Kronig penney

EEE 352: LECTURE 13

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

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IONIONION

POSITION

PO

TE

NT

IAL

EN

ER

GY

V = 0

V

We simplify the potential, in order to be able to solve the problem in any simple manner.

Potential core around theatom.

X=0 X=a

X=d

V

Potential barrier between the atoms.

We will eventually letV and d 0 in the problem.

The Kronig-Penney Model


We now solve the time-independent Schrödinger equation.

ax 0 0 xd

0

2

12

21

2

121

22

dx

d

Edx

d

m

22 2

mE

0

2

22

22

2

2222

22

dx

d

EVdx

d

m

22 )(2

EVm

An energy band has coherent transport over the entire region.So, we seek a general solution of the form:

)()( xuex iikx

i )()( xudaxu ii ,

NOTE


0)(2 1221

21

2

kdx

duik

dx

ud0)(2 2

22222

2

ukdx

duik

dx

ud

xi eu

0)(2 221

21 kik 0)(2 22

222 kik

iik 1 ik2

xiikxxiikx BeAeu 1

xikxxikx DeCeu 2


)0()0( 21 uu DCBA

0

2

0

1

xx dx

du

dx

du

DikCik

AkiAki

)()(

)()(

)()()( 221 duauau

dikdik

akiaki

DeCe

BeAe

)()(

)()(

dxax dx

du

dx

du

21

dikdik

akiaki

DeikCeik

BekiAeki

)()(

)()(

)()(

)()(

This simple b.c. enforces the periodicity onto the solution.


0

)()(

)()(

1111

)()()()(

)()()()(

D

C

B

A

eikeikekieki

eeee

ikikkiki

dikdikakiaki

dikdikakiaki

Since the RHS is 0, there must be an intrinsic solution that arises without any forcing functions.

This requires the determinant of the large square matrix to vanish:

0.det coeff


)](cos[)cos()cosh()sin()sinh(2

22

dakadad

To simplify this, we take the limit V , d 0, in such a manner that Vd = Q.

1)sinh(

, 1)cosh( d

dd

)cos()cos()sin(2

2

kaaad

Function of the energy E Depends only upon theWavevector k

The wavevector k is real only for certain allowed ranges of E,which we illustrate by a graphical solution.


a

)cos()sin(

aa

aP

1

-1

Regions where the equation is satisfied, hence wherethe solution exists.

In general, as the energy increases (a increases), each successive band gets wider, and each successive gap gets narrower.

Boundaries are for ka = n.

No solutionexists, k2 < 0

a

1

-1

d

d

2

d

3

d

4

d

d

2

d

3

d

4

The Kronig-Penney model gives us DETAILED solutions for the bands, which are almost, but not, cosinusoidal in nature.

d

d

2

d

3

d

4

d

d

2

d

3

d

4

Extended zone scheme

d

d

2

d

3

d

4

d

d

2

d

3

d

4

Reduced zone scheme

As energy increases, the bands get WIDER and the gaps get NARROWER

[web]

http://fermi.la.asu.edu/schmidt/applets/kp/plugkp.html



3D Band Structure of Silicon

(100)(111) (110)

Valence band—this is the M shell, and is FULL. Thewave functions here are the symmetric combinationsof the sp3 hybrids.

EG=1.08 eV

Conduction band—thisis composed of the anti-symmetric combinationsof the sp3 hybrids.

Because the minimumof the conduction bandis away from 0, along the(100) direction, thereare 6 equivalent minima.

k = 0

3D Band Structure of Silicon

kx

kz

ky

There are 3 values of the mass in each of the six valleys.In the direction along the valley,we have the longitudinal mass,while the other two directionshave the transverse mass.

However, in each valley, wecan make a coordinate transformationso that we can treat the transportas isotropic.

Surfaces of CONSTANT ENERGY

The Electron’s Effective Mass

The energy bands are closer to cosines than to a free

electron parabola.

Hence, we must be very careful about defining an

effective mass, as the mass will vary with energy!

As a result, we must return to our basic connection for

momentum:

v*mk

We introduce our effective mass through this defining equation, which relates the crystal momentum to the real momentum.


