Weeks10-11HWSolutions

F m x f0 1 1 e f0 f0 f k bt 0

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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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