HW7Sol

# HW7Sol - 10.2 Sketch 10.3 eD A n(a llsl n 32 30...

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Unformatted text preview: 10.2 Sketch 10.3 eD A n (a) llsl : n 32 30 x3 (1.6x10'” )(2o)(10“ )(10‘) - 10“ or 15 = 3.2x10‘" A (b) _ . 44 05 (1) 10 = 3.2x10 ex => 0.0259 ic = 7.75 [.IA .. _ -14 0.6 (11) to = 3.2x10 exp => 0.0259 ic = 0.368 mA . 44 0.7 (111) 1c =3.2x10 ex => 0.0259 ic = 17.5 mA 10.4 a 0.9920 (8) 3=—=—=> 3:124 1— a 1- 0.9920 (b) From 10.3b (i) For ic =7.75uA; i3 :§=E= is = 0.0625 M, ii = (ﬂ) - ic = (gyms) => 3 124 is = 7.81 ,uA (ii) For ic = 0.368 mA, i8 = 2.97 ”A, is = 0.371 mA (iii) For ic = 17.5 mA , is = 0.141 mA , is = 17.64 mA 10.5 is 510 (a) [3: —=—=> B: 85 is 6 B 85 a=—=—=> (1:0.9884 1+B 86 — is =is+is =510+6=> is =516uA (b) 2.65 ﬁ=—=> B=53 0.05 53 a=—=> (1:0.9815 54 — is = 2.65+ 0.05 = is = 2.70 mA 10.22 (a) We have V exp i —1 EDEN” 1 K Jﬂ. = — +— L I I 3 ml A] ”ml 4:] L3 L3 "We ﬁnd that n? (sznmzlm)1 3 _] NM 2 —' = —M = 4.5110 cm N 511D .E‘ and L3 = #33130 = .. f(15)(5x10‘”) = 3.551104 cm Then (1.6x10"°)(15)(4.5x103) 8.66x10“ J : nE 0 0259 ) Slnh(070)+e x[0.70) 8 66 8 66 J“. = 1.79 A /cm2 We also have eD V l JpE = EpEO exp i _ l . — LE V; tanh x5 LE 01‘ pm=7: 8 —=225x102cm'3 and L2 = 4ng = ,/(8)(10“) = 2.83x10“ cm I] (1.6x10"’)(8)(2.25x102) J = ’5 2.83x10“ [ p( 0.60 ) ] 1 X ex —1 '— 0.0259 1( 0.8) 2.83 or 2 JPE = 0.0425 A/cm We can ﬁnd exp —1 J _ eDan 0.0259 1 (1.6x10"°)(15)(4.5x10’) — 8.66x10“ p( 0.60 ) ex 0.0259 1 —+— , 0.7 0.7 smh(—) tanh(—) 8.66 8.66 Jnc = 1.78 A / cm2 The recombination current is X 01‘ J J [EVE J R '0 2H" 2(31194)exp[ﬂ] 2(ﬂ.[}259) JR 2 3.22::10'} Afcmz (b) Using the calculated cun'ents, we ﬁnd I 1.79 T——ﬂ ——=;, JﬂE + JPE 1.?9 + [1.9425 1" = 9.97? We ﬁnd J”: 1.73 D: = 2—: HT=ﬂ994 I J 1.?9 RE and 6 _ JEE + Jﬁ. _ 139 + [1.0425 Jﬂ + JR + JPE _ 1.?9 + {1.00322 + 0.0425 5 = [1.993 Then {I 2 Mg 2 (ﬂ.97?)(ﬂ.994)(ﬂ.993) 2} ﬂ: = 0.969 Haw a [1.959 ﬁ=—=—=..~ 13:31.3 1—4:: 1—0359 — ...
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