lecture 12

lecture 12 - EEE 352: LECTURE 12 Last time, we explored the...

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Unformatted text preview: EEE 352: LECTURE 12 Last time, we explored the free electron model and then justified the shape of the cosinusoidal energy bands by a simple nearest neighbor model. Now, we want to look at a more formal treatment. The Kronig-Penney ModelA Unified Model * Solving for tunneling through the potentials between the atoms * Introducing periodicity into the wave solutions Electron bands Energy gaps Effective Mass + + + ION ION ION POSITION POTENTIAL ENERGY V = 0 V We simplify the potential, in order to be able to solve the problem in any simple manner. Periodic potentialsReal and Simplified Now, simply, what we are going to do is find the Fourier series for the periodic potential. However, the straight-forward Fourier series will have a set of discrete levels, at given values of the Fourier coefficient k (which correspond to energies). This would be the result for static standing waves. We are going to force traveling waves; which requires tunneling between the different wells of the potential, and this broadens the discrete levels into bands, which are ranges of allowed values for k , and hence for energy. The values of energy for which we have allowed values of k are the bands , and the ranges of energy for which we do not have allowed values of k are the gaps . Potential core around the atom. X=0 X=a X= d V Potential barrier between the atoms. We will eventually let V and d 0 in the problem. The Kronig-Penney Model Ralph Kronig We will force a wave onto the solution. Since, k is associated with the wave, we cannot use it as a constant in the equation. The Kronig-Penney Model We now solve the time-independent Schrdinger equation. An energy band has coherent transport over the entire region. So, we seek a general solution of the form: , NOTE Between the barriers In the barrier k comes with the forced wave: The Kronig-Penney Model d dx e ikx u ( x ) ( ) = ike ikx u ( x ) + e ikx du dx d 2 dx 2 e ikx u ( x ) ( ) = d dx ike ikx u ( x ) + e ikx du dx = k 2 e ikx u ( x ) + 2 ike ikx du dx + e ikx d 2 u dx 2 Now, we put this result into the two differential equations for the two regions (between the barriers and within the barrier) d 2 1 dx 2 + 2...
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lecture 12 - EEE 352: LECTURE 12 Last time, we explored the...

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