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Fφ x ≡ ∑ ⎜ m*υ x ⎟ υ x f x, p, t = n m* υ x . Ω p ⎝2
2
⎠
The associated generation rate is nφ ( x , t ) = ECE
656 ( ) ( ) 16 Fall 2013 Mark Lundstrom ⎧1
∂φ ⎫
⎪
⎪
Gφ = − qE x ⎨ ∑
f ⎬ . ⎪ Ω p ∂px ⎪
⎩
⎭
The term inside the sum is ⎛1
2⎞
∂ ⎜ m *υ x ⎟
2
⎝2
⎠
∂φ
1 ∂ px
=
=
= υ x , ∂px
∂px
2 m * ∂px
so we find ⎧
⎫
⎪1
⎪
Gφ = − qE i ⎨ ∑ υ x f ⎬ = J nxE x ⎪
⎪
⎩Ω p
⎭
Finally, the recombination term is: 0
nφ − nφ Wxx − Wx0
x
Rφ ≡
=
τφ
τW (v) (vi) (vii) () (viii) xx Now we can put this all together beginning with eqn. (i) and using eqns. (iii), (iv), (vii) and (viii) to find the Wxx balance equation as ∂Wxx
Wxx − Wx0
d ⎛ nm* 3 ⎞
x
=− ⎜
υ x ⎟ + J nxE x −
. ∂t
dx ⎝ 2
⎠
τW (ix) xx 3
To terminate the equations, w would need to develop an approximation for υ x . 8) Derive a balance equation for nk BTe , where Te is the carrier temperature (not the carrier energy). Solution: The general balance equation is ∂nφ = −∇ • Fφ + Gφ − Rφ . ∂t (i) () For nk BTe , we need the corresponding φ p . From eqn. (5.45) in Fundamentals of Carrier Transport, 3
1
nk BTe = nm* c 2 2
2
nk BTe = ECE
656 nm* 2
1
m*c 2
1 c=∑
f x , p, t = ∑ φ ( p ) f x , p, t 3
Ωp 3
Ωp
( ) ( 17 ) Fall 2013 Mark Lundstrom We conclude that
φ ( p ) = m*c 2 3 . (ii) (iii) (iv) (v) The associated quantity is nk BTE 1 ∑ φ ( p) f x, p, t = nk BTe . Ωp
The associated flux is 1 ⎛ m*c 2 ⎞
1
m*c 2 Fφ x ≡ ∑ ⎜
υ x f x , p, t = ∑
υ dx + cx f Ω p⎝ 3 ⎟
Ωp 3
⎠ ( nφ ( x , t ) = ( ) ( ) ⎧1
⎫
m*c 2 ⎫ 2 ⎧ 1
1
⎪
⎪
⎪
⎪
Fφ x = υ dx ⎨ ∑
f ⎬ + ⎨ ∑ m*c 2 cx f ⎬ 2
⎪Ω p 3
⎪ 3 ⎪Ω p
⎪ ⎩
⎭
⎩
⎭
2
Fφ x = nk BTeυ dx + Qx 3
The associated generation rate is ⎧
⎫
∂φ ⎪
⎪1
Gφ = − qE x ⎨ ∑
f ⎬ . ⎪
⎪
⎩ Ω p ∂px ⎭
The term inside the sum is 2
2
2
*2
∂φ ∂ m c 3 m * ∂ cx + cy + cz
2
∂c
=
=
= m *cx x
∂px
∂px
3
∂px
3
∂px
so we find ⎧1 ⎛2 ⎞ ⎫
⎪
⎪
Gφ = − qE i ⎨ ∑ ⎜ cx ⎟ f ⎬ = 0 ⎝3
⎠⎪
⎪Ω p
⎩
⎭ ( ) ( ) ) = 2 * ∂ (υ x − υ dx ) 2
m cx
= cx , (vi) 3
∂px
3
(vii) because cx = 0 . The electric field does not generate temperature because temperature is random kinetic energy. Finally, the recombination term is 0
nφ − nφ nk B Te − TL
. (viii) Rφ ≡
=
τφ
τT ( ) e Now we can put this all together beginning with eqn. (i) and using eqns. (iii), (iv), (vii) and (viii) to find the nk BTe balance equation as ( ∂ nk BT
∂t ) e =− ( ) d⎛
2 ⎞ nk T − T
nk BTeυ dx + Qx ⎟ − B e L . dx ⎜
3⎠
⎝
τT (ix) e We could proceed to derive a balance equation for the heat flux, Qx by similar procedures. ECE
656 18 Fall 2013 Mark Lundstrom 9) Using the balance equation approach, derive a set of coupled current equations for near
equilibrium transport and compare them to the results obtain from the Landauer approach (or the near
equilibrium BTE). Solution: The coupled current equations are: dT
E x = ρJx + S
(A) dx
dT
J Qx = π J x − κ e
(B) dx
We’ll do the first equation and leave the second one as an exercise. It is most convenient to derive the equation: J x = σ E x − Sσ dT
dx The appropriate function of momentum is φ ( p ) = ( − q )υ ( p )...
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 Fall '14

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