HW-5 solutions

HW-5 solutions - .. I I 971.3964” >xu E CE 305 mm”)...

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Unformatted text preview: .. I I 971.3964” >xu E CE 305 mm”) SOMCEIOVLS _ I_ 0 N wweofl 192m _-;?c_v\‘>2cs. 2:6: Lwlibfim) w-eloi‘m SL412 6-?) .g. f; "‘ , ‘ a ‘ ‘4‘. “I I t | :7. g . M; .M5T§'-';DV€J};—,-.,\i1 V) (fir): .. ._ _ kw :1. “Per .M give-rm _;::-_1i§;;_____}jii*tldy'xwg‘“ ,,,)o,~xr'e‘-"r§f\§e{ “ P 4' ‘4 “35°” , nomdégéwu‘éléfa “fie/9% afapife» In. ‘ _ ‘ \ j. ’ J ., “.7 Li]: L". NW», w; -- . .3.nCcfi-‘féxgfiw & vchonvo is: ,defvzla‘H-b‘f'h Wyvewiimacfimm i v-. 1. fir.) Gin)- Eym Uta ékie'rm‘na E ammw‘w. Hwy-ma w...» unnww w".:=im'.r.7.:~_:.k5c .u..:. asm‘aazagaa‘u A a - [akin-14 sch-l: nan-nu aus‘;,.-::<».ax;. m: u:..-.~:;_-._~..-:-. . ' L a 7 Am... .. . . n, _ ‘ " , .3. " V? I'— , vi," H ‘ s .w quD________;_u;i___ . 7, O < 5; , _ a m . .. .. .. .. . 10‘ is ,, ., , ._ __ . Yatmn , Ivy. C93. " , (.36 W _ wT3t€.. .. I 1%? [Ks ‘ 4c? gm. g ~m/L 7 - m Mu}; S M Xilz " N's '/Ksao “ ~ M1 \< “at” .1“ J L; 6% above , ' 4w. Aw jwfiofiwfla. 91mm __,z__.———v-' _ 2K5?» l . ‘ _ ‘-, Ifi=1j“_7 : w * OW FE : E?“ ‘5 who“ NW): mac ND " ' I $Nn IVA—W0 'Tflflwfiw . , -' . 'Wb1~VH>J*‘ . " .W__.tr‘r._'_rree=bls M efi...xg.._..3202?fiu‘e..i 2 Z \ Alf; 9L; o _ .. , ,. gga aar- MH‘ 2“ 2. (a) Sketch the band-diagram of the device shown (below along the line indicated) using the four-step method procedure discussed in the class. Assume that kBT=0.025 eV and the statistics is non—degenerate. Also use the following parameters InP 1-5 0-75 — NV intrinsic intrinsic — (b) What is the valence band discontinuity at the n—InP and i—InGaAs junction? Egz-Ecl-(Xl-X2)= 6V (0) What is the built-in potential? You should determine this by inspection of the band-diagram. 1.5 eV.... ...
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This note was uploaded on 02/05/2012 for the course ECE 305 taught by Professor Melloch during the Spring '08 term at Purdue.

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HW-5 solutions - .. I I 971.3964” >xu E CE 305 mm”)...

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