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( From the chaits, we find that Db and rp, in the ,-type region, are about 4.5 cm2/s and 1.1 ps, respectively. Similarly, in thel,',typ€ regior, D" and rn are about 25 cm2ls and 2.2 ps, respectively. This gives Lh = 22.2 W aud L.= 74.2 w. T\e minority concentrations are found to be ? | " lo2o n-4 = l00cm-3 , 'n" No l0l8c,,3 nJ I to2ocn 6 ,-o n^o-: - ,."".-'lU cm'. '- Na 10'ocm-' Hencq the satuIation current is - n. -nI,.^,,,]
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  • Spring '06
  • effw
  • Trigraph, Orders of magnitude, jtmaion deplaion width, Si p-n juttction, junotion built-in yoltage, width ond length

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