q1_soln - Fall 2005 6.012 Microelectronic Devices and...

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Unformatted text preview: Fall 2005 6.012 Microelectronic Devices and Circuits Prof. J. A. del Alamo October 13, 2005 — Quiz #1 problem grade joLAJ'HsA/S Name: Recitation: General guidelines (please read carefully before starting): 0 Make sure to write your name on the space designated above. 0 Open book: you can use any material you wish. 0 All answers should be given in the space provided. Please do not turn in any extra material. If you need more space, use the back page. 0 You have 120 minutes to complete your quiz. 0 Make reasonable approximations and state them, i.e. quasi-neutrality, depletion approxima— 7 tion, etc. 0 Partial credit will be given for setting up problems without calculations. N0 credit will be given for answers without reasons. 0 Use the symbolsutilized in class for the various physical parameters, i.e. pm, I D, E, etc. 0 Every numerical answer must have the proper units next to it. Points will be subtracted for answers without units or with wrong units. 0 Use d) = 0 at no = 1)., = m as potential reference. 0 Use the following fundamental constants and physical parameters for silicon and silicon dioxide at room temperature: 71,- = 1 x 1010 cm—3 kT/q = 0.025 V q = 1.60 x 10-19 0 es = 1.05 X 10‘1?F/cm (503 = 3.45 X 10‘13 F/cm 1. ( 25 points) A bar of silicon is doped with acceptors as shown below. The doping density variers smoothly and motononically in the a: direction from N A << 71,- at :5 = 0 to N A >> 71,- at :v = L. Na(L/ 2) = 71,-. Around :5 = L, the acceptor profile becomes uniform. The donor density is zero everywhere. This is a thermal equilibrium situation. log Na V uniform , [ doping L 2 On the basis of this description, answer the following questions by circling the correct answer. Write a brief justification for your choice below. (1a) (2 points) Where is the hole concentration the greatest? $20 0<$<L/2 :v:L/2 L/2<$<L uniform TR kbig wmwlmk‘pc i! areal-Ml NM Na i; whim}- (1b) (2 points) Where is the electron concentration the greatest? $=0 0<$<L/2 :v=L/2 L/2<$<L $=L uniform \ TN dukes Lovzmuln’klek ‘1 'ynci‘ml «ame “M kak uwwirzfiw u \s-wvd' (wwnbx L who“. NA \‘J Iowa}. (1c) (2 points) In whichdirection does the hole diffusion current flow? Q) no current +:’v’ lieu?» abikbaob 010w ‘l'hk WauUuJ‘J (\‘(ial 4“ \L'fi‘_ Md’4N we Lilla/iot— WmIL pom [mm by} 1% L4}. (1d) (2 points) In which direction does the hole drift current flow? —:L' no current ( £5 ) 1y iWMA eXQl/ILYIVW h‘u (yr/Lu” km ‘l‘a LC BMMah‘h 'l’o \N—u $11VM\°\ Lyryd‘ (1e) ( 2 points) In which direction does the electron diffusion current flow? @ nocurrent +:i" ammo aunts“ 1% ML l~ fikk 7% , m mum“ cvrer—J’ 110w; {M (7‘th 1L: (1f) (2 points) In which direction does the electron drift current flow? —a: no current GED curthl‘ lab“ in, 5L IY‘MQHi‘k l‘a aLfMuw‘m uA’rA “I , I (1g) ( 2 points) In which direction does the electric field point? __55 . no field +5 7v» 7, n1 9%. +k4~ vb quJ l» Afill +. rfikL «Kl ckctwwlt‘ am} b 4m {4% (1h) ( 2 points) Where is the electrostatic potential the greatest? a:=0 0<a:<L/2 a:=L/2 . L/2<a:<L a:=L uniform TN' yu'lih‘lx‘d }, h}er 0w “4L “w! uL 4m /awLy1\4 92kg“ AILLV>L I; ' w-MkBIVi CVL7wN’r’K. (1i) (3 points) In the axis provided below, sketch the volume charge density along x. MgrL Aakh luck I « Lam F’O cal X; ti C :K i”; I ;+ I‘) 4" “\lk’ aL X ; cn’AAQ “A. NWJJ‘HB M M-rL AtLLWLN-O Mme “WA, (lj) (3 points ) In the axis provided below, sketch the electric field distribution along 1:. E(x) (1k) (3 points) In the axis provided below, sketch the electrostatic potential distribution along 11:. Use as reference (I) = 0 at no 2 p0 = 11,-. ¢(X) ,{gr Malt) 3k lul'dmic Wv-n-Lvoi‘ry ’2. (10 points) Consider an MOS structure on an n-type substrate: VGB semiconductor contact (nrrtype) contact The doping level in the substrate (or body) is N D = 1017 0771—3. The doping level in the gate is N E = 1020 cm‘3. - (2a) (5 points) In the axis below, qualitatively sketch the volume charge density across this structure at zero bias. Explain your result. 900 1N Lul\}— :V‘ 9L *M’TJ J J‘NL;Wk 3; “,1 4 2 4 -4 ., expanse- on- ms» [3 Sr y. 01' m ' [BIB - 'o.15v 71421 F1 yt-M‘KVQN 7M), Amends a )hf‘w’w‘u 1m} ‘F U4"? ‘J‘ H“ mm . «A. warm w ~ WW my a m Mamet”; m \ @Ml) 63) h .0.» '1 +5 lilch an GCLuHA'Lcfi‘Qs {AW .fi ‘ pl” 'U\J'L-’MLDV\ ivfilfl‘r ihlf—r‘fihf. (2b) { 5 points) Calculate the flatband voltage of this structure {numerical answer with appropriate sign and units expected). \ (he lam 3*: and) at ngvl VAlML *L V543 -l—o mya ‘J- 1N cud—mu dLmevkL ‘bCou‘ ,(4 3. (30 points) Consider a pn junction at zero bias with an electric field distribution as sketched below. The metallurgical junction is placed at a: = 0. ‘E(x) [V/cm] 105 0 50 100 x [nm] (3a) (10 points ) Calculate the depletion capacitance at zero bias ( numerical answer with appropriate sign and units expected). ' A'l' W RM, DQJJJJJFBH. erbgk 9} [93 MWK‘ TN aavmu'lmkk TI 4%? 6 l' L F/LW “L CJ\ ; ; ’ loSy° : L’xh-? p/UM XA /co,|=’1cw\ (3b) (5 points ) Calculate the built-in potential (numerical answer with appropriate sign and units expected). TM/ {I QMMI) 4N lull-VJ all 4-“ ckb‘l'"\c 1AM A‘l’ Um 9:111: (3c) { 5 points) Estimate the doping type and doping level of the region between 50 < a: < 100 nm {numerical answer with appropriate sign and units expected). :Vl‘l. T) Y\‘ V;va ‘(W X 7 o ) 4R) ‘2 I) rzfou‘ .L -\-\--k P ~v\ SuthV’p". TWL LI ’ (cm 3-“ WVVK Lyn/M YLLVCYLJ H,pr ‘u JtNmrhA L) ,VALéMWJAlw “Luphfl_ .t,‘ \ T,w HM roex ¢ (6: MW r%a\ -HN/YL u a rawyx. ’TI’V?’ 4"ka a [,de C‘AAYK Leawutragol. y‘vw L3 Anupucvlém "-47192 oLq/QB 16.4. “m mama we? Mil, 7w. 6“ w a p we G's — __ 2 ' 5—— -' ——-. 3‘ \ 0”“ e; x cm S°VID—}LW (3d) (5 points) What can you say about the doping type and doping level of the region between 0 < a: < 50 nm? 1“ mg, a?“ MN influx LN“ 7; 5le Pull“ (Wt "L kw MAL Wi‘allar \A ‘mL’R'M-x T) lun «i 51:. | [n HAW I \ah 4R w J r 3 w / (ALpl’ylg V‘LLJ ohuuli‘ CVAV. l’KUl/Hk M""W VOL/0N. char“ dunkivl> \ 3} 4H [Lek *l NAM/L Wt jg]; Mh‘MLi‘Lr/K \IK 9N.- perx\ovJ Allie“. )‘WML, ‘H‘L Lml‘v («a chp ya) {D - \ ll '2 N: (4 /.'3,L|= cw (3e) (5 points) What can you say about the doping type and doping level of the region defined as a: < 0? TV) \I( V\'—‘ifir,( lfl.&>.~._ CXLLLJ‘ 'L AW finkfie‘ va‘ok qylwucr‘l WMLL +\I-k is/QVB levuL ll qu “(L MK 4N JuAL «L +14» cL.y\;.~> Lava .‘x +W IOLKL1°0 aw— \ ruvhvfi l’UMLL 4% L9,} (NL LCM J4) u: ) NA 3) '_'3,‘(’ OWL 4. (40 points) Consider a MOS structure as sketched below: VGB FW‘VT—V’WT 3 semiconductor contaCt ‘» A l. contact The oxide thickness os to: = 50 nm and the doping level in the substrate is Na = 1016 cm’3. This problem is about calculating the hole concentration at a: = 0 (the oxide-semiconductor interface) under the following conditions: (4a) (10 points) At flatband (numerical answer with appropriate sign and units expected). 18+ “sud, L Mwl¥~>v' ML wk «wirz’cim at x» r» 3 _ ciuak "l, “M Awe.“ la 4N bl); (4b) {10 points ) At threshold { numerical answer with appropriate sign and units expected). M— wwwu, L) «MW—HQ- ‘+N swim Loud“ ‘ox A n, ‘51 Dual 4" “AL :3“ ‘ \de "P 4-H kn: : “W > n (4d) (10 points) At a condition when the capacitance per unit area of the MOS structure is 50 nF/cm2 (numerical answer with appropriate sign and units expected). dI‘T-LEYH}, ‘4 L‘V‘A V¥k QC gk’fi‘hmw Kl I fifiokg I I” F -lg '69 .9 P W -? - Cox: .1 —_ 2 " h ._ eff“. f-lcw‘L : “MP/cw t3" Sox‘la'} cw TW} 7/ a WJM AKA/5g” J-km- 4H Unfit/3M4“ yw‘vw 43o“ [ZN/14% tlA/‘u [#‘Nl/“VVL 1‘! ,‘c‘ VLLrtubCPL' KWQ YICR camnyL" flag “La/kl ,rf Muk‘OL Y’W\o\ rLW\u\M3 WL 14ch M LcythA‘J-‘orj ,‘vx' Jzn‘fi: I _ I ' A , __ + _ C can C; _ ) [NM 1 : I 1 - i: J, _ i I I X4 C— C°* ga - 66 TM CI‘L x" ’ 5i _ /.c>S:l9 F/LM : XTx‘O—é cm 1 gg “M ' -‘z ‘ lilxl" F/L’V'x —I\I\L Suf\+,‘\_ yi-*lutig\ acre.” Ahuggk “Livy; h ycvw 1 ,_ ._ i ’1. Va firm a the"? c “Mm 3‘ (mm ‘ m) ,1 ' P“ ’ *- 2.4x)” V .Lx Lbrxlo'rl’ F/LH" WQ “9” d"- e\ CmeLll—utcoh- Hm!" +- ‘H’LaL aL I‘m/F 4C: L, 3" w 2.6 o' #3 MW») ’— PUG?) "h - urge. ’9 Q>=Y\- __:—1—§— : Zine"; UM ! 9.9 L (4c) {10 points) At a condition in which the potential build up from the quasi-neutral body of the semiconductor to a: = 0 is 0.5 V {numerical answer with appropriate sign and units expected). ’WL (an v11 4%. R¢HZIMM~IA I'L\c}7:a\. L. rLiA‘i‘k cur/9&1» (‘Mlh)'r:.tt‘°kq “L“” H“ Al/tLth‘P“ rufiou -L A M‘J I'LNL-iwrg pVLbLLV L43“; ‘,/.'. V!g ._ $904“ ) 'l‘—'.-.Ii~+«\(‘x; ya) ; L1)“ ROI“) 5 No) !~\V\‘Wj (or r1(x:9\ _ 1 a , _ c .. I A Mx») ~ P04») 6*: -1? [ ¢Cx_-) 4 RAM} ] - t Z 2hr) — : LZ'V’Iy‘o-J- (“fl—2 9°15 ...
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q1_soln - Fall 2005 6.012 Microelectronic Devices and...

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