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# Notes14 - ECE215B/Materials206B Fundamentals of Solids for...

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Unformatted text preview: ECE215B/Materials206B Fundamentals of Solids for Electronics E.R. Brown/Spring 2008 1 Transport Theory #8 Canonical Examples of Semiclassical Scattering Theory 1. Ionized Impurity Scattering Ionized point-like defects, such as donor and acceptors in semiconductors, are fundamentally important and common in solids. And because they are not charge-neutral, they create an electrostatic potential that can scatter “free” carriers such as electrons or holes. But when the solid is a metal, semi-metal or semiconductor, there is generally a high enough concentration of such carriers that the electrostatic potential is not the simple Coulomb potential from electrostatics. There is a collective effect of the charge carriers as a whole to “screen” the impurity potential such that the normal 1/r variation of the Coulomb potential is multiplied by another term – a decaying exponential - that causes the potential to decay away much faster at great distance from the impurity. This is the screened Coulomb electrostatic potential (for impurity charge q S , ionization number Z) ( ) ( ) ( ) 4 e x p S r D V r q r r L πε ε = − (1) At low carrier densities, L D is the Debye length, L D = [(k B T ε r ε /(ne 2 )] 1/2 (2) and at higher densities it becomes the Thomas-Fermi screening length. To calculate the scattering effects with (1) and (2), we first need the perturbation Hamiltonian from a Bloch state k 1 to a Bloch state k 2 . In the spatial representation this is given by k 1 k 2 θ /2 |k 1 |sin( θ /2) k 1 k 1 k 1 k 2 θ /2 |k 1 |sin( θ /2) k 1 k 1 Fig. 1. ECE215B/Materials206B Fundamentals of Solids for Electronics E.R. Brown/Spring 2008 2 [ ] ( ) 2, 1 3 exp p k k q H V r j d r V = − ⋅ ∫ G G G 1 2 k k r (3) where q P is the charge of the incident particle. ( ) 3 exp( / ) exp 4 P S D r q q r L j d r V r πε ε − = − ⋅ ∫ G G G 1 2 k k r (4) ( ) ) 2 exp( / exp sin 4 D P S r r L q q j r d r d d V r θ θ φ πε ε − = − ⋅ ∫ ∫ ∫ G G G 1 2 k k r (5) where the last step is just re-expression in spherical coordinates. This is a classic integral of science and can be solved simply by first defining the direction of the vector − G G 1 2 k k (which is fixed during the integration) arbitrarily along the z axis of a same spherical coordinate system. Hence, ( ) | | | c o s c o s k r θ θ − − ≡ ∆ ⋅ ⋅ G G G G G G i 1 2 1 2 k k r = k k r | , and there is no dependence on azimuthal angle, so that (5) becomes, ( ) 2, 1 ) 2 exp( / exp cos sin 2 D P S k k r r L q q H j k r r d r d V r θ θ θ ε ε − = ∆ ⋅ ∫ ∫ (6) The θ integral is straightforward and yields ( ) 2, 1 ) 2 exp cos exp( / 2 ( ) D P S k k r j k r r L q q H r d r V r j k r π θ ε ε − ∆ ⋅ − = ⋅ ∆ ⋅ ∫ sin( ) exp( / ) P S D r q q k r r L dr V k ε ε ∞ ∆ ⋅ = − ∆ ∫ From standard integral tables or a good symbolic integration tool, this can be evaluated as...
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## This note was uploaded on 01/17/2012 for the course ECE 215B taught by Professor Brown during the Spring '08 term at UCSB.

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Notes14 - ECE215B/Materials206B Fundamentals of Solids for...

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