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
http://www.fulton.asu.edu/~ferry/EEE 352.htm
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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
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
The Kronig-Penney Model
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
The Kronig-Penney Model
)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.
The Kronig-Penney Model
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
The Kronig-Penney Model
)](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.
The Kronig-Penney Model
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.
The Electron’s Effective Mass
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.
The Electron’s 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
The Electron’s Effective Mass
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
The Electron’s Effective Mass
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=
The Electron’s Effective Mass
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=
The Electron’s Effective Mass
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
The Electron’s Effective Mass
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