Chapter 2 Semiconductor Heterostructures - Cornell Universitysemiconductor (1) to semiconductor (2) and hole flow from semiconductor (2) to semiconductor (1) continues until the electric

1

ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Chapter 2

Semiconductor Heterostructures

In this lecture you will learn:

• Energy band diagrams in real space• Semiconductor heterostructures and heterojunctions• Electron affinity and work function• Heterojunctions in equilibrium• Electrons at Heterojunctions• Semiconductor Quantum wells Herbert Kroemer

(1920-)Nobel Prize 2000 for the Semiconductor Heterostructure Laser


Band Diagrams in Real Space - I

For devices, it is useful to draw the conduction and valence band edges in real space:

cE

vE

fE

x

cE

vEfE

x

N-type semiconductor

Energy

k

fE

cE

vE

Energy

k

fEcE

vE

P-type semiconductor

KTEEc

fceNn

KTEEv

vfeNp

2


Band Diagrams in Real Space - II

Electrostatic potential and electric field:

cE

vE

fE

xN-type semiconductor

An electrostatic potential (and an electric field) can be present in a crystal:

The total energy of an electron in a crystal is then given not just by the energy band dispersion but also includes the potential energy coming from the potential:

Therefore, the conduction and valence band edges also become position dependent:

rrEr

and

kEn

rekEkE nn

reEEreEE vvcc

Example: Uniform x-directed electric field

xeExExE

xExr

xErE

xcc

x

x

0

0

ˆ

xErE x ˆ


Electron Affinity and Work Function

Electron affinity “” is the energy required to remove an electron from the bottom of the conduction band to outside the crystal, i.e. to the vacuum level

0 x

0

Vacuum level

Potential in a crystal

Conduction band

Energy

V

cE

vE

fE

x

W

Work function “W ” is the energy required to remove an electron from the Fermi level to the vacuum level

• Work function changes with doping but affinity is a constant for a given material

3


Semiconductor N-N Heterostructure: Electron Affinity Rule

Heterostructure: A semiconductor structure in which more than one semiconductor material is used and the structure contains interfaces or junctions between two different semiconductors

Consider the following heterostructure interface between a wide bandgap and a narrow bandgap semiconductor (both n-type):

1gE 2gE

1 2

1cE

1vE

1fE 2cE

2vE

2fE

21

V

1gE2gE

The electron affinity ruletells how the energy band edges of the two semiconductors line up at a hetero-interface


Semiconductor N-N Heterojunction

1cE

1vE

1fE 2cE

2vE

2fE

21

V

1gE2gE

Something is wrong here:the Fermi level (the chemical potential) has to be the same everywhere in equilibrium (i.e. a flat line)

• Once a junction is made, electrons will flow from the side with higher Fermi level (1) to the side with lower Fermi level (2)

Electrons

4


• Electrons will flow from the side with higher Fermi level (1) to the side with lower Fermi level (2)

• Electron flow away from semiconductor (1) will result in a region at the interface which is depleted of electrons (depletion region). Because of positively charged donor atoms, the depletion region has net positive charge density

• Electron flow into semiconductor (2) will result in a region at the interface which has an accumulation of electrons (accumulation region). The accumulation region has net negative charge density

1cE

1vE

1fE2cE

2vE

2fE

2

1

V

1gE2gE

Depletion region Accumulation

region

1gE 2gE

1 2+++++++++++++++

---------------

Note: the vacuum level follows the electrostatic potential:

00 xxexVxV

Semiconductor N-N Heterojunction: Equilibrium


1cE

1vE

1fE2cE

2vE

2fE

2

1

V

1gE2gE

• Electron flow from semiconductor (1) to semiconductor (2) continues until the electric field due to the formation of depletion and accumulation regions becomes so large that the Fermi levels on both sides become the same

• In equilibrium, because of the electric field at the interface, there is a potential difference between the two sides – called the built-in voltage

• The built-in voltage is related to the difference in the Fermi levels before the equilibrium was established:

