 2-lflmcm" * Nd: l-lfl'scm" " Ki: [)5 tun - L5 = 0.5 pm * I” = I105 pm - VI": {13 v - Channel stop doping = 16.0 Jr. (pg-pa substrate doping) Fmdthe effecfive drain parasitic capacitance when the drain nod: voltage changes from 5V IDEJV. Wn=10urn SOLUTION: L :5. m 4,0 =£Cm Na IND wrest” M ={13961'Wl q n.- [L45xlfl'°} -‘ 15‘ an rpm zfltn NA 3"” =flfl26£n M =D.9688[V] q n: (“sun”) 3,9. N .N 1 C. = 4._ —d—P— - I" 2 [NA+ND] x7410 _|4 —I5| '5 : W = HISXID'QIFJCHII] C = efl-q[Nfl'~NE]_ 1 F“ 2 NA'+ND :me -l4 -l9 IS = {lLi‘xassalxm xLfixlfi xlfixlc- =3fir94xmgpfm2] ZXQQfiBE C 2X6,”=u5x104xaag4x1fi4=w4irprcm J-fh' J Jr A=wa=fixlfl=60[um3] P=2[r+w}=2(5+m]=32[um] K _2‘/_[J¢0+5-1x¢n+2.5] q- ¢'u _ 5-2.5 = 2J1]. 395? ——H—‘dmnz'5 W] = {1. 44 K '_2\/¢—[\||¢Iflm+5- m _ 05w 5-15 1'5. 9633 - 1.1'3. 4633 y'flgfifis —--—-— [ 15 =2 =fl.4fi Cmi-l'l' :Kw 'Cja'A'Fqu'pr'P =D.44><9.fixlfl'9xfiflxlfl'“ +fi.46xls4?xm'”><32xm" =5.ES[fF] 3.12 The: following parameters an: given for an [IMOS process. - x” = 500A ' substrate doping NA = l- lfll‘cm3 - polysilicon gate doping ND = 1* 10m crud ' ofide-htterfaoe fixed-chargt: density N“ = 210‘”ch {3) Calculate VT for an unimpianted transistor. Cb} What type and what concentration of impurities must be implanted u:- achjcvc V, = +2v andVT = -2V'? SOLUTION: (a) For unimplantod transistm‘. k? n,- L45 3-: mm thl’lsubsmtte) = ?ln—N—; = 0.02611: W = -fl.35[ V] kt" ND 1x101” r =—tn—-&=o.mm ——-—=u.5 v Mg“) 4 nj "usual" 9[ } (DEC = ¢F(substmte) — ¢F{gate} = 4135 - {159 = —0. 94[V] em = 1' qua.m£n'|2¢ r1 = «Jana >< 10'” x10“ x1 L?x3.35><1fi"‘ x2><l135 = 4.32xln—“[cxcm=] C _§fl_ 3.9x385x1fl‘“ M t 2 500x104 =a9x10'3[wcm3] =4} m-zxr-u 35J——M32X1W - zflflmxifixm—w wig”? ‘ ngxm" 6.9x1fl" (b)FoI V1. =2v: Negative chat-gm needed inthis case, so it must hep-type implant in the amount pf Qt: =9”: =f'1’rr -VTII]JCm flgxlfl'“ N, =f2—fl.4l}>< L6 x mm = 1135 x1fl“[cm'3] For VT = —2V,positive charges mod. must be n-type implant, lump-5 .3 N: =f2+fi.41}K-W=LMXIUL1[€M I 3.16 An enhancement—firm nMDS transistor has the fuliewing parameters: t 'r'm = [LEV - y: 0.21.?“ s l = DflE-V'l - |2¢-F| = {153V '- k': 2D um: {a} When the transistor is biased with VG = 2.3 V, VD: 5 V, V5 2 l V. Va = fl V. the drain current ID: 1:124 mA. Determine W31. {b} Calculate In fer Va: 5 V, Va: 4 V, V5: 2 V and V5: [2' V. {c} If Pf 50"] cmEN-s and (1“: CW; W1: = L-flxltfr” F, find W and L. SDLUTIDN: {a} For enhancement transistor and V“ :r t}, it must be rIMDS. VT{FSBJ = IL’rrru +"I'(‘I.I'|2¢rl+ V33 “142411!” =t}.S+IJ.2-[JI153+1—~.."t153)=fl.899{t’] VD; =4 2»- VGS — VT = LE —fl.399 zflflfll nMDS transistor is in saturation region. rpm” = kEilr-[vm - VT}2(1+AVM} + E: 2~IEfsatj L Ii"r'il'fuss ‘VT)2{1+J”VM] 2-0.24xlfl'3 =————__=14 54 zuxlufirs—usssfir+0.05“) {h} virer = vm +~r(\f|2¢FI+ V33 4PM) =G.S+t12-[n.l"fl.53+2 ~Ju.ss')=usss[v] VDS=2cVGS—VT =3—t].969=2.031 HMOS trausister is in linear region. . k' W rpm"; = E? [2wm — t»; )VM u VD5=][1+ Mm] =lfixlfl"£24.64[2K[3—fl.9159)x2—4ll+fl.fl5x2] =1.lZ[m.-4] {c} _£_ Efixlfl'fi c- “I p“ Stilt) =4xln"[Frcm2] C lD—IE L=—¥—= : 4::(1IIJI'E W~ 2.5xw'“[cm*} 1;: 24.64 Solve for W and L {W = 1'. Sim] L = fl.32[|.1m] ...
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This homework help was uploaded on 04/10/2008 for the course ECE 1238 taught by Professor Kim during the Spring '08 term at Pittsburgh.

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Solution3 - 3.15 Consider a layout of an nMDS transistor...

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