1-SEMICONDUCTOR CRYSTAL STRUCTURE
1.1INTRODUCTION
While studying semiconductors we are interested in behaviour of a large number of electrons moving through positively charged ions.
To appreciate the complexity of the problem, we can examine the number density of electrons involved.
"Free" valance electrons
Electrons in a solid
valance electrons (-eZc)
core electrons (-e(Za - Zc))
nucleus with charge (eZa)
where Za = atomic number
Fig: 1.1 Electrons in a solid
Consider now an element with density, , having a unit volume.
Then mass = , . Let the atomic mass be A , and use Avagadro's Number (6.022 1023 atoms per mole),
Then the number of moles = /A, and
n = (6.022 1023 atoms per mole)* ( /A)*Zc------1.1
where,
Zc = number of electronsin the outermost shell, and
n = electron density for conduction electrons.
For most metals n 1023 electrons.
1.2BASIC LATTICE TYPES
Crystal is formed by identical building blocks that consist of an atom or a group of atoms.
Natural crystals have a fixed symmetry, BUT artificial crystals are produced with a modified crystalline structure. These lead to "superlattices". Altering crystal structure also changes the electronic properties.
The building block which is repeated infinitely to produce the crystal may be :
- very simple (single atom for many metals, such as copper, gold, aluminium etc.) OR
- complex ( as in some proteins)
In superlattices the number of the building block can be arbitrary, but in natural crystals it is mainly two.
Now we shall have some definitions:
- Lattice is a set of points in space which form a periodic structure. It is a mathematical abstraction.
- Basis is a building block of atoms attached to each lattice point.
Lattice point R', can be obtained from any other lattice point R by a translation:
R' = R +m1 a1 + m2 a2 + m3 a3 ------1.2
where a1, a2,a3 are vectors and m1, m2, m3, are integers. Such a lattice is called Bravais Lattice and the lattice structure together with the basis form the Crystal Structure, as indicated below in Fig. 1.2.
******
******
a-Space lattice
0o
b-Basis with two different ions
0o0o0o0o 0o 0o
0o0o0o0o 0o 0o
c-Crystal structure
Fig. 1.2 Crystal structure
The translation vectors, a1, a2,a3 , are called primitive if the volume of the cell formed is smallest possible.
One choice would be to take,
a1as the shortest period of the lattice,
a2 as the shortest period of the lattice, and not parallel to a1
a3 as the shortest period of the lattice, and notcoplanar witha1 anda2
The basic building block of the crystal is called the unit cell.
The lattice separation is very small, so in practical terms one deals with almost infinite lattices. Boundaries, interfaces play an important role, as will be seen later.
Basic lattice types: Lattice types are important in the sense that the structure of lattice also describes the symmetry which effects the electronic properties. We will only consider cubic lattices, which is structure taken by all semiconductors. Basically there are THREE types:
- Simple cubic
- Body centred cubic ( abbreviated to bcc)
- Face centred cubic ( abbreviated to fcc)
- Simple cubic lattice
This is generated by the primitive vectors
a x, a y, az ------1.3
where x, y, z are unit vectors, and a is the cube side.
Fig. 1.3Simple cubic structure.
- Body centred cubic ( abbreviated to bcc)
This is generated from the simple cubic structure by placing an extra atom at the Centre of the cube
Fig. 1.4.Body centred cubic structure
- Face centred cubic ( abbreviated to fcc)
This is generated from the simple cubic structure by adding atoms in the Centre of each square face.
Fig. 1.5 Face centred cubic structure.
"fcc" is the most important lattice for semiconductors. The primitive vector set is given by:
a1 = a/2 (y + z ), a2 = a/2 (z + x ), a3 = a/2 (x + y )------1.4
where a is the cube edge and is called lattice constant of the semiconductor.
The diamond and zinc-blende structures:
Essentially all semiconductors of interest for electronics and optoelectronics have "fcc" structure with two atoms per basis.
The two basis atoms have coordinates relative to each other as, (0,0,0), and (a/4, a/4, a/4) where "a" is the lattice constant.
Fig 1.6 Two basis atoms of a diamond structure, and two diamond structures interpenetrating.
Each atom lies on its own "fcc" , and the structure may be thought of two interpenetrating "fcc" structures, where one is displaced by (a/4, a/4, a/4) along the body diagonal of a cube with sides "a".
Now, if the two atoms of the basis are identical the structure is called DIAMOND (eg. as in Si, Ge, C, etc.)
But, if the two atoms are different, it is called ZINC BLENDE (GaAs, CdS, AlAs, etc.)
Usually, semiconductors with a diamond structure is called ELEMENTAL semiconductors, whereas those having Zinc Blende structure are called COMPOUND semiconductors. As an example, in a compound semiconductor of GaAs there are 4 Ga and 4 As atoms in each cube of volume a3.
Example 1.1
Lattice constant of silicon is 5.43 A. Calculate the number of silicone atoms in a cubic centimetre. Also calculate the number density of Ga atoms in GaAs which has a lattice constant of 5.65 A.
Solution
Since we have silicone we should already know that we have a n "fcc" diamond structure where we also have two identical atoms per basis.
"fcc" gives 8 lattice points at the cube edges
"fcc" gives 6 points at the face centres
Lattice points at the edges are shared by 8 other cubes, while the faces are shared by 6 other cubes, as indicated below.
Hence the number of point per cube of volume a3 is:
N(a3) = 8*(1/8) + 6*(1/2) = 1 + 3 = 4 points
In silicone there are two atoms per lattice point (basis), hence the number density is then, (Angstrom = 10-10 meters, 10-8 cm):
N Si = (4*2 ) / a3 = 8 / a3 = 8 /(5.43 10-8 cm)3 = 4.997 10 22 atoms/cm3.
In GaAs there are only one Ga atom and only one As atom at each lattice point. Hence the number density of Ga atoms is given by:
N Ga= 4 / a3 = 4 / (5.65 10-8 cm)3 = 2.22 1022 atoms / cm3
Also there are equal number of As atoms.
Example 1. 2
A Si device on a VLSI chip represents one of the smallest devices and GaAs laser one of the largest. If Si device has dimensions of (5*2*1)(m)3 and GaAs has (200*10*5)( m)3 , calculate number of atoms in each device.
Solution
Using the results obtained in Example 1, for the number of atoms per cm3 , Si (5 1022 atoms / cm3 ) and GaAs (2.22 1022 atoms /cm3 ) .
N Si = (5 1022 atoms / cm3 ) * 10 (10-6 102 cm)3 = 5 1011 atoms.
N Ga = (2.22 1022 atoms /cm3 ) * 104 (10-6 102 cm)3 = 2.22 1014 atoms.
Also,
N As = (2.22 1022 atoms /cm3 ) * 104 (10-6 102 cm)3 = 2.22 1014 atoms.
1.3MILLER INDICES
Miller Indices are used to simplify description of lattice planes.
There is a simple procedure to follow when a drown plane is to be indexed. This is:
- Define the x, y, z, axes.
- Take the intercepts of the plane along the axes in units of lattice constant.
- Take reciprocals of the intercepts and reduce them to smallest integers.
A family of parallel planes are indicated by a notation h k l .
Fig. 1.7 Miller Indices for various planes
Example 1.3
Calculate the surface density of Ga atoms on a Ga terminated (001) GaAs surface.
Solution
The atoms at the top of (001) surface are either Ga or As. If it is Ga terminated then the atom at the top are Ga. The (001) plane has an area of a 2 , 4 corner atoms and 1 Centre atom. Each atom of the corner atoms are shared by 4 similar planes, so that the contribution is 1/4, but the Centre atom is contributed wholly.
Hence the number of atoms per a2 = 4*(1/4) + 1 = 2 atoms
The surface density is then, S Ga = 2 / a2 = 2 / (5.65 10-8 cm)2 = 6.26 1014 atoms cm-2.
Example 1. 4
Plot the following planes for "bcc" and "fcc" structures.
(010),(110),(200),(123).
2-QUANTUM MECHANICS AND STATISTICAL PHYSICS OF ELECTRONS
2.1 INTRODUCTION
The physical world can be viewed in two ways using:
- Classical physics
- Quantum mechanics
The interaction of experimental observations and theoretical descriptions in 20th century defied the classical explanations. Experimental observations such as ;
black-body radiation, with
Energy (E) = hv ------2.1
where "v" is the frequency and "h" is the Planck's constant;
photoelectric effect with,
Energy (E) = h' w ------2.2
where h' = h / 2, and w = 2v;
and atomic spectra are all explained by the quantum mechanics approach.
The quantum mechanics approach can be described by either,
- Heisenberg uncertainty relation between momentum "p" and position "x", and between energy "E" and time "t". That is
p*x h' /2 and E*t h' /2
- Schrodinger Equation ( This is what we will choose to study).
2.2 SCHRODINGER EQUATION
The quantities that are physically seen are represented by mathematical operators in Schrodinger approach. Some important operators are represented by following operations.
Momentum :px -i h' ( / x)
py -i h' ( / y) ------2.3
EnergyE i h' ( / t)
Now consider kinetic energy (KE) and potential energy (PE) of an electron with mass mo. Then,
KE + PE = E or,
p2 /2mo + U(r,t) = E ------2.4
where U(r,t) is the potential in which the electron moves. If we represent observables "p" and "E" by their operators, and operate on a general state (r,t), we get Schrodinger equation;
(-h'2/2mo) 2(r,t) + U(r,t) (r,t) = i h' ( / t)(r,t) = E (r,t) --2.5
where "t" is the time and "r" all the spatial variables (x,y,z), and (r,t) is called the wave function. The wave function has the meaning that;
(r,t)2 dr ------2.6
gives the probability that the electron can be found at time "t" in the volume element "dr".
As an example take a hypothetical (not true) case and plot (x)2
in one dimension as in Fig. 2.1 and assume that it is time independent.
Fig. 2.1 The electron is always between "xo" and "x4" , (i.e. probability is zero for xo x x4 ).
Since, in Fig. 2.1, the electron must be somewhere, the probability of finding it between xo" and "x4" must be unity. That is;
(x)2 dx = 1 ------2.7
This equation only helps to illustrate meaning of (x)2 .
If the potential energy term U(r,t) has no time dependence, then we can rewrite Eq. 2.5 with i) time dependent and ii) time independent parts.
We may also rewrite the wave function with the time dependent and time independent components as;
(r,t) = (r) f(t) ------2.8
If we substitute Eq. 2.8 into Eq. 2.5, divide by "(r) f(t)" and equate each side to "E" , we get ;
- one time independent equation as,
(-h'2/2mo) 2(r) + U(r) (r) = E (r) ------2.9
- one time dependent equation as;
i h' ( / t)f(t) = E f(t) ------2.10
The solutions for the two cases are;
i) Time dependent case;
The time dependent part of the solution has a general form,
f(t) = exp(iEt / h') ------2.11a
and if we set E = h' then we get;
f(t) = exp(it ) ------2.11b
where "E" is the energy of the electron that has to be solved for by solving Eq. 2.9.
In Eq.2.9 the unknowns are both "(r)" and "E". In general, there will be a series of solutions "(r)" with associated energies En.
It is clear that the associated wave functions are complex, and the physical interpretation according to which the probability to find the electron in a region of space Vo is given by;
P(Vo) = d3r *(r )(r ) ------2.12
The allowed wave functions may be normalised to a particular volume "V" (ie. unit volume) signifying that there is oneelectron in the volume "V".
ii) time independent case;
The time independent case can be considered under two different conditions;
- condition 1, when there is no potential, U(r) = 0
- condition 2, when the potential changes uniformly in space
Condition 1
This is the case when we set U(r) = 0, in Eq. 2.9.
This concept deals with quantum mechanics of "free" electrons moving in absence of any potential.
So the Schrodinger (time independent ) equation for free electrons is ; ( Eq. 2.9 with no potential energy term),
(-h'2/2mo) (2 / x2 + 2 / y2 + 2 / z2 )(r) = E (r) ------2.14
with a general solution;
(r) = (1 / V) exp(ik.r) ------2.15
and the corresponding energy (only the kinetic energy since we already had no potential energy) is given by;
E = (h'2 k2 )/2mo ------2.16
where "k" is a wave vector and the factor, "(1 / V)" is used such that "V" is a volume having one electron and obeys the relation,
d3r (r) 2 = 1 ------2.17
The energy of the electron is,
E = (h'2 k2 )/2mo p2 /2mo ------2.18
and the momentum is;
-ih' ( / r ) h' k ------2.19
while the velocity is;
v = h' k / mo ------2.20
Assume now that the volume "V" is a cube of sides "L". To correlate with physical conditions, two boundary conditions are imposed on the wave function.