We justify this approach by beginning with the de Broglie relation:

groupvm

h

k

p

h

*

2

kmvmk group

**

Since k only has significance within the first Brillouin zone (periodicity in 2/d), we say that this is an “effective” mass.


k

E

kgroup

1

vv

k

E

km

2

1

*

1

This is the basic definition of the effective mass.

Then, what is wrong with the definition given in many books?

F

k

Em

t

k

km

tmF

2

21*

v*

v*

2

2

2

1

*

1

k

E

m


F

k

Em

t

k

km

tmF

2

21*

v*

v*

This is NOT Newton’s Law!

Instead, we must write:

F

k

Emm

t

mm

dt

md

dt

dpF

2

21

v

*v*

v

v

*v*

)v*(

2

2

2

11

v

*v*

k

Emm


If we have a parabolic energy surface (free electrons), both give the same result!

On the other hand, for our cosinusoidal energy bands, the second derivative only gives an approximate result that is valid very near the minimum (maximum) of the appropriate energy band. It gives a very wrong answer as one moves away from the edge of the band.

/a0

k

vgroup

/a0

k

E

0 /a0/a0

k

Ehole

EMPTYELECTRON

STATES

FILLEDHOLE

STATES=


We find the group velocity by taking the first derivative of the cosinusoidal energy band:

Slope and velocity are > 0

Slope and velocity are < 0

Velocity goes to zero at k = /d.

*holem

*m

WAVENUMBER k

0/a0 /a0

k

E

0 /a0/a0

k

Ehole

EMPTYELECTRON

STATES

FILLEDHOLE

STATES=


k

E

km

2

1

*

1

Both k and dE/dk go to zero, but ratio is non-zero as E goes to

zero as k2.

Only dE/dk goes to zero here.

Divergence of the mass is required for the velocity to go to zero when k 0.

[web]http://www.benfold.com/sse/kp.html

http://www.benfold.com/sse/kp.html

http://www.benfold.com/sse/kp.html


Some values of electron effective mass:

GaAs 0.067

InAs 0.22

InSb 0.13

Si 0.19,0.91*

* Minima are not at center of zone, but are ellipsoids.

O.K. We have energy bands and we have gaps. How do we know whether the material is an insulator, a metal, or a semiconductor?

Well, let us reconsider some of the things we have learned so far:

1. The crystal potential and the wave functions are periodic functions. If the crystal has length L, then we require

2. Hence, we have that the exponential part of the wave function must satisfy

There are N, where N is the number of atoms, values of n.

dx

d

dx

Ld

L

)0()(

)0()(

NnL

nk

nkL

ee ikikL

,...,2,12

2

10

3. This means that each band can hold 2N electrons (the factor of 2 is for spin).

4. Thus, a material with only 1 (outer shell) electron per atom, such as Li, K, Cu, Au, Ag, etc., will be a metal, since only one-half of the available states are filled. The highest band (which we will call the conduction band) is one-half filled. (We assume that, in 3D, the material has a comparable band structure to the simple cubic.)

5. In Si, however, there are 8 atoms per FCC cell: 8 corner atoms, shared between 8 cubes, gives 1; 6 face atoms, shared between 2 cubes each, give 3; and 4 internal atoms, which are not shared with any other cube, gives 4. But, this is considering the basis. The basic FCC cell has only 4 atoms, and each can contribute 8 states, so that there are 32 states per unit cell in the band. Now, we have 8 atoms, each with 4 electrons, and this means 32 electrons. Hence, all the states in the band are filled, and Si should be an insulator!

6. An insulator has all the states in the topmost occupied band FULL. Si is therefore an insulator.

7. But, most insulators have a band gap of 4-10 eV. Silicon has a gap of only 1.0 eV, so electrons can be excited over the gap, and we call Si a semiconductor.

8. In a metal, the number of electrons does not change with temperature. The scattering does increase with temperature, so that the conductance goes down with T.

9. In a semiconductor, the number of electrons increases exponentially with temperature, so that even though the scattering increases, the conductance increases with T.

Size of the Unit Cell

Interatomic distance = twice the atomic radii = 2 X 0.117 nm = 0.234 nm

Unit cell edge length =

nm4.53

234.04

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