Depletion region Accumulation

region

1cE

1vE

1fE 2cE

2vE

2fE

21

V

1gE2gE

beV

21 ffb EEeV

beV

Semiconductor N-N Heterojunction: Equilibrium

5


1cE

1vE

1fE 2cE

2vE2fE

21

V

1gE 2gE

Once a junction is made:

• Electrons will flow from the side with higher Fermi level (1) to the side with lower Fermi level (2)

• Holes will flow from the side with lower Fermi level (2) to the side with higher Fermi level (1)

beV

Electrons

Holes

Semiconductor P-N Heterojunction


1cE

1vE

1fE

2cE

2vE2fE

2

1

V

1gE

2gE

Depletion region Depletion

region

beV• Electron flow away from semiconductor (1) will result in a region at the interface which is depleted of electrons (depletion region). Because of positively charged donor atoms, the depletion region has net positive charge density

• Hole flow away from semiconductor (2) will result in a region at the interface which is depleted of holes (depletion region). Because of negatively charged acceptor atoms, the depletion region has net negative charge density

1gE 2gE

1 2+++++++++++++++

---------------

Note: the vacuum level follows the electrostatic potential:

00 xxexVxV

Semiconductor P-N Heterojunction: Equilibrium

6


1cE

1vE

1fE

2cE

2vE2fE

2

1

V

1gE

2gE

Depletion region Depletion

region

1cE

1vE

1fE2cE

2vE2fE

21

V

1gE 2gE

beV

beV

• Electron flow from semiconductor (1) to semiconductor (2) and hole flow from semiconductor (2) to semiconductor (1) continues until the electric field due to the formation of depletion regions becomes so large that the Fermi levels on both sides become the same

• The built-in voltage is related to the difference in the Fermi levels before the equilibrium was established:

21 ffb EEeV

Semiconductor P-N Heterojunction: Equilibrium


Types of Semiconductor Heterojunctions

Type-I: Straddling gap

Type-II: Staggered gap

1cE

1vE

2cE

2vE

21

V

1gE 2gE

1cE

1vE

2cE

2vE

21V

1gE

2gE

1cE

1vE

2cE

2vE

2

1

V

1gE

2gE

Type-III: Broken gap

7


Band Offsets in Heterojunctions

1cE

1vE

2cE

2vE

21

V

1gE2gE

cE

vE

The conduction and valence band offsets are determined as follows:

cggcgv

c

EEEEEE

E

21

12


A PN Heterojunction

Ec1

Ef1

Ev1

Ec2Ef2

Ev2

q1q2

Eg1 Eg2

Vacuum level

Ec

Ev

x

x

1 (p-doped) 2 (n-doped)

+ +

+ +

+ +

+ +

- -

- -

- -

- -

-xpxn

12 ffbi EEqV

222 ln

c

dcf N

NKTEE

111 ln

v

afv N

NKTEE

.ln12

2

vc

davgbi NN

NNKTEEqV

a

ipo N

nnn

21

apo Np dno Nn

d

ino N

npp

22

8


A PN Heterojunction in Equilibrium

Ec1

Ef

Ev1

Ec2

Ef

Ev2

q1

q2Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

x

+

-

-xp

xn

(x)

+qNd

-qNa

The Depletion Approximation:


A PN Heterojunction in Equilibrium

x

-xp

xn

E(x)

elsewhere0

0

0

1

2

xxxxqN

xxxxqN

xE ppa

nnd

xxExdxd

Electric Field:

x-xp

xn

(x)

VbiElectrostatic Potential:

2

2

2nd xN

q

1

2

2paxN

q

pand xqNxqN

1

2

2

2

22 pand

bixN

qxN

qV

21

2121

21

2121

2

2

da

bi

d

an

da

bi

a

dp

NN

V

N

N

qx

NN

V

N

N

qx

QxqNxqN pand

Charge per unit area:

9


Ec1

Ef1Ev1 Ec2

Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

-qV

A PN Heterojunction in Reverse Bias

x-xp xn

1 (p-doped) 2 (n-doped)+ ++ ++ +

- -- -- -

V+ -

Wn-Wp

V<0

Quasi Fermi Levels and their Splitting:



x-xpxn

1 (p-doped) 2 (n-doped)+ ++ ++ +

- -- -- -

Wn-Wp

V<0

21

2121

21

2121

2

2

da

bi

d

ap

da

bi

a

dn

NN

VV

N

N

qVx

NN

VV

N

N

qVx

Depletion regions grow in width:

V+ -

10


A PN Heterojunction in Forward Bias

x-xpxn

1 (p-doped) 2 (n-doped)+ +-

V+ -

Wn-Wp

V>0

Ec1

Ef1Ev1

Ec2Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

qV

--

+ ++ +

--

-

Now diffusion exceeds drift!!Minority carrier injection………

Electrons

Holes



Ec1

Ef1Ev1

Ec2Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

qV

Electrons

Holes

Assumption:

Main bottleneck for current flow are the quasineutral regions and not the depletion regions

a

KTxExEvp

KTqVpo

KTqV

a

i

KTxExEKTxExEc

KTxExEcp

NeNxp

eneNn

eeNeNxn

pfpv

pfpfpcpfpcpf

)()(1

21

)()()()(1

)()(1

1

1212

)(

)(

Electron concentration on the p-side:

11



Ec1

Ef1Ev1

Ec2Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

qV

Electrons

Holes

Assumption:

Main bottleneck for current flow are the quasineutral regions and not the depletion regions

Hole concentration on the n-side:

KTqVno

KTqV

d

i

KTxExEKTxExEv

KTxExEvn

dKTxExE

cn

epeNn

eeNeNxp

NeNxn

nfnfnfnvnfnv

ncnf

22

)()()()(2

)()(2

)()(2

1221

2

)(

)(



x-xp xn Wn-Wp

p(x)

n(x)

Minority carrier concentrations:

Electrons on the p-side:

1e

poee

nxnxGxR

Excess electrons injected in the p-side will recombine with the holes

xxn

DqxJ ee

1 Diffusion current

xRxGxJxqt

neee

1

Need to solve: 0

1

2

2

1e

poe

nxn

x

xnD

12


A PN Heterojunction in Forward BiasMinority carrier concentrations:

111

21

2

2

12

2

1

eee

e

po

e

poe

DL

L

nxn

x

xn

nxn

x

xnD

222

22

2

2

hhh

h

no

DL

L

pxp

x

xp

P-side: N-side:

Boundary conditions:KTqV

pop enxn )(

pop nWn )(

Boundary conditions:KTqV

non epxp )(

non pWp )(?? ??


A PN Heterojunction in Forward BiasMinority carrier concentrations:

ppKT

qV

e

pp

e

p

popo xxWe

L

xW

L

xW

nnxn

1

sinh

sinh

1

1

nnKT

qV

h

nn

h

n

nono Wxxe

LxW

LxW

ppxp

1

sinh

sinh

2

2

13



Majority carrier concentrations and charge neutrality:

P-side:

One must have: poa nxnxnNxpxp

Excess majority carrier density must balance the excess minority carrier density

N-side:

One must have: nod pxpxpNxnxn


A PN Heterojunction in Forward BiasMinority carrier current:

ppKTqv

e

pp

e

p

e

e

a

i

eee

xxWe

L

xW

L

xW

LD

Nn

q

xxn

DqxEqnxJ

1

sinh

cosh

1

1

1

121

11

P-side:~0

nnKT

qv

n

nn

n

n

h

h

d

i

nhhh

Wxxe

LxW

LxW

LD

Nn

q

xxp

DqxEqpxJ

1

sinh

cosh

2

2

2

222

2

N-side:~0

14


x-xp xn Wn-Wp

Jh(x)Je(x)


Since there is no obstacle to current flow in the depletion regions, and if we ignore electron-hole recombination in the depletion region, we must have:

x-xp xn Wn-Wp

Jh(x)Je(x)

Total current:

xJxJJ heT Must be constant throughout the device

JT

Minority carrier current:



x-xp xn Wn-Wp

Jh(x)Je(x)

JT

Total current:

1cothcoth

22

222

11

121 KT

qv

h

nn

h

h

d

i

e

pp

e

e

a

iT e

LxW

LD

Nn

L

xW

L

D

Nn

qJ

1KT

qv

oT eIAJI

22

222

11

121 cothcoth

h

nn

h

h

d

i

e

pp

e

e

a

io L

xWLD

Nn

L

xW

L

D

Nn

qAI

V+ -

I

A

15



x-xp xn Wn-Wp

Jh(x)Je(x)

JT

Majority carrier current:

P-side:

N-side:

xJJxJ eTh

xJJxJ hTe



Ec1

Ef1

Ev1

Ec2

Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

qV

Quasi Fermi Levels:

1KT

qv

oT eIAJI

16



Ec1

Ef1Ev1 Ec2

Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

-qV

1KT

qv

oT eIAJI

Reverse bias current: I -Io

Quasi Fermi Levels:

Why?



Electron-Hole Recombination in the Depletion Region:

Ec1

Ef1

Ev1

Ec2

Ef2

Ev2

q1

q2

Eg1

Eg2

Vacuum level

Ec

Ev

xxn-xp

qV

xRxGxJxq eee

1

dxxGxRqxJxJn

p

x

xeepene

17




x-xp xn Wn-Wp

Jh(x)Je(x)

JT

dxxGxRqxJxJn

p

x

xeepene

xRxGxJxq eee

1

dxxGxRqxJxJn

p

x

xhhnhph

Similarly:




x-xp xn Wn-Wp

Jh(x)Je(x)

JT

dxxGxRqeL

xWLD

Nn

L

xW

L

D

Nn

qJn

p

x

xee

KTqv

h

nn

h

h

d

i

e

pp

e

e

a

iT

1cothcoth

22

222

11

121

2iee nnpxRxG

KTEEi

ffennp 122

1KT

qv

ee exRxG

18


Effective Mass Schrodinger Equation

Consider a semiconductor with energy band dispersion:

oeocc kkMkkEkE

..2

12

Energy

k

fEcE

vE

The Bloch functions are solutions of the equation:

rkErrVm kcckcLattice

,,

22

2

ruV

er kc

rki

kc

,

.

,

What if one needs to solve the equation:

rErrUrVm Lattice

2

22

Some extra potential (perhaps due to some crystal impurity, defect, or external electric field)

ok


Effective Mass Schrodinger EquationEnergy

k

fEcE

vE

One can in most cases write the solution as:

rrrokc

,

Envelope function

Where the envelope function satisfies the “effective mass Schrodinger equation”:

rErrUikE oc

ˆ

ok

rErrUrVm Lattice

2

22

19


The Envelope Function

rErrUikE oc

ˆ

Energy

k

rrrokc

,

r

rokc

,

Slowly varying envelope function Bloch function

ok

Electron wavefunction


The Effective Mass Schrodinger Equation: An Example Energy

k

ok

Consider a conduction energy band with the dispersion:

zz

ozz

yy

oyy

xx

oxxcc m

kkm

kk

mkk

EkE222

222222

rErrUikE oc

ˆ

Note that one has to make the following replacements in the energy dispersion relation:

z

ikky

ikkx

ikkikEkE ozzoyyoxxocc

ˆ

The operator is then: ikE oc

ˆ

2

22

2

22

2

22

222 zmymxmEikE

zzyyxxcoc

The effective mass Schrodinger equation becomes:

rErrUEzmymxm c

zzyyxx

2

22

2

22

2

22

222

What is this equation:

20


Electrons at Heterojunctions

1cE

1vE

2cE

2vE

1gE 2gE

cE

vE

Question: What happens to the electron that approaches the interface (as shown)? How does it see the band offset? Does it bounce back? Does it go on the under side?

The effective mass equation can be used to answer all the above questions

In semiconductor 1:

In semiconductor 2:

rErrUikE oc

111ˆ

rrrokc

,111

rErrUikE oc

222ˆ

rrrokc

,222


Electrons at Heterojunctions; Effect of Band Offsets

1cE

1vE

2cE

2vE

1gE 2gE

cE

vE

1

22

11 2 ecc m

kEkE

rrr

okc

0,111

Assume for the electron in the conduction band of semiconductor 1:

rErEm c

e

111

2

1

2

2

2

22

22 2 ecc m

kEkE

rrr

okc

0,222

And for the electron in semiconductor 2:

rErEm c

e

222

2

2

2

2

Notice that the conduction band edge energy (i.e. Ec1 or Ec2) appears as a constant potential in the effective mass Schrodinger equation

Conduction band offset at the heterojunction therefore appears like a potential step to the electron

0rU

21


Electrons at Heterojunctions: Boundary Conditions

(1) Continuity of the wavefunction at the boundary:

0201 xx rr

(2) Continuity of the normal component of the probability current at the boundary:

In text book quantum mechanics the probability current is defined as:

rim

rrim

rccrim

rrJ ***

22..

2

..2

* ccrim

rrJ

Or in shorter component notation:

Probability current is always continuous across a boundaryWe need an expression for the probability current in terms of the envelope function

0201 xx rr

If one assumes: rr

oo kckc

,2,1



Probability Current: In a material with energy band dispersion given by:

oonoonn kkkkm

EkkMkkEkE

,

21

2

2..

2

The expression for the electron probability current (in terms of the envelope function) is:

..2

* ccrim

rrJ

Continuity of the probability current:The continuity of the normal component of the probability current across a heterojunction gives another boundary condition for the envelope function:

02

2011

11xxxx

rm

rm

For:

zz

yy

xx

m

m

m

M

1

1

11

0

2

20

1

1

11

xxxxxx xr

mxr

m

22



(1) Continuity of the envelope function at the boundary:

(2) Continuity of the normal component of the probability current at the boundary:

0x x

Semiconductor 1 Semiconductor 2

0201 xx rr

02

2011

11xxxx

rm

rm

zz

yy

xx

m

m

m

M

1

1

11

0

2

20

1

1

11

xxxxxx xr

mxr

m

If in both the materials the inverse effective mass matrix is diagonal then this boundary condition becomes:


The Effective Mass Theory for Heterojunctions

1cE2cE

cE

x0

Assume in semiconductor (1):

rErEzmymxm

rEriE

rErrUikE

czyx

c

oc

1112

2

1

2

2

2

1

2

2

2

1

2

111

111

222

ˆ

ˆ

1

22

1

22

1

22

11 222 z

z

y

y

x

xcc m

km

k

mk

EkE

Assume in semiconductor (2):

2

22

2

22

2

22

22 222 z

z

y

y

x

xcc m

km

k

mk

EkE

In semiconductor (1):

0ok

0ok

23



1cE2cE

cE

x0

rErEzmymxm c

zyx

1112

2

1

2

2

2

1

2

2

2

1

2

222

Assume a plane wave solution: zkykxki zyxer 1

1

Plug it in to get:1

22

1

22

1

21

2

1 222 z

z

y

y

x

xc m

km

k

mk

EE

We expect a reflected wave also so we write the total solution in semiconductor (1) as:

zkykxkizkykxki zyxzyx erer 11

1

A plane wave solution works


r t r

1 r

2



rErEzmymxm c

zyx

2222

2

2

2

2

2

2

2

2

2

2

2

222

Assume a plane wave solution: zkykxki zyxetr 2

2

Plug it in to get:2

22

2

22

2

22

2

2 222 z

z

y

y

x

xc m

km

k

mk

EE

A plane wave solution works here also


1cE2cE

cE

x0

r t r

1 r

2

24


Boundary Conditions at Heterojunctions

zkykxkizkykxki zyxzyx erer

111

zkykxki zyxetr

22

(1) Envelope functions must be continuous at the interface:

1

22

1

22

1

21

2

1 222 z

z

y

y

x

xc m

km

k

mk

EE

2

22

2

22

2

22

2

2 222 z

z

y

y

x

xc m

km

k

mk

EE

tr

etere

xxzkykizkykizkyki zyzyzy

1

00 21

Note that this boundary condition can only be satisfied if the components of the wavevector parallel to the interface are the same on both sides

1cE2cE

cE

x0

r t r

1 r

2




111

zkykxki zyxetr

22

Energy conservation:

1

22

1

22

1

21

2

1 222 z

z

y

y

x

xc m

km

k

mk

EE

2

22

2

22

2

22

2

2 222 z

z

y

y

x

xc m

km

k

mk

EE

zyeffx

x

x

x

zz

z

yy

yc

x

x

x

x

z

z

y

y

x

xc

z

z

y

y

x

xc

kkVmk

mk

mmk

mm

kE

mk

mk

mk

m

k

mk

Emk

m

k

mk

EE

,22

112

11222

222222

1

21

2

2

22

2

12

22

12

22

1

21

2

2

22

2

2

22

2

22

2

22

2

21

22

1

22

1

21

2

1

Note that the effective barrier height depends on the band offset as well as the parallel components of the wavevector

1cE2cE

cE

x0

r t r

1 r

2

25




111

zkykxki zyxetr

22

(2) Probability current must be continuous at the interface:

1

22

1

22

1

21

2

1 222 z

z

y

y

x

xc m

km

k

mk

EE

2

22

2

22

2

22

2

2 222 z

z

y

y

x

xc m

km

k

mk

EE

tmk

rmk

etmik

eremik

xmxm

x

x

x

x

zkyki

x

xzkykizkyki

x

x

xxxx

zyzyzy

2

2

1

1

2

2

1

1

0

2

20

1

1

1

11

Conservation of probability current at the interface

1cE2cE

cE

x0

r t r

1 r

2


Transmission and Reflection at Heterojunctions

tr 1 tmk

rmk

x

x

x

x

2

2

1

1 1

We have two equations in two unknowns:

The solution is:

1221

1221

1221 11

12

xxxx

xxxx

xxxx kmkmkmkm

rkmkm

t

zyeffx

x

x

x kkVmk

mk

,22 1

21

2

2

22

2

Where:

Special case: If the RHS in the above equation is negative, then kx2 becomes imaginary and the wavefunction decays exponentially for x>0 (in semiconductor 2). In this case:

and the electron is completely reflected from the hetero-interface

1r

1cE2cE

cE

x0

r t

26


Semiconductor Quantum Wells

1cE2cE

cE2cE

1vE

2vE2vE

AlGaAs AlGaAsGaAsA thin (~1-10 nm) narrow bandgap material sandwiched between two wide bandgap materials

GaAs

GaAsInGaAs quantum well (1-10 nm)

Semiconductor quantum wells can be composed of pretty much any semiconductor from the groups II, III, IV, V, and VI of the periodic table

TEM micrograph

GaAs

GaAs

InG

aA

s


Semiconductor Quantum Well: Conduction Band Solution

1cE

2cE

cE2cE

x

0Assumptions and solutions:

e

cc mk

EkE2

22

11

e

cc mk

EkE2

22

22

rErEm

rEriE

ce

c

111

22

111

2

ˆ

rErEm

rEriE

ce

c

222

22

222

2

ˆ

zkykix

zkykix

zy

zy

exk

exkAr

sin

cos1

2

2

2

2 Lxee

eeBr zkykiLx

zkykiLx

zy

zy

2

2

2

2 Lxee

eeBr zkykiLx

zkykiLx

zy

zy

L

Symmetric

Anti-symmetric

27


22

2||

22

2

2||

22

1

2

22

xce

ec

e

xc

kEm

m

kE

m

kkEE

1cE2cE

cE2cE

x

0

L

Energy conservation condition:

The two unknowns A and B can be found by imposing the continuity of the wavefunction condition and the probability current continuity condition to get the following conditions for the wavevector kx:

x

xce

x

x

x

xce

x

x

k

kEm

kLk

k

kEm

kLk

22

22

2

2cot

2

2tan

Wavevector kx cannot be arbitrary!Its value must satisfy these transcendental equations

222|| zy kkk

Semiconductor Quantum Well: Conduction Band Solution


1cE

2cE

cE2cE

x

0

L

x

xce

x

x

x

xce

x

x

k

kEm

kLk

k

kEm

kLk

22

22

2

2cot

2

2tan

2Lkx2

23 2

250

Different red curves for Increasing Ec values

Graphical solution:

• Values of kx are quantized• Only a finite number of solutions are possible – depending on the value of Ec

In the limit Ec ∞ the values of kxare:

Lpkx ( p = 1,2,3……..

