Week7HW Solutions

From these numbers we find ece 606 5 spring 2013

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online