Unformatted text preview: (v) (vi) (vii) ⎤
dm* dFn ⎥
−
2
dx
dx ⎥
⎥ ⎦
1 d ⎛ 1⎞
+ ⎡ EC x + E k , x − Fn x ⎤
⎣
⎦ k dx ⎜ T ⎟
⎝⎠
B (viii) (ix) ! f = "# m$ x
f0 = 1
1 + e! ! f0 ! f0 !"
! f !"
=
= k BT 0
! x !" ! x
! E !x { } !" !
$ E ( x ) + E ( k , x ) # Fn ( x ) & k BT =
'
!x !x % C () ⎡
∂Θ
1 ⎢ ∂EC x 2 k 2 1
=
−
∂x k BT ⎢ ∂x
2 m*
⎢
⎣ () () () () Using (vii) in (vi), we find () ⎧ ⎡ ∂E x
⎛ 1 dm* ⎞ dFn ⎤ ⎫
⎪⎢ C
− E k, x ⎜ *
⎥⎪
⎟−
∂f0 ∂f0 ⎪ ⎢ ∂x
⎝ m dx ⎠ dx ⎥ ⎪
⎦ ⎬ =
⎨⎣
∂x ∂E ⎪
d ⎛ 1⎞⎪
⎪+T ⎡ EC x + E k , x − Fn x ⎤ dx ⎜ T ⎟ ⎪
⎣
⎦ ⎝⎠
⎩
⎭ () () () () ECE
656 2 Fall 2013 Mark Lundstrom 10/27/13 Now consider the derivative in momentum space: ! f0 ! f0 !"
! f !"
=
= k BT 0
! px !" ! px
! E ! px (x) !"
= #x
! px ( k T ) (xi) B (xii) so (x) becomes ! f0 ! f0
=
" ! px ! E x Using (ix) and (xii) in (iv), we write the solution to the BTE as δ f = −τ mυ x ⎧
⎛ 1 dm* ⎞ ⎫ ∂f0
∂f0
⎪ dE
⎪
+ τ m ⎨ C − E k, x ⎜ *
⎟ ⎬ ∂p ∂x
⎝ m dx ⎠ ⎪ x
⎪ dx
⎩
⎭ () (xiii) After simplifying the algebra, the final result: % $ f ( 0 dF
d % 1(3
! f = "# m ' " 0 * + x 1" n + T , EC ( x ) + E ( k , x ) " Fn ( x ) . ' * 4 / dx & T )
& $ E ) 2 dx
5
is exactly the same as the result for a position
independent bandtructure. Accordingly, the current equation for a semiconductor with a position
dependent bandstructure is identical to that of a uniform semiconductor: J nx = σ
3) ( d Fn q
dx ) − σ S dT dx In Lecture 24 (Lecture 15, Fall 2011) we stated that for power law scattering in 2D !m = !0 ( " s+2 () "2 ) , where s is the characteristic exponent in the expression ! ( E " EC ) = ! 0 #( E &...
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