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ECE 874: Physical Electronics. Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu. Lecture 16, 23 Oct 12. Effective mass: How: practical discussion:. Reminder: how you got the E-k curves: Kronig-Penney model allowed energy levels, Chp. 03:. LHS.
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ECE 874:Physical Electronics Prof. Virginia Ayres Electrical & Computer Engineering Michigan State University ayresv@msu.edu
Lecture 16, 23 Oct 12 VM Ayres, ECE874, F12
Effective mass: How: practical discussion: VM Ayres, ECE874, F12
Reminder: how you got the E-k curves:Kronig-Penney model allowed energy levels, Chp. 03: LHS RHS Graphical solution for number and values of energy levels E1, E2,…in eV. a = width of well, b = width of barrier, a + b = Block periodicity aBl VM Ayres, ECE874, F12
k = ± p a + b k = 0 VM Ayres, ECE874, F12
(b) VM Ayres, ECE874, F12
(b) VM Ayres, ECE874, F12
Get: the E-k curves. Matlab can do numerical derivatives Note that the effective mass m* isn’t a single number. Note also that a + b = aBl varies depending on what direction you move in, so there are more curves than are on this single ± direction chart. VM Ayres, ECE874, F12
Get: the E-k curves. Region of biggest change of tangent = greatest curvature: the parabolas shown. Example problem: Which band has the sharpest curvature d2E/dk2? Which band has the lightest effective mass? Which band has the heaviest effective mass? Where in k-space, for both? VM Ayres, ECE874, F12
Get: the E-k curves. Region of biggest change of tangent = greatest curvature: the parabolas shown. Example problem: Which band has the sharpest curvature d2E/dk2? Band 4 Which band has the lightest effective mass? Which band has the heaviest effective mass? Band 1: broadest = least curvature divide by smallest number = heaviest m* Where in k-space, for both? At k= 0 called the G point VM Ayres, ECE874, F12
Where in k-space, for both? VM Ayres, ECE874, F12
Where in k-space, for both? m*A at G k = 0 m*B at about ½ way between G and X in [100] direction: k = 0 VM Ayres, ECE874, F12
a + b = aBl aBl for [100] = aLC k = p/aBl = p/aLC at end of Zone 1 This is X for [100] VM Ayres, ECE874, F12
Where in k-space, for both? m*A at G k = 0 m*B at about ½ way between G and X in [100] direction at k = p/2 aLC k = 0 VM Ayres, ECE874, F12
Assume T = 300K and it doesn’t change Ec = Egap = constant at a given T Hint: compare the answers for b = 0 and b ≠ 0 in (a) VM Ayres, ECE874, F12
Pick correct curve: VM Ayres, ECE874, F12
Pick conduction or valence bands:: E – Ec (eV) VM Ayres, ECE874, F12
Pick conduction minima. Where in k-space are they? E – EV (eV) L G X <111> <100> VM Ayres, ECE874, F12
Pick conduction minima. Where in k-space are they? G at k = 0 L at k = p/aBl for <111> Could work out the aBl distance between atomic cores in a <111> direction if needed. Not needed to finish answering the question. E – EV (eV) L G X <111> <100> VM Ayres, ECE874, F12
Go back to here: Note that the effective mass m* isn’t a single number. Note also that a + b = aBl varies depending on what direction you move in, so there are more curves than are on this single ± direction chart. VM Ayres, ECE874, F12
(From practical to fundamental!) VM Ayres, ECE874, F12
In 3 D: VM Ayres, ECE874, F12
Write this in 2D: all three parts. Integrate a -> v -> r. Vector r (t) is the direction. The final answer contains time t. VM Ayres, ECE874, F12
Start with [m*ij] Then F = qE Then a = dv/dt for dvx/dt and dvy/dt Integrate with respect to time, 2x’s, to get x(t) and y(t). VM Ayres, ECE874, F12
k = 0 VM Ayres, ECE874, F12
Region of biggest change of tangent = greatest curvature: the parabolas shown. 1D: Any one of these parabolas could be modelled as: VM Ayres, ECE874, F12
Region of biggest change of tangent = greatest curvature: the parabolas shown. 3D: <111> + <100> E – EV (eV) L G X <111> <100> For any of these parabolas: There’s a major axis but also two minor ones VM Ayres, ECE874, F12
E – EV (eV) Same: truncate 1/2 L G X <111> <100> VM Ayres, ECE874, F12
k = 0 VM Ayres, ECE874, F12