HW 8 Solution

# HW 8 Solution - 15.2{a For I = BGDK Silicon N gal-2r = In a H 10” = —{3.3259 111 = 43.34 V l'II 1.5.113 Now 4 E \$55'— l—H = ewe

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Unformatted text preview: 15.2 {a} For I = BGDK Silicon: N gal-2r: = In a H]. 10” = — {3.3259 111 = 43.34? V [ l'II 1.5.113 Now 4 E \$55 '— l—H' = . ewe 4(11.?)(8.35x1n'”)(0347} I; _[ (1.5x1ﬂ'”)(1ﬂ""‘) ] DI" IQ. = 13.39pm Ala-0 — lgmhnaxll = eNﬂxﬁ = (1.6111349111]”)(ﬂjﬂxlﬂ4) e11" lﬂgﬂhnaﬂl = 4.8.1::1Cr—E C cm: Gaﬂs: lUlﬁ g” = —{D.D259)ln 5 = 43.5811?" ’F LEIID and . - ] I. 4(13.1}(3.85I10'1'M11581) ‘1"--.'1' : “‘ (l.61‘1ﬂ"g_](1ﬂ'ﬁ_] or .th = 13.4111] gm; Then IQ;D(1naX}| = 6.:‘3453'13'E C‘ run-a3 Gennannnn .1615 gala, = —(D.D259)ln[—] = 43.156 V 2.4.110” Then 4(16](3.35x10‘1")[nlsel " ‘11-“- : “‘ (1.6.1'10'”_](10m] 01" Id: = 13.235 hm Then lggﬂhnaxﬂ = 3.?61‘10'3 C en;2 {'3} 1301‘ I = 21393:. EDD P; = {13.132519} — = QUITE? V 313E] Silicon: it: = 168.110" ("m—3 W e obtain \$55? = 43.442 V and If! 2 {1383 Fran . |Q;D[tnax)| = 5.4.rltit'E C I ran;1 Gariia: a: = 1.33 eat—3 We obtain a5? = 43.53] I? and I5! = {1423;31:13 . |Q;D[tnax}| = 5.351111]?3 C I can1 Gennanintn: a]. = 2.16.1101” cm"! We obtain \$51,; = 43.225 JV and .‘t' = {llﬁﬁtan . |Q;D[tnax)| = 4.5.1134 C em2 a"! 45.4 p-type silieon. Na = orli}lj ear—3 Then. from Figure 6.21: {a} Alinninnni gate gm 2 —o.93 V {b} n— polysilieon gate: :33” 1- —1.D8 V {C} p+ polysilioon gate: on” a +0.31 I? 15.45 n-type silicon. N“; = 113'” (“HF—3. H— poly gate. Froin Figure 6.21. 1.95:“ e 43.33 If Now. V =05 — ﬂ and CM: F3 ' m; {a} For rm. = 400 .4‘ _ (3.9)(8.35x10'“ '] CH _3 ' =3.63xm'3 Firm: ' 400x10 (1') (1.6xlﬂ'” )(1n”‘ ] F}? = 4133 — —-3- 3.63m)" 01" :33 = 43.349 V (in — (1.5x10'”](1n“] P}? = 41.33 — 8.631‘1D‘ 01" F3 I7 = 43.515 V {b} For rm, = 200 A" (3.9 1(3351113'“ ] CM. _3 ' = 1.?31‘10'7 F x cm] 200x10 {i} _ _ (1.5x10'”)(1n”'] VFE' = 4133 _ LTBxlD 01" :35 = 43.339 V {ii} (Lax-10'” K10“) VF; = 4:133 — ‘ 1.73m)" 01" V = —D.422 V FB me axle” Imam ] = 43.3294 V eff = —(ﬂ.ﬂ259}ln[ Surface potential: 59; = Z‘Iﬁﬁ, = 2((13294) = [1,559 V We have 95}, yrs 2 g‘ﬁn _ = —U.9ﬂ V Ever New 0' max VT=—|‘SD( )l+¢,+Vm ' Cf ' d 1' W e obtain 4 E‘gbﬁ '— I-ﬂ' : — ' eNﬂ _ [4(11.7')(e..e:<m;1r::r”)(rarjae),J:l (1.51-1n'lg](5.«:1n"5] er I” = 13.413 gm: Then Then |Q;D(max)| = (1.6111349)(ixlﬂlj)(UAISIIW'i] IDI' IQ;D(1113X]| = 3.3IIIJIIICJ—3 C I win"? We also ﬁnd (3.9}(33 5110'” _] En. [far 2 A = _E rm 400x10 CIT Car = 8.63.1155 cnf Then ' max VI = lei )| +6: 4p —C _ - SO 3.3DIIU'E VT = —g + 0.555;: — 0.90 ' 3.63.1113" 01" V} = +n.141 I7 5.19 From Problem 6.10. Na = 5x10]: rm:—3 . V =—ﬂ.QUV;F01‘Q;==U.VH =05 F3 - W Now lggnomxn . . Vn = C + Fig + 2 gr? D: and I ixlﬂj .95” = —(0.0259)1n —],I = 41329 V Lleu“ Also _ _ 4(11.7}(3353510—14](G-329)1LE 1dr — (1.6.151U_;9:](5Ilﬂ;s) J CIT Id: = [1413 “um Then |Q§D+1maxj+| = (1.45.1194;)(51‘1015)(94131'104] or |Q;E[max)| = 3#.3CJ.*I.'1I[III4j Cr" cm: And E {3.9)(8.35x10‘”] Cot 2 DJ: 2 - I f9: 3‘51: D. For rt = 2D Ac; C71 = 1331-104 F mf a . SO Ayn-11:)"E V“. = —ﬁ — 0.90 + 2(0329) ' L?3x10' 01" Pa. = 41.223 V For r =5GDA° I}: CI = ﬁﬂﬂxlﬂ'E Fm”: a . ‘50 3.301163 = —E — 0.9:: + 2(0329} ' 5.9m—10‘ 01' F. = +£12.36 V 1'2". ...
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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 8 Solution - 15.2{a For I = BGDK Silicon N gal-2r = In a H 10” = —{3.3259 111 = 43.34 V l'II 1.5.113 Now 4 E \$55'— l—H = ewe

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