4 when we write the resistance as r rball 1 l 0

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Unformatted text preview: 2 1D 1 M1D ( E F ) ≡ σ ball h L 1D σ ball = 2q 2 M1D ( E F ) L h 2D Gball = ECE- 656 2q 2 2D W WM 2 D ( E F ) ≡ σ ball h L 3 Fall 2013 Mark Lundstrom 2D σ ball = 10/5/13 2q 2 M 2 D ( E F ) L h 3D Gball = 2q 2 3D A AM 3 D ( E F ) ≡ σ ball h L 3D σ ball = 2q 2 M 3D ( EF ) L h So the results are: 2q 2 M1 D ( E F ) L h 2q 2 = M 2 D ( EF ) L h 2q 2 = M 3D ( EF ) L h 1D σ ball = 2D σ ball 3D σ ball To go further, we need to specify M E F . Let’s assume parabolic energy bands: () M1D ( E F ) = H ( E F − EC ) M 2 D ( EF ) = 2 m* ( E F − EC ) π H ( E F − EC ) m* M 3D ( E F ) = A ( E − EC ) H ( EF − EC ) 2π 2 F As an exercise, you might now want to derive the ballistic mobilities in 1D, 2D, and 3D. 4) When we write the resistance as R = Rball 1 + L λ0 , we assume a constant (energy- ( ) independent) mean- free- path. What is the corresponding expression for an energy dependent mean- free- path, λ E ? () Solution: Begin with the expression for the conductance: ⎛ ∂f ⎞ 2q 2 G= ∫T ( E ) M ( E ) ⎜ − ∂ E0 ⎟ dE h ⎝ ⎠ ECE- 656 4 Fall 2013 Mark Lundstrom G= 10/5/13 λ ( E) ⎛...
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This document was uploaded on 01/15/2014.

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