ecee302hw5 - ’11 Na Nd fl=Km I H I where Ii 2 0.0259 V...

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Unformatted text preview: ’11 Na Nd fl=Km I H. I where Ii; 2 0.0259 V and n}. 21.51101" cm‘i We find (a) 15 (a) n—side: Nd EF — En = k?" In ”1' 5x10” 2 (0.0259) 111 m 1.51:“) 01' HF — EE 2 0.3294 EV p-side: NH En —EF 2 H111 H = (0.0259)111[ 101? J 1.5110” 01' ER — EF 2 0.40m all (h) V“ = 0.3294 + 0.4070 01' VM 2 03354 V (c) NE NE g=Km 2 H. I (10” )(5x10”) 2 (0.0259)1n (15100”): I'L- = 0.?363 V 1» {2:121me _ [201513331044 )(0336) 1.611049 101-3 1 ] 1.:‘2 X — 5x10” 10” + 5110” x" = 0.426 ,um DI' Now _ [2(11.?)(3.35x10'” )(0.?36) P 1.51104” 5x10” 1 j m K — 10” 10” + 51:10“ Dr xp 2 0.0213 m We have EN 1 IEMI = d " E (1.61:1 0'” )(5x1015 )(n.425x10‘* ) (11.?)(3.35x10‘“) or |Em| = 3.29;:10‘ w cm 129 (101‘)(5x10”)] (a) I; = (0.0259)111|: (1.50:101“)2 Dr I; = 0.?96 V U2 , ee N‘Nfl, (32 AC 2 A — [2(0— +0)(Na + 30)] (1.6::10'19 )(11.7)(s.35110‘” ) 2(0— + 0.) X (10‘*)(5x10“)]“ (10“ + 5110”) 2(500—5)[ or 4121 10‘” "2 _ . I C: (5110 5)[m] 51' I For I; =0, CzllélpF For 173 23V, €20.521pF F01" 1‘; =6V, 020339;»? We can write [LT :;[W] C A2 eeNflNd For the p+n junction 61 —[‘—)1 sothat fiU/Cf _ 1 2 £1}; —E 661W; Wehave 1 2 11 For V11 2 0, — = 7.6911510 C 2 1 24 For VI = 6 V, [—] = 6.61.1:10 C T111911= for iii/R 26V, 6( 1/ C): = 5.34110“ Wefind 2 1 N1: 2 '—2 A65 611/13) 61/]! 2 (5110—5)2(1.6110_19)(l 1.?)(365110‘1‘) 1 564x10“ 6 so that N, = 4961:1015 = 51:10” amt New, for a straight line y = mx + E: X 6(1/ Cf 5.8411024 m I — Z — 6V 6 I l 2 At VE = 0, [—j = 1691:1023 = E: C Then 1 2 5.34110“ 23 — = — -1;+T.691:10 C 6 1 2 New, at [—] = D , (’3 [5.3411024 0 = — 6 which fields VR 2 4}; = —0.790 V J-VR + 1691:1023 ...
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