POWERPOINT SLIDES
, Chapter
1
STRUCTURAL
PROPERTIES OF
SEMICONDUCTORS
, SEMICONDUCTORS: STRUCTURAL ISSUES
Different states of matter and a classification based
on the order present in the material.
MATTER
SOLID LIQUID GASEOUS LIQUID CRYSTAL
Crystalline: No short- or No short- or Long-range order
Long-range order. long-range long-range and flow of
order. order. atoms/molecules.
Polycrystalline:
Long-range order over Matter can
several microns. "flow" and
take the
shape of the
Amorphous/Glasses: container.
Good short-range
order, but no long-
range order.
Semiconductors used in most technologies are high
quality crystalline materials (some exceptions are
amorphous silicon, used for thin film transistor and
solar cells).
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
, CRYSTALLINE MATERIALS: SOME DEFINITIONS
BRAVAIS LATTICE: Collection of points that fill up space.
Every point has the same environment
around it.
TRANSLATION VECTORS: A translation of the crystal by a
vector T that takes a point R to R+T and
leaves the entire crystal invariant.
PRIMITIVE TRANSLATION VECTORS: Starting at any
particular lattice point, we can construct
3 vectors that take us to 3 nearest
neighbors points (non-coplanar).
The smallest such vectors are called
primitive vectors a1, a2, a3.
BASIS: A crystal is produced by attaching a basis to every
lattice point. The basis consists of one
or more atoms.
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
,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.
Alternative primitive cell for 2D lattices.
The shape of primitive cells can be defined by a set of
parameters
a1 = |a1|; a2 = |a2|; a3 = |a3|
a •a a •a a •a
α1 = cos–1 2 3 ; α2 = cos–1 1 3 ; α3 = cos–1 1 2
a2 a3 a1 a3 a1 a 2
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
, a3
Notation for a
primitive cell. α1
a2
α2
α3 a1
Bravais lattices can be formed only if
α = 0; π or 5π ; π or 3π ; 2π or 4π ; π
3 3 2 2 3 3
Number of Restrictions on conventional
System lattices cell axes and angles
Triclinic 1 a1 = a2 = a3
α=β=γ
Monoclinic 2 a1 = a2 = a3
α = γ = 90° = β
Orthorhombic 4 a1 = a2 = a3
α = β = γ = 90°
Tetragonal 2 a1 = a2 = a3
α = β = γ = 90°
Cubic 3 a1 = a2 = a3
α = β = γ = 90°
Trigonal 1 a1 = a2 = a3
α = β = γ < 120°, = 90°
Hexagonal 1 a1 = a2 a3
α = β = 90°
γ = 120°
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
, c
c
b
β α
γ b
a a
triclinic monoclinic base-centered
c
b
a
orthorhombic base-centered body-centered face-centered
c
b
a
tetragonal body-centered
c
a a
a
rhomboherdral hexagonal
a
a
a
cubic body-cubic face-centered cubic
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
,Most semiconductors have an underlying lattice that is
either face-centered cubic (fcc) or hexagonal closed pack
(hcp).
FACE CENTERED CUBIC: The lattice sites at the edges of a
cube and at the center of tis faces. The edge of the
cube a is called the lattice constant.
z
a a2
a1
a3 y
x
Primitive basis vectors for the face-centered cubic lattice.
a a a
a1 = (y + z), a2 = (z + x), a3 = (x + y)
2 2 2
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
, DIAMOND AND ZINC BLENDE STRUCTURES
a
The zinc blende crystal structure. The structure consists of
the interpenetrating fcc lattices, one displaced from the other
by a vector ( 4a , 4a ,4a ) along the body diagonal. The underlying
Bravais lattice is fcc with a two-atom basis. The positions of
the two atoms is (000) and ( 4a , 4a ,4a ).
Positions of the atoms
A = (n ,n ,n ), (n + 1 ,n + 1 ,n + 1 ),
1 2 3 1 4 2 4 3 4
(n1 + 12 ,n2 + 12 ,n3),(n1 + 34 ,n2 + 34 ,n3 + ), 1
4
(n1 + 12 ,n2,n3 + 12 ),(n1 + 34 ,n2 + 14 ,n3 + ), 3
4
(n1,n2 + 12 ,n3 + 12 ),(n1 + 14 ,n2 + 34 ,n3 + ) 3
4
(0, 12 ,12 )
( 12 ,0, 12 ) Nature of
chemical bonds
( 14, 14, 14 ,) in a diamond
or zinc blende
(0,0,0) ( 12 , 12 ,0) structure
Tetrahedral bonding
© Prof. Jasprit Singh www.eecs.umich.edu/~singh
, IMPORTANT PLANES IN ZINC BLENDE
OR DIAMOND STRUCTURES
ATOMS ON THE (110) PLANE
Each atom has 4 bonds:
• 2 bonds in the (110) plane
• 1 bond connects each atom to adjacent (110)
planes
Cleaving adjacent planes requires
breaking 1 bond per atom.
ATOMS ON THE (001) PLANE
2 bonds connect each atom to adjacent (001)
plane.
Atoms are either Ga or As in a GaAs crystal.
Cleaving adjacent planes requires
breaking 2 bonds per atom.
ATOMS ON THE (111) PLANE
Could be either Ga or As.
1 bond connecting an adjacent plane on
one side.
3 bonds connecting an adjacent plane on
the other side.
Some important planes in the cubic system with their Miller indices. This figure
also shows how many bonds connect adjacent planes. This number determines
how easy or difficult it is to cleave the crystal along these planes by cutting the
bonds joining the adjacent planes.
© Prof. Jasprit Singh www.eecs.umich.edu/~singh