Semiconductor Quantum Well: Conduction Band Solution r

28


Electrons in Quantum Wells: A 2D Fermi Gas

1cE

2cE

cE2cE

x

0

L

epc

ee

xc

m

kEE

m

k

mk

EE

2

222||

2

1

2||

222

1

Since values of kx are quantized, the energy dispersion can be written as:

p = 1,2,3……..

• We say that the motion in the x-direction is quantized (the energy associated with that motion can only take a discrete set of values)• The freedom of motion is now available only in the y and z directions (i.e. in directions that are in the plane of the quantum well)• Electrons in the quantum well are essentially a two dimensional Fermi gas!

1E

2E

In the limit Ec ∞ the values of Ep are:22

2

Lp

mE

ep

p = 1,2,3……..

222|| zy kkk


1cE

2cE

cE2cE

x

0

L1E

2E

Energy Subbands in Quantum Wells

e

pcc m

kEEkpE

2,

2||

2

1||

p =1,2,3……..

The energy dispersion for electrons in the quantum wells can be plotted as shown

It consists of energy subbands (i.e. subbands of the conduction band)

Electrons in each subband constitute a 2D Fermi gas||k

1cE11 EEc

21 EEc

E

kz

ky

31 EEc

E

222|| zy kkk

29


1cE

2cE

cE2cE

x

0

L1E

2E

Density of States in Quantum Wells

Suppose, given a Fermi level position Ef , we need to find the electron density:We can add the electron present in each subband as follows:

pfc EkpEf

kdn ||2

||2

,2

2

fEIf we want to write the above as:

fQWE

EEfEgdEnc

1

Then the question is what is the density of states gQW(E ) ?||k

1cE

11 EEc

21 EEc 31 EEc

||k

1cE

11 EEc

21 EEc 31 EEc


Density of States in Quantum Wells

fE

||k

1cE

11 EEc

21 EEc 31 EEc

||k

1cE

11 EEc

21 EEc 31 EEc

pfc EkpEf

kdn ||2

||2

,2

2

Start from:

e

pcc m

kEEkpE

2,

2||

2

1||

And convert the k-space integral to energy space:

fp

pce

E

pf

e

EE

EEfEEEm

dE

EEfm

dEn

c

pc

12

2

1

1

This implies:

ppc

eQW EEE

mEg 12

EgQW

1cE 11 EEc 21 EEc 31 EEc

2em

22

em

23

em

30


Density of States: From Bulk (3D) to QW (2D)

The modification of the density of states by quantum confinement in nanostructures can be used to:

i) Control and design custom energy levels for laser and optoelectronic applicationsii) Control and design carrier scattering rates, recombination rates, mobilities, for electronic applications iii) Achieve ultra low-power electronic and optoelectronic devices

Eg D3k

1cE

E E E

Eg D2

2em

22

em

23

em

||k

1cE

11 EEc

21 EEc

31 EEc

EE

Eg D2

2em

22

em

23

em

||k

1cE

11 EEc

21 EEc

31 EEc

E

EgQW


Semiconductor Quantum Well: Valence Band Solution1vE

2vEvE2vE

x

0Assumptions and solutions:

h

vv mk

EkE2

22

11

v

vv mk

EkE2

22

22

rErEm

rErEm

rEriE

vh

vh

v

111

22

111

22

111

2

2

ˆ

rErEm

rErEm

rEriE

vh

vh

v

222

22

222

22

222

2

2

ˆ

zkykix

zkykix

zy

zy

exk

exkAr

sin

cos1

2

2

2

2 Lxee

eeBr zkykiLx

zkykiLx

zy

zy

2

2

2

2 Lxee

eeBr zkykiLx

zkykiLx

zy

zy

L

Symmetric

Anti-symmetric

31


22

2||

22

2

2||

22

1

2

22

xvh

ev

xv

kEm

m

kE

mh

kkEE

Energy conservation condition:

The two unknowns A and B can be found by imposing the continuity of the wavefunction condition and the probability current conservation condition to get the following conditions for the wavevector kx:

x

xvh

x

x

x

xvh

x

x

k

kEm

kLk

k

kEm

kLk

22

22

2

2cot

2

2tan

Wavevector kx cannot be arbitrary!

