VLSI_Class_Exam_1_Soln_2002

VLSI_Class_Exam_1_Soln_2002 - Exam 1 Solution Set 2” 2....

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Unformatted text preview: Exam 1 Solution Set 2” 2. Problem 1 i) The implant goes in from the wafer surface through 40 nm (Oxide thickness) and then into the Silicon. Also, the best implant condition that gives roughly 200nm into the wafer is 609V with a Rp = 192 nm and a ARp = 55.6nm. The corrected background doping is 5x10“13. Ndose := 5 x ION-cm— 2 Nbackground := 5-1013-cm_ 3 RP := 0.1921-1-10‘ 4-cm Dcl_Rp:= 0.0566-1-10_ 4-cm ._ _..__Nd°56 — x — R 2 u 19 l Npeak“ 2 N“) :Z N akflp ( p) Npeak— 3.524x 10 —3 (2- ) D l R pa 2 ' cm W ( 6— p) 2-Del_Rp - 4 l7 1 := R-p + _2 NbaCkground ‘ -cm) — X '—3 Npeak 5 “"1 Xj = 4.858 x 10— cm 19 3.54-10 1-102” : 0 0.1 0.2 0.3 0.4 x, x, 0.04 Depth in microns w»— tracel ........... .. trace-2 ii) Diffusion of implant Using Class notes values for Diffusion coefficients for Boron. Assume profile redistributes with a Gaussian from the Silicon surface. With the peak at the Oxide surface (40nm). Do := 0.76 EA:= 3.46 Q:= 5.10l4 T:= 273 +1000 "(5150 -— - — := 1.5 3600 D.— Do ex 0.026”. D 51.823 x 10 14 ttotal ( ) 300 Q-exp - 4 D- NHX) := ttotal Jimmtotalj ~ 4 — 3 - N . -1. . xj := .04-1-10 4 + 2 D-ttotal-ln I 0 04 10 cm Nbackgmund —4 19 --?623x10_5 N1(.04-1-10 )= 2.843x 10 "J “ ' 1 102“ 1-11)19 E 140'8 D .E a 1-10” g N1(x10‘4) E ——— 1 1016 fl 8 1-1015 1‘10l4 13 ' - 1.10 0 0.2 0.4 0.6 0.3 1 X Alternative solution Depth in microns For Jaeger Book Diffusion Values D0 := 10.5 EA := 3.69 D :: Do-ex {EA} 0.026-T _4 )2 300 0.04-1.10 — _ Q'exp 1"; D=3..131x 10 14 4 ' 13' l:tota] “at-D-tmlj N2(.04-1-10'4) =2.169x 1019 N2(x) := N2 0.04-1-10“ Nbackground xj := .04110' 4 + 2 D-ttotal-ln xj =9.??x10‘S DIE .— 0 N2( - Concentration in cm-3 Problem 2. Oxidation, oxidation proceeds from all sides and the pillar oxides from the sides fastest. The correct equation to find the final oxide thickness is below. 0.2 0.4 x Depth in microns 0.6 0.8 Part a) 2 ~0.78 B := 3.86-10 - ex —-—— B =0.328 “m p 0.026"? “m 300 2 — 4 cm B E Bum'[(1'm ) BoverAumz= 0.97-108- exp BoverA ;= BoverAum- 1 -10‘ 4-51-13 BoverAum = 0.826 r B A .= BoverA A = X 10— 5 cm x0 := (0.4-1-10" 4-0111) tau := 0hr x02 x o t . t . -= — + — tau ox1del ox1del ‘ -— = 0,972 (hours) B 130va 3600-sec Part b) xinit := 0.05-1-10" 4-cm 2 o o x. - tau := Xm" + ""t tau = 245.295 sec B BoverA -2.05 0.026-T 300 The polyslicon region oxides fastest from the sides. Xox is 0.25um from each side. Assume that the vertical oxidation on Oxide 1 does not interfere. Assume the oxidation will finish when the bottom of the region is oxidized from the sides. The final oxide thickness is 50nm plus the polysilcon thickness 250nm!0.46 .25 _ xox := [0.05 + LJ-l-IO 4-cm —- S 0.46 Km = 5.935 x 10 cm 2 t .__ xox + xox t t 01 ex polyex" B BoverA an «L- : 1.723 hours 3600-sec If you assume a vertical solution (not right) the Xox is 550nm - 4 0.500 _ 4 x := 0.05-10 cm + —-—-10 -cm — 4 0X _45 xox 2 1.137 x 10 cm 2 t '- xox + xox ta tpol ox polyox " B Bovem u ——-—y = 5.246 hours 3600-sec Problem #3 Worst case etch condtion is for thick oxide and slow etch. Part a) “m := 1,10‘ 7_cm = 233.333 sec [am-(100% — 10%)] Timeovwch := 233.3-(100% + 10%) sec n 1 Timeoveretch = 256.63 m -256.6sec-— OxideetchJateraLtOP z: 2 sec 4 . - 5 Ox1deetchwlatemLtop = 1.283 x 10 cm The bottom of the oxide starts lateral etching in the nominal case after 400n mf(2nmlsec). This is 200 seconds. . nm 1 Ox'deetchmlateraLbottom 3: 2 ;'[(256-6'sec - 200866;] 6 OXideetchJatcraLbottom “ 2‘83 x 10 cm . . 2 nm . . - 7 Polysfllconetch := fi-—-256.6sec Polysfllconctch = 5.132 x 10 cm sec " -— 2 “m 2566 200 ) —7 SlhconetCh '_ E a.“ ‘ Isec _ sec ) Siliconetch = 1.132 x 10 cm ...
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VLSI_Class_Exam_1_Soln_2002 - Exam 1 Solution Set 2” 2....

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