Lecture 7 3D Crystals and Band Structure...

ECE 6451 - Dr. Alan DoolittleGeorgia Tech

Lecture 7

3D Crystals and Band Structure

Reading:

Notes and Brennan Chapter 7.4, 8.0-8.2


Importance of k-Space Boundaries at k=(+/-)π/aCrystal Structures, Brillouin Zones and Bragg Reflection

After Neudeck and Peirret Fig 4.1

The crystal lattice consists of a periodic array of atoms. Unit Cell Concept

While crystals have rotational symmetry, we restrict ourselves to methods of reconstructing the entire crystal (every lattice point) using translation of a unit cell (a special type known as a Bravaiscell) only – no rotation.


Unit Cell Concept

A “building block” that can be periodically duplicated to result in the crystal lattice is known as the “unit cell”.


The crystal lattice consists of a periodic array of atoms.

The smallest “building block”that can be periodically

duplicated to result in the crystal lattice is known as the

“primitive unit cell”.

The unit cell may not be

unique.

Unit Cell Concept

A “building block” that can be periodically duplicated to result in the crystal lattice is known as the “unit cell”.


Unit Cell Concept: Translation of a 3D Bravais LatticeDeconstructing a Hexagonal Crystal From a Trigonal P Bravais Lattice

Side View


Unit Cell Concept: Translation of a 3D Bravais LatticeDeconstructing a Hexagonal Crystal From a Trigonal P Bravais Lattice

Top View with Trigonal Lattice

Apparent

The crystal is reconstructed by translating the Bravais Lattice along vectors with 60 degree symmetry.

!orthogonal be tohavenot do c and b,a :Note

clblalR vector lattice space Real 321∧∧∧

∧∧∧

++=

∧

a

∧

b

plane. theofout pointed is c ∧


Unit Cell Concept

Lattice Constant: A length that describes the unit cell. It is normally given in Å, angstroms = 1e-10 meters.

Diamond Structure: Constructed by 2

“inter-penetrating”FCC Lattices

Zincblende is a diamond structure with every other atom a different

element. Example: Ga only bonds to As and As only bonds

to Ga.

Note: In class show DiamondTM

examples.


Unit Cell Concept

Some unit cells have hexagonal symmetry.

Rocksalt unit cells are one of the simplest practical unit cells.


k-space or Reciprocal Space Description of a Crystal

Since the Bloch wavefunction is distributed throughout the crystal, the position of the electron is highly delocalized. Thus, from the uncertainty principle, the momentum of the electron in the crystal should be well defined*.

Therefore, it is convenient for us to consider the k-space equivalent of the crystal and in particular the Reciprocal Lattice.

* Strictly speaking the Bloch wavefunction is not an eigenfunction of the momentum operator and thus, the momentum is not exactly known. However, due to the uncertainty principle, the vast delocalization of the electron in the crystal (in Bloch states) will result in well defined but not singular value of momentum. Thus, to a good approximation, the electrons in the crystal will be treated as nearly free electrons with well defined momentum.

Cell Unit Space Realof Volume

ba2c


ac2b


cb2a

integer. an is n wheren2πRG

by, clbkahG vector lattice reciprocala define can wethen

clblalR vector lattice space reala Given

***

***

321

=

=

=

=•

++=

++=

∧∧

∧

∧∧

∧

∧∧

∧

∧∧∧

∧∧∧

xxxπππ

The RS unit cell fully reconstructs the entire reciprocal space via only translations just like the real space unit cell reconstructs the entire crystal via only translations.

Note that the Reciprocal space unit cell maintains the same symmetry as the real space unit cell because it was derived from the real space unit cell.

∧∧∧

∧∧

∧

∧∧∧

∧∧

∧

∧∧∧

∧∧

∧

×•

=×•

=×•

=cba

ba2c

cba

ac2b

cba

cb2a ***

xxxπππ


An example for a Body Centered Cubic (BCC) RECIPROCAL LATTICE material*.


*It can be shown that the Face centered cubic and Body centered cubic structures are Fourier analogs so the above example is the reciprocal lattice equivalent of an FCC crystal.

Fourier Transform


Connect an arbitrary lattice point to all of its nearest neighbors (green lines)...



...construct the perpendicular bisectors to all of these lines. The 1st

Brillouin Zone is the volume enclosed within this region.



kz

kykx

http://britneyspears.ac/physics/dos/dos.htm

Cubic GaN

Real SpaceCrystal Momentum Space

•Different potentials exist in different directions

•Electron wavelength and crystal momentum, k=2π/λ, differs with direction

•Plots of E-k are 4D plots, thus have to be represented in other ways (as slices along certain directions).

•Many different parabolic E-k relationships exist depending on our crystalline momentum

Now consider the 3D periodic potential in a cubic crystal


•All equivalent directions give redundant information and thus are not repeated

•Most important k-space points

• Γ-point is the center of crystal momentum space (k-space) at k=0

• X-point is the edge of the first Brillouin zone (π/L edge) of crystal momentum space (k-space) in the <100> direction

• L-point is the edge of the first Brillouin zone (π/L edge) of crystal momentum space (k-space) in the <111> direction

Cubic GaN

Now consider the 3D periodic potential in a cubic crystal

Neudeck and Peirret Fig 3.13

ECE 6451 - Dr. Alan DoolittleGeorgia TechSuzuki, M, T. Uenoyama, A. Yanase, First-principles calculations of effective-mass parameters of AlN and GaN, Phys. Rev. B 52, 11 (1995), 8132-8139.

kx

ky

kz

Now consider the 3D periodic potential in a hexagonal crystal

Full Band Diagram

Real SpaceCrystal Momentum Space

ECE 6451 - Dr. Alan DoolittleGeorgia TechNeudeck and Peirret Fig 3.14

Where are the electron trajectories/momentum vectors in the crystal?

Neudeck and Peirret Fig 3.13

Valence Band E-k and constant

energy surfaces all look similar

Si EC minima occurs at ~0.8(π/a)

In these figures, the Energy is fixed near the band edges (E>ECfor example) and al values of k having this fixed energy are plotted to give a 3D E-k representation.


k3

k1

k2

k3

k1

k2

For Valence Bands and EC of Direct gap materials,

For EC of Indirect gap materials,

( ) ( )masses effective e transversand allongitudin theare and where

22

E-E 2

E-E

z andy x,ji,for 1 and 1 where1dtvd

crystals 3Dfor So Fam

**

23

22*

221*

2

C23

22

21*

2

C

2

21

111

111

111

**

*

tl

tle

jiij

yyzyzx

yzyyyx

xxxyxx

mm

kkm

km

kkkm

kkEm

mmmmmmmmm

mF

m

++=++=

=∂∂

∂=

==

=

−

−−−

−−−

−−−

hhh

h

3D Crystalline Effects on Effective Mass


k3

k1

k2

k3

k1

k2

For Valence Bands and EC of Direct gap materials,

For EC of Indirect gap materials,

( ) ( )

( )

( )

( ) ( ) ( )

( ) ( )( )Gefor )8

21(4 and Sifor 6N

N

so N

to,leads this34

34N

Thus, ellipsoid. of volume the toequal is spherenew theof volume that thesuch 2mk radius of volume

sphericala define weso, do To averaged.y effectivel are crystal theof properties canisotropi thethus, andoccur events scatteringmany that being validis This .properties canisotropi theaverages

thatm electronsfor mass effective average an define can wecase, ellispsoid For the

masses effective e transversand allongitudin theare and where

22

E-E 2

E-E

Zone Brillioun1 in ellipsoids

312**3

2

Zone Brillioun1 in ellipsoids*

23*2**

Zone Brillioun1 in ellipsoids

3321Zone Brillioun1 in ellipsoids

2

*n

eff

*n

**

23

22*

221*

2

C23

22

21*

2

C

st

st

st

st

==

=

=

=

−=

++=++=

tln

ntl

eff

C

tl

tle

mmm

mmm

kkkk

EE

mm

kkm

km

kkkm

ππ

h

hhh



k3

k1

k2

For Valence Bands

( ) ( )masses effective holeheavy and holelight theare and where

band, valencefor theSimilarly

**

32

23*2

3**

hhlh

lhhhp

mm

mmm

+=


After Neudeck and Pierret Tables 3.1 and 4.1


How do electrons and holes populate the bands?Derivation of Density of States Concept

We can use the idea quantum confinement of a set of states ( 1D well region) to derive the number of states in a given volume (volume of our crystal).

Consider the surfaces of a volume of semiconductor to be infinite potential barriers (i.e. the electron can not leave the crystal). Thus, the electron is contained in a 3D box.

After Neudeck and Pierret Figure 4.1 and 4.2

mkmE

kzyx

Exm

EVm

2or E 22k where

cz0 b,y0 a,x0...for

0

02

2

22

2

22

2

2

2

2

2

2

22

22

h

h

h

h

===

<<<<<<

=Ψ+∂Ψ∂

+∂Ψ∂

+∂Ψ∂

=Ψ−∂Ψ∂

−

Ψ=Ψ

+∇−

λπ



Using separation of variables...

