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Unformatted text preview: Homework 4 1. Problem: Find the resistivity ρ (in ohm-cm) 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: μ n = 232 + 1180 1 + N A + N D 8 × 10 16 . 9 , (1) μ 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 p-type 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 ρ = 1 / ( epμ 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 μ p ≈ 51 cm 2 /Vs), we get ρ ≈ . 0122 Ω cm....
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- Spring '10