HW SOLN 6

# HW SOLN 6 - ECE331 Homework#6 Solution 1(a Since n0 < ni so...

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1/4 page ECE331 Homework #6 Solution 1. (a) Since n0 << ni, so this is the minority carrier concentration. This is a p-type semiconductor. (b) The minority carrier concentration n 0 = 4.5×10 4 cm -3 . The majority carrier concentration p 0 , 3 15 4 2 10 10 0 . 5 10 5 . 4 ) 10 5 . 1 ( 0 2 0 × = × × = = cm n ni p (c) We have both donors and acceptors in the material. The donor concentration is given. Based on the space charge neutrality, the acceptor concentration, Na = p0 + Nd – n0 = 1.0×10 16 cm -3 . (d) The carrier mobilities for electrons and holes are 1330 cm 2 /Vs and 495 cm 2 /Vs, respectively (Anderson Text p. 116). So the conductivity is cm S Vs cm cm C qp qp qn p p n / 396 . 0 495 10 5 10 6 . 1 2 3 15 19 = × × × × × = + = μ σ 2. (a) The intrinsic carrier concentration at 300 K is 3 9 10 473 . 8 ) 2 / exp( × = = cm kT E N N n g v c i The current density at 300K, 2 14 19 n 1 / 6 . 1 100 1000 10 10 6 . 1 E ) ( cm A E qN qn E qp qn E J n d p n = × × × × = = + = = (b) Since J 2 = 1.05J 1 , if we assume the mobility is independent of temperature, then n 02 = 1.05n

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HW SOLN 6 - ECE331 Homework#6 Solution 1(a Since n0 < ni so...

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