EE7603 Lecture Notes FWJ 2013

EE7603

Advanced Semiconductor Physics

Dr. Fan Wei Jun Associate Professor

Office: S2.2-B2-20 Phone: 6790 4359 email: [email protected]

1

2

Jasprit Singh, Physics of Semiconductors

and Their Heterostructures, McGraw-Hill, 1993.

J. Singh, Electronic and Optoelectronic Properties of Semiconductor Structures, Cambridge 2003.

S. L. Chuang, Physics of Photonic Devices, Wiley 2009.

References

3

Crystal Structure

Outline

Semiconductor Band Theory

Band Gap Engineering

Doping and Carrier Concentrations

Crystal Structure

4

Different states of matter and a classification based on the order present in the material.

5 Strong bonds Weak bonds No bonds

CRYSTALLINE MATERIALS: SOME DEFINITIONS

BRAVAIS LATTICE TRANSLATION VECTORS PRIMITIVE TRANSLATION VECTORS BASIS PRIMITIVE CELL UNIT CELL WIGNER-SEITZ CELL

7

BRAVAIS LATTICE: Collection of points that fill up space. Every point has the same environment around it. An array of discrete points that appears

exactly the same from any point the array is viewed.

A kind of 2D Bravais Lattice 8

Auguste Bravais (1811-1863) was a French physicist, well known for his work in crystallography.

TRANSLATION VECTOR: A translation of the crystal by a vector R that takes a point R0 to R0+R and leaves the entire crystal invariant.

PRIMITIVE TRANSLATION VECTORS: The a1, a2, a3 are called primitive if the volume of the cell formed by them is the smallest possible. Primitive translation vector also called as primitive vector.

The choice of primitive vectors is not unique.

a1

a2

9

R = n1a1 + n2a2 + n3a3 For 3D

Primitive vectors are used to define a translation vector, R

For 2D R = n1a1 + n2a2

10

BASIS: A physical crystal can be described by giving its underlying Bravais lattice, together with a description of the arrangement of atoms within a particular primitive cell. This arrangement within the primitive cell is referred to as the Basis The basis can consist of one or more atoms.

A Basis consisted of 2 atoms

11

Lattice + Basis = Crystal Structure

+

||

2D lattice

Basis

Crystal Structure

12

PRIMITIVE CELL: The primitive vectors define a parallelopiped of volume a1 a2 x a3 which is called the primitive cell.

There are many different ways of selecting a primitive cell.

Primitive cell is not unique.

Each primitive cell contains the equivalent of one atom.

Generally, primitive cell can not display the full symmetry of the crystal.

a1

a2

13

UNIT CELL: a smallest building block of crystal, which can display the full symmetry of the crystal. It can have more than one atoms.

primitive vector and primitive cell for the face centered cubic (FCC) unit cell

14

WignerSeitz Cell A very useful way to define a primitive cell is the

procedure given by Wigner and Seitz. The procedure involves:

1. Drawing lines to connect a given lattice point to all neighboring points;

2. Drawing bisecting lines or planes to the previous lines.

The smallest volume enclosed within the bisections is called as the WignerSeitz cell.

15

Eugene Paul "E. P." Wigner (1902 1995) was a Hungarian American physicist and mathematician. He received a share of the Nobel Prize in Physics in 1963 "for his contributions to the theory of the atomic nucleus and the elementary particles, particularly through the discovery and application of fundamental symmetry principles";

Frederick Seitz (1911 2008) was an American physicist and a pioneer of solid state physics. Wigners Ph.D student.

From wikipedia

Construction of a Wigner-Seitz primitive cell

16

of FCC and BCC

17

A state in which parts on opposite sides of a plane, line, or point display arrangements that are related to one another via a symmetry operation such as translation, rotation, reflection or inversion.

and www.bioc.rice.edu/~bios482/Xtal_PPT/lecture2.ppt http://www.fkkt.uni-lj.si/attachments/dsk2423/ 18

Application of the symmetry operators leaves the entire crystal unchanged.

Symmetry

Symmetry Elements

Translation

moves all the points in the asymmetric unit the same distance in the same direction. This has no effect on the handedness of figures in the plane. There are no invariant points (points that map onto themselves) under a translation.

In geometry, a translation "slides" an object by a a: Ta(p) = p + a.

19

Symmetry Elements

Rotation

Turns all the points in the asymmetric unit around one axis, the center of rotation. A rotation does not change the handedness of figures. The center of rotation is the only invariant point (point that maps onto itself).

20

Also called as proper rotation.

n-fold rotational symmetry n-fold rotational symmetry with respect to a particular point (in 2D) or axis (in 3D) means that rotation by an angle of 360/n (180, 120, 90, 72, 60, 51 3/7 , etc.) does not change the object. Note that "1-fold" symmetry is no symmetry, and "2-fold" is the simplest symmetry. n is the number of the repeat times after rotating 360.

The notation for n-fold symmetry is Cn or n.

C2 C3 C4

Wiki 21

MIT OCW 3.012 22

23 From: http://www.fkkt.uni-lj.si/attachments/dsk2423/3d-symmetry-2007_2008.pdf

Symmetry Elements

Inversion, or center of symmetry

26

Symmetry Elements

Mirror plane or Reflection

29

Symmetry elements: mirror plane and inversion center

The handedness is changed.

32

Improper rotation

An improper rotation (Sn) is performed by rotating the object 360/n followed by reflection through the plane perpendicular to the rotation axis.

33

n=? http://symmetry.otterbein.edu/tutorial/improper.html

Symmetry Elements

Ethane C2H6

S3 S6 ?

Crystal system

Crystals are grouped into seven crystal systems, according to characteristic symmetry of their unit cell.

The characteristic symmetry of a crystal is a combination of one or more rotations and inversions.

36

French physicist and mineralogist, best remembered for his work on the lattice theory of crystals. Bravais lattices are named for him. In 1850, he showed that crystals could be divided into 14 unit cells These unit cells fall into seven geometrical categories, which differ in their relative edge lengths and internal angles. In 1866, he elaborated the relationships between the ideal lattice and the material crystal. Sixty years later, Bravais' work provided the mathematical and conceptual basis for the determination of crystal structures after Laue's discovery of X-ray diffraction in 1911.

Auguste Bravais (1811-1863)

37

5 two-dimensional Lattices

38 graphene

There are 14 three-dimensional Lattices classified into 7 crystal systems

The volume of the unit cell = where a , b and c are the primitive vectors

a b c

Q: Which are unit cells and also primitive cells?

14 three-dimensional Lattices

end face centered

Simple; Body centered; Face centered, respectively

http://jas.eng.buffalo.edu/

39

40

S- Simple lattice has only a lattice point at each corner of the three-dimensional unit cell B - Body-centered lattice contains not only lattice points at each corner of the unit cell but also contains a lattice point at the center of the three-dimensional unit cell F - Face-centered lattice possesses not only lattice points at the corners of the unit cell but also at the centers of all three pairs of faces E End face centered lattice possesses not only lattice points at the corners of the unit cell but also at the centers of just one pair of faces

The Bravais lattices come in 4 different types:

Lattice types of Bravais lattices

7 crystal systems and 14 Bravais lattices Distance

Parameters (a, b, c)

Angle Parameters (,,) Name

No. of Lattices / Types Example

a = b = c = = = 900 Cubic 3 S,B,F CsCl, NaCl

a = b c = = = 900

Tetragonal 2

S,B TiO2 (Rutile), SnO2 (Cassiterite)

a b c = = = 900

Orthorhombic 4

S,B,E,F KNO3, BaSO4 (Baryte)

a = b = c = =

42

Dendrytic ice crystals imaged with a scanning electron microscope. The colors are computer generated.

NaCl

Ice crystal

TiO2

Ice

Ice structure

Cubic Crystal System

Primitive centering (PCC) Body centered (BCC) Face centered (FCC)

eight atoms (at corners). one atom/unit cell.

eight atoms at corners. one atom at body centre. two atoms/unit cell.

eight atoms at corners. six atoms at face centres. four atoms/unit cell.

43 http://www.youtube.com/watch?v=Rm-i1c7zr6Q&feature=player_detailpage

Primitive vectors of BCC lattice

a, Lattice Constant: the cube dimension

( )( )( )

11 2

12 2

13 2

a x y z a x y z

a x y z

aaa

= + = + + = +

44

12

3

( )2

( )2

( )2

aa y z

aa z x

aa x y

= +

= +

= +

Primitive vectors of FCC lattice

Volume of primitive cell, 3

1 2 3 / 4V a a a a= =

45

Primitive vectors of simple hexagonal lattice

zca

yaxaa

yaxaa

23

2

23

2

3

2

1

=

+=

=

x

46

a1

a2

a3

y

z

Total of two atoms per unit cell (shaded) Net 1 atom at corners Plus 1 atom inside cell

Hexagonal Close-Packed (HCP)

ENGR 145, Chemistry of Materials Case Western Reserve University

83

ca

=

47

48

c/2

22 2 2

2 3 3c x xy x = = =

a a/2

x

y

32

ax =2 223

ac y = =

2 2 1.6333

ca

= =

Show for Ideal HCP: 83

ca

=

49

HCP: AB. CCP: ABC

CCP (Cubic Close Pack) = FCC

Lattice video

diamond

Diamond & Zincblende Most semiconductors adopt a cubic lattice, but the

cubic structure is complicated, known as the diamond (such as Si) or zincblende (GaAs) structure.

zincblende

50 http://www.youtube.com/watch?v=u2-0y6oxmSk&feature=player_detailpage http://www.youtube.com/watch?v=MD79L2W9sp4&feature=player_detailpage

Diamond & Zincblende

Taking a block of dimension a/2, we can see how the diamond structure relates to the FCC structure.

a/2

Diamond lattice: two overlapping FCC sublattices, offset from one another. Offset is one quarter of a body diagonal.

Zincblende structure: the two offset lattices are of different atoms. Each group III site is surrounded by 4 group V sites, and vice-versa.

51

Packing Density

Packing density = No. of sphere x volume of spherevolume of unit cell

BCC: 68% FCC: 74% Diamond Structure: 34% HCP:74%

52

Example. BCC packing density Relationship between atomic radius r

and lattice constant a: Number of atoms/unit cell: 2 Volume of atoms: Volume of unit cell:

4 3r a=

Vatoms = 2

4r3

3

=

38

a3

Vcell = a3

3 0.688

atoms

cell

VPDV

= = = a

a

b

Examples: Alkali metals (Li, Na, K, Rb), Cr,Mo, W, Mn, -Fe (< 912 C), -Ti (> 882 C).

Body centered cubic (BCC) crystal structure. (a) A BCC unit cellwith closely packed hard spheres representing the Fe atoms. (b) Areduced-sphere unit cell.

ENGR 145, Chemistry of Materials Case Western Reserve University

53

The reciprocal (space) lattice of a Bravais lattice is the set of all vectors K such that

for all lattice point position vectors R. Reciprocal space is also called Fourier space, k- space, or momentum space.

Reciprocal Lattice

1=RiKe

54

For an infinite three dimensional lattice, defined by its primitive vectors, a1, a2, a3 , its reciprocal lattice can be determined by generating its three reciprocal primitive vectors, through the formulae

= 321 2aab

= 132 2aab

= 213 2aab

|)(| 321 aaa =

Note: ijji ba 2=55

The real lattice is described at the left, the reciprocal lattice is described at the right. g is the reciprocal lattice vector. The absolute value of g is equal to 2pi/d and the direction is that of the normal N to the appropriate set of parallel atomic planes of the real lattice separated by distance d. Two parallel planes of the three-dimensional lattice are shown below.

http://www.chembio.uoguelph.ca/educmat/chm729/recip/3vis.htm

Vlad Zamlynny 56

The reciprocal lattice of an FCC lattice is the body-centred cubic (BCC) lattice.

Primitive lattice vectors Reciprocal lattice vectors a1 = ( 0 , 1 , 1)a/2 b1 = ( -1, 1, 1 ) 2 /a a2 = ( 1 , 0 , 1)a/2 b2 = ( 1, -1, 1 ) 2 /a a3 = ( 1 , 1 , 0)a/2 b3 = ( 1, 1, -1 ) 2 /a The reciprocal lattice of a BCC lattice is the

FCC lattice. a1 = ( -1, 1, 1)a/2 b1 = ( 0, 1, 1 ) 2 /a a2 = ( 1, -1, 1)a/2 b2 = ( 1, 0, 1 ) 2 /a a3 = ( 1 , 1, -1)a/2 b3 = ( 1, 1, 0 ) 2 /a

57

Brillouin zone The first Brillouin

zone is a uniquely defined primitive cell of the reciprocal lattice. It is found by the same method as for the WignerSeitz cell in the Bravais lattice. First Brillouin zones of (a) square

reciprocal lattice and (b) hexagonal reciprocal lattice.

http://en.wikipedia.org/wiki/Brillouin_zone

The first Brillouin zone = WignerSeitz cell of reciprocal lattice 58

Lon Nicolas Brillouin (1889 1969) was a French physicist. He made contributions to quantum mechanics, radio wave propagation in the atmosphere, solid state physics, and information theory.

First Four Brillouin Zones: Square Lattice

www.lehigh.edu/~jdg4/classwork/reciprocallattice.ppt 59

All Brillouin Zones: Square Lattice

www.lehigh.edu/~jdg4/classwork/reciprocallattice.ppt 60

15th Brillouin Zones of the simple cubic crystal:

61 The first 20 Brillouin Zones of the simple cubic crystal video

First Brillouin Zone of BCC Reciprocal Lattice

(FCC Lattice) with high symmetry k points marked

http://en.wikipedia.org/wiki/Brillouin_zone

(0, 0, 0) Center

L (0.5, 0.5, 0.5) 2/a

X (1, 0, 0) 2/a

U,K (0.25,0.25,1) 2/a

W (1, 0.5, 0) 2/a

K Middle of an edge joining two hexagonal faces

L Center of a hexagonal face

U Middle of an edge joining a hexagonal and a square face

W Corner point X Center of a square face

Si,

GaAs,

AlAs,

InP,

c-GaN, etc 62

First Brillouin Zone of FCC Reciprocal Lattice (BCC lattice), with high symmetry k points marked

http://cst-www.nrl.navy.mil/bind/kpts/

(0, 0, 0) Center

H (1, 0, 0) 2/a

P (0.5, 0.5, 0.5) 2/a

N (0.5, 0.5, 0) 2/a

H Corner point joining four edges N Center of a face P Corner point joining three edges CsCl

63

The First Brillouin zone of a hexagonal lattice, with high symmetry k points marked

3 3

(0, 0, 0) Center

M (0, 1/ , 0) 2/a

L (0, 1/ , 0.5a/c) 2/a

A (0, 0, 0.5a/c) 2/a

K (2/3, 0, 0) 2/a

3

A Center of a hexagonal face H Corner point K Middle of an edge joining two rectangular faces

L Middle of an edge joining a hexagonal and a rectangular face

M Center of a rectangular face

ZnO, MgO, w-GaN, w-AlN, etc

64

Directions and Planes

65

William Hallowes Miller (1801 1880), British mineralogist and crystallographer

From: http://www.myspace.com/alexjvecchio

Directions

66

Planes

x

z

y (0,0,0)

(hkl) [hkl]

reciprocals are 1/2,1/4,1/3

Miller indices of plane are (6 3 4) Planes lie in the direction [6 3 4]

- Select an atom/point in your lattice as the origin, and define coordinate axes.

- For a given plane, determine the intercept with the axes (integral number of basis vectors).

- Take the reciprocal values, and convert to the smallest integer values with same ratio (hkl)

- Example: along the x, y, and z axes,

our intercepts occur at 2, 4 and 3 basic units respectively

Note: [hkl](hkl) for cubic crystal. (How to show?)

68

Common Crystal Planes

x y

z

a

a

a

Intercepts are 1, , and (parallel to y and z), so the plane is the (100) plane.

Depending on our reference point, any one of the six faces could be the (100) plane. Here, we have six planes in total (100), (010), (001), and . The bar indicates a minus direction. The six planes are equivalent, and called as {100}.

(100), (010)(001)

(100)

Planes equivalent are denoted in curly brackets around the indices {hkl}. 69

Common Crystal Planes

Intercepts are 1, 1, and (parallel to z). Plane is the (110) plane.

x y

z

a

a

a

(110)

70

The orientation of a crystal plane is determined by three points in the plane that are not collinear to each other.

Example: Three points in a plane are: P1(0,2,2), P2(2,0,2) and P3(2,1,0).

Calculate the Miller indices of the plane.

P3

P2

P1

71

We define the following vectors: r1=0i+2j+2k, r2=2i+0j+2k, r3=2i+j+0k and calculate the following differences: r - r1=xi + (y-2)j + (z-2)k; r2 - r1=2i - 2j + 0k; r3-r1 = 2i j - 2k We then use the fact that: (r-r1) [(r2-r1) (r3-r1)] =A(BC)= 0

We now use the following matrix representation, that gives The end result of this manipulation is an equation of the form: 4x+4y+2z=12 The intercepts are located at: x=3, y=3, z=6 The Miller indices of this plane are then: (221)

0212

02222

)(

321

321

321

=

==zyx

CCCBBBAAA

CBA

Solution:

72

Bravais-Miller indices for hexagonal lattice

In the case of an hexagonal lattice, one uses four axes, a1, a2, a3, c and four indices, (hkil), called Bravais-Miller indices, where h, k, i, l are again inversely proportional to the intercepts of a plane of the family with the four axes. The indices h, k, i are cyclically permutable and are related by

h + k + i = 0 or i = - h - k

73

Bravais-Miller indices example

(112) (1122)

i= 1/S

S is the intercept of the plane with the axis [110]

i = - h k=-(1+1)= 2

74

Symmetry and band structure

75

The equal-energy contour plots of the first valence subband (HH1) of InGaAsN/GaAs [11N] QWs.

We can see that all the patterns have mirror plane symmetry. And, the (110), (113) and (001) cases have the C4 rotation symmetry and the (111) case has the C6 rotation symmetry. The patterns just reflect the symmetry of (11N)-orientation in real space.

76

Semiconductor Band Theory

Band Structure

Also called as Electronic Structure

E-K Relations or Energy Dispersion Curves

77

ELECTRONS IN AN ATOM

Wavefunction:

78

the Bohr radius, the generalized Laguerre polynomials of degree n 1, a spherical harmonic function of degree and order m.

a0

3D Image of the eigenstate wavefunction 4,3,1. The solid body contains 45% of the electron's probability.

NATURE OF ATOMIC FUNCTIONS: l =0: called s-state; l =1: called p-state

79

Origin of Bands : Electron in vacuum

Single electron (in vacuum) Schrodinger Equation provides the solution : Plane waves as eigen vectors. E = (k)2/2m as eigen energy.

Eigen energy can take continuous values for every value of k. E-k relationship produces continuous energy bands.

E-k relationship

E = (k)2/2m

E

k

(k) = Aexp(-ik.r)

= |)(|)( kEkH

mkkH 2

)()(2=

Schrodinger Equation

Plane Waves (Eigen vectors)

Eigen Energy

Free electron kinetic energy Hamiltonian

Continuous energy band

k = Momentum vector E = Kinetic energy

Abhijeet Paul 80

Origin of Bands : Electron in Crystal

Atoms

Periodic potential due to crystal V(r)

k = Crystal momentum vector E = Kinetic energy Schrodinger Equation

| ( ) |H E k = 2( ) ( )2

kH V rm= +

GAP

GAP

GAP

E

k Discontinuous energy bands

E-k Relationship

Electron travelling in a crystal sees an extra crystal potential , V(r)=V(r+R) Eigen vectors are no longer simple plane waves. Eigen energies cannot take all the values. Energy bands become discontinuous, thereby producing the BAND-GAPS.

Electron Hamiltonian in a periodic crystal

Abhijeet Paul 81

Energy bands, Bandgap and effective mass

Vacuum electron E-k relationship

E

k

ek m

kE2

22=

Electron mass in vacuum = 9.1e-31kg

Continuous bands

E

k

Band Gap

E

k

Band Gap

/a -/a

Energy bands

E-k relationship in periodic potential

Similar E-k relationship. Now free electron mass is replaced by effective mass (m*) . Effective mass provides the energy band curvature.

More details on effective mass here: http://en.wikipedia.org/wiki/Effective_mass_(solid-state_physics)

okk

k

oe kE

km=

= 2

2

2*

1)(

1

*

22

2 ek

mkE =

Lattice constant = a. -/a k /a is the first BRILLOUIN ZONE E-k relation in this zone is called reduced E-k relation.

Abhijeet Paul

82

83

Effective mass

Example

Example

Note that the wavefunction is not periodic. k is called the crystal momentum. uk(r) is called the cell periodic part of the wavefunction, Bloch function.

BLOCH THEOREM

84

Felix Bloch (1905 1983) Swiss physicist, awarded the 1952 Nobel Prize

Different forms of Bloch theorem

85 BL-Bravais Lattice

Show k has discrete values in 1D crystal Consider 1 D crystal with length of L=Na, N is the number of unit cell, a is the lattice constant. Boundary condition requires

(0) ( )k k L =

Applying Blochs theorem, we have

0 (0) ( )2 2 21

ik ikNak k

ikNa

e u e u Nane k kN a Na L

=

= = = =

k has discrete values, so will E(k)!

86

Kronig-Penny Model

87

Real case

Kroning-Penny Model a

88

Ralph Kronig (1904 1995) German-American physicist. Known for Discovery of particle spin; Kronig-Penney model; Coster-Kronig transition; KramersKronig relation.

William Penney, (1909 1991) British physicist who was responsible for the development of British nuclear technology

sinh(x)=(exex)/2

cosh(x)=(ex+ex)/2 89

GAP

GAP

GAP

E

k

Solving the Kronig-Penney Model

Which is best solved graphically

RHS

d=a+b

90

Check out the Kronig Penney applet at http://fermi.la.asu.edu/schmidt/applets/kp/plugkp.html

91

Kronig-Penney Model Summary

some values of E imaginary k physically unacceptable these Es are forbidden allowed and forbidden energy bands created

band discontinuities occur at k = n / d Problems of Kronig-Penney model : (1) not much physical insight (2) does not give the # of energy states in a band

92

93

Tight-Binding Approximation

from NCTU OCW

94 from NCTU OCW

95 from NCTU OCW

96 from NCTU OCW

Example

Example

FROM ATOMIC LEVELS TO ENERGY BANDS

As atoms are brought closer and closer to each other to form a crystal, the discrete atomic levels start to broaden to form bands of allowed energies separated by gaps. The electronic states in the allowed bands are Bloch states.

Low lying core levels are relatively unaffected. Higher levels are broadened significantly to form bands. 97

98

Metal, Semiconductor and Insulator

from NCTU OCW

99

Example

from NCTU OCW

100

Example

from NCTU OCW

101

Example

from NCTU OCW

102

C : 2s 2p Si : 3s 3p Ge : 4s 4p Sn : 5s 5p Pb : 6s 6p

4N 6N 4N 6N

C ( diamond ), Si, and Ge have similar band structures

4N 6N

4N

2N

from NCTU OCW

103

Eg: C (diamond)~6 eV Si ~ 1.1 eV Ge ~ 0.7 eV

from NCTU OCW

104 from NCTU OCW

Classification of Materials

0 < Eg < ~3.5 eVs 105

SEMICONDUCTOR BAND STRUCTURE

In semiconductors we are pimarily interested in the valence band and conduction band. Moreover, for most applications we are interested in what happens near the top of the valence band and the bottom of the conduction band. These states originate from the atomic levels of the valence shell in the elements making up the semiconductor.

Outermost atomic levels are either s-type or p-type.

106

Direct and Indirect Bandgap Semiconductors The top of the valence band usually occurs at an effective

momentum value of zero (k=0), but the bottom of the conduction band may not coincide.

conduction band

valence band

E

k

Eg

conduction band

valence band

E

k Eg

Direct bandgap Semiconductors : Both CBM and VBM are at k=0 points. Such as, GaAs, InP, InAs

Indirect bandgap semiconductors: CBM and VBM are not at the same k point. Such as, Si, Ge, AlAs

107

BAND STRUCTURE OF SEMICONDUCTORS

First Brillouin Zone of Si

108

BANDSTRUCTURE NEAR BANDEDGES

Behavior of electrons near the bandedges determines most device properties. Near the bandedges the electrons can be described by simple effective mass pictures, i.e., the electrons behave as if they are in free space except their masses are m*.

Schematic of the valence band, direct bandgap, and indirect bandgap conduction bands. The conduction band of the direct gap semiconductor is shown in the solid line, while the conduction band of the indirect semiconductor is shown in the dashed line. The curves I, II, and III in the valence band are called heavy hole, light hole, and split-off hole states, respectively.

109

EFFECTIVE MASS DESCRIPTION

110

ELECTRON EFFECTIVE MASS OF MOST IMPORTANT SEMICONDUCTORS

111

CHARACTER OF THE WAVEFUNCTIONS NEAR THE BANDEDGES

The wavefunction (central cell) determines the nature of optical transistion in optoelectronic devices.

Top of the valence band is made from p-type states. Combining spin, the total angular momentum of the states is 3/2

112

BANDSTRUCTURE: Si

113

y

BANDSTRUCTURE: GaAs

114

115

Form UCSB

BANDSTRUCTURE: Ge AlAs, InAs, InP

116

ELECTRONIC PROPERTIES OF SOME SEMICONDUCTORS

Properties of some semiconductors. D and I stand for direct and indirect gap, respectively. The data are at 300 K. Note that Si has six conduction band valleys, while Ge has four. 117

BANDSTRUCTURE OF InN, GaN, AlN

Bandstructure of InN, GaN, and AlN. Also shown is the Brillioun zone. These materials are important for blue light emission and high power/high temperature electronics. 118

Density Of State

119

The general form of DOS of a system with volume V is given as

120 Considering spin up and down, the DOS must time 2

Alternative Derivation of DOS g=3D Consider a volume in k-space (sphere) The occupied volume of one state is The number of state in the sphere is then Considering spin up and down, So, DOS in k-space is where, The DOS in energy space is For electron in bulk semiconductors, So,

3

34 kVk =

zyx LLL 222

zyx LLLkN 2

3

6=

zyx LLLkNN 2

3

32'

==

2

33

3')(

k

VNkD == zyx LLLV =

dEdk

dEkdED

3

23

31)()(

==

*

22

2mkE =

EmEmdE

dEmED 23

2

*

221

23

2

*

2

23

23

2

*

23 )2(

21

23)2(

31)2(

31)(

===

121

How to Calculate Band Structure?

