{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

VLSI_Class_Exam_1_Soln_2002

VLSI_Class_Exam_1_Soln_2002 - Exam 1 Solution Set 22979 2...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern