ee7603 lecture notes fwj 2013
DESCRIPTION
Advanced Semiconductor Physics Lecture Notes, part ITRANSCRIPT
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EE7603
Advanced Semiconductor Physics
Dr. Fan Wei Jun Associate Professor
Office: S2.2-B2-20 Phone: 6790 4359 email: [email protected]
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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
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3
Crystal Structure
Outline
Semiconductor Band Theory
Band Gap Engineering
Doping and Carrier Concentrations
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Crystal Structure
4
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Different states of matter and a classification based on the order present in the material.
5 Strong bonds Weak bonds No bonds
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CRYSTALLINE MATERIALS: SOME DEFINITIONS
BRAVAIS LATTICE TRANSLATION VECTORS PRIMITIVE TRANSLATION VECTORS BASIS PRIMITIVE CELL UNIT CELL WIGNER-SEITZ CELL
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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.
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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
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R = n1a1 + n2a2 + n3a3 For 3D
Primitive vectors are used to define a translation vector, R
For 2D R = n1a1 + n2a2
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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
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Lattice + Basis = Crystal Structure
+
||
2D lattice
Basis
Crystal Structure
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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
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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
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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.
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From wikipedia
Construction of a Wigner-Seitz primitive cell
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of FCC and BCC
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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
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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.
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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).
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Also called as proper rotation.
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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
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MIT OCW 3.012 22
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23 From: http://www.fkkt.uni-lj.si/attachments/dsk2423/3d-symmetry-2007_2008.pdf
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25
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Symmetry Elements
Inversion, or center of symmetry
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Symmetry Elements
Mirror plane or Reflection
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31
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Symmetry elements: mirror plane and inversion center
The handedness is changed.
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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.
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n=? http://symmetry.otterbein.edu/tutorial/improper.html
Symmetry Elements
Ethane C2H6
S3 S6 ?
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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.
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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)
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5 two-dimensional Lattices
38 graphene
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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/
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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
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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 = =
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42
Dendrytic ice crystals imaged with a scanning electron microscope. The colors are computer generated.
NaCl
Ice crystal
TiO2
Ice
Ice structure
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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
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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
= + = + + = +
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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= =
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Primitive vectors of simple hexagonal lattice
zca
yaxaa
yaxaa
23
2
23
2
3
2
1
=
+=
=
x
46
a1
a2
a3
y
z
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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
=
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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
=
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HCP: AB. CCP: ABC
CCP (Cubic Close Pack) = FCC
Lattice video
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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
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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.
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Packing Density
Packing density = No. of sphere x volume of spherevolume of unit cell
BCC: 68% FCC: 74% Diamond Structure: 34% HCP:74%
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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
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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
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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
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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
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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
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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.
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First Four Brillouin Zones: Square Lattice
www.lehigh.edu/~jdg4/classwork/reciprocallattice.ppt 59
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All Brillouin Zones: Square Lattice
www.lehigh.edu/~jdg4/classwork/reciprocallattice.ppt 60
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15th Brillouin Zones of the simple cubic crystal:
61 The first 20 Brillouin Zones of the simple cubic crystal video
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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
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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
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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
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Directions and Planes
65
William Hallowes Miller (1801 1880), British mineralogist and crystallographer
From: http://www.myspace.com/alexjvecchio
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Directions
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67
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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
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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
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Common Crystal Planes
Intercepts are 1, 1, and (parallel to z). Plane is the (110) plane.
x y
z
a
a
a
(110)
70
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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
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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:
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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
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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
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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.
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Semiconductor Band Theory
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Band Structure
Also called as Electronic Structure
E-K Relations or Energy Dispersion Curves
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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.
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NATURE OF ATOMIC FUNCTIONS: l =0: called s-state; l =1: called p-state
79
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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
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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
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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
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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
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Different forms of Bloch theorem
85 BL-Bravais Lattice
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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
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Kronig-Penny Model
87
Real case
Kroning-Penny Model a
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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
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sinh(x)=(exex)/2
cosh(x)=(ex+ex)/2 89
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GAP
GAP
GAP
E
k
Solving the Kronig-Penney Model
Which is best solved graphically
RHS
d=a+b
90
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Check out the Kronig Penney applet at http://fermi.la.asu.edu/schmidt/applets/kp/plugkp.html
91
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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
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93
Tight-Binding Approximation
from NCTU OCW
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94 from NCTU OCW
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95 from NCTU OCW
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96 from NCTU OCW
Example
Example
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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
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98
Metal, Semiconductor and Insulator
from NCTU OCW
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99
Example
from NCTU OCW
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100
Example
from NCTU OCW
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101
Example
from NCTU OCW
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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
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103
Eg: C (diamond)~6 eV Si ~ 1.1 eV Ge ~ 0.7 eV
from NCTU OCW
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104 from NCTU OCW
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Classification of Materials
0 < Eg < ~3.5 eVs 105
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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
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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
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BAND STRUCTURE OF SEMICONDUCTORS
First Brillouin Zone of Si
108
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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
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EFFECTIVE MASS DESCRIPTION
110
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ELECTRON EFFECTIVE MASS OF MOST IMPORTANT SEMICONDUCTORS
111
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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
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BANDSTRUCTURE: Si
113
y
-
BANDSTRUCTURE: GaAs
114
-
115
Form UCSB
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BANDSTRUCTURE: Ge AlAs, InAs, InP
116
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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
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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
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Density Of State
119
The general form of DOS of a system with volume V is given as
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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
-
122
-
123
-
124
-
125
-
126
-
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
-
145
-
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
-
170
-
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.
Lattice matching and strain
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.
Lattice matching and strain
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
Lattice matching and strain
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
-
226
-
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