1vE

2vEvE2vE

x

0

L

Semiconductor Quantum Well: Valence Band Solution


1vE

2vE

vE

2vE

x

0

L

x

xvh

x

x

x

xvh

x

x

k

kEm

kLk

k

kEm

kLk

22

22

2

2cot

2

2tan

2Lkx2

23 2

250

Different red curves for Increasing Ev values

Graphical solution:

• Values of kx are quantized• Only a finite number of solutions are possible – depending on the value of Ev

In the limit Ev ∞ the values of kxare:

Lpkx ( p = 1,2,3……..

Semiconductor Quantum Well: Valence Band Solution

32


1vE

2vE

vE

2vEx

0

L

hpv

hh

xv

m

kEE

m

k

mk

EE

2

222||

2

1

2||

222

1

Since values of kx are quantized, the energy dispersion can be written as:

p = 1,2,3……..

• We say that the motion in the x-direction is quantized (the energy associated with that motion can only take a discrete set of values)• The freedom of motion is now available only in the y and z directions (i.e. in directions that are in the plane of the quantum well)• Electrons (or holes) in the quantum well are essentially a two dimensional Fermi gas!

1E

2E

In the limit Ev ∞ the values of Ep are:22

2

Lp

mE

hp

p = 1,2,3……..

Semiconductor Quantum Wells: A 2D Fermi Gas

Light-hole/heavy-hole degeneracy breaks!


Density of States in Quantum Wells: Valence Band

fE

pfv EkpEf

kdp ||2

||2

,12

2

Start from:

h

pvv m

kEEkpE

2,

2||

2

1||

And convert the k-space integral to energy space:

fp

pvh

E

pf

hEE

EEfEEEm

dE

EEfm

dEp

v

pv

1

1

12

2

1

1

This implies:

ppv

hQW EEE

mEg 12

EgQW

1vE11 EEv 21 EEv 31 EEv

2hm

22

hm

23hm

||k

1vE

11 EEv

21 EEv

31 EEv

E

33


Growth of Semiconductor Heterostructures: MBE

Low pressure (10-11 Torr), near-equilibrium, chemical reaction free, layer-by-layer growth


Growth of Semiconductor Heterostructures: MOCVD or MOVPE

Adsorption

TM-In

PH3

CH4

Growth of InP by MOCVD

Atm pressure (760 Torr) growth, involves gas flow and chemical reactions

34


Epitaxial Growth and Lattice Mismatch

A lattice mismatch between the epitaxial layer and the substrate means that the layer grown will be strained (biaxial strain):

sub

sub

a

aa

0

0

Tensile strain

Compressive strain

if the thickness h of the coherently strained layer exceeds a certain critical thickness hc the coherent strain relaxes and this process generates crystal dislocations (crystal defects). Critical thickness is given by:

b

hbh c

c lncos1

cos14

2

2ab

a

asub

Poisson ratio

and are both equal to 60-degrees for diamond and zinc-blende lattices

for diamond and zinc-blende lattices

Matthews-Blakeslee Formula


Strain Compensation

b

hbh c

c lncos1

cos14

2

How does one calculate the critical thickness for a multiple layer stack?

321

332211

hhh

hhhavg

Strain compensation can be used to grow much thicker dislocation-free layers!

Substrate

h1

h2

Documents

Chapter 2 Semiconductor Heterostructures - Cornell Universitysemiconductor (1) to semiconductor (2) and hole flow from semiconductor (2) to semiconductor (1) continues until the electric