Week13HWSolutionsV2

Dx 3 t ix e we could proceed to derive

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Unformatted text preview: 3 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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