127

Many Body Hamiltonian Many body Hamiltonian in the adiabatic approximation and fix ion charge

Too difficult, we need approximations

= EH

{ }( )ir=

iieiee VVVTH +++=

The total energy of the many body system is given by

Where the many body wave function is

Sir William Rowan Hamilton (1805 1865) Irish physicist, astronomer, and mathematician, who made important contributions to classical mechanics, optics, and algebra.

128

One electron Hamiltonian Typical approximations: Hartree, Hartree-Fock, Local Density approximation (LDA) . These approximations reduce the many body system to the problem of one electron moving in an effective field

)(2

2

rVm

pH += V has the crystal symmetry V(r+R)=V(r)

( ) ( )rr kkk nnn EH =

To find the one electron energy En, one has to solve the Schroedinger equation

129

The first-principles method: very accurate for valence band, but under-estimated band gap due to local density approximation (LDA) and very time-consuming. Usually, not used for QW calculation.

Tight Binding Method: describe the band structure over the entire Brillouin zone; computational cost is low; Can calculate bulk material and quantum wells.

Empirical Pseudopotential Method: reasonable results; quick. Widely used for bulk material calculations.

k.p method: accurate results within a small region near center of the Brillouin zone; quick. Widely used for QW calculation.

130

Tight Binding Method We attempt to solve the one electron Hamiltonian in terms of a Linear Combination of Atomic Orbitals (LCAO)

( ) ( ) = i

iiiC site,

atomic orbitals,

Rrr

Ci= coefficients, i= atomic orbitals (s,p,d) Ci

= orbital

ijjiji S == ,( ) ( )rr EH =

matrix notation:

HC=EC

[ ] 0 jsite,

atomic orbitals,, =

jnji CEH

131

where the Hamiltonian matrix element

( ) ( )ijji RrHRrdH = *, r

The calculation of these integrals requires the knowledge of both basis function and potentials even though they do not appear separately in the final matrix element

Empirical tight binding develops approximations only for the Hamiltonian matrix elements Hi,j themselves without attempting to model the potential and the explicit form of the basis functions

132

( ) ( )ijji RrHRrdH = *, r

H =k k k ( ) ( ) ( )3,

1 *l mi m ll m

e d HN

= k R R r r R r R

( ) ( ) ( )3,

1 *l mi m ll m

e d HN

= + k R R r r R R r

( ) ( )3,

1 *ni nn m

e d HN

= k R r r R r

( ) ( )3 *ni nn

e d H = k R r r R rKeep only on site & nearest neighbor terms:

( ) ( )3 *d H = r r r ( ) ( )3 1*d H = r r R r

. .

l

l

i

n ne

= k RkR

Band structure arising from a single atomic s-level

Considering only s orbital,

( ) ( ) =>= i

iiNk

site,atomic orbitals,

i ie 1|

Rrrk.R

( ) ( )i i

Nk Rrr k.R =>=

site,atomic

i ie 1|

133

Fcc lattice: 12 N.N. at ( ) ( ) ( ){ }1, 1,0 , 1,0, 1 , 0, 1, 12a

=

( ) ( ) ( ) ( ) ( ) ( )( )/2 /2 /2 /2 , ,x y x y x y x yi k k a i k k a i k k a i k k ae e e e x y y z x z y x + += + + + + + k

( ) ( )

( ) ( )

( ) ( )

1 12 cos cos2 21 1cos cos2 21 1cos cos2 2

x y x y

y z y z

z x z x

k k a k k a

k k a k k a

k k a k k a

= + +

+ + +

+ + +

1 1 1 1 1 14 cos cos cos cos cos cos2 2 2 2 2 2x y y z z x

k a k a k a k a k a k a = + +

. .

l

l

i

n ne

= k RkR

134

)()cos( 21 ixix eex +=apply

135

Band width = 16

Along X

2

2* 2m

a=

(0, 0, 0)

L (0.5, 0.5, 0.5) 2/a

X (1, 0, 0) 2/a

U,K (0.25,0.25,1) 2/a

W (1, 0.5, 0) 2/a

kx=2v/a

ky=kz

(k)=--4 (1+2cosv)

0 v1

For =0, =1 eV, a=4 , m*=0.2m0 along [100].

Models: Atomic Basis Set

NN-sp3 vs. NN-sp3s*

NN-sp3 model captures key features of valence band (VB), but fails for conduction band (CB) in indirect bandgap materials. NN-sp3s* reproduces indirect conduction bandgap but with wrong effective masses.

s xp *szpyp

sp3

sp3s*

NB=4

NB=5

136

137

Models: Atomic Basis Set sp3s* Nb=5

sp3s*d5

Nb=10 2 2x yd zxdyzdxyd

s xp *szpyp

2 2 /3z rd

2NN-sp3s* vs. NN-sp3s*d5

Both models offer correct effective mass. 2NN-sp3s* has smaller basis size compared to NN-sp3s*d5. Accurate modeling strain is difficult for distant neighbor interactions.

Why Tight-Binding ? Allows us to describe the band structure over the entire Brillouin zone

Relaxes all the approximations of Envelope Function approaches

Allows us to describe thin layer perturbation (few )

Describes correctly band mixing

Gives atomic details

The computational cost is low

It is a real space approach

138

EPM The pseudopotential Hamiltonian for a semiconductor crystal is given by

),(2

22

rVm

H +=

where V(r) is the pseudopotential of the crystal and is the sum of all the atomic local pseudopotential. The V(r) can be expanded in reciprocal lattice vectors G. For zinc-blende structure, this yields

,)sin)(cos)(()( rGGGr += iAG

S eGiVGVV

where = 1 = -2 = a(1,1,1)/8, and a is the lattice constant. VS and VA are the symmetric and antisymmetric pseudopotential form factors of a binary compound respectively, and can be written as

V V VS ( ) [ ( ) ( )],G G G= +12 1 2

V V VA ( ) [ ( ) ( )],G G G= 12 1 2 139

( ) ( )( )

, ,i

n n

i n i

e u

e C e

=

=

k rk k

k r G rG

G

r r

k

The expansion of the wave-function in Bloch states:

Substitute into the Schrdinger equation, we can show

( ) ( ) ( ) ( )22

02

E C V Cm

+ + = G G

k GG G G G

Thus implying a set of simultaneous equations for C(G). The E can be obtained by solve a secular equation.

140

In terms of the unit Cartesian vectors, nearest neighbor sites in reciprocal space correspond to:

G group (units 2/a)

Number of permutations

Total number of elements

|G|2 (2/a)2 units

(0,0,0) 1 1 0

(1,1,1) 8 9 3

(2,0,0) 6 15 4

(2,2,0) 12 27 8

(3,1,1) 24 51 11

(2,2,2) 8 59 12

(4,0,0) 6 65 16

(3,3,1) 24 89 19

The pseudo-potential form factors are typically given up to G2 = 11.

141

Pseudopotential in real space

142

Pseudopotential is not unique.

143

Pseudopotential in reciprocal space

144

Flow Chart for EPM Input Vs and VA

H = p2 + V(r)

H = E

Get E and

Compare with experimental results, Eg(), Eg(L) and Eg(X)

Agree with experiment?

Plot band structure. End. Yes

No

146

Si Band structure calculated by EPM

-14

-12

-10

-8

-6

-4

-2

0

2

4

6

8

10

Si

L X K

E (e

V)

147

Band structure of GaN and AlN calculated by EPM

148

k p Method

2 2 2

0 0 0

( ) ( ) ( ) ( )2 2nk n nkp kk p V r u r E k u rm m m

+ + =

2

0

( ) ( ) ( ) ( )2 nk n nkp V r r E k rm

+ =

The single particle Schrdinger equation reads

When written in terms of unk (r), it becauses

Using second order perturbation theory we find

Perturbation term is related k.p, so called as k.p method

'

'0 '

22 2 2

20 0

| |( ) (0)

2 (0) (0)nn

n n nnn n n n

k pkE k E k pm m m E E

= + + +

Note pnn=0 149

2*, ,

1( ) (0)2n n

E k E D k k k km

= =

We have

and

2 2 2' ' ' '

2 *'0 0 '

12 2 (0) (0) 2

nn n n nn n n

n n n n

p p p pDm m E E m

+ = + =

,=x,y, and z.

