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Weeks10-11HWSolutions

# 0 0 s 5 2 f s3 2 1 f 2 0 0

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Unformatted text preview: &! 2 ) 3/ 2 # 0 . ( s + 3 / 2 )F s+1/ 2 (/ F ) # 0 f0 dk \$% * m (k T 2B = ) 3/ 2 ( ( 3 / 2 )F 1/ 2 () F ) ! # ( s + 3 / 2 ) F s+1/ 2 (\$ F ) num =0 denom # (3 / 2) F 1/ 2 (\$ F ) For non ­degenerate conditions power law scattering, and 1D, !m ECE ­656 = !0 " ( s + 3 / 2) " (3 / 2) 5 (xiii) Fall 2013 Mark Lundstrom 10/27/13 3c) Work out the corresponding result in 3D. Solution: The result proceeds as in 3a) and 3b) num = "! # f = 2 xm0 k 2% % ) 0 0 () & d\$ & d' & (! sin' cos\$ ) * = gv 2m ( k BT 4! 3m* denom = "! f = 2 x0 k s ) 3/ 2 2 * E ( EC #0 , / f0 dk + k BT . # 0 0 ( s + 5 / 2 )F s+3/ 2 (1 F ) 2\$ \$ ( 0 0 '( % d# % d& % (! sin& cos# ) 2 f0 dk * gv 2m 3/ 2 k T ) ) (5 / 2 )F 3/ 2 (* F ) 3*(B 4! m 2 " x! m ! # ( s + 5 / 2 ) F s+5/ 2 (\$ F ) num = = =0 2 denom # (5 / 2 ) F 3/ 2 (\$ F ) "x = !m For non ­degenerate conditions power law scattering, and 1D, !m = !0 " ( s + 5 / 2) " (5 / 2 ) (xiii) Problem 3) summary: For power law scattering, parabolic energy bands, and non ­degenerate carrier statistics. the “transport average” momentum relaxation time in 1D, 2D, and 3D are: 1D : !m = !0 2D : !m = !0 3D : !m = !0 " ( s + 3 / 2) " (3 / 2) " ( s + 2) " ( 2) " ( s + 5 / 2) " (5 / 2 ) Knowing these times, we get the mobility from q τm µ= m* ECE ­656 6 Fall 2013 Mark Lundstrom 10/27/13 The analogous procedure in the Landauer approach is to relate the mobility to the mean ­ free ­path according to υT λ Dn µ= = k BT q 2 k BT q To compute λ () , we assume power law scattering ( λ E = λ0 ⎡ E − EC ⎣ ) k BT ⎤ ⎦ r where the characteristic exponent for the mfp is “r” rather than “s”. From the definition of the average mean ­free ­path ⎛ ∂f0 ⎞ ∫ λ E M E ⎜ − ∂E ⎟ dE ⎝ ⎠ λ≡ ⎛ ∂f0 ⎞ ∫ M E ⎜ − ∂E ⎟ dE ⎝ ⎠ () () () assuming parabolic bands and nondegenerate conditions, we could express λ in terms of Gamma f...
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