ECE606: Solid State Devices Lecture 4 - Purdue...

Klimeck – ECE606 Fall 2012 – notes adopted from Alam

ECE606: Solid State Devices

Lecture 4Gerhard Klimeck

[email protected]

Catchup from last lecture

1


Reference: Vol. 6, Ch. 3

Presentation Outline

•Schrodinger equation in periodic U(x)•Bloch theorem•Band structure•Properties of electronic bands•E-k diagram and constant energy surfaces •Conclusions

2


Three Critical Conceptual Steps

• Basis State Selection»Physical problem is based on a 1D/2D/3D periodic array of atoms»Desired system/signal response is periodic in space»Physical space is infinitely extended

� Can the physical space be collapsed into a different representation?»Chose a basis system of plane waves»New finite reciprocal space is representative of the original system»In 2D and 3D the reciprocal space may have critical axes/symmetries

• Solution of the Schroedinger Equation»Solution with plane waves in reciprocal space»1D solution performed in Kroenig-Penney model (last lecture)

� Band formation»2D/3D solution along critical paths in the reciprocal space

• Filling of the states»States are filled from the bottom up.»Thermal distribution will cause some disorder

3


Fourier Transform Reminders

finite <=> infinite

infinite <=> infinite

iaxe

Space Mapping

Periodic => discrete1

( )cos ax4


2..... 1,0,1,....

2 2

n N Nk n

NL

π= ± = − −

k

L

πL

π−2

kNL

π∆ =

E

4 states per atom, N atoms=> 4 bands, N states in each bandAll states are included in the first zoneInvariant to shift by

1D Brillouin Zone and Number of States

[ ] ( ) ikLNx NL x eψ ψ+ =

E

1 2 3

N-1

U(x)

max min2

2kStates L N

band kNL

k π

π−= = =

∆

2( ) ikLN i mx e e πψ=( ) ikLN imkLx e eψ=

21 im imkLe eπ= =5


L

xReal

2

L

π 4

L

π2

L

π−4

L

π−

kK-lattice

L

πL

π−

k

L

πL

π−2

kNL

π∆ =

E

Solution Space: Brillouin Zone

4 states per atom, N atoms=> 4 bands, N states in each bandAll states are included in the first zoneInvariant to shift byCollapse of an infinite space into a discrete space

21 im imkLe eπ= =6


Fourier Transform Reminders

finite <=> infinite

infinite <=> infinite

iaxe

Space Mapping

Periodic => discrete1

( )cos ax7


A 1D periodic function:

can be expanded in a Fourier series:

The Fourier components are defined on a discrete set of periodically arranged points (analogy: frequencies) in a reciprocal space to coordinate space.

3D Generalization:

( ) ( ); f x f x l l nL= + =

2 / 2( ) i nx L igx

n gn g

nf x A e A e g

Lπ π= = =∑ ∑

( ) ( )a

;r, 321G

G

⊥

++=∑= ⋅

G

bbbGkk rG lkhefu inn

Where hkl are integers. G=Reciprocal lattice vector

Reciprocal Space

Notes adopted from Dragica Vasileska, ASU 8


2 2 2x y z

b c c a a bk k k

a b c a b c a b cπ π π× × ×= = =

× × ×i i i

1) Define reciprocal lattice with the following vectors ….

2) Use Wigner Seitz algorithm to find the unit cell in the wave-vector (reciprocal) space.

Brillouin Zone –

Allowed States in a Reciprocal Lattice

( )31( ) ( )exp2

f r d k f k ikrπ

= ∫� � � �� Fourier transform:

