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Weeks10-11HWSolutions

# In 2d we have 1 num f 2 2 k 2 xm0 2 s 2 2 s

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Unformatted text preview: quot; EC ) k BT % , \$ & s and nondegenerate conditions are assumed. ECE ­656 3 Fall 2013 Mark Lundstrom 10/27/13 3a) Prove this result. Solution: Begin with the definition of average momentum relaxation time: τm ≡ 2 υ xτ m 2 υx (i) Recall the definition of the average, X X ∑ X ( E) f ( E) . = ∑ f ( E) 0 k τm = (ii) (iii) ) E(E , & d% & ! cos % # 0 + k BT C . f0 kdk * 0 0 (iv) 0 k 2 υ xτ m υ = 2 x num denom Consider the numerator first. In 2D, we have: 1 num = "! # f = 2\$ 2 k 2 xm0 2\$ s ' 2 2 s ! 0 \$ 2 ' E & EC * num = # f kdk 2" % ) k BT , 0 ( + 0 Assume parabolic energy bands: 2 ( E # EC ) m* kdk = gV dE " 2 = ! !2 m* so (v) becomes g! num = V 20 "! # \$ EC & E % EC ) ( kT + 'B* ECE ­656 ) (v) (vi) (vii) (viii) (ix) (x) s+1 f0 dE Change variables to # E " EC & # E F " EC & !=% ( ! F = % k T ( \$ k BT ' \$ ' B so % g! # s+1 num = V 20 k BT & d# # \$# F "! 0 1+ e g! num = V 20 k BT # s + 2 F s+1 \$ F "! ( () 4 Fall 2013 Mark Lundstrom 10/27/13 Now work on the denominator: 2# & 1 2 denom = "! x f0 = d\$ % ! 2 cos 2 \$ f0 kdk 2# 2 % k 0 0 & gv ' = 2 k BT % d' ' (' F #! 0 1+ e = (xi) (xii) gv k T ) ( 2 )F 1 (' F ) # !2 B Using (x) and (xi), we find 2 " x! m ! # ( s + 2 ) F s+1 (\$ F ) num !m = = =0 2 denom # ( 2) F 1 (\$ F ) "x Assuming non ­degenerate conditions, we find !m = !0 " ( s + 2) " ( 2) QED. 3b) Work out the corresponding result in 1D. Solution: The solution proceeds much as in problem 3a). We will not show all the details. s ' E \$ EC * num = "! # f = & ! # 0 ) , f0 dk ( k BT + k \$% % 2 xm0 = 2 * 4 gv - !m* m (k T 2B denom = "! f = 2 x0 k = !m = 4 gv ' !m* 2 " x! m " 2 x %...
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