MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 1 -

Problem Set 6 Solutions

From the Kronig-Penney model as covered in Robert F. Pierret “Advanced Semiconductor Fundamentals”, 2nd Ed., Vol. 6 (Addison Weseley, 2003), do the following. 1. From the Kronig-Penney model, we found that:

22

2 ( )om E Vβ −=

.

Prove that β β+= for E >Vo and iβ β−= for 0 < E < Vo. Show all steps for full credit.

Answer:

We know that:

2 ( )om E Vβ

−= ±

thus 2 ( )om E V

β β+

−= + =

for E >Vo.

Additionally:

1 i− = thus 1 1 1i− = − = − .

For 0 < E < Vo., letting β be the minus of the plus or minus, we have:

2

2 ( )

2 ( )1

2 ( )

2 ( )

2 ( )1

2 ( )

o

o

o

o

o

o

m E V

m E V

m E Vi

m E Vii

m E Vi

m V Ei

β−

= −

−= −

−=

−=

−= −

−=

Let:

2 ( )om V Eβ−

−=

and substitute this into the last equation, we have:

iβ β−=

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 2 -

2. Using the solutions to the Schödinger equation in the Kronig-Penney Model given by:

( ) ( )( )a a ax A Sin x B Cos xψ α α= +

and

( ) ( )( )b b bx A Sin x B Cos xψ β β= +

and using the boundary conditions for any potential barrier quantum mechanics problem (as we defined in class), show that one can obtain the following two equations (show all work):

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

0

0

ik a b ik a ba a

ik a b ik a ba a

A Sin a e Sin b B Cos a e Cos b

A Cos a e Cos b B Sin a e Sin b

αα β α ββ

α α α β α α β β

+ +

+ +

+ + − = − + − − =

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 3 -

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 4 -

3. In the Kronig-Penney Model, the determinant formed from the coefficients of the unknowns, Aa and Ba, is zero. Show that by taking the determinant, the equation that results is:

( ) ( ) ( ) ( ) ( )2 2

2Sin a Sin b Cos a Cos b Cosk a bα β α β α β

αβ+

− + = +

Show all work.

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 5 -

4. Using your class notes, the notes from the website, and/or the handout of scanned pages from Pierret’s book titled “Advanced Semiconductor Fundamentals”, sketch a plot of the solution to the Kronig-Penney model. That is, sketch the dispersion relation ( ) ( )f f kξ = . Fully label the sketch of the plot. Show what parts of the plot are the allowed states and the forbidden states (i.e., no solutions). If you want to actually plot the solutions using a mathematical program, that is fine (a few points of extra credit), but it is not required. Provide 3 to 4 sentences of commentary relative to the significance of this finding.

Solution: Your sketch should follow that of the Kronig-Penney Model solution shown in class. It is shown in the figure below:

Note that the allowed energy bands, ξ, are the gray areas as they fall within -1 ≤f(ξ) ≤1. For f(ξ) outside this range, gaps are produced in ξ.

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 6 -

The significance of the solution to the Kronig-Penney Model is demonstrated by the alternating allowed and forbidden energy bands. The solutions show, when a periodic arrangement of atoms modeled simply by a periodicity combination of quantum wells and barriers of constant potential energy , Vo, that energy bands result. The periodicity allow the band to be mapped into the first Brillouin zone. The valence and conduction band can then be identified and we become very close to unlocking the secret to the difference between nonconductors, semiconductors and conductors. With the Fermi-Dirac distribution function used to define the Fermi Energy level, the conductivity/resistivity differences between nonconductors, semiconductors and conductors (over 30 orders of magnitude!) can be explained with the combination of quantum mechanics and statistical thermodynamics. [1] Robert F. Pierret, "Advanced Semiconductor Fundamentals", 2nd Ed., Vol. 4 of Modular Series on Solid State Devices, Editors G. Neudeck, R. Pierret (Prentice Hall, 2003)

Figure Caption: The region of f(ξ) between -1 and 1 that shows the allowed energy states (pink bands) and the forbidden energy band gaps (white). The plot was generated in Mathematica.

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 7 -

5. In class, we reviewed several dispersion relations (i.e., E-k diagrams or E-vs-k diagrams) of electrons in various semiconductors and a metal. Find a dispersion relation that differs from the ones we covered in class. Label all the important features including Eg, CB, VB, heavy and light holes. Describe which, in general, is lighter and faster carrier, the electron or hole, using the effective mass and group velocity.

1g

Evk kω∂ ∂ = = ∂ ∂

while the effective mass is given by: 12 2

222

2

effEm

kEk

− ∂

== = ∂ ∂ ∂

.

Solution: The students can use any dispersion relation they find off the internet. It just needs to be label with Eg, CB, VB, heavy and light holes and whether the electron mass is lighter or heavier than the hole mass. The lighter carrier will lead to a greater group velocity.

6. Draw energy band diagrams in real space for the following: a. Conductor b. Semiconductor c. Nonconductor Thoroughly label the diagrams at least including the valence band top, conduction band bottom, vacuum energy, energy band gap, electron affinity, work function and the energy axis. Include the energy band gap conditions for semiconductor and nonconductor.

MSE 410/ECE 340 Dept. of Materials Science and Engineering Electrical Properties Of Materials Fall 2016/Bill Knowlton

- 8 -

Solution: