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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 Fall '14

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