D is the inverse effective mass in matrix form multiplied by 2/2

150

Kanes model and Luttinger-Kohns model

151

152

Kane model of [001]-oriented Hamiltonian under |S>, |X>, |Y>, |Z> basis is given by

+

++

+

++

+

++

++

=

0222

222'''

'

0222

222''

''

0222

222'02222'

int

)(

)(

)(

mk

yxzzyzxz

zymk

zxyyxy

zxyxmk

zyxx

zyxmk

g

kkMkLkkNkkNiPk

kkNkkMkL

kkNiPk

kkNkkNkkMkL

iPk

iPkiPkiPkkAE

H

Considering spin up and down, the 8-band Kane Hamiltonian under |S>, |X>, |Y>, |Z>, and |S>, |X>, |Y>, |Z>, basis can be written as

153

=

int

int

00

HH

H

The relationships between L, M, N and 1, 2, 3 are

'20

31

3

'20

31

2

'20

32

1

)(

1)2(

N

ML

ML

m

m

m

=

=

+=

3022'

21022'

21022

6

)14(

)12(

m

m

m

N

L

M

=

++=

+=

Note: here r1, r2, r3 are the modified L-K effective mass parameters

154

[001]-oriented L-K Hamiltonian under |X>, |Y>, |Z> basis is given by

+

++

+

++

+

++

=

0222

2220222

2220222

222

int

)(

)(

)(

mk

yxzzyzx

zymk

zxyyx

zxyxmk

zyx

kkMkLkkNkkN

kkNkkMkL

kkN

kkNkkNkkMkL

H

Lm

LLm

LLm

N

L

M

3022

21022

21022

6

)14(

)12(

=

++=

+=Note: here r1L, r2L, r3L are L-K effective mass parameters

The modified Luttinger parameters are related to Luttinger parameters in the following manner:

155

G

PL

EE

31

11 =

G

PL

EE

61

22 =

G

PL

EE

61

33 =

Note: in the following, we will focus on L-K mode, and use r replace rL parameter

New Band Edge Basis Functions

156

c

so

lh

hh

Eg

0

32

32

32

12

32

12

32

32

12

12

12

12

12

62

16

2

213

3

, ( ) ,

, [( ) ] ,

, [( ) ] ,

, ( ) ,

, [( ) ] .

, [ ( ) ] ,

= +

= +

= +

=

= + +

= +

X iY

i X iY Z

X iY Z

i X iY

X iY Z

i X iY Z

|s >; |s >;

21,

23,

21,

23

23,

23,

23,

23

21,

21,

21,

21

6-band L-K Hamiltonian

157

|3/2,3/2>, |3/2,1/2>, |3/2,-1/2>, |3/2,-3/2>, |1/2,1/2>, |1/2,-1/2>.

=

SiDiii

SiiDii

iiH

DiiL

iDiL

iiH

Hv

02/2

2/32

022/32

2/

2/20

]2

[2/30

2/3]2

[0

22/0

**

*

****

**

*

6

For unstrained bulk semiconductor

158

,])[(2

],4)(2[2

],)(2[32

],)([322

)],2())([(2

)],2())([(2

01222

0

2

22

222

0

2

222

30

2

30

2

212

2122

0

2

212

2122

0

2

+++=

+=

=

=

+++=

+++=

zyx

zyx

yxyx

xyz

zyx

zyx

kkkm

S

kkkm

D

kkkikm

kikkm

kkkm

L

kkkm

H

Where, 0 is the spin-orbit split-off energy, m0 is the free-electron mass, and 1, 2, 3 are the Luttinger parameters, kx, ky and kz are the wavevectors Note: we take hole energy positive

Heavy hole term

Light hole term

SO hole term

,, D are the interaction terms among hh, lh and so holes.

The conduction band Hamiltonian of QWs

22 2 2

* ( ) ( )2c x y z cH k k k V z

m= + + +

Where m* is the electron effective mass, Vc(z) is the conduction band periodic potential of MQWs after hydrostatic strain.

The electron wave functions of QWs is

>++= )(|])2(exp[1)](exp[1 , szLmkiLgykxkiLL zm mnyxyxn cc

L=l+d is the period of the QW, where l and d are the widths of the wells and the barriers, respectively.

159

HC Hamiltonian Matrix Elements

mnzk Lnk

mnHm 2*

2

)2(2

|| +>=<

mnV

mnV

dzezVL

nzVm

cmnLlmn

Ld

c

zmniL

L

L

==

>=, |3/2,1/2>, |3/2,-1/2>, |3/2,-3/2>, |1/2,1/2>, |1/2,-1/2>.

161

V(z) is the valence band periodic potential of MQWs after hydrostatic strain, b=-2Du /3 is the valence-band shear strain deformation potential, 0 and B are the spin-orbit split-off energy in well and barrier, respectively, m0 is the free-electron mass, and 1, 2, 3 are the Luttinger parameters. The in-plane strain exx is equal to (as-a0)/a0, where as is the substrate lattice constant and a0 is the unstrained well material lattice constant. C11 and C12 are the elastic stiffness constants.

22 2 2

1 2 1 20

22 2 2

1 2 1 20

2

30

22 2

3 20

22 2 2

2 20

22 2 2

1 00

[( )( ) ( 2 )] ( ),2

[( )( ) ( 2 )] ( ),2

2 3[ ( ) ],2

3[2 ( ) ],2

[2( ) 4 ],2

[( ) ] ,2

(

x y z

x y z

z y x

x y x y

x y z

x y z

H k k k E zm

L k k k E zm

k ik km

ik k k km

D k k km

S k k km

E

= + + +

= + + + +

=

=

= +

= + + +

12 11

0

2 (1 2 / ) , .) 3

0, .

, .( )

0 , .

u xxD c c e in wellzin barrier

V in barrierV z

in well

+=

=

162

+

=well

barrierVzV B

0)(' 00

The six dimensional hole envelope wave function for the QWs can be expanded as :

Hole wave functions of QWs

{ } nv nvj j= =, ( , ,... .),1 2 6

nvj

x y nv mj

mzi k x k y a L

i k mL

z= + +exp[ ( )] exp[ ( ) ] ,,1 2

Where,

163

yxLL1

How good is effective mass aprox. ?

EC

AlGaAs GaAs

E1

E1

d [nm]

d

1 2 3 4 5 6 7 8 9 10 11

Effective mass

Exact

164

165 -4 -3 -2 -1 0 1 2 3 4

-100

-80

-60

-40

-20

0

HH3

HH2

LH1

HH1

S.L. Chuang IEEE JQE 26, 13(1990)

GaAs/Al0.3Ga0.7As 100/200 AE

(meV

)

Band structure calculated by 6 band k.p method

0.0

0.2

0.4

0.6

0.8

1.0

GeGe

Ge0.55Si0.15Sn0.3

Ge0.60Si0.15Sn0.25Ge

Ge0.65Si0.15Sn0.20

E (e

V)

The band lineups of strained Ge/GeSiSn.

166

-4 -2 0 2 4-300

-250

-200

-150

-100

-50

0

-4 -2 0 2 4 -4 -2 0 2 4

Sn=0.3

LH3

LH2HH2

HH1

LH1

E (m

eV)

[100] k (2/L) [110]

LH3LH2

HH2

HH1

LH1

Sn=0.25

LH2LH3

HH2

HH1

LH1

Sn=0.2

The hole energy dispersion curves of 80 /200 Ge/Ge1-x-ySiySnx QWs. y=0.15, take hole energy positive.

167

Valence band structure of Gan/AlGaN QW calculated by 6-band k.p method

168

Table. The band parameters at 300K for bulk -Sn, Ge, and Si Parameter Sn Ge Si a () 6.4892 5.6573 5.4307 mc (m0) -0.058 0.038 0.528 1 -15 13.38 4.22 2 -11.45 4.24 0.39 3 -8.55 5.69 1.44 Ep (eV) 24.0 26.3 21.6 Eg (eV) -0.413 0.7985 4.185 EgL (eV) 0.092 0.664 1.65 (eV) 0.8 0.29 0.044 ac (eV) -5.33 -8.24 1.98 av (eV) 1.55 1.24 2.46 aL (eV) -0.342 -1.54 -0.66 b (eV) -2.7 -2.9 -2.1 C11 (10 GPa) 6.9 12.853 16.577 C12 (10 GPa) 2.9 4.826 6.393 nr 4.90 4.02 3.45

169

Band Gap Engineering

171

Modification of band structure: Band Gap Engineering

Band Gap Engineering by Semiconductor alloy Band Gap Engineering by Low-dimensional

semiconductor Band Gap Engineering by Strained semiconductor

172

Semiconductors Alloys

A semiconductor alloy - a combination of binary, or elemental, semiconductors

Binary alloy (an alloy of group IV) SixGe1-x = (x)Si + (1-x)Ge

Ternary alloy (two binaries mix) AlxGa1-xAs = (x)AlAs + (1-x)GaAs Note: x fraction (< =unity)

173

Quarternary alloys

1. 2 elements from one column, 2 from the other: (4 binaries mix) AIII(1-x)BIII(x)CV(1-y)DV(y) = (1-x)(1-y)[AIIICV] + (1-x)y[AIIIDV] + x(1-y)[BIIICV] + xy[BIIIDV] For example: Ga1-xInxAs1-yPy 2. 3 elements from one column, 1 from the other: (3 binaries mix) AIII(1-x-y)BIII(x)CIII(y)DV = (1-x-y) [AIIIDV] + (x) [BIIIDV] + (y)[CIIIDV] For example: Ga1-x-yAlx InyAs AIIIBV(1-x-y)CV(x)DV(y) = (1-x-y) [AIIIBV] + (x)[AIIICV] + (y)[AIIIDV] For example: GaAs1-x-ySbxNy

174

Bandgap of alloys Bandgaps of the ternary semiconductor approximately follow: Eg(AxB1-xC) = x Eg AC + (1-x)Eg BC

(very rough approximation!)

