00_exam1spr_sol

# 00_exam1spr_sol - 6.012 Electronic Devices and Circuits...

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6.012 Electronic Devices and Circuits Answers for Exam No. 1 - Spring 2000 (Ave = 75, s = 13.5, n = 130) Problem 1: a) The sample is n-type so n o ≈ 10 15 cm -3 and p o ≈ n i 2 /n o = -3 10 5 cm . b) ø = (kT/q) ln (n o /n i ) = 15 x 0.06 V = 0.30 V n g) c) f (x) [V] d) 0.3 V Straight and flat 0.5 5.0 -5.0 x [µm] x [µm] 0.5 1.0 +5.0 -5.0 f (x) [V] 0.8 V 0.7 V 0.3 V 0.55 V Straight 0.5 V e) The electric field is 0.5 V/10 -3 cm = 500 V/cm, so dr J = q µ e n E e o = 1.6 x 10 -19 x 1.6 x 10 3 x 10 15 x 5 x 10 2 = 128 A/cm 2 dr J h ≈ 0 (i.e., negligibly small) f) x [µm] 0.5 -0.5 f (x) [V] 5.0 -5.0 -x no x po Parabolic 0.3 V 0.6 V 0.0 V x [µm] 0.5 -0.5 f (x) [V] 5.0 -5.0 -xno xpo Parabolic 0.25 V 1.1 V 0.5 V 1.0 Flat Flat 0.8 V Problem 2: a) D = kT µ e /q = 1600/40 = 40 cm 2 /s e L = (D e t ) 1/2 = (4 x 10 1 x 10 -5 ) = 2 x 10 -2 cm = 200 µm e e b) With uniform constant illumination, G, the excess minority carrier population is G t e ; thus n’(x) = 10 13 cm . c) The population decays as exp(-t/ t e ); thus -3 n’(x,t) = 10 13 exp (-t/ t ) cm e d) n’(x) will be essentially the same as it was in Part b. Prove this to yourself if it isn’t yet intuitively obvious.

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00_exam1spr_sol - 6.012 Electronic Devices and Circuits...

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