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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 pointlike defects, such as donor and acceptors in semiconductors, are fundamentally important and common in solids. And because they are not chargeneutral, they create an electrostatic potential that can scatter “free” carriers such as electrons or holes. But when the solid is a metal, semimetal 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 ThomasFermi 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 reexpression 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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