3080 Exam 2_Spring08

# 3080 Exam 2_Spring08 - ECE 3080 Semiconductor Devices...

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Unformatted text preview: ECE 3080 Semiconductor Devices Spring 2008 Dr. Klein Exam 2 Instructions: (1) Closed book, closed notes. You are only permitted a calculator and a writing instrument. (2) Show all work to receive full credit. Proctor will provide additional paper on request. (3) You have 80 minutes to work. Don‘t get hung up on one problem. If you get stuck, move on and come back later. (4) Each problem is worth 20 points. (5) Circle, box, or otherwise indicate your ﬁnal answer. (6) Answers must be legible. (7) Calculate numerical values where possible. I. The doping proﬁle in a semiconductor p-n junction is given by: No = 0. NA = B forx < -m’2 cm ND - NA = B :sin(x). 4:12 cm < x < 11/2 cm No: B. NA=0 forx> 11/2 cm This doping proﬁle is sketched below. Assume that —xp > 4:12 cm and xn < n12 cm in the depletion approximation in equilibrium. ND — NA (Ema) *9 x (cm) _ FJQ (Pun +m0v WA) ;: (fl/o- WA) - ' - (a) Sketch the equilibrium electron concentration It as a function of x in 31 — gsinu) regions of the device. Use the axes provided below. n (cm‘3) ("I l 2 n : f NA (b) Repeat part (a) for the equilibriun} hole concentration p. p (cm'3) ' B i (c) Using the depletion approximation, derive an expression for the built-in electric ﬁeld as a function of x, E(x). :Dx PQS‘DR @: xé—xf, ilz‘zzsfj 2. A bipolarjunction transistor (BJT) consists of two back-to—back pn junctions. Consider the pnp BJT sketched below: A P A r) P ollector ‘3. . f E l , 3 Grth 0 ii[1 — rn cat" ‘ “m, (am 31“ “W Ssh—912% Jessa/zen; V? agent \J-sleen that the collector-base applied voltage VCB = 0 V, derive an expression for n'i‘“ “ii the hole diffusion current at x=0 in terms of the emitter-base applied voltage VEB. «WM Sblo'lllow bi: m‘uwrt boarflif AJQQOSW‘ ‘ 'in 352/" 3. The band edge diagram of a P-I-N heterojunction photodiode is sketched below. The cross-sectional area of the photodiode is A (cmz). N (AlAs) WI The bandgap of GaAs is 1.425 eV, and the bandgap of AlAs is 2.2 eV. If the diode is not biased, and light of wavelength A = 655 nm is incident on the photodiode from the left, calculate the current in the diode in terms of the incident intensity Linc. Assume that the surface is antireflection coated, so we can take R = 0. Also, W; = 1.0 pm, and the absorption coefﬁcient of GaAs at 655 nm is a=3.40e4cm'1. in): r/L: awn”) j: 1" qASO €16de GLCXB 1’61?) LOW OAK—:6 L4,er [>5 1 LAN. [SAX ("W thx 0L..—_o ‘m RQRS/becho “(Q 99%:- e'\/ :- L-Qi‘CY—J—e ‘30 A i— 04 ‘ \A‘T’ I: 0‘ me, a 3.x .. _L_ . _. "‘"3‘: _ qA [ma LMD e, W O.‘7(o:t 4. Below is a band edge diagram ofa biased pn junction, zoomed in on the depletion region: 1.2 1 0.8‘ 0.6 E 0.4 Lu 0.2 0» -D.2* —0.4 — .6 0-6 (a) Estimate the magnitude of the applied bias |VA|. \VM': 0.2V VA 0.8 urn-counu-u-----.n-un------------—.----- «spun-un-ca F": - .............. ...._... .......... .._._.. .... -4 2 o x (m) (b) Is this diode forward or reverse biased? How can you tell? (213ch (c) Estimate the magnitude of the built—in voltage |Vbi|. COILJN‘A, — mom, Lift/Chins 0L Kw (5&0; -' \$iC‘LQ/ oi— 5. Short answer (a) For an edge-emitting laser diode with cavity length L = 10 pm and facet reﬂectance R = 0.3, ﬁnd the threshold gain per uni len th gm. ’-'- J” l C 2}? 3% 2L K c a —-\ (b) Why is silicon not used as the active region material in semiconductor laser diedes‘? £w\$iftc+ 36 WWDl' AOLS nD"— ‘ ln‘t‘ PJK‘LE c,le W; 3" ‘Q (0) Why are laser diodes generally capable of higher direct switching speeds than LEDs? R+ 5 Jr} «hols-RA m. c o m6xn>4§ow I in bcﬁvu “maﬁa UL [w J‘O‘LQS W‘l—‘ité WCCSS CNrtQ/(G (d) For the laser diode of part (a), calculate the gain per unit length 9 when the pump current is 10 times the threshold current, I = 10 Im. Sm msaqf - via \C. Ch‘M-«ié +0 WWW: oz ‘4on : LZOKpLZ arm—1 fl ...
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## This note was uploaded on 10/12/2010 for the course ECE 3080 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.

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3080 Exam 2_Spring08 - ECE 3080 Semiconductor Devices...

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