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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 Fall '14

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