Week13HWSolutionsV2

2 when deriving balance equations in 3d we obtained a

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Unformatted text preview: which can be written as dn ! J ni ( r , t ) = nqµnE i + qDn dxi k BTe Dn = µn q We have derived the familiar drift ­diffusion equation from the first moment of the BTE. 2) When deriving balance equations in 3D, we obtained a tensor 1 υi p j Wij = ∑ f ( r , p, t ) . Ωp 2 In general, ! Wxx Wxy Wxz # Wij = # Wyx Wyy Wyz # # Wzx Wzy Wzz " $ & & . & & % Evaluate this tensor in equilibrium for a non ­degenerate semiconductor and show that: !1 0 0$ W0 # & W= #0 1 0& 3 #0 0 1& " % 0 ij What is W0 ? Solution: In equilibrium: 1 υi p j Wij = ∑ f0 ( r , p, t ) Ωp 2 !! The equilibrium distribution function, f0 ( r , p, t ) , is an even function of momentum. ECE ­656 3 Fall 2013 Mark Lundstrom Consider: 1 υp Wzz = ∑ z z f0 ( r , p, t ) Ωp 2 The quantity, ! z , is odd in momentum ( pz ), and so is pz . The product, pz! z f0 2 , is, therefore even, so the integral is finite. pυ pυ 1 Wzz = ∑ z z f0 ( r , p, t ) = n0 z z > 0 Ωp 2 2 We get similar results for Wxx and Wyy . Now consider an off ­diagonal component: 1 υ x pz ∑ 2 f0 ( r , p, t ) Ωp When integrating over υ x , we are integrating and odd function of px times f0 , which is an even function of px , so the integral is zero. (Same argument for pz .) We conclude that all off ­diagonal components are zero. The result is ⎡ υp ⎤ xx 0 0 ⎢ ⎥ 2 ⎢ ⎥ ⎢ ⎥ υ y py 0 ⎢ ⎥ . 0 0 Wij = n0 2 ⎢ ⎥ ⎢ ⎥ υ z pz ⎥ ⎢ 0 0 ⎢ ⎥ 2 ⎣ ⎦ For parabolic energy bands and non ­degenerate conditions, υ z pz m *υ z2 1 kT = = υ z2 = B L , 2 2 2 2 where TL is the lattice temperature, which is also the electron temperature in equilibrium. Similar results apply in the x and y directions, so ! kT $ # BL 0 0& #2 & # & kBTL Wij0 = n0 # 0 0 & . 2 # & # kBTL & 0 #0 & 2& # " % Wxz = ECE ­656 4 Fall 2013 Mark Lundstrom In equilibrium, the kinetic energy is equally distributed between the three degrees of freedom, so the equilibrium kinetic energy density, a scalar, is m *υ 2 1 3 ⎛1 2 2 2⎞ W0 = n0 = n0 m * υ x + υ y + υ z2 = n0 3 × ⎜ m * υ x ⎟ = n0 kBTL . ⎝2 ⎠ 2 2 2 Therefore, we can write the tensor as ⎡1 0 0⎤ W0 ⎢ ⎥ W= ⎢0 1 0⎥ 3 ⎢0 0 1⎥ ⎣ ⎦ 0 ij 3) In Lecture 32 (L31 Fall 2011), we developed a 1D energy flux equation. In 3D with spatial variations along the x ­direction alone, the analogous result is ∂FWx x , t dX F 5q (A) = − xx − WE x − Wx , * ∂t dx 3m τF () W which can be simplified to: 5 10 d (W k BTe / q ) FW = ! µ EWE x ! µ E 3 3 dx (B) Verify that (i) and (ii) are correct under the appropriate simplifying assumptions. Note that there is an error in the course text, Fundamentals of Carrier Transport, problem 5.7, p. 244. Solution: i) The general balance equation is ∂nφ = −∇ • Fφ + Gφ − Rφ . ∂t The x ­directed...
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