Week13HWSolutionsV2

Ap write the velocity as 2 2 2 f x 2 y x

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Unformatted text preview: W = −∇i FW + J n iE − ∂t τE (ix) But now we need a balance equation for the energy flux. ECE ­656 11 Fall 2013 Mark Lundstrom For the energy flux: φ ( p ) = E p υ = E p υ i () () The associated quantity is 1 nφ ( x, t ) = ∑ φ ( p ) f x, p, t = FW Ap The associated flux in the jth direction is 1 Fφ j ≡ ∑ E ( p )υ iυ j f x , p, t ≡ X ij Ap The associated generation rate is ⎧1 ⎧1 ∂ ( Eυ i ) ∂φ ⎫ ⎪ ⎪ ⎪ Gφ = − qE j ⎨ ∑ f ⎬ = − qE j ⎨ ∑ ⎪ ⎪ ⎪ ⎩ A p ∂p j ⎭ ⎩ A p ∂p j ( (x) (xi) (xii) ⎫ ⎪ f ⎬ ⎪ ⎭ (xiii) ) ( ) Need to evaluate the quantity ∂ ( Eυ i ) . ∂p j From the dispersion of graphene 2 E = υ F k x2 + k y υx = 1 ∂E = υF ∂k x kx 2 k x2 + k y ki → υi = υ F 2 k x2 + k y = ki () 2 E υ F 2 Eυ i = υ F pi ∂ ( Eυ i ) 2 = υ Fδ i , j ∂p j (xix) Using (xix) in (xiii), we find ⎧1 ⎫ ⎪ 2⎪ 2 Gφ = − qE i ⎨ ∑ υ F f ⎬ = − qnSυ FE i . ⎪ ⎪ ⎩A p ⎭ (xx) (xxi) Finally, the recombination term is: 0 nφ − nφ F Rφ ≡ = Wi . τφ τF W Now we can put this all together beginning with eqn. (i) and using eqns. (xi), (xii), (xx) and (xxi) to find: ∂FWi F ∂ 2 (xxii) =− X ij − qnSυ FE i − Wi ∂t ∂x j τ FW Now let’s simplify this equation. Assume “slow” time variations to write (xxii) as ECE ­656 12 Fall 2013 Mark Lundstrom FWi = − q τ F W ∂X ij 2 nSυ FE i − τ F ∂x j W . Assume that the tensor, 1 X ij ≡ ∑ E ( p )υ iυ j f x , p, t , Ap ( ) is diagonal so that X xx = X yy and that X xy = X yx = 0 . The diagonal component is 1 ∑ E ( p)υ x2 f x, p, t . Ap Write the velocity as 2 2 υ 2 = υ F = υ x + υ 2 y ( X xx = ) and assume an isotropic distribution, so that on average 2 υ2 υF 2 υx = = . 2 2 The term, X xx , becomes υ2 1 υ2 1 2 E ( p )υ x f x , p, t = F ∑ E ( p ) f x , p, t = F W . ∑ Ap 2Ap 2 ( X xx = ) ( ) Putting this all together, we find 2 ∂X ij υ F ∂W 2 2 FWi = − q τ F nSυ FE i − τ F = − q τ F nSυ FE i − τ F . W W W W ∂x j 2 ∂x j Now define the “energy diffusion coefficient” as DE ≡ 2 υF τ F W 2 to write FWi = − q τ F W 2 nSυ FE i − DE ∂W . ∂x j Now let’s work on the drift term. 2 nSυ FE i = − −q τ F W q τF W W 2 nSυ FWE i = − q τF W (W n ) υ 2 F WE i Define the average effective mass for electrons in graphene as W nS m* = 2 υF ( S ) ( ) Note that the quantity W nS is the average kinetic energy per electron. Finally, we can define the “energy mobility” as ECE ­656 13 Fall 2013 Mark Lundstrom µE ≡ q τF = W (W n ) υ S 2 F q τF W * m to write the final result as ∂W FWi = − µ EWE i − DE ∂xi We have developed a closed system of equations that can be solved: W −W 0 ∂W = −∇i FW + J n iE − ∂t τE FWi = − µ EWE i − DE µE = DE = ∂W...
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