{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution3

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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] ...
View Full Document

{[ snackBarMessage ]}