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VLSI_Class_Exam_1_Soln_2002

# VLSI_Class_Exam_1_Soln_2002 - Exam 1 Solution Set 22979 2...

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Unformatted text preview: Exam 1 Solution Set 22979 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 = 55.6mm. The corrected background doping is 5x1ot‘13. Ndose := 5 x ION-cm— 2 Nbackground := 5-1013-cm_ 3 RP :2 0.1921-1-10‘ 4-cm Dc1_Rp:= 005664.10' 4-cm ._ Ndose — x — R 2 u 19—1 Npeak ‘_ —_‘—_7 Neg ;: Npeakexp ( P) Npeak _ 3.524 X 10 3 (2-7:)(Del_Rp) 2.061 sz ' cm 2 Nbackground N(.04-1-10_ 4-cm) = 9.52? x 1017-1— Xj := R1) + —2 (De1_Rp) -ln —--— 3 Npeak cm Xj = 4.858 x 10‘ 5 cm 20 3.54-1019 Concentration in cm-3 0 0.1 0.2 0.3 0.4 x,x,0.04 Depth in microns _ trace 1 ----------- 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). D0 := 0.76 EA:= 3.46 Q:= 5.10l4 T:= 273 +1000 "(BM _ 0.026-T D = 1.823 x 10 300 D := Do-e‘x 14 ttotal := (1.5)3600 Concentration in cm—3 Alternative solution 2i .— f'—"‘h ¥ p... D 1 ~45 ._. (0.04-1-10‘4 — x)2 4 ‘D' ttotal \I (“'D'ttotalj 1mm, = 5.4 x 103 Q-exp - N100 := _.4 _3 xj ;= _04.].10‘4+ 2 D'ttotal'ln NI 0.04-1-10 -cm Nbackgmund ‘4 19 '=7623x10_5 N1 .04-1-10 =2_g43X 10 xJ . 1-102“ 1-10I9 1-1018 1-10” _1016 1-1015 1.11:1l4 1 .10” ' ' 0 0.2 0.4 0.6 0.3 1 X 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 —x) -14 Gem A D: 3.131 x 10 4'D'ttotal N200 I: —'-_'—"—‘— _ 4 19 HE'D‘ttotali N2(.04-1-10 )= 2.169 x 10 _ N2 0.04-1-1 “ X]- := .04-lo10 4 + 2 D‘ttotal'l“ 0 Nbackground xj =9.'1?x10‘S Concentration in cm-3 Z N A bl! E 0 0.2 0.4 0.6 0.8 l x Depth in microns 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. Part a) ~0.78 2 B := 3.86-10 - ex B =0.328 “m p 0.026"? “m 300 2 2 ._ — 4) ﬂ _ B 'h Bum-[(l-IO ' hr :| BoverAum := 0.97-108- exp ﬂi- 0.026-T 300 BoverA ;= BoverAum- 1 -10‘ 4.51;"! BoverAum = 0.826 r B A .= Bevel-A A = 3.973 X 10— 5 cm ( “4 ) x0 := 0.4-1-10 -cm tau := 0hr x02 x o t . t . -= — + — tau ox1del oxtdel ‘ -— = 0,972 (hours) B 130va 3600-sec Part b) xinit := 0.05-1-10" 4-cm 2 o o x. - tau := Xm" + ""t tau= 245.29Ssec B BoverA 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 0.25 _ Xox := [0.05 + -——]-1-10 4-(:m S 0.46 Xox = 5.935 x 10* cm 2 t '- Xox + Xox t t 0] ex p01y0X'_ B BoverA an "i— : 1.723 hours 3600-sec If you assume a vertical solution (not right) the Xox is 550nm - 4 0.500 _ 4 D x := 0.05-10 cm + —-—-10 -cm — 4 0X (( _45 Xox = 1.137 x 10 cm 2 xox xox 4. — tau tpolyox --—-— = 5.246 hours BOW“ 3600-sec t := polyox B Problem #3 Worst case etch condtion is for thick oxide and slow etch. Part a) “m := 1.10— 7_cm W)— : 233.333 sec [2.ﬂ.(100% — 10%)] Timeovemch := 233.3-(100% + 10%) sec n 1 Timcoveretch = 256.63 m -256.6sec-— OxideetchJateraLtOP := 2 sec 4 . — 5 OXIdeetchwlateraLtop = 1.283 x 10 cm The bottom of the oxide starts lateral etching in the nominal case after 400n mi(2nmisec). This is 200 seconds. IIITI OXideetchmlateraLbottom := 2 l [(256.6‘scc — 2003cc)—:| sec 4 6 OXideetchJatcraLbottom z 2‘83 x 10 cm ' ‘ “m ' ‘ — 132 10" 7 Polysﬂlconetch := E-Q-ﬁﬁﬁsec Polysﬂlconctch — 5. x om ‘ ' - 2 “m 200 7 Siliconetch .= — —--((256.6-sec — sec)) 100 sec Siliconetch = 1.132x 10— cm ...
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