ECE3080_Homework2_Solution

# ECE3080_Homework2_Solution - GEORGIA INSTITUTE OF...

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ECE 3080: Homework #2, Solution June 4, 2009 Page 1 of 8 G EORGIA I NSTITUTE OF T ECHNOLOGY S CHOOL OF E LECTRICAL AND C OMPUTER E NGINEERING ECE 3080: Semiconductor Devices for Computer Engineering and Telecommunication Systems Summer Semester 2009, Homework #2 SOLUTION 1. Carrier Generation: Continuity Equation (50 points) Assume a 1-dimensional silicon crystal, which has been doped with 10 17 cm -3 phosphorous atoms. Using a high-energy light source, 10 14 cm -3 additional electron-hole pairs are generated at x = 0 in the stationary state (dp/dt = dn/dt = 0). (a) Can we consider low-level injection or do we have to assume high-level injection? Justify your answer. (b) Calculate analytical expressions for the minority carrier distribution p n (x) in the stationary state for x > 0 for the following two cases: (i) Infinitely long semiconductor with p n (x= ) = p n0 (see sketch (i)) (ii) Semiconductor with length W and p n (x=W) = p n0 (see sketch (ii)) (c) In case of the semiconductor with length W: give an approximate solution of the minority carrier distribution in the case L p = (D p τ p ) 1/2 >> W (hint: do a Taylor expansion of the resulting sinh function, i.e. sinh(x) x, e x 1 + x, and e -x 1 – x for small x) Case (i): Case (ii): Comment: Use the 1-D continuity equation, L p = (D p τ p ) 1/2 , and sinh(x) = (e x - e -x )/2.

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ECE 3080: Homework #2, Solution June 4, 2009 Page 2 of 8 Solution: (a) Low-level injection can be considered, because the number of injected carriers Δ p = Δ n = 10 14 cm -3 is far smaller than the majority carrier density n n0 = N D = 10 17 cm -3 . This way, the majority carrier concentration remains basically unchanged, while the minority carrier
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## This note was uploaded on 07/30/2009 for the course ECE 3080 taught by Professor Staff during the Spring '08 term at Georgia Tech.

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ECE3080_Homework2_Solution - GEORGIA INSTITUTE OF...

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