However, in most alloys, there is a bowing effect arising from the increasing disorder due to the alloying.

The energy bandgap of alloy can be better described by the following expression:

Eg(alloy)=Eg0 + cx +bx2 Ego is the bandgap of the

lower bandgap binary. b is bowing factor. 175

Bandgap data for some III-V ternary compounds at 300K

0.18 - 0.41x + 0.58x2 InAsxSb1-x 0.36 + 0.891x + 0.101x2 InPxAs1-x 0.726 0.502x + 1.2x2 GaAsxSb1-x 1.424 + 1.15x + 0.176x2 GaPxAs1-x 0.172 + 0.139x +0.415x2 GaxIn1-xSb 0.36 + 0.628x + 0.438x2 GaxIn1-xAs 1.351 + 0.643x +0.786x2 GaxIn1-xP

0.8+0.746x+0.334x2 1.02+0.492x+0.077x2 0.726 + 1.129x + 0.368x2 AlxGa1-xSb 0.36 + 2.012x + 0.698x2 AlxIn1-xAs

1.708 + 0.642x 1.9 + 0.125x + 0.143x2 1.424+1.247x+1.147(x-0.45)2 (x>0.45) 1.424 + 1.247x (x

Vegards Law Many properties of semiconductor alloys are determined by

the parameters of the constituent binaries, and are found to vary roughly linearly with composition - Vegards Law

For example, the lattice constant a of GaAs1-xPx is given by

For quaternaries (AxB1-xCyD1-y) (A, B and C, D are group III and V,

respectively). An arbitrary parameter Q may be expressed by

GaAsGaPPGaAs axxaa xx )1(1 +=

BDAD

BCAC

QyxQyx

QyxxyQyxQ

)1)(1()1(

))(1(),(

++

+=

177

Semiconductor heterojunctions Formed when two semiconductors with different bandgaps

(and lattice constants) are brought together. Three types of band edge lineups in heterostructures

178

To estimate the band alignment, we should know the

work function s (energy required to excite an electron from the Fermi level to vacuum) and the electron affinity s (energy to excite an electron from the bottom of the conduction band to vacuum).

Conduction band offset:

Valence band offset:

Semiconductor heterojunctions

12 = CE

)( 12 = gV EE

)( 12 >

+

EC1

EV1 EF

vacuum

1 1

EC2

EV2

EF

2 2

Eg1 Eg2

179

Bandgap and Electron Affinities,

Bandgaps and electron affinities for some elemental and binary semiconductors.

E g ( eV) ( eV) GaP 2.21 (I) 4.3 InP 1.35 (D) 4.35 AlAs 2.16 (I) 2.62 GaAs 1.424 (D) 4.07 InAs 0.36 (D) 4.9 AlSb 1.65 (I) 3.65 GaSb 0.73 (D) 4.06 InSb 0.17 (D) 4.59 Ge 0.66 (I) 4.13 Si 1.12 (I) 4.01

Sometimes, offsets will be calculated using the 60:40 rule,

e.g. The bandgap difference between GaAs and Ga0.7Al0.3As is ~ 0.374 eV EC = 0.6 x 0.374 eV = 0.224 eV EV = 0.4 x 0.374 eV = 0.150 eV 180

Band offset calculation: Model Solid Theory

EV = Evav + /3

Evav is the valence band average, is the SO splitting energy

A B

Ec

Ev

Evav

Ec

Ev

Evav /3

Eg=Ec-Ev

Ec=Ec(B)-Ec(A)

Ev=-[EV(B)-EV(A)]

Eg(B)

181

Anderson Model: Rules for construction:

- Fermi levels align under equilibrium. - Electrons move from the semiconductor with higher

EF to that with lower EF. - Bands bend up in carrier motion direction so as to

provide a barrier when equilibrium is reached.

Semiconductor heterojunctions

182

Isotype heterojunctions First, we will consider an isotype heterojunction (e.g n-type,

denoted n-N) N is the wide bandgap material.

(a) before junction formation (b) equilibrium

junction Vbi is the built-in potential 183

For a p-P isotype structure, this figure becomes:-

(a) before junction formation (b) equilibrium junction.

Isotype heterojunctions

EC1

EV1

EF

1 1

Eg1 EC2

EV2

EF

2 2

Eg2

184

Note in Isotype heterojunctions

An accumulation of charge will occur at the discontinuities - we get a spike due to the bands moving in opposite directions at the junction.

In forming an n-N junction, electrons move from N to n material until the fermi levels are equal. The formation of a potential barrier EC prevents further electron motion.

The accumulation region forms a narrow quantum well, forming a 2-D electron gas. Thus heterojunctions can provide confinement.

185

Anisotype heterojunctions Anisotype heterojunctions are formed when we bring two

materials with different doping together. (e.g. P-n)

+

EC1

EV1 EF

P n

vacuum

1 1

EC2

EV2

EF

2 2

Eg1 Eg2

We use the Anderson model to construct the equilibrium

band diagram. 1 2 1 1 1 2 2( ) ( )bi g F FqV E E E = = + +

186

So the P-n equilibrium structure becomes: Q: How about N-p structure??

EC

EV

EF

EC

EV

EF2 EF1

Anisotype heterojunctions

187

Double heterostructures Bandgap discontinuities are

good for confining charge to a specific region. Can also make use of two adjacent heterostructures for two types of carrier!

Other material properties, such as dielectric constant, will also show discontinuities (good for optical confinement).

Example: P-AlGaAs/n-GaAs/N-AlGaAs

Equilibrium (zero bias)

Forward biased

Refractive index profile

188

Quantum well (QW)

The QW is a sandwich made of a thin (e.g., 10 nm) layer of a narrower-bandgap semiconductor, surrounded by two wider-bandgap semiconductor layers, and showing type I band line-up, i.e., with the energy minimum for electrons and holes confined in narrower bandgap semiconductor.

189

The energy is quantized due to the electrons confined in QW, the energy is called sub-band energy level.

The DOS is step function for the two-dimensional

system.

Effective bandgap, Eg Eg = Eg0 + E1 + E1(HH)

Subbands energy levels and density of state (DOS) of QW

190

Energy levels and wavefunction in infinite deep QW

The Schrodinger equation for the electron states in the quantum well can be written as:

=+

EzVm

)](2

[ *22

Assuming the potential is infinite outside the well, we have

++== *22

2* 2

)(2

1),,(mk

Wn

mkknEE xyx

*

22

2mk y

n=1,2,3,W is the well width, m* is the effective mass, k is wavevector

0sin2 =

= katz

Wn

W

191

LOW-DIMENSIONAL STRUCTURES: DENSITY OF STATES

192

Derive DOS 2D Consider an area in k-space The occupied volume of one state is The number of state in the area is then Considering spin up and down, So, DOS in k-space is where, The DOS in energy space is For electron in 2D semiconductors, So, for one En subband For all subband subband

2kAk =

yx LL 22

yxLLkN4

2

=

yxLLkNN2

2'2

==

2')(

22 k

ANkD == yxLLA =

dEdk

dEkdE

DD

222

21)()(

==

222*

22

,2 yxn

kkkEmkE +=+=

2

*2 )(

mED =

=n

nD

t EEmE )()( 2

*2

is step function 193

The lattice constant of the substrate determines the alloy compositions which may be grown.

If the lattice constant of the epitaxial layer (af) is different from that of the substrate (as), defects in the crystal structure can occur.

Lattice matching and strain

isolated lattices

combined lattices

Misfit dislocation

194

For perfect epitaxial growth, there is no mismatch, so the layer is unstrained

However, a small mismatch can be accepted, causing an elastically strained layer, but the layer must be thin for defects not to occur.


isolated lattices

combined lattices

Strained

x

z (growth direction)

y

195

In-plane strain =// = xx = yy = (as-af) /af

where as is the lattice constant of substrate, af

is epi-layer lattice constant without strain. Vertical strain = zz = - 2(C12/C11)// where C11 and C12 are elastic stiffness

constants. For most of the III-V semiconductors, C12 0.5 C11.


196

Binary AlAs AlSb GaP GaAs

Lattice constant () 5.6605 6.1355 5.4512 5.6533

Binary GaSb InP InAs InSb

Lattice constant () 6.0959 5.86875 6.0584 6.47937


Consider growing AlxGa1-xAs on GaAs (x = 40%), from Vegards law, lattice constant is given by a1 = 0.4aAlAs + 0.6aGaAs = 5.6562

Lattice mismatch (strain) =(5.6533-5.6562)/5.6562 = -0.05% The negative sign indicates compressive strain (larger lattice

constant than the substrate)

Tensile strain: when the lattice constant is smaller than that of the substrate

197

Lattice matching and strain Strain energy - will accumulate

and is linear with thickness.