2222

22

22

2

22

2

2

22

2

2

2

2

2

22

2

2

2

2

2

0)()(

1 ,0)(

)(1 ,0)(

)(1

0)()(

1)()(

1)()(

1

get... we(2)by dividing and (1) into (2) Inserting)()()(),,( (2)

0 (1)

zyx

zz

zy

y

yx

x

x

z

z

y

y

x

x

zyx

kkkkwhere

kzz

zk

yy

yk

xx

x

kzz

zyy

yxx

x

zyxzyx

kzyx

++=

=+∂Ψ∂

Ψ=+

∂Ψ∂

Ψ=+

∂Ψ∂

Ψ

=+∂Ψ∂

Ψ+

∂Ψ∂

Ψ+

∂Ψ∂

Ψ

ΨΨΨ=Ψ

=Ψ+∂Ψ∂

+∂Ψ∂

+∂Ψ∂

Since k is a constant for a given energy, each of the three terms on the left side must individually be equal to a constant.

So this is just 3 equivalent 1D solutions which we have already done...



Cont’d...( ) ( ) ( )

++=

±±±====

=Ψ

ΨΨΨ=Ψ

2

2

2

2

2

222

nz ny, nx,

zz

yy

xx

2 Eand

3...,2 1,nfor ,a

nk ,a

nk ,

ank where

sinsinsin),,()()()(),,(

cn

bn

an

m

zkykxkAzyxzyxzyx

zyx

zyx

zyx

hπ

πππ

0

aπ

bπ

cπ

kx

ky

kzEach solution (i.e. each combination of nx, ny, nz) results in a volume of “k-space”. If we add up all possible combinations, we would have an infinite solution. Thus, we will only consider states contained in a “fermi-sphere” (see next page). See Pierret and Neudeck for a

more general approach that does not require the fermisphere argument. Tie this into the GaAs equi-energy surface in lecture 7.



Cont’d...f

22

f energy E electron averagefor valuemomentuma defines 2

E ⇒=mk fh

Volume of a single state “cube”: V3

state single

=

=

Vcbaππππ

Volume of a “fermi-sphere”: 34V 3

sphere-fermi

= fkπ

A “Fermi-Sphere” is defined by the number of states in k-space necessary to hold all the electrons needed to add up to the average energy of the crystal (known as the fermienergy).

“V” is the physical volume of the crystal where as all other volumes used here refer to volume in k-space. Note that: Vsingle-state is the smallest unit in k-space. Vsingle-state is required to “hold” a single electron.



Cont’d...k-space volume of a single state “cube”: V

3

state single

=

=

Vcbaππππ

k-space volume of a “fermi-sphere”: 34V 3

sphere-fermi

= fkπ

Number of filled states in a fermi-sphere: 2

3

3

3

sin 3

4

34

21

21

212N

ππ

πf

f

stategle

spherefermi kV

V

kxxx

VV

=

=

=≡

−

−

Correction for allowing 2 electrons per state (+/- spin)

Correction for redundancy in counting identical states resulting

from +/- nx, +/- ny, +/- nz. Specifically, sin(-π)=sin(+π) so the state would be the same. Same as

counting only the positive octant in fermi-sphere.



Cont’d...

Number of filled states in a fermi-sphere:

31

2

f2

3 3k 3

N

=⇒=≡

VNkV f π

π

Ef varies in Si from 0 to ~1.1 eV as n varies from 0 to ~5e21cm-3

( ) density electron theis n where2

3 E 2

3

2

E3

222

f

32

22

22

f mn

mVN

mk f π

πh

hh

=⇒

==

23

222

3 233

N

==

h

mEVkV f

ππThus,



Cont’d...

E

V

mmEV

32

2

21

22

2mm G(E)

223

23

G(E)

VdEdN

per volumeenergy per states of # G(E)

h

hh

π

π

=

=

=≡

23

222

3 233

N

==

h

mEVkV f

ππ

Applying to the semiconductor we must recognize m m* and since we have only considered kinetic energy (not the potential energy) we have E E-Ec

cEE −= 32

** 2mm G(E)hπ

Finally, we can define the density of states function:

Lecture 7 3D Crystals and Band Structure...

Documents

Transcript of Lecture 7 3D Crystals and Band Structure...