Represented real-spacewith plane waves

( )( )31( ) ( )exp2

f r R d k f k ik r Rπ

+ = +∫� ��

Impose periodicity in R( ) ( )f r R f r+ =� ��

( )exp 1ik R =��

2k R nπ=��

x x zk G hk kk lk= = + +� ��

Reciprocal vector G

9


Wigner-Seitz Method for Reciprocal Space

Primitive cell in real space Unit-cell in reciprocal lattice

a

b ky

kx

ˆ ˆ2 2

ˆ ˆx y

b z z ak k

a b z a b zπ π× ×= =

× ×i i

10


Brillouin Zone for One-dimensional Solids

Real-space

1st B-Z

L

L

πL

π−

E-k diagram

kx2

L

π2

L

π−

L

πL

π−kx

0

Replacing

(a+b) by L …

11


E-k diagram in 2D solids

Real-space

1st B-Z

E-k diagram

a

b

a

πb

πkx

0

a

πa

π−

kx

b

π

b

π−

ky

ky

kxky

E

12


Constant Energy-surface in 2D

1st B-Z

E-k diagram

Const. EnergySurface

a

πb

πkx

0

kx

a

π

a

π−

kx

b

π

b

π−

ky

ky

kxky

E

ky

E1

k1

E1

13


Conclusions

• Solution of Schrodinger equation is relatively easy for systems with well-defined periodicity.

• Electrons can only sit in-specific energy bands. Effective masses and band gaps summarize information about possible electronic states.

• Effective mass is not a fundamental concept. There are systems for which effective mass can not be defined.

• Kronig-Penney model is analytically solvable. Real band-structures are solved on computer. Such solutions are relatively easy – we will do HW problems on nanohub.org on this topic.

• Effective mass is not a fundamental concept. There are systems for which effective mass can not be defined.

• Of all the possible bands, only a few contribute to conduction. These are often called conduction and valence bands.

• For 2D/3D systems, energy-bands are often difficult to visualize. E-k diagrams along specific direction and constant energy surfaces for specific bands summarize such information.

• Most of the practical problems can only be analyzed by numerical solution.

14




[email protected]

15




•Reminder – bandstructure in 1D – Brillouin Zone•E-k diagram/constant energy surfaces in 3D solids

•Definition of a density of states•Density of States for specific materials•Conclusions

16


Original Problem

PeriodicStructure

How do electrons move through a real crystal?

17


N Wells => 2N States => 2 Bands

Vb=110meV, W=6nm, B=2nm

Vb=400meV, W=6nm, B=2nm18


N Wells => 2N States => 2 Bands



1 state/well => 1band

2 states/well => 2bands19


GaAs Well Comparison – 80 Barriers

A GaAs structure with6nm wells, 2nm barriersand 0.4eV barrier heightis modeled as follows,• PPL-Periodic structure

repeated indefinitely.• TB: 80 barriers using

tight-binding.• TM: 80 barriers using

transfer matrices.It can be seen that theresults of these threeapproaches agree well.

20


2..... 1,0,1,....

2 2

n N Nk n

NL

π= ± = − −

k

L

πL

π−2

kNL

π∆ =

E

4 states per atom, N atoms=> 4 bands, N states in each bandAll states are included in the first zoneInvariant to shift by

Brillouin Zone and Number of States

[ ] ( ) ikLNx NL x eψ ψ+ =

E

1 2 3

N-1

U(x)

max min2

2kStates L N

band kNL

k π

π−= = =

∆

2( ) ikLN i mx e e πψ=( ) ikLN imkLx e eψ=

21 im imkLe eπ= =21


A 1D periodic function:

can be expanded in a Fourier series:

The Fourier components are defined on a discrete set of periodically arranged points (analogy: frequencies) in a reciprocal space to coordinate space.

3D Generalization:

( ) ( ); f x f x l l nL= + =

2 / 2( ) i nx L igx

n gn g

nf x A e A e g

Lπ π= = =∑ ∑

( ) ( )a

;r, 321G

G

⊥

++=∑= ⋅

G

bbbGkk rG lkhefu inn

Where hkl are integers. G=Reciprocal lattice vector

Reciprocal Space

Notes adopted from Dragica Vasileska, ASU 22


2 2 2x y z

b c c a a bk k k

a b c a b c a b cπ π π× × ×= = =

× × ×i i i

1) Define reciprocal lattice with the following vectors ….

2) Use Wigner Seitz algorithm to find the unit cell in the wave-vector (reciprocal) space.

Brillouin Zone –

Allowed States in a Reciprocal Lattice

( )31( ) ( )exp2

f r d k f k ikrπ

= ∫� � � �� Fourier transform:

Represented real-spacewith plane waves

( )( )31( ) ( )exp2

f r R d k f k ik r Rπ

+ = +∫� ��

Impose periodicity in R( ) ( )f r R f r+ =� ��

( )exp 1ik R =��

2k R nπ=��

x x zk G hk kk lk= = + +� ��

Reciprocal vector G

23


Brillouin Zone in Cubic Lattice …

Real Space Cubic Lattice Reciprocal Lattice Brillouin Zone

2 aπa

aπ

aπ−0

Follow W-S algorithm, but now for reciprocal lattice

24


Brillouin Zone in Real FCC Lattices …

Real Space FCC(for Si, Ge, GaAs)

Brillouin Zone of

Reciprocal LatticeReciprocal Lattice

Diamond lattice and bcc are Fourier Transforms of each other!

25


Brillouin Zone in Real FCC Lattices …

Real Space FCC(for Si, Ge, GaAs)

Brillouin Zone of

Reciprocal LatticeReciprocal Lattice

2π/a

0.87∗2π/a

Note unlike cubic lattice, zone edge is not at π/a

Diamond lattice and bcc are Fourier Transforms of each other!

26


Brillouin Zone of

Reciprocal Lattice2π/a

0.87∗2π/a

Note unlike cubic lattice, zone edge is not at π/a

Brillouine Zone is Space Filling

27






28


Analogy for E-k Diagram: 4D info thiough 2D Plots

Density (x,y,z)4D information

Cut along (θ1, φ1) …

r

ρ

θ1φ1

A series of line-sections canRepresent the 4D info in 2D plots

29


E-k along ΓΓΓΓ-X Direction

Ge

30


E-k along ΓΓΓΓ-L direction

Ge

31


E-k Diagram

� 3 valence bands (light hole, heavy hole, split-off)valence bands near k=0 is essentially E ~ k2

� Minima may not be at zone center �(Ge: 8 L valleys, Si: 6 X valleys, and GaAs: Γ valleys)

32


E-k diagram for GaAs

Direct bandgap material

Zone-edge gaps (L6-Γ8,X6-Γ8) close to direct gap

Has important implications For transport

33


Analogy for E-k Diagram

Density (x,y,z)

Contours of density ….

34


Constant-E surface for Conduction Band

)( 23

22

21 kkBAkEE c +++= )( 2

322

21 kkkAEE c +++=

35


Constant E-surface …

2

2

1 1

ij i j

E

m k k

∂=∂ ∂ℏ

2 2 21 2 3( )c BAE E k k k= + + + 2 2 2

1 2 3( )cE k k kAE= + + +

11 22 33

1 1 1 1;

(2 0

)ij

Am m m m i j

= = = =≠2 2

11 22 33

1 1 1 1; ; 0

2

)

2

(ij

B

m m i

A

m m j= = = =

≠ℏ ℏ

Offdiagonal Elements=0=> Force and movement aligned 36


Four valleys inside BZ for Germanium

37


Constant E-surface for Valence Band

)]([ 2222222422xzzyyxv kkkkkkCkBAkEE +++−= ∓

Si: A=4.29, B=0.68, C=4.87; Ge: A=13.38, B=8.48, C=13.1538






39


Density of States

A single band has total of

N-states

Only a fraction of states

are occupied

How many states are

occupied up to E?

Or equivalently…

How many states per unit

energy ? (DOS)kk-k

E

40


Density of States in 1-D Semiconductors

k

a

π2k

Na

πδ =

E

1unit enerStates/ @ gyNa

E

kE

π∆=∆

1 1States between + 2 & k

Ek

E Eδ∆∆ = ×

E1+∆E

E1

k1+∆kk1

22

k

Naπ∆= ×

a

N atoms

41


1D-DOS

( )

( )

2 20

0 2

20

2

2

2

*

*

*

m E

dk

d

EkE E k

m

m

E EE

−− = ⇒ =

=−

ℏ

ℏ

ℏ

States/unit energy @ N ka

EEπ

= ∆∆

( )20

States/unit energy @ 2

*m

E

LE

Eπ=

−ℏ

( )20

unit lengthStates/unit energy/ @

1

2

*

E

mDOS

E Eπ≡ =

−ℏ

k

a

π2k

Na

πδ =

E

E1+∆E

E1

k1+∆kk1

a

N atomsL

42


1D-DOS

k

a

π2

kNa

π∆ =

E E

DOS

( )20

1

2

*mDOS

E Eπ=

−ℏ

Conservation of DOS

43


Density of States in 2D Semiconductors

ab

Show that 2D DOS is a constant independent of energy!

44


Density of States in 3D Semiconductors

( )( )

2 20

0 2 20

2

2 2

* *

*

m E E mk dkE E k

m dE E E

−− = ⇒ = ⇒ =

−ℏ

ℏ ℏ

( )1 1

3 3

22

States between + &

4 43 3

2 2 2 2

E E E

k dk k V

H

k k

L W

π π

π π π π

∆

+ −= = ∆

22

unit energStates/ @ 2

yV k

EE k

dπ∆=

( )1

02 3

States/unit energy/unit volume @

22

**

E

mDOS m E E

π= −

ℏ

kK+dk

2π/L

2π/W2π/H

L

W

H

Macroscopic Sample

45


3D-DOS

k

a

πa

π−2

kNa

π∆ =

E E

DOS

Conservation of DOS

( )3 3 0

2 3

2* *m m E EDOS

π−

=ℏ

46





•Definition of a density of states•Density of States for specific materials

47


Density of States of GaAs: Conduction/Valence Bands

( )

( )

( )

2 3

2 3

2

2

2

2

* *hh hh

* *lh lh

m m E E

g E

m m E E

υ

υ

υ

π

π

−=

−

ℏ

ℏ

( ) ( )2 3

2

2

* *n n c

c

m m E Eg E

π−

=ℏ

48


Four valleys inside BZ for Germanium

49


Ellipsoidal Bands and DOS Effective Mass

2 22 2 2 231 2

2 2 2C * *tl*

t

kk k

m mE

mE − = + + ℏℏ ℏ

( ) ( ) ( )22 2

31 2

2 2 2

12 2 2* *

t Cl t*

C CE E E

k

E E E

k

m m

k

m − −= + +

−ℏ ℏ ℏ

24

3k elN παβ =

V

( ) ( ) ( ) ( )3

2 2 2 2

22 2 24 4

3 3

* * *cl c t c t

*ec ff

el

E Em E E m E E m E EN

mπ π

−− − − ≡

ℏ ℏ ℏ ℏ

( )1 32 3 2* * *eff el l tm N m m=

Transform into …

k1

k2

k3

E=const. ellipsoid

const. E

34

3 effkπ≡

2α 2β

Nel is number of equivalent ellipsoids50


DOS Effective Mass for Conduction Band

( )1 32 3 26* * *eff l tm m m=

( ) ( )2 3

2

2

* *eff eff c

c

m m E Eg E

π−

=ℏ

( )1 32 3 24* * *eff l tm m m=

( ) ( )2 3

2

2

* *eff eff c

c

m m E Eg E

π−

=ℏ

51


Conclusions

1) E-k diagram/constant energy surfaces are simple ways to

represent the locations where electrons can sit. They

arise from the solution of Schrodinger equation in

periodic lattice.

2) E-k diagram and energy bands contain equivalent

information. In principle, any one can be used to

construct the other.

3) Only a fraction of the available states are occupied. The

number of available states change with energy. DOS

captures this variation.

4) DOS is an important and useful characteristic of a

material that should be understood carefully.

52




[email protected]


Reference: Vol. 6, Ch. 3 & 4


•Reminder – Density of states»Possible states as a function of Energy

•Reality check - Measurements of Bandgaps•Reality check - Measurements of Effective Mass•Rules of filling electronic states •Derivation of Fermi-Dirac Statistics: three techniques

•Intrinsic carrier concentration•Conclusions


k

a

πa

π− 2k

Na

π∆ =

E E

DOS

( )3 3 0

2 3

2* *m m E EDOS

π−

=ℏ

Reminder: Momentum vs. DOS

Important things to remember:• Momentum k entered our thinking as a quantum number• Each quantum number is creating ONE state• Often “just” need the number of available states in an energy range

=> Density of States => appears to be solely determined by »1) band edge, »2) effective mass


Measurement of Band Gap

k

E

1

2

3

4

Photons are only absorbed between bands that have filled and empty states

E23

abso

rptio

n


Measurement of Energy Gap

k

hν

Iout

Iin

E23

I out -I in

E23+∆

23E E−∼

( )2

23 phE E E− −∼

k

E

1

2

3

4


Temperature-dependent Band Gap

k

E

1

2

3

4

( ) ( )2

0G G

TE T E

T

αβ

= −+


Reference: Vol. 6, Ch. 3 & 4


•Reminder – Density of states»Possible states as a function of Energy

•Reality check - Measurements of Bandgaps•Reality check - Measurements of Effective Mass•Rules of filling electronic states •Derivation of Fermi-Dirac Statistics: three techniques

•Intrinsic carrier concentration•Conclusions


Measurements of Effective Masses

Important things to remember:• Full bands do not conduct –

electrons have no space to go• Empty bands to not conduct –

there are no electrons to go aroundQuestion:• We are interested in the top-most

valence band holes and the bottom-most electron states

• We want to figure out the slope of the bands

• How can we probe just one particular species of electrons/holes?

• We do not want to transfer them from one band to the next!

=> can we rotate the electrons around in a single band?

k

E

3

2

1


(kx,0)

(0,-ky)

(0,ky)

(-kx,0)

kx

Energy=constant.

kxky

kz

ky

Liquid He temperature …x

y

Motion in Real Space and Phase Space


Derive the Cyclotron Formula

2

0

0 0

z z

m*q B q B

r

qB r

m*

υ υ υ

υ

= × =

=

B

For an particle in (x-y) plane with B-field in z-direction, the Lorentz force is …

0

0

00

00 0

2 2

12

2

r m*

qB

qB

m*qB

m*

π πτυ

ντ π

ω πν

= =

≡ =

= =

0

02

qm*

B

vπ=


Measurement of Effective Mass

ν0=24 GHz(fixed)

IoutIin

B field variable …

00

0

0

2 2

q qm

B B*

vm*πν

π= =

I out -I in

BBcon

kxky

kz

Bval kxky

kz


Effective mass in Ge

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ]

111 111 111 111

111 111 111 111

4 angles between B field and the ellipsoids …Recall the HW1

B


Derivation for the Cyclotron Formula

B

Show that 2 2

2 2

1

c t l t

cos sin

m m m m

θ θ= +

[ ] dF q B M

dt

υυ= × =

The Lorentz force on electrons in a B-field

( )

( )

( )

* xx y z z y t

y*y z x x z t

* zz x y y x l

dF q B B m

dtd

F q B B mdt

dF q B B m

dt

υυ υ

υυ υ

υυ υ

= − =

= − =

= − =

In other words,

Given three mc and three θ,we will Find mt, and ml


Continued …

22 2 2 2 2

2

0 0 00

0 with l

l

yy t t

t* * *c t l

dsin cos

dtqB qB qB

m m m

wω ω

ω

υυ

ω

ω ω θ θ

ω

+ = ≡ +

≡ ≡ ≡

Differentiate (vy) and use other equations to find …

Let (B) make an angle (θ) with longitudinal axis of the ellipsoid (ellipsoids oriented along kz)

( )2 2

2 2

1*

l t tc

sin cos

m m mm

θ θ= +so that …

B0

ky

kx

kz

( ) ( )0 00x y zB B cos , B , B B sin ,θ θ= = =


Measurement of Effective Mass

Three peaks B1, B2, B3Three masses mc1,mc2,mc3Three unique angles: 7, 65, 73

2 2

2 2

1

tc l t

cos sin

m mm m

θ θ= +

Known θ and mc allows calculation of mt and ml.

[110]

B=[0.61, 0.61, 0.5]

1

02cmqB

vπ=


Valence Band Effective Mass

Which peaks relate to valence band?Why are there two valence band peaks?


Conclusions

1) Only a fraction of the available states are occupied. The

number of available states change with energy. DOS

captures this variation.

2) DOS is an important and useful characteristic of a

material that should be understood carefully.

3) Experimental measurements are key to making sure that

the theoretical calculations are correct. We will discuss

them in the next class.

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