Unformatted text preview: that E. The bands flatten out as shown below. 4b) For this bandstructure, derive an expression for the velocity, υ x ( kx ) as a function of kx . Solution: Velocity is related to bandstructure by: 1 dE
υx = dk x
ECE 656 7 Fall 2013 Mark Lundstrom 8/24/2013 ECE 656 Homework 2 (Week 2) (continued) 2 kx2
2
E +αE = 2 m* ( 0 ) dE
dE 2 kx
+ 2α E
= dk x
dk x m* ( 0 ) dE
2 k
(1 + 2α E ) = * x dk x
m (0) 1 dE 1 2 kx
1
=
= υ x * dk x m ( 0 ) (1 + 2α E ) υx = kx
1 *
m ( 0 ) (1 + 2α E ) alternatively, we could define an energy dependent effective mass by: m* ( E ) = m* ( 0 ) (1 + 2α E ) and write the velocity as k
υ x = * x m (E) 5) For parabolic energy bands, the 2D density of states is D2 D m*
E=
Θ E − ε1
π 2 . () ( ) Assume a non parabolic band described by the so called Kane dispersion, 2k 2
E k ⎡1 + α E k ⎤ =
⎣
⎦ 2 m* 0 () () ( ) , and derive the density of states. ECE 656 8 Fall 2013 Mark Lundstrom 8/24/2013 ECE 656 Homework 2 (Week 2) (continued) Solution: First, find the number of states in 2D k space: ⎛A
⎞
A
A
N ( k ) dk x dk y = ⎜
× 2⎟ 2π kdk =
2π kdk = kdk 2
2
π
2π
⎜ ( 2π )
⎟
⎝
⎠ Note that N ( k ) dk x dk y...
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