Fall04_Test2

Fall04_Test2 - E CE 3 52 F all 2 004 Test2 S olutions 2...

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ECE 352 - Fall 2004 Test 2 - Solutions October 29, 2004 A Si p-n juttction is measured to have a built-in l)tential of 0.9542 I/ and a jtmaion deplaion width of 0.113 pn. Whal are ND and NA in the n- and p-type rcgrons. respectivelv'! The n-ttpe r"gion is th" h"ai?r lop?d oftheJwo. From the junotion built-in yoltage equation, we can find the product NDNA as: /-. -. \ -, -, lNnNnl e v bi = Kat tnl --;- | \ tti ) ,1 Ne" "t ^ t , 0'4*/ "i 'n.tt3Y4 2eVbiIND+NA'| " \ NrN" ) = 1.0 x l036cm-6 From the junction width, we can find the sum ofthe two dopants as: "w2 LD I NA-NDNal'1, -t.o^toto"^ t -> "tbi We can now solve these two equations together to find: No = l.oxtol9czr-3 ly'; = l.0x10r/cm-: Anp-njunction is doped with ND .10'3 tm' ancl \,lA 1016 cm'3. The cross- secircnal area of the device is 2 cm'. llith o hias of V*a - 0.J y whot is th(
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This note was uploaded on 01/23/2011 for the course ECE 352 taught by Professor Effw during the Spring '06 term at ASU.

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Fall04_Test2 - E CE 3 52 F all 2 004 Test2 S olutions 2...

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