- Firstly, for electrons confined in a finite volume; the wave function goes to zero at the boundaries of the volume. The wave solutions are of the form sin(kxx) or cos(kxx) and k-values are restricted to the positive values;
kx = / L, 2 / L, 3 / L, nx / L ------2.21
where "nx" is an integer, and the allowed values of energy are;
E = (h'2 k2 )/2mo = (h2 / 8m0L2 )(nx2 + ny2 + nz2 ) ------2.22
This standing wave solution is used to describe stationary electrons (e.g. electrons in quantum wells) as shown in Fig. 2.2 below.
- Secondly, for moving electrons; the boundary conditions is known as periodic boundary conditions. The wave can be spread in all space (although finite in volume, V,) made up of cubes of side "L".
Then,
(x, y, z+L)= (x,y,z)
(x, y+L, z) = (x,y,z) ------2.23
(x+L, y, z) = (x,y,z)
Fig.2.2 a) Standing waves(stationary boundary).
b) Exponential wave function with electron probability
equal in all regions (periodic boundary).
Because of boundary conditions, the allowed values of "k" are;
kx = 2 nx / L, ky = 2 ny / L, kz = 2 nz / L, ------2.24
where "n" are integers, positive or negative.
If "L" is large, spacing between "k" values is very small.
Now consider volume in k-space that each electron occupies. As shown in Fig. 2.3 this volume in 3-dimensions is;
(2 / L)3 = 83 / V ------2.25
where the volume in real space V= L3
Fig.2.3 k-space , the allowed values are separated by 2 / L.
Now, if "" is any volume in k-space, the number of electronic states in this volume are;
() / (83 / V) = V / 83 ------2.26
Condition 2
This is the case when the electrons are in a periodic potential as in a crystal lattice.
Normally, the equation of motion for "free" electrons may be written as;
mo dv / dt = Fint + Fext ------2.27
where " Fint " and " Fext " are the internal and external forces respectively.
The internal forces, " Fint " may be very complex. However, the periodic nature of the crystal potential allow an equation like, "Eq. 2.27" to be written except that the effect of the internal forces are accounted for by using "m*" (effective mass) rather than the free electron mass "mO".
Hence;dpeff / dt = Fext ------2.28
The Schrodinger equation for an electron in a semiconductor is;
(-h'2/2mo) 2(r) + U(r) (r) = E (r) ------2.29
where U(r) is the background potential. Due to crystalline nature of material, U(r) has the same periodicity "R" as the crystal lattice.
Hence;U(r) = U( r + R ) ------2.30
and the electron probability is the same in all unit cell.
If the potential was random, this would not be the case as shown in Fig. 2.4.
Fig. 2.4 : a) non-periodic potential. b) periodic potential.
Through the Bloch's Theorem, Fig. 2.4b, shows that the probability (not the wave function) is periodic through the crystal.
2.3 FROM ATOMIC LEVELS TO BANDS
There are different models representing the back-ground potential in the crystal. One of them, the Kronig-Penny model represents the back-ground potential (periodic) seen by electrons in the crystal as a simple potential as shown in Fig. 2.4b above.
Graphical solution of Eq.2.29 to obtain allowed energy levels is shown in Fig. 2.5 below, where f(E) is a function of energy that must lie between the values of 1.0 and -1.0.
Fig. 2.5 a) f(E) against electron energy. b) electron energy, E, against wave vector kx.
A simpler representation of energy states is shown in Fig. 2.6, where the allowed bands are infinitesimally narrow and the forbidden gaps are relatively wide.
Vacuum Energy Level Electrons in free space
Electrons are bound in the material
Crystal Isolated atoms
(allowed bands) (atomic levels)
spacing between atoms------
spacing 1-2 A spacing 1000 A
Fig. 2.6 Allowed and forbidden bands.
If the atoms are brought together to form a crