{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ecee302hw6

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

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}