Week1HWSolutions

1 and write the answer as i1 w ece 656 3 2 m 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online