Week13HWSolutionsV2

5 we find 1 2 e g qe x x f m

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: energy flux is φ ( p ) = E p υ x , where E p is the kinetic energy. The associated quantity is () () 1 ∑ φ ( p) f x, p, t = FWx Ωp The associated flux is nφ ( x , t ) = ECE ­656 ( ) 5 (i) (ii) (iii) Fall 2013 Mark Lundstrom 1 ∑ E ( p)υ x υ x f x, p, t ≡ n Eυ x2 ≡ X xx Ωp The associated generation rate is ⎧1 ∂φ ⎫ ⎪ ⎪ Gφ = − qE x ⎨ ∑ f ⎬ , ⎪ ⎪ ⎩ Ω p ∂px ⎭ ( Fφ x ≡ ) ( ) (iv) (v) and the derivative inside the sum is ∂φ ∂ ∂E ∂υ = ( Eυ x ) = ∂p υ x + E ∂p x ∂px ∂px x x (vi) Now assume parabolic energy bands ∂φ ⎛ 2 E ⎞ = ⎜ υ x + * ⎟ ∂px ⎝ m⎠ (vii) Inserting this in eqn. (5), we find ⎧ ⎫ ⎪1 ⎛ 2 E ⎞ ⎪ Gφ = − qE x ⎨ ∑ ⎜ υ x + * ⎟ f ⎬ . m⎠ ⎭ ⎪ ⎪ ⎩Ω p ⎝ (viii) (ix) (x) (xi) (xii) (xiii) Recall that 1 ⎛1 2⎞ Wxx = ∑ ⎜ m *υ x ⎟ f ⎝2 ⎠ Ωp and 1 W = ∑ E ( p ) f , Ωp so the associated generation rate, eqn. (8), becomes qE Gφ = − *x {2Wxx + W } m Finally, the recombination term, 0 nφ − nφ Rφ ≡ τφ is simply, F Rφ = Wx τF W Now we can put this all together beginning with eqn. (i) and using eqns. (iii), (iv), (xi) and (xii) to find ∂FWx dX F q (xiv) = − xx − * 2Wxx + W E x − Wx . ∂t dx m τ ( ) FW Now, we make one more simplifying assumption (in addition to the parabolic energy band assumption). We assume that the kinetic energy is equally distributed among the three degrees of freedom, so ECE ­656 6 Fall 2013 Mark Lundstrom W 3 We finally obtain ∂FWx dX F 5q = − xx − WE x − Wx * ∂t dx 3m τF Wxx = (xv) (xvi) W which is eqn. (A). ii)) Simplify the result to a drift diffusion equation for the energy current. Solve (xvi) for the energy flux ∂FWx dX xx 5 q τ FW + FWx = − τ F − WE x (xvii) W W ∂t dx 3 m* Assume that time variations are slow on the scale of the energy flux relaxation time, so (xvii) becomes τF dX xx dX xx 5 q τ FW 5 WE x − τ F = − µ EWE x − τ F , * W W 3m dx 3 dx where we have defined the “energy mobility” as FWx = − µE ≡ (xviii) q τF . (xix) m We almost have a DD equation for the energy flux, but we need an expression for X xx . To terminate the hierarchy, assume f p = f S p to evaluate this moment. W * () () The distribution function has an equilibrium shape and is given by 1 fS r , p = E− F k T 1 + e( n ) B () From (iv), we find 1 2 X xx ≈ ∑ Eυ x f...
View Full Document

This document was uploaded on 01/15/2014.

Ask a homework question - tutors are online