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ecee302hw6 - 8.4 The cross-sectional area is I 10x10"...

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Unformatted text preview: 8.4 The cross-sectional area is I 10x10" A=—=—=5xlO_4 cm2 J 20 We have VD 0.65 J=Jsexp— => 20=J5ex K 0.0259 so that J5 = 2.52x10"° A/cm’ We can write We want 1 D" N 1' " "" = 0.10 1 D" + 1 D, Na THO Nd 1"po 01' 1 25 N, 5x10‘7 7.07x10 = + = 0.10 7.07x10’ + ‘ (4.47x10’) Nd which yields N —' = 14.24 Nd Now 2 J5 = 2.52x10"° = (1.6x10'")(1.5x10‘° X[1.25+i.1°] (14.24)Nd 5x10" N, 5x10‘7 Wefind N‘' = 7.1x10" cm-3 and N =1.01x1016 cm" 8.6 For a silicon p+n junction, I - —Aen,.2 -— —'D 1 12 =(10( (1.6x10'” )(L5x101°)2- .5 1 10 10‘ 01' Is = 3.94x10‘” A Then V 0.50 ID = Is exp —” =(3.94x10“’)exp( ) V, 0.0259 01‘ ID = 9.54x10" A 8.7 We want J " = 0.95 J + J n p eDflnPo Dn _ Ln _ LnNa ean0+erpn0 D" + DP L" L LnNa L N, DI] _ L" Dn + DP Na L" L N, We obtain Ln = Jar“, = ,/(25)(0.1x10*‘) = L" = 15.8 pm -6 =1/Dp1’po =1/(1o)(o.1x10 ) => LP = 10 pm Then 25 15.8 0.95:— 25 10 N _+_. _" 15.8 10 Nd whichyields N‘— 8.8 . NE (a) p-sxde: En — E; = kT ” 5x10" = (0.0259)1n 10 => 1.5x10 Efi —EF = 0.329 eV Also . Nd n-51de: E, —E,, = len — n i =(0.0259)1n[ 10” )=> 1.5x10‘° E; — En. = 0.407 eV (b) We can find D" = (1250)(o.0259) = 32.4 cm2 ls D = (420)(o.0259) = 10.9 cm2 /s N E J5 :(enH: 32.4+10.9 x5x110" —*’0 +10" 0" or (.l6x10 J5 = 4.48x10'“ A / cm’ Then 15 = AJS = (10*)(4.48x10'“) or 49) (115x10 -15 [5 =4. 48x10 A We find VD [=15 exp — V, =(4.48x10"s)exp( 05 ) 0.0259 01' I = 1.08 ,uA (C) The hole current is proportional to [Piecen -A—(:: -1.) 1 10.9 (=1. 6x10 1)(.5)x10(10"17 _, 10 10 or -16 I oc3.76xlO A Then I, _ 3.76x10'“ 1, _ —-—.u=> ——0.0839 1 4.48x10 1 ...
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