Week7HW Solutions

# I sf 0 cms which implies that there is no

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: nm wide near x = 0 where the lifetime is 0.1 nsec. Find the steady state excess minority carrier concentration and QFL’s vs. position. You may assume that the sample extends to x = +∞ . HINT: treat the thin layer at the surface as a boundary condition – do not try to resolve Δn ( x ) inside this thin layer. Approach the problem as follows. 7a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn d 2 Δn Δn Begin with: = Dn − + GL ∂t dx 2 τn d 2 Δn Δn Simplify for steady- state conditions: 0 = Dn − + GL dx 2 τn The simplified MDE equation is: d 2 Δn Δn d 2 Δn Δn GL Dn − + GL = 0 − 2+ = 0 Ln = Dnτ n dx 2 τ n dx 2 Ln Dn d 2 Δn Δn GL − 2+ = 0 where Ln = Dnτ n is the minority carrier “diffusion dx 2 Ln Dn length.” 7b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady- state problem, there is no initial condition. As x → ∞ , we have a uniform semiconductor with a uniform generation rate. In a uniform semiconductor under illumination, Δn = GLτ n , so Δn ( x → ∞ ) = GLτ n ECE- 606 9 Spring 2013 Mark Lundstrom 2/24/2013 At the surface, the total number of e- h pairs recombining per cm2 per second is Δn ( 0 ) Δx Δx cm/s is the “front RS = Δx = Δn ( 0 ) = S F Δn ( 0 ) cm- 2- s- 1 where S F = τS τS τS surface recombination velocity. Δx 10−6 SF = = −10 = 104 cm/s τ S 10 In steady- state, carriers must diffuse to the surface at the same rate that there are recombining there so that the excess minority carrier concentra...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online