semisolpr03 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 3 Solutions Manual Problem Solutions 23 Chapter 3 Problem Solutions 3.1 If a o were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If a o were to decrease, the bandgap energy would increase and the material would begin to behave more like an insulator. 3.2 Schrodinger°s wave equation + = ∂Ψ ( ) ( ) ( ) ( ) h h 2 2 2 2 m x t x V x x t j x t t Ψ Ψ , , , Let the solution be of the form Ψ x t u x j kx E t , exp ( ) ( ) F H I K F H I K L N M O Q P = h Region I, V x ( ) = 0 , so substituting the proposed solution into the wave equation, we obtain ( ) F H I K F H I K L N M O Q P h h 2 2 m x jku x j kx E t { exp + ( ) F H I K F H I K L N M O Q P U V W u x x j kx E t exp h = F H I K ( ) F H I K F H I K L N M O Q P j jE u x j kx E t h h h exp which becomes ( ) ( ) F H I K F H I K L N M O Q P h h 2 2 2 m jk u x j kx E t { exp + ( ) F H I K F H I K L N M O Q P 2 jk u x x j kx E t exp h + ( ) F H I K F H I K L N M O Q P U V W 2 2 u x x j kx E t exp h = + ( ) F H I K F H I K L N M O Q P Eu x j kx E t exp h This equation can then be written as + + + = ( ) ( ) ( ) ( ) k u x jk u x x u x x mE u x 2 2 2 2 2 2 0 h Setting u x u x ( ) ( ) = 1 for region I, this equation becomes d u x dx jk du x dx k u x 2 1 2 1 2 2 1 2 0 ( ) ( ) ( ) + = α b g where α 2 2 2 = mE h Q.E.D. In region II, V x V O ( ) = . Assume the same form of the solution Ψ x t u x j kx E t , exp ( ) ( ) F H I K F H I K L N M O Q P = h Substituting into Schrodinger°s wave equation, we obtain ( ) ( ) F H I K F H I K L N M O Q P h h 2 2 2 m jk u x j kx E t { exp + ( ) F H I K F H I K L N M O Q P 2 jk u x x j kx E t exp h + ( ) F H I K F H I K L N M O Q P U V W 2 2 u x x j kx E t exp h + ( ) F H I K F H I K L N M O Q P V u x j kx E t O exp h = ( ) F H I K F H I K L N M O Q P Eu x j kx E t exp h This equation can be written as + + ( ) ( ) ( ) k u x jk u x x u x x 2 2 2 2 + = ( ) ( ) 2 2 0 2 2 mV u x mE u x O h h Setting u x u x ( ) ( ) = 2 for region II, this equation becomes d u x dx jk du x dx 2 2 2 2 2 ( ) ( ) + + = F H I K ( ) k mV u x O 2 2 2 2 2 0 α h where α 2 2 2 = mE h Q.E.D.
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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 3 Solutions Manual Problem Solutions 24 3.3 We have d u x dx jk du x dx k u x 2 1 2 1 2 2 1 2 0 ( ) ( ) ( ) + = α b g The proposed solution is u x A j k x B j k x 1 ( ) ( ) ( ) = + + exp exp α α The first derivative is du x dx j k A j k x 1 ( ) ( ) ( ) = α α exp + + ( ) ( ) j k B j k x α α exp and the second derivative becomes d u x dx j k A j k x 2 1 2 2 ( ) ( ) ( ) = α α exp + + + ( ) ( ) j k B j k x α α 2 exp Substituting these equations into the differential equation, we find ( ) ( ) α α k A j k x 2 exp + + ( ) ( ) α α k B j k x 2 exp + ( ) { ( ) 2 jk j k A j k x α α exp + + ( ) ( ) } j k B j k x α α exp ( ) { k A j k x 2 2 α α b g exp + + = ( ) } B j k x exp α 0 Combining terms, we have + ( ) α α α 2 2 2 2 k k k k b g l ( ) k A j k x 2 2 α α b gq exp + − + + + + α α α
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