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Week2HWSolutions

# Ky mdos kx hint you

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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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