HW 9 Solution - 6.2!} {a} For f = 1H: We have EH...

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Unformatted text preview: 6.2!} {a} For f = 1H: We have EH (3.9}(3.35x10'“] L“! .- d. r 4001-1 0'3 WI 01' CV = 3.531-10'5 Ff mf I}... and C“ _ _ (3.9}(3351-1 0'“) I’I‘I _E [3.9) IIIWJIESQHI1.?)(835xlU'H) 400x10 + — |—_ 11.? \I {1.5x10'19){10”) OT (7;? = 6.43.1104 cnf I 13 4 W] 9N u! L". n. "I II l—I Now . 10"5 9!; = —(0.0259)1n —_fi = 4134—; V Lfixlfl” Then Then 4(11.?)(8351‘1fl—H)(D.34T} '1 A“ _ (1.6x10'”)(10”] CIT rm. 2 0.33 pm so That Cr (3.9](835110‘1‘) m 4UUI1D_S —(£](n.30x1u'4] 11.? ' CIT (it = 2.4?1-10'3 F r‘mf Also C'fi'm'} = Cg. = 8.I:53.14:1Ifl_E F r" cm: {'3} For f = 1 MH: . we have u CW = 3.533.1{1'3 Ff mf 6;? = 6.43:10'“ cnf C“. =2.4?I1D'3 F :‘cnf :IIIJL and C(nn') 2 Erin = 2.4T1'1D'3 F fem] 1301' the ideal MOS CHpE’ICiTOl‘. 95's = D . then VIE = = 3.2 — [3.25 + 0.56 + 034?} 01" 55 V = 43.95? V Also — |Q;p(11133)l = (1.1511049 :](1”:](DISUI1fl—4 or |Q;D(1113X)| = 4.311(3—3 C rm: Haw r536 K“. = IQ;[IHHH}|[ ]+ gar"m: + 2‘65? E 0.1: (4.311134 )(4001-1 0‘9 I] = — 0.957 — 201.34? 1 [3.9)(8.35xlfl ' ] 01' pin. = 43.293 V 15.21 {a} At f 21H: E _ (3.9}(3.85x10‘“ ] [jar = i = g rm 49mm 01' CM = 3.15mi?g Eff-m: A130 'v' En: £5.13 2 I ED, TH") E, rm. + ‘ | — —‘ E3 V e efiffl [3.9)(8.851‘1[}'H] = I - '— - —1_-- _S 3.9 1(0.0359][11.,=(3.83.110 ] 40mm — I'fi 11.? \I (1.5m) _](5x10 _] UT 6;? = 3.4mm"? Frcm‘ Now I Eat Cm = F:1'+ IJI-a'i' E: We find I. 2 4 E and If: 2 5b!" eNd 4(1 1.?)(8351-10'” )(mm) H _ (1.5.110‘” )(51-10“ ] CIT I” = 1.13 pm Then [3.9 1(335x10‘“ ] [if = . IL‘IIII. 3'9 I I arming-11$S —( )(umm‘) 11.7 - GI" 63;: = man-mg F; rm] Also C'O'J'n'} = C3. = 8.631(1ij5 F x" cm] {13} At CW = 13.53::1crE Ff cm] C}? = 3.42.110” Firm; 311 c”: = n.?9?x1n'3 F; rm] and C'Gm') = {Trim 2 DJQTHD'E F r" em] {C} For the ideal oxide. VH3 = gbm = 3.2 — [3.25 + 0.56 — (1.2?) or V = 43.34 I7 F5 We find I I I IQ;B(111£13]| = (1.1511049KSIIUI;I](1.18.151C|'_1I] CIT IQ;B(1113X]| = H.9441'1D_3 C." em] Then ITI r f? pr}? 2 _|__OSD(111aX}|[ D" ]+ £5“... — 25‘s»: n'J 1' {0.9443163 Katmai-10's ] = — 0.34 — 203.2?) (3.9)(8fiiflfl _] 01" K = 43.939 V JP 15.2? {a} 11-type {b} We have 200x104; 3 ? Cm, = —_, = 1.1112} F cm" 2110 ’ Also Err E [3.9}(3.85x10'”] Cm. = 4:3 rm = =—, ' rm: I C5: or {C} Ql‘ p23 : gbw — 55 {301' or Q; —fl8fl==—fl5fl— _7 1'3 01' Q" = 31113—3 CECE”: = LETSxID“ cm"; {(1} C" E: = — (3.9}(335x10‘H ] l— _5 39 I' (11.7}(335x10‘1’) 3.45310 + ||(0.0259) —.- 11.? (lfixlfl'm_](2.1'1fl15:] or 6;? = 1823:1'3'5 cnfi and CE = 156 pF 6.23 {a} Point 1: 111156131011 2: Threshold 3. Depletion 4: Platband 5: Accumulation 45.39 {a} We have Wu C V ‘3 Iflfim] = ‘ “ ”" (VG= 45.) Eli ' ' New fixlfid —1+' 525](3.9}(8.35x1fl ]( )1 = — —.' 5— CLTS L 2 (40mm?) which yields W — = 14.7 L {b} 6x104 _ 3cm )M ' ' (5— [3135}2 L (4fl0x1fl4] which YiElds '3‘ — = 25.? L ...
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This note was uploaded on 10/03/2010 for the course ESE 231 taught by Professor Kamoua,r during the Spring '08 term at SUNY Stony Brook.

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HW 9 Solution - 6.2!} {a} For f = 1H: We have EH...

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