hw1soln

hw1soln - mV K 3.37 j = V T ln N A N D n i 2 = 0.025 V ( 29...

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UNIVERSITY OF CALIFORNIA, SANTA BARBARA Department of Electrical and Computer Engineering ECE 137A WINTER 2009 Instructor: Luke Theogarajan HOMEWORK ASSIGNMENT #1 SOLUTION 3.1 φ j = V T ln N A N D n i 2 = 0.025 V ( 29 ln 10 19 cm - 3 ( 29 10 18 cm - 3 ( 29 10 20 cm - 6 = 0.979 V w do = 2 ε s q 1 N A + 1 N D j = 2 11.7 8.854 x 10 - 14 F cm - 1 ( 29 1.602 x 10 - 19 C 1 10 19 cm - 3 + 1 10 18 cm - 3 0.979V ( 29 w do = 3.73 x 10 - 6 cm = 0.0373 μ m x n = w do 1 + N D N A = 0.0373 m 1 + 10 18 cm - 3 10 19 cm - 3 = 0.0339 m | x p = w do 1 + N A N D = 0.0373 m 1 + 10 19 cm - 3 10 18 cm - 3 = 3.39 x 10 -3 m E MAX = qN A x p s = 1.60 x 10 - 19 C ( 29 10 19 cm - 3 ( 29 3.39 x 10 - 7 cm ( 29 11.7 8.854 x 10 - 14 F / cm = 5.24 x 10 5 V cm 3.6 w d = w do 1 + V R j | (a) w d = 2 w do requires V R = 3 j = 2.55 V | w d = 0.4 m 1 + 5 0.85 = 1.05 m 3.12 j p = q p pE - qD p dp dx = 0 E = - D p p 1 p dp dx = - kT q 1 p dp dx p ( x ) = N o exp - x L | 1 p dp dx = 1 L | E = - V T L = - 0.025 V 10 - 4 cm = - 250 V cm The exponential doping results in a constant electric field.
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3.13 j p = qD n dn dx = q μ n V T dn dx | dn dx = 2000 A / cm 2 1.60 x 10 - 19 C ( 29 500 cm 2 / V - s ( 29 0.025 V ( 29 = 1.00 x 10 21 cm 4 3.34 dv D dT = v D - V G - 3 V T T = 0.7 - 1.21 - 3 0.0259 ( 29 300 = - 1.96
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Unformatted text preview: mV K 3.37 j = V T ln N A N D n i 2 = 0.025 V ( 29 ln 10 16 cm-3 ( 29 10 15 cm-3 ( 29 10 20 cm-6 = 0.633 V w do = 2 s q 1 N A + 1 N D j = 2 11.7 8.854 x 10-14 F cm-1 ( 29 1.602 x 10-19 C 1 10 16 cm-3 + 1 10 15 cm-3 0.633V ( 29 w do = 0.949 m | w d = w do 1 + V R j w d = 0.949 m 1 + 10 V 0.633 V = 3.89 m | w d = 0.949 m 1 + 100 V 0.633 V = 12.0 m 3.39 E max = 2 j + V R ( 29 w d = 2 j + V R ( 29 w do 1 + V R j = 2 j w do 1 + V R j 3 x 10 5 V cm = 2 0.6 V ( 29 10-4 cm 1 + V R 0.6 V R = 374 V...
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hw1soln - mV K 3.37 j = V T ln N A N D n i 2 = 0.025 V ( 29...

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