semisolpr03

semisolpr03 - Semiconductor Physics and Devices: Basic...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 22 2 2 m xt x Vx j t Ψ Ψ , , , Let the solution be of the form Ψ ux j k x E t ,e x p () ( ) F H I K F H I K L N M O Q P =− h Region I, () = 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 mx jku x j kx E t { exp + F H I K F H I K L N M O Q P U V W x jk x E t exp h = ⋅− F H I K F H I K F H I K L N M O Q P j jE x E t h hh 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 x x E t exp h + F H I K F H I K L N M O Q P U V W 2 2 x x 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 −+ + = kux j k x x mE 2 2 2 2 0 h Setting u x = 1 for region I, this equation becomes du x dx jk du x dx ku x 2 1 2 1 1 20 +− = α bg where 2 2 2 = mE h Q.E.D. In region II, Vx V O . Assume the same form of the solution Ψ x E t x p ) 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 x x E t exp h + F H I K F H I K L N M O Q P U V W 2 2 x x E t exp h F H I K F H I K L N M O Q P Vux 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 + j k x x 2 2 2 2 = 0 mV mE O Setting = 2 for region II, this equation becomes dx jk du x dx 2 2 2 2 2 + −− + = F H I K k mV O 2 2 2 0 h where 2 2 2 = mE h Q.E.D.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 3 Solutions Manual Problem Solutions 24 3.3 We have du x dx jk du x dx ku x 2 1 2 1 22 1 20 () +− = α bg The proposed solution is u x A j kx B j kx 1 ( ) ( ) =− + + exp exp αα The first derivative is du x dx jk Ajk x 1 exp −+ B x exp and the second derivative becomes dx x 2 1 2 2 exp ++ + B x 2 exp Substituting these equations into the differential equation, we find −− kA j k x 2 exp − + kB j k x 2 exp {( ) 2 jk j k A j k x exp } B x exp { j k x exp += } Bj k x exp 0 Combining terms, we have kk k k l j k x q exp + + + + k k ch af m + = j k x 0 q exp We find that 00 = Q.E.D. For the differential equation in ux 2 and the proposed solution, the procedure is exactly the same as above.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 05/07/2008 for the course ECE E-302 taught by Professor Nabet during the Spring '08 term at Drexel.

Page1 / 12

semisolpr03 - Semiconductor Physics and Devices: Basic...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online