Unformatted text preview: (viii) nH z The (v) becomes E x = ! n J x + ! n ( µ n rH Bz ) J y Similarly for E y : So the final result is: E x = ! n J x + ! n ( µn rH Bz ) J y E y = " ! n ( µn rH Bz ) J x + ! n J y 6) But see problem 6d). We can actually get this result without assuming a small magnetic field. This homework exercise will help you become familiar with how B –fields affect transport Consider the equation of motion for an average electron, dp Fe = − qE  qυ × B =
. dt (i) Assume that the electron moves for a time, ! m , then scatters, returning the average momentum to zero, so dp
p
= ! . (ii) dt
"m
Assuming that p = m *υ , we find an equation for the average velocity as qτ qτ υ = − * E  * υ × B . (iii) m
m
This equation can be solved exactly for the velocity (see prob. 4.18, Lundstrom, Fundamentals of Carrier Transport, 2000), but let’s take a different approach. ECE
656 9 Fall 2013 Mark Lundstrom 10/27/13 6a) Assume carriers move in 2D and that only a z
directed B
field is present. Evaluate eqn. (iii) and find two coupled equations for ! x and ! y . Solution: Assuming a z
directed B
field, it is straightforward to show from (iii) that q# m
q#
E x  *m ! y Bz
*
m
m
q# m
q# m
! y = " * E y + * ! x Bz
m
m !x = " (*) 6b) Solve the two equations for ! x and ! y in terms of the electric field and the B
field. Solution: From (*), it is straightforward to show: !x =
!y = 2
" µnE x + µnE y Bz 1 + (# c$ ) 2 2
" µnE y " µnE x Bz 1 + (# c$ ) Where ! c = (**) 2 qBz
is the so
called cyclotron frequency. m* 6c) Write the current densities as J x = ! nS q" x...
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