HW2 solution

# HW2 solution - (b W[2 Eu P[NJ Nd 1]” e NHNJ 0mm...

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Unformatted text preview: (b) W: [2 Eu; +P;}[NJ +Nd 1]” e NHNJ 0mm? _ 2(11.?)(3.35x1n'“)(;; + Va) 0113453 ={ 1.5xm‘” . I.I'2 lxlﬂ” +2x1ﬂ” x — NM 5x19” 5x10” 2x19” 2x10” m£ , 02M? ( m ( t } III W: [112x134 (I; #3)] Fur Ix; =3, W=t11591x1f am For F; = 3 V, W: 2.4mm“ cm (C) E 22091:) max W 5.15 For I; = {1, Em = 1.94x1ﬂ‘ Wm MN 4 (a) ﬁzﬁln[ “1”] FDIFZ=3V,EM=TDJCID Vim n! (23:10” (23:19” =(n_0259)1n — (1.51:10” 2 01' I»; = ﬂﬁTl V 5.42 For the Schntﬂiy diode. J3. = 3514:?3 .4 Leaf .4 = 53:14)“ cmI For I =1mA . in" 5x145“ We have 1;: “5L; 5,]: (0.41259) 5145] {31' J: =2Afcm2 VD = {145? V (Schnttky diode} Fur the pnjuncti-zm, JS 2 3:4:1‘3—1I A I cm: Then 2 I; = (ﬂ_ﬂ259)lr( u ] 351-5 01' VD = 0.?{15 V (pn junction diode) 6.2 (a) For T 2 300K Silicon: 4—4] ”I. 1016 = —(0.0259)1n = 41.34? V 1.5.410m 01' 01' m;[4—E‘¢Fp ]1I2 m:[4.—(11(.(7) 85510—148)(0.34T):|” 1.551049 )(10“‘) xii? = 0.30 m A150 Ignmax)‘ = = (1.5454149[101510305104) IQ;D(Inax)l = 4.3::10‘“ Cf c442 GaAs: lﬂm = —(0.0259)1n a = 41.531 V 1.8x1ﬂ and (1.5510"”)(10”) xﬂ. = 0.419 m _|:4(13.1)(8.8551"4)(U.581):|M Then IQ;D(Inax)l = 6.561104 Cicmz Germanium mm 0.10 4” = —(0.0259)1n —u = —0.155 V 5x10” 2.4110 45\$ 2 —(0.0259)1n m = 41.3294 V 1.5110 Then Surf te t' 1 -14 m ace po 11 1a : x = [W] 4:; = 2‘¢Fp| = 2(03294) = 0.559 V at!" —]9 16 (1'61“) )(10 ) We have or _ ers _ x” = 0.235 m F33 — 45m, — c: — —0.90 V Then New 9' max = 3.761004 Cfcmz ' I SD( )| V: ZIQSD(ﬂ13-X)‘+¢ +Vm (b) C“ ‘ For 2" 2 200K, we 0mm 200 m p; = (0.0259) — = 0.0122? F 4 5‘44 ‘ 300 I” = 3N FF Silicon: r11. = 1".682194 (rm—3 " —14 “2 we obtain (3514 = \$442 .v and _ [4(11-2)(3.35x10 )(0.329)] _ —19 15 x” = 0.388 m, Q;D(max)l = 5.4100“9 01"ch (153510 X53510 ) _ _ _3 or (331313. 71’. —1.33 cm 3‘0: : 0-413 .19” We obtain my 2 —0.631 V and Then x” = 0.428 m, Q;D(max)l = 0353:1093 Cfcmz \Qgﬂ(max)| = (1.6x10'")(5x1015)(0.413x10“) 01' Germanium: Hr. : 2.] 6.23101" (:10?—3 We obtain 05"?! = —0.225 V and x” = 0.232144, QMM)‘ = 4.51104 cram-2 lQQJmax)‘ = 3.309000“ Cfcm VVEEﬂmJEHd e (3.9)(8.35x10‘“) C; 2.41. 4 f 400x10 M 01' CM = 0:531:10“? chmz Then I max VI : erﬂ +173? Car SO 3.30100“a VI = —4 + 0.659 — 0.90 8.63x10 Of VI = +0141 V ...
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