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Lecture%2013

# Lecture%2013 - EEE 352 LECTURE 13 Last time we explored the...

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1 EEE 352: LECTURE 13 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 Model—A 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 potentials—Real and Simplified 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 The Kronig-Penney Model We now solve the time-independent Schrödinger equation. a x < < 0 0 < < x d 0 2 1 2 2 1 2 1 2 1 2 2 = + = ψ α dx d E dx d m h 2 2 2 h mE = 0 2 2 2 2 2 2 2 2 2 2 2 2 = = + γ dx d E V dx d m h 2 2 ) ( 2 h E V m = An energy band has coherent transport over the entire region. So, we seek a general solution of the form: ) ( ) ( x u e x i ikx i = ) ( ) ( x u d a x u i i = + + , NOTE Between the barriers In the barrier

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2 The Kronig-Penney Model ( ) ( ) 2 2 2 2 2 2 ) ( ) ( ) ( ) ( ) ( dx u d e dx du ike x u e k dx du e x u ike dx d x u e dx d dx du e x u ike x u e dx d ikx ikx ikx ikx ikx ikx ikx ikx ikx + + = + = + = Now, we put this result into the two differential equations for the two regions (between the barriers and within the barrier) ( ) 0 2 ) ( 2 ) ( 1 2 2 1 2 1 2 1 2 2 1 2 1 1 2 1 2 2 1 2 = + + = + + + = + u k dx du ik dx u d e x u e dx u d e dx du ike x u e k dx d ikx ikx ikx ikx ikx α ψ ( ) 0 2 ) ( 2 ) ( 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 = + + = + + = u k dx du ik dx u d e x u e dx u d e dx du ike x u e k dx d ikx ikx ikx ikx ikx γ (3.4) (3.8) The Kronig-Penney Model 0 ) ( 2 1 2 2 1 2 1 2 = + + u k dx du ik dx u d 0 ) ( 2 2 2 2 2 2 2 2 = + + u k dx du ik dx u d x i e u δ = 0 ) ( 2 2 2 1 2 1 = + + k ik 0 ) ( 2 2 2 2 2 2 = + + k ik i ik ± = 1 ± = ik 2 x i ikx x i ikx Be Ae u + + = 1 x ikx x ikx De Ce u + + = 2 The Kronig-Penney Model ) 0 ( ) 0 ( 2 1 u u = D C B A + = + 0 2 0 1 = = = x x dx du dx du D ik C ik A k i A k i ) ( ) ( ) ( ) ( + = = + ) ( ) ( ) ( 2 2 1 d u a u a u = = d ik d ik a k i a k i De Ce Be Ae ) ( ) ( ) ( ) ( + + + = = + d x a x dx du dx du = = = 2 1 d ik d ik a k i a k i De ik Ce ik Be k i Ae k i ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( + + + = = + This simple b.c. enforces the periodicity onto the solution.
3 The Kronig-Penney Model () 0 ) ( ) ( ) ( ) ( 1 1 1 1 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( = +

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Lecture%2013 - EEE 352 LECTURE 13 Last time we explored the...

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