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Unformatted text preview: Homework 4 1. Problem: Find the resistivity (in ohmcm) for a piece of Si doped with both acceptors ( N A = 10 19 cm 3 ) and donors ( N D = 10 16 cm 3 ). Since the electron and hole mobilities depend on the concentration of the dopants, use the following emprirical expressions to evaluate them: n = 232 + 1180 1 + N A + N D 8 10 16 . 9 , (1) p = 48 + 447 1 + N A + N D 1 . 3 10 16 . 76 , (2) where the mobilities are expressed in cm 2 /V s and N A and N D in cm 3 . Solution: At equilibrium and steady state, a semiconductor with both acceptors and donors will behave like a semiconductor with net acceptor doping N A N D if N A > N D or net donor doping N D N A if N D > N A . Thus, in our case our sample will be ptype with net doping concentration N A N D = 10 19 10 16 cm 3 (recall the concept of doping compensation). We can safely assume N A N D 10 19 cm 3 , since the error will be only of 0.1%. Thus the resistivity of the sample will be = 1 / ( ep p ) . Assuming p = N A N D and using Eq. (2) (assuming also N A + N D 10 19 cm 3 ) to compute the hole mobility (obtaining p 51 cm 2 /Vs), we get . 0122 cm....
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 Spring '10
 Polizinni

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