Week13HWSolutionsV2

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

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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...
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