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eng 45 homework5_sol

# eng 45 homework5_sol - 15.43 Calculate the fraction of G...

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Unformatted text preview: 15.43. Calculate the fraction of G: atoms that pmvidcs a con- ducﬁon electron at room-tempcmmm. any ﬂaw 1, J fqm‘i: 53-7—3 XML-611 xmxobazM/a’H-m cm 1‘ -»-- 3 72.5? 3 = 44.!xta 174M/* 3 - 72,461 15.5 am 41,; 23m”»~'3 .-. ﬂag a,“ may M9741“ mm : 3x a” -3 _ ﬁaiﬁbq': i;— = 5.23910 m «Jana—7.53 _ 15.44. Wm: motion of the conducivin of inn-insic silicon at room—mpmm is due to (a) electrons and (b) clan-on m r = We”: “gr/amh “Jﬂﬁvﬁ r . "I —:= L: #ﬁ _. ‘7 “MW/“h? ,uu/a. u»; Tunas, 0-!- 0..“— 0.140 0,140+ 0.03: a 0.737 Satin]? , U— h - 59‘947111‘) = o—'.°_§_'————-= 0.140 4-0.0}! 15.41. Using the data in Table 15.5. calculate the room tem- pcrature conductivity of intrinsic lnSb. WWW I = (I3-5’“0’5—"5X0.Ibonxxo"’c)(&ao40.a4s)% = [.74 #0431»?! Section 17.1 - Intrinsic, Elemental Semiconductors PP 17 ,1 In Sample Problem 17.1. we calculate the density of conduction electrons in silicon. The result is in agreement with data in Ta— bles 15.5 and 17.5. In Problem 15.43, the fraction of germanium atoms contributing conduction electrons at room temperature was calculated. Make a similar calculation for germanium at 150°C. (Ignore the effect of thermal expansion of germanium.) Them“ FAA-#é-m/5143. a4. ﬂﬁht= 4-4.1x1027 Mme/M3 Aim ale-769%“ [7.2, “mm 7442 Mull-‘1. a7La-n Mala, layman: “J MW) mew. - £3 /2.£T e -{0.6b eV/2)/_A (Horns)?— W— Mlso'ﬂ : a my“ - {macaw/aw] (awn-Inn e. - (Oiéch/Jk) (TIE; ' fie-ax.) e. or .. MKO'C. ‘ “dng . I .- J— =(13xfouh'3) e: (ﬁ%mj 4J5:- Jaw.) : ‘7. 4noum'3 300K 10 -3 ' _ how: 0"4xw 1'" = 2.13610 Y “.iKJOL-Jt—‘s (Mile! TL}. advert/ugly Sax/0’” 03‘3WK, ML W114 PM 11.5. Starting From an ambient temperature of 300 K. what temperature increase is necessary to double the conduc- tivity of puxe silicon? f J... 2—1—2, Gig—E’A‘AT = e-Egjg 7'_3mn> 0'5... - qe-Eaizibwz] or E ___f____J_ LQ" '37-:(7' 300:. W .1. _ _I__ z _ L2 ' 7 sum (53/91) or I _ ﬂ é.- - 49‘: T J 3"” 1 _ 21.4.1 '1 _1_ _ 2(K.leo"¢V/K!AZ " ‘ [3“K Ala7ev : 310K .'. 47': 310K - 200K = 10K 17.10. An As-doped silicon has a conductivity of 2.00 x 10'25'2" - m'l at room temperature. (a) What is the predominant charge carrier in this material? {b} What is the density of these charge carriers? (c) What is the drift velocity of these carriers under an electrical ﬁeld strength of 200 Wm? (The n, and p... values given in Table 15.5 also apply for an extrinsic material with low impurity levels.) (a) Arsavn'c 6401. 7e». Guy-II. oF 2‘4 Mind/i: >454:- ' Mid, d-u ‘h-bﬂ ans-krill! of!“ #1 Jch a... M 0L4}: Mfr/1. Mb+ﬂﬂy ﬁllet—e Eth‘h-i 0356/4:- or- vn1= '2:— 3/: Tm 15:5- .5..— Iw) 691%.. /q,_=o.:4a~‘/c’ws) or 2.09 1:10' '12"- h." (a .m: mi *"c)(o. I40 :— V5451) ' (CL) 5'"— %VA#M _ I. 1b.?“ 5: 0.140%. Rang:- 2865/5 h: I7 _3 = 3.72::[0 H 17.11. Repeal Problem 17.10 for the case of a Ga-doped silicon with a conductivity of 2.00 x 10‘19" - m" at room temperature. (a) G'IHI'u-n (h); rem... Gnu,- m2: Juana Lf-fyfe. “an mu m (Adm Me, a #4 Palm...” any; Mm. all-’4‘)». 0': Jag/(4A or m: 37“: Twas: (or/7.1) JIM/4A: clonal/(155)“. 2.00119“ (2.": g..." (0.169: x/a”'C )(D‘CBEAVEV-SJ) h: =3.:szo"a—.‘3 1::/‘E= 0.03:;; , Rang: '24 ,_/5 42] 17.14. CIICulatc ch: conducdvity for the cxhnustion range of silicon doped with lO—ppb antimony. 17.14 A; MM». mm... mm, \$37 = m G710“..be 12175251, x Ian’s.- x 15.3 NJ SI. 1...": O.‘013ﬁl°uaﬁMSL S," lot-5H1- ar) m: L/Sxmu ThkﬁfJLm/aﬁgdwﬂ. ‘ a” Jer‘léms A 11%;. M-ﬁbpe mmﬁacfor’, A} ‘5‘.“ [5ij I74 ﬂaw-IL 0L7: 94;. “mi/“e -_- (I.ISNOMm'3)(0-/6’Vodrcﬁ"40 = 2.574%,“4' ...
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