A critical thickness occurs at which strain energy is higher than dislocation energy - defects occur.

This happens at the critical thickness: dc as / |2| .

Accurate calculation should use Matthews equation.

Misfit dislocation defect

198

Strain influence on band structure

ac and av are the hydrostatic strain deformation potential constants for conduction and valence bands, respectively. b is the shear strain deformation potential constant 199

The consequence of pseudomorphic strain on the bandedges of a direct bandgap semiconductor

200

[001]-oriented Strained Hamiltonian

201

22

2

2

0

22

0

12

0

2

bm

am

am

v

cc

Based on Hk, the strained Hamiltonian Hs can be obtained using the following correspondence relations,

),,,(''' zyxjikk ijji =

xx =yy and zz = C12/C11

The final Hamiltonian for bulk semiconductor can be obtained H= Hk,+Hs. For QW, H= Hk,+Hs+V. Q: How about the H

under electric field?

For strained semiconductor layer, the conduction band edge is shifted by

Ec = ac(xx +yy +zz ) = 2ac(1-C12/C11) and the valence bands are shifted by EHH = -P - Q (see S. L. Chuangs book)

ELH = -P + Q P = - av(xx +yy +zz ) = -2av(1-C12/C11) Q = -b(xx +yy -2zz )/2 = -b(1+2C12/C11)

202

So, the Strained Bandgaps: For compressive strain: EC-HH = Eg + EC - EHH = Eg + Ec + P + Q = Eg + 2(ac -av )(1-C12/C11) -b(1 + 2C12/C11) Eg + a -2b (if a= ac -av and c11=2c12) For tensile strain: EC-LH= Eg + Ec - ELH = Eg + Ec + P - Q = Eg + 2(ac -av ) (1-C12/C11) + b(1+2C12/C11) Eg + a + 2b (if a= ac -av and c11=2c12)

203

Unstrained and strained valence band structures

-4 -2 0 2 4-100

-80

-60

-40

-20

-0

LH2

HH3

HH2

LH1

HH16 band

E (m

eV)

[100] [110]-4 -2 0 2 4

100

150

200

250

300

[100] [110]

LH3

LH2HH2

HH1

LH1

6 Band

E (me

V)

7/13 nm Ge/Ge0.75Si0.05Sn0.2 QW 10/15 nm GaAs/Al0.2Ga0.8As QW

Strain effect on band lineup of Ge-Ge0.986Si0.014 QW

well width of 112 and barrier width of 85

0 50 100 150 200-300

-200

-100

0

100

200

300

400

500

600

700

800

900

1000

Strain=0

E (m

eV)

Z (A)

SO LH HH EC

0 50 100 150 200-300

-200

-100

0

100

200

300

400

500

600

700

800

900

1000

Strain=0.5%

E (m

eV)

Z (A)

SO LH HH EC

0 50 100 150 200-300

-200

-100

0

100

200

300

400

500

600

700

800

900

1000

Strain=1%

E (m

eV)

Z (A)

SO LH HH EC

Strain effect on band structure of the Ge/GeSi QW.

0

50

100800

900

1000

1100

Ge/Ge0.986Si0.014Strain=0

6 4 2 0 2 4 6 [110] k (2/L) [100]

LH3

LH1

LH2HH3HH2

HH1

E1

E2

E3

E (m

eV)

0

50

100800

900

1000

1100

Ge/Ge0.986Si0.014Strain=0.5%

6 4 2 0 2 4 6 [110] k (2/L) [100]

LH3

LH1

LH2HH3HH2HH1

E1E2

E3

E (m

eV)

0

50

100

800

900

1000

1100

Ge/Ge0.986Si0.014Strain=1%

6 4 2 0 2 4 6 [110] k (2/L) [100]

LH3

LH1

LH2HH3

HH2HH1

E1

E2

E3

E (m

eV)

Doping and Carrier Concentrations

207

208 from NCTU OCW

HOLES IN SEMICONDUCTORS

The empty states in the valence band are called holes.

209

HOW DO HOLES MOVE?

210

Semiconductors can be classified into Intrinsic and extrinsic semiconductors

211

Intrinsic semiconductors:

from NCTU OCW

Carrier Density: n(E) and p(E) and Carrier Concentration: n and p

The density of electrons with energy between E and E+dE is equal to the density of energy states, gC(E), between E and E+dE times the probability that the energy states are occupied.

The density of holes with energy between E and E+dE is equal to the density of energy states gV(E), between E and E+dE times the probability that the energy states are not occupied

( ) ( ) ( )

( )Cdn n E dE f E g E dE

n n E dE

= =

=

( ) [1 ( )] ( )

( )Vdp p E dE f E g E dE

p p E dE

= =

=

212

For electrons (and holes), use Fermi-Dirac statistics to determine occupation of the states. The probability of an energy level E being occupied is

TkEE BFe

Ef /)(11)( +

=

Ef is the Fermi level, kB is Boltzmanns constant.

Energy density of energy states: Obtained by considering an electron moving in a part of crystal momentum space (so it is related to k).

dEEEm

dEEg CeC 32* )(2

)(3

= dEEEm

dEEg vhv 32* )(2

)(3

=

= h/2 is reduced Plancks constant m* is the effective mass

f(E)

E EF 0

T = 0

T > 0

1

1/2

Eg

213

Enrico Fermi (1901 1954) Italian-American physicist. Known for development of the first nuclear reactor, Chicago Pile-1. Awarded the 1938 Nobel Prize in Physics for his work on induced radioactivity.

Paul Adrien Maurice Dirac, (1902 1984) English theoretical physicist who made fundamental contributions to the early development of both quantum mechanics and quantum electrodynamics. He shared the Nobel Prize in physics for 1933 with Erwin Schrdinger

E

EC

EV

gC(E)

gV(E)

0 1

E

f(E)

1 - f(E)

EF

Total electron concentration E

n(E)

p(E)

EC

EV

214

( ) ( ) ( )

( )Cdn n E dE f E g E dE

n n E dE

= =

=

( ) [1 ( )] ( )

( )Vdp p E dE f E g E dE

p p E dE

= =

= Total hole concentration

])(exp[TkEENn

B

Fcc

= ])(exp[

TkEENp

B

vFv

=

Intrinsic carrier concentration, ni

For intrinsic semiconductor, we can use f exp[-(E-Ef)/kBT] to derive:

2)exp( iB

gvc nTk

ENNnp ==

Nc, Nv: the effective density of state at the CB and VB edges, respectively.

215

Example

Example

Nc: effective density of state at the CB edge.

1.08m0 1.5x1016

from NCTU OCW

Intrinsic carrier concentration

Intrinsic semiconductor: no doping ni2 = n0p0, n0=p0 where ni is the intrinsic carrier concentration

Temperature dependence of ni in Si, Ge, GaAs

Effective densities and intrinsic carrier concentrations of Si, Ge and GaAs.

216

Fermi level in intrinsic semiconductor

Fermi level lies at about the mid-bandgap for intrinsic semiconductor

1 12 2 ln( )

cFi v g B

v

NE E E k TN

= +

Nc, Nv: the effective density of state at the CB and VB edges.

217

])(exp[TkEENn

B

Fcc

=

])(exp[TkEENp

B

vFv

=

For intrinsic semiconductors, we have

For intrinsic semiconductors, n = p, we can dervive

Example GaAs has an effective density of states at the conduction CB Nc of

4.7X1017 cm-3 and an effective density of states at the VB edge Nv of 7X1018 cm-3. Given its bandgap Eg of 1.42 eV calculate the intrinsic concentration and the intrinsic resistivity at room temperature (take as 300 K). Where is the Fermi level?

160 220 310 380 400 h (cm2 V-1 s-1)

2400 4000 7000 8000 8500 e (cm2 V-1 s-1)

1018 1017 1016 1015 0 Dopant concentration (cm-3)

Dopant impurities scatter carriers and reduce the drift mobility (e for electrons and h for holes).

218

ni = Nc Nv( )1/ 2

exp Eg

2kBT

( )( ) ( )( )1/ 217 3 18 3

5 1

6 3

1.42 eV4.7 10 cm 7 10 cm exp2 8.6174 10 eV K 300 K

2.146 10 cm

in

= =

( )( )( )( )19 6 3 2 1 1 2 1 1

9 1 1

1.608 10 C 2.146 10 cm 8500 cm V s 400 cm V s

3.06 10 cm

i e hen

= +

= +

=

( )8

9 1 1

1 1 3.27 10 cm3.06 10 cm

= = =

Soulution:

Ec EFi = kBT lnNcni

( )( ) ( )( )17 3

5 16 3

4.7 10 cm8.617 10 eVK 300 K ln 0.675 eV

2.146 10 cmc FiE E

= =

Why the Fermi level is above the middle band gap of the intrinsic semiconductor?

