ECE 453 Assignment 2 Solutions0[1]

ECE 453 Assignment 2 Solutions0[1] - ECE 453 Assignment 2...

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ECE 453 Assignment 2 Solutions 2.1 From the attached equations: N ( p ) = K N p h d Also, ( 29 ( 29 ( 29 2 1 ( ) 2 2 d N d dN E dN p dp D E dE dp dE p E m dp m dE E K dp and D E dp h dE - = = = = = Density of States: Note: Density of states is usually expressed as per unit dimension (length, area or volume). Therefore, for final expressions, we divide out the dimension (L, WL, AL) from both the sides. We can then compare the result with any standard textbook expression. Substituting values for each case and replacing p with E using the parabolic dispersion relation: 1 D: ( 29 1 2 2 N K m m D E h E h E = = 2 D: ( 29 2 2 2 2 2 2 2 2 2 N K m m m D E p mE h E h E h π = = = 3 D: ( 29 ( 29 2 3 3 3 4 4 3 3 2 2 2 3 2 N K m m m D E p mE mE h E h E h = = = (You can replace E by E-Ec. The above derivation assumes Ec = 0)
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Density of Modes: For modes, we can just use the expression given in the attached equations and substitute for p as we did for the density of modes: 1 D: ( 29 1 M E = 2 D: ( 29 2 2 2 2 W Wp W mE M E h p h h = = = 3 D: ( 29 ( 29 2 2 2 A A mE M E h h p π = =
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2.2
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ECE 453 Assignment 2 Solutions0[1] - ECE 453 Assignment 2...

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