Unformatted text preview: ntegral ∞ I1 = ∫W EC 2 m* ( E − EC ) π ( E − EF ) 1+ e 2 m* ( E − EC ) ( E − E )
dE ≈ ∫ W
eF
π
E
∞ 1 k BT C k BT dE but it is easier to recognize that “non degenerate” means E F << E or η F << 0 and that F 1/ 2 (η F ) → exp (η F ) for η F << 0 so we can use the result of prob. 1) and write the answer as I1 ≈ W
ECE 656 3/ 2
m* 2π
( k BT ) exp (η F ) 2 Fall 2013 Mark Lundstrom 8/24/2013 ECE 656 Homework 1: Week 1 (continued) 3) For still more practice, work out this integral: ∞
⎛ ∂f ⎞
I 2 = ∫ M ( E ) ⎜ − 0 ⎟ dE , ⎝ ∂E ⎠
E
C where M ( E ) is as given in problem 1). Solution: From the form of the Fermi function, we see that ⎛ ∂ f0 ⎞ ⎛ ∂ f0 ⎞
⎜ − ∂E ⎟ = ⎜ + ∂E ⎟
⎝
⎠⎝
F ⎠ so the integral becomes ∞
⎛ ∂f ⎞
I 2 = ∫ M ( E ) ⎜ + 0 ⎟ dE . ⎝ ∂ EF ⎠
EC Since we are integrating with respect to energy, not Fermi energy, we can move the derivative outside of the integral to write ∞
∞
∞
∂
1
∂
1∂
I2 =
∫ M ( E ) f0 ( E ) dE = k BT ∂( E k T ) E∫ M ( E ) f0 ( E ) dE = k BT ∂η F E∫ M ( E ) f0 ( E ) dE
∂ EF E
F
B
C C C The integral can be recognized as the one we worked out in prob. 1), so ⎫
m*k BT 2π ∂
3/ 2
m* 2π
1∂
1 ∂⎧
⎪
⎪
I2 =
I=
F (η )
( k BT ) F 1/2 (η F )⎬ = W
⎨W
k BT ∂η...
View
Full
Document
This document was uploaded on 01/15/2014.
 Fall '14

Click to edit the document details