219

220

: Doped semiconductors

from NCTU OCW

221 from NCTU OCW

222 from NCTU OCW

223 from NCTU OCW

P- & N-type Dopants in Semiconductors

B C N

Al Si P

Ga Ge As

In Sn Sb

Zn

Cd

Hg

S

Se

Te

IIB IIIA IVA VA VIA 5 6 7

13 14 15 16

30 31 32 33 34

48 49 50 51 52

80

Mg 12

Be 4 P-type dopant N-type dopant

Si: B, In P, As GaAs: Be, Mg, Zn, C S, Si

Si is the most popular n-type dopants for III-V compound semiconductors. Be, Zn and C are used for p-type dopants for III-V semiconductors.

224

FREE CARRIERS IN DOPED SEMICONDUCTORS

If electron (hole) concentration is measured as a function of temperature in a doped semiconductor, one observes three regimes:

Freezeout: Temperature is too small to ionize the donors (acceptors), i.e.,

kBT < EC ED (kBT< EA EV). Saturation: Most of the donors (acceptors) are

ionzed.

Intrinsic: Temperature is so high that ni > doping concentration.

225

Fermi level in extrinsic semiconductor In a doped semiconductor, the Fermi level will lie closer to one of

the band edges, and the electron and hole densities will adjust accordingly.

Extrinsic semiconductor (e.g. n-type): n = ND, but number of holes decreases, as np = ni2 (constant!) (lightly doped only!)

E

EC EV

gC(E)

gV(E)

0 1

E

f(E)

1 - f(E)

EF n E

n(E)

p(E)

EC

EV p

227

Notes on degenerate and non-degenerate semiconductors

A semiconductor is said to be 'non-degenerate' if only a small number of donor and/or acceptor atoms have been introduced into it, such that the concentration of these impurity atoms are small in comparison to that of the host atoms. In an n-type non-degenerate semiconductor, the small number of donor atoms are spaced far apart from each other, such that there is no interaction among their donor electrons. In a p-type non-degenerate semiconductor, the small number of acceptor atoms are also far enough from each other to prevent interaction among the acceptor holes. The product of the electron and hole density of a non-degenerate semiconductor is always equal to the square of its intrinsic carrier density, whether the semiconductor is intrinsic or extrinsic.

A 'degenerate' semiconductor, on the other hand, is one that has been doped to such high levels that the

dopant atoms are an appreciable fraction of the host atoms. In a degenerate semiconductor, the donor (or acceptor) atoms become close enough to each other to allow their donor electrons (or acceptor holes) to interact.

In an n-type degenerate semiconductor, the single discrete donor energy level will split into a band of

energies as the donor electrons begin to interact. As donor impurity concentration is increased, this donor energy band widens, and may reach a point that it overlaps the bottom of the conduction band. This overlap occurs when the donor concentration becomes comparable with the effective density of states, Nc. When the concentration of electrons in the conduction band exceeds Nc, then the Fermi level will lie within the conduction band.

The same is true for p-type degenerate semiconductors - an increase in the acceptor atom concentration will result in the widening of the acceptor energy band to the point that it may overlap with the top of the valence band. When the concentration of holes in the valence band exceeds the effective density of states Nv, then the Fermi level will lie within the valence band.

The overlapping of the donor band with the conduction band in an n-type degenerate semiconductor will

make it behave more like a conductor than a semiconductor. Likewise, a p-type degenerate semiconductor whose acceptor energy band overlaps with the valence band will behave more like a conductor than a semiconductor.

228

Modulation-doped QW: self-consistent solution

Modulation doping: dopants introduced to barrier regions; liberated free carriers migrate to well region (less scattering)

The illustration is for n-type doping only; for generality, consider both types

Ref: S L Chuang

e/h envelope fns within single-band approx.

heavy or light or SO band holes

e

229

The total potential profiles for e and h

( ) ( ) | | ( )ee bi HV z V z e Fz V z= + +

( ) ( ) | | ( )hh bi HV z V z e Fz V z= + +

( ) | | ( )HV z e z= Hartree potential

built-in potentials

External electric field

(z) is the electrostatic potential

230

Hartree potential

Start with Gauss law:

QW

Charge distribution is given by:

Free carriers concentrations

ionized donor acceptor concentrations 231

Free carrier (e/h) concentrations

We use:

Finally, the Fermi level is determined from the charge neutrality condition:

No external injection of carriers: Fc=Fv=EF

Subband surface e/h concentrations

232

Self-consistent calculation flow chart

Start with a (guess) Hartree potential

Solve EMA equations for e/h envelope fns and subband energies

Generate the e/h concentrations n(z)/p(z) and solve Poissons equation

Potentials converged ?

End Yes

No

233

Modulation doped heterojunctions (2DEG)

Very similar to modulation doped-QW, but w/o 2nd top barrier region Modulation doping again separates charge carriers from ions This reduces the ionized impurity scattering, leading to very high mobility A 2D electron gas (2DEG) is formed ; backbone of the HEMTs

Ref: Yu-Cardona 234

Optical Transition Momentum Matrix Element

The optical transition matrix elements of momentum for transitions between the valence hole subbands and the conduction electron subbands are given by

235

zyxipPvc

vcnin

nni ,,,|| =>==

= + + +

| |

( ), , , , ,*

02 3 5 62

626

13 3

*,

6,

5,

4,

3,

2,

1,0 )3

132

166

12

(

||

mnmnmnmnmnmnm

mn

nynnn

y

cvvvvvv

vc

vc

gaaiaaiaaiP

pP

++=

>=, |X>, |Y>, |Z> basis is given byConsidering spin up and down, the 8-band Kane Hamiltonian under |S>, |X>, |Y>, |Z>, and |S>, |X>, |Y>, |Z>, basis can be written as[001]-oriented L-K Hamiltonian under |X>, |Y>, |Z> basis is given byThe modified Luttinger parameters are related to Luttinger parameters in the following manner: New Band Edge Basis Functions6-band L-K HamiltonianSlide Number 158The conduction band Hamiltonian of QWsSlide Number 160Slide Number 161Slide Number 162Slide Number 163How good is effective mass aprox. ?Band structure calculated by 6 band k.p method Slide Number 166Slide Number 167Valence band structure of Gan/AlGaN QW calculated by 6-band k.p methodSlide Number 169Slide Number 170Band Gap EngineeringModification of band structure:Band Gap EngineeringSlide Number 173Quarternary alloysSlide Number 175Slide Number 176Slide Number 177Slide Number 178Semiconductor heterojunctionsSlide Number 180Band offset calculation: Model Solid TheorySemiconductor heterojunctionsSlide Number 183Slide Number 184Slide Number 185Slide Number 186Slide Number 187Slide Number 188Quantum well (QW)The energy is quantized due to the electrons confined in QW, the energy is called sub-band energy level.The DOS is step function for the two-dimensional system. Effective bandgap, EgEg = Eg0 + E1 + E1(HH)Energy levels and wavefunction in infinite deep QWLOW-DIMENSIONAL STRUCTURES: DENSITY OF STATESDerive DOS 2DSlide Number 194Slide Number 195Lattice matching and strainSlide Number 197Slide Number 198Strain influence on band structureThe consequence of pseudomorphic strain on the bandedges of a direct bandgap semiconductor[001]-oriented Strained HamiltonianSlide Number 202Slide Number 203Unstrained and strained valence band structuresStrain effect on band lineup of Ge-Ge0.986Si0.014 QWStrain effect on band structure of the Ge/GeSi QW.Doping and Carrier ConcentrationsSlide Number 208HOLES IN SEMICONDUCTORS HOW DO HOLES MOVE?Slide Number 211Slide Number 212Slide Number 213Slide Number 214Slide Number 215Slide Number 216Slide Number 217ExampleSlide Number 219Slide Number 220Slide Number 221Slide Number 222Slide Number 223Slide Number 224FREE CARRIERS IN DOPED SEMICONDUCTORSSlide Number 226Slide Number 227Notes on degenerate and non-degenerate semiconductorsSlide Number 229Slide Number 230Slide Number 231Slide Number 232Slide Number 233Slide Number 234Optical Transition Momentum Matrix ElementExample: explicit 6 band k.p Slide Number 237Squared optical transition matrix elements of 10/15 nm GaAs/Al0.2Ga0.8As QWOptical transition ruleSpontaneous emission rate and optical gain Slide Number 241Band lineup of strained Ge-Ge0.679Si0.1Sn0.221 QWSquared optical transition matrix element of strained Ge-Ge0.679Si0.1Sn0.221 QWTM mode optical gain of Ge/Ge0.679Si0.1Sn0.221 QW TM mode spontaneous emission rate of Ge/Ge0.679Si0.1Sn0.221 QW

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EE7603 Lecture Notes FWJ 2013

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