A simple scheme is used to describe lattice planes, directions and points. For a plane, we use the following procedure:
(1) Define the x, y, z axes (primitive vectors).
(2) Take the intercepts of the plane along the axes in units of lattice con- stants.
1.2. Crystalline materials 13
Zinc blende and wurtzite
MATERIAL
C Si SiC Ge BN BN BP BAs A1N A1P AlAs AlSb GaN GaP GaAs GaSb InN InP InAs InSb ZnO ZnS ZnS ZnSe ZnTe CdO CdS CdS CdSe CdSe CdTe PbS PbSe PbTe
CRYSTAL STRUCTURE
DI DI ZB DI HEX
ZB ZB ZB W ZB ZB ZB W ZB ZB ZB W ZB ZB ZB W ZB W ZB ZB R W ZB W ZB ZB R R R
BANDGAP (EV) 5.50,1 1.1242,1 2.416,1 0.664,1 5.2,1 6.4,1 2.4,1
— 6.2.D 2.45,1 2.153,1 1.615,1 3.44,D 2.272,1 1.4241.D 0.75,D
1.89JD 1.344,D 0.354,D 0.230,D 3.44,D 3.68,D 3.9107JD 2.8215.D 2.3941,D 0.84,1 2.501,D 2.50,D 1.751,D 1.475,D 0.41, D*
0.278,D*
0.310,D*
STATIC DIELECTRIC
CONSTANT 5.570 11.9
9.72 16.2 ell = 5.06 e-L = 6.85 7.1 11.
—
£ = 9.14 9.8 10.06 12.04 eil=10.4
8i=9.5 11.11 13.18 15.69 12.56 15.15 16.8 ell= 8.75
£i=7.8 8.9
£ = 9.6 9.1 8.7 21.9
£ = 9.83
—
£11=10.16
£i= 9.29 10.2 169.
210.
414.
LATTICE CONSTANT
(A)
3.56683 5.431073 4.3596 5.6579060 a =6.6612 c = 2.5040 3.6157 4.5383 4.777 0=3.111 c= 4.981 5.4635 5.660 6.1355 0 = 3 . 1 7 5 c= 5.158 5.4505 5.65325 6.09593 a = 3.5446 c = 8.7034 5.8687 6.0583 6.47937 a = 3.253 c= 5.213 5.4102 a = 3.8226 c = 6.2605 5.6676 6.1037 4.689 0=4.1362 c= 6.714
5.818 a = 4.2999 c= 7.0109 6.052 6.482 5.936 6.117 6.462
DENSITY (gm-crrr3)
3.51525 2.329002 3.166 5.3234 2.18 3.4870 2.97 5.22 3.255 2.401 3.760 4.26 6.095 4.138 5.3176 5.6137 6.81 4.81 5.667 5.7747 5.67526 4.079 4.084 5.266 5.636 8.15 4.82
— 5.81 5.87 7.597 8.26 8.219 Data are given at room temperature values (300 K).
Key: DI: diamond; HEX: hexagonal; R: rocksalt; W: wurtzite; ZB: zinc blende;
*: gap at L point; D: direct; I: indirect ell: parallel to c-axis; e i : perpendicular to c-axis.
Table 1.2: Structural properties of some important semiconductors.
Material c/a Be
Cd Mg Ti Zn Zr
2.29 2.98 3.21 2.95 2.66 3.23
3.58 5.62 5.21 4.69 4.95 5.15
1.56 1.89 1.62 1.59 1.86 1.59
Table 1.3: Materials with hep closed-packed structure. The "ideal" c/a ratio is 1.6.
(3) Take the reciprocal of the intercepts and reduce them to the smallest integers.
The notation (hkl) denotes a family of parallel planes.
The notation (hkl) denotes a family of equivalent planes.
To denote directions, we use the smallest set of integers having the same ratio as the direction cosines of the direction.
In a cubic system, the Miller indices of a plane are the same as the direction perpendicular to the plane. The notation [ ] is for a set of parallel directions; < > is for a set of equivalent direction. Fig. 1.12 shows some examples of the use of the Miller indices to define planes.
EXAMPLE 1.1 The lattice constant of silicon is 5.43 A. Calculate the number of silicon atoms in a cubic centimeter. Also calculate the number density of Ga atoms in GaAs which has a lattice constant of 5.65 A.
Silicon has a diamond structure, which is made up of the fee lattice with two atoms on each lattice point. The fee unit cube has a volume a3. The cube has eight lattice sites at the cube edges. However, each of these points is shared with eight other cubes. In addition, there are six lattice points on the cube face centers. Each of these points is shared by two adjacent cubes. Thus the number of lattice points per cube of volume a3 are
In silicon, there are two silicon atoms per lattice point. The number density is, therefore
AT 4 X 2
_ 4 X 2 = 4 9 9 ? x 1Q22 a t o m s/c m3 b l a3 (5.43 x 10-8)3
In GaAs, there is one Ga atom and one As atom per lattice point. The Ga atom density is, therefore
NG* = i = (5.65 x410-*)3 = 2'2 2 X ^ a t°™/cm3 There are an equal number of As atoms.
EXAMPLE 1.2 In semiconductor technology, a Si device on a VLSI chip represents one of the smallest devices, while a GaAs laser represents one of the larger devices. Consider a
1.2. Crystalline materials 15
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
= u > 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
F i g u r e 1.12: Some important planes in the zinc blende or diamond structure along 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.
Si device with dimensions ( 5 x 2 x 1 ) /im3 and a GaAs semiconductor laser with dimensions (200 X 10 x 5) ^m3. Calculate the number of atoms in each device.
From Example 1.1 the number of Si atoms in the Si transistor are ATgi = (5 x 1022 atoms/cm3)(10 x 10~12 cm3) = 5 x 1011 atoms The number of Ga atoms in the GaAs laser are
7VGa = (2.22 x 1022)(104 x 10"12) = 2.22 x 1014 atoms An equal number of As atoms are also present in the laser.
E X A M P L E 1.3 Calculate the surface density of Ga atoms on a Ga terminated (001) GaAs surface.
In the (001) surfaces, the top atoms are either Ga or As leading to the terminology Ga terminated (or Ga stabilized) and As terminated (or As stabilized), respectively. A square of area a2 has four atoms on the edges of the square and one atom at the center of the square.
The atoms on the square edges are shared by a total of four squares. The total number of atoms per square is
7V(a2) = 1 + 1 = 2 The surface density is then
"Via ~ a2 ~ (5 t 6 5 x 10-8)2 - — - - —
E X A M P L E 1.4 Calculate the height of a GaAs monolayer in the (001) direction.
In the case of GaAs, a monolayer is defined as the combination of a Ga and As atomic layer. The monolayer distance in the (001) direction is simply
_ a _ 5.65 _ o
m l ~ 2 ~~ 2 ~