Week7HW Solutions

# From these numbers we find ece 606 5 spring 2013

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Unformatted text preview: − en nT In equilibrium: rN 0 = cn0 pT 0 n0 − en0 nT 0 = 0 From eqn. (5.11a): en0 = cn0 n1 so rN 0 = cn0 pT 0 n0 − cn0 n1nT 0 = 0 pT 0 n0 = n1nT 0 ECE- 606 4 Spring 2013 Mark Lundstrom (N T 2/24/2013 − nT 0 ) n0 = n1nT 0 N T n0 = nT 0 ( n1 + n0 ) nT 0 n0 1 = fT = = NT n1 + n0 1 + n1 n0 n1 ni e( T i ) = n0 ni e( EF − Ei ) E′ − E k BT k BT = e( ET − Ei ) k BT ′ fT ( ET ) = ′ 1 ( ET′ − EF ) 1+ e k BT ET = ET ± k BT ln gT (plus sign for acceptor- like defects and minus sign for donor- like ′ defects) For the plus sign (acceptor- like states): fT ( ET ) = 1 + gT e( 1 ET − E F ) k BT and for the minus sign (donor- like states): fT ( ET ) = 1 1+ 1 ( ET − EF ) e gT k BT These are almost Fermi functions – except for the trap degeneracy factor (like the degeneracy factors we saw for donors and acceptors). - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Problems 5) – 13): The following problems concern the Minority Carrier Diffusion equation for electrons as follows: ∂ Δn d 2 Δn Δn = Dn − + GL ∂t dx 2 τn For all the following problems, assume silicon at room temperature, uniformly doped with N A = 1017 cm- 3, µ n = 300 cm2/V sec, τ n = 10−6 s. From these numbers, we find: ECE- 606 5 Spring 2013 Mark Lundstrom Dn = kBT µn = 7.8 cm 2 s q 2/24/2013 Ln = Dnτ n = 27.9 µm Unless otherwise stated, these parameters apply to all of the problems below. 5) The sample is uniformly illuminated with light, resulting in an optical generation rate GL = 1020 cm- 3 sec- 1. Find the...
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## This note was uploaded on 01/15/2014 for the course ECE 606 taught by Professor Staff during the Fall '08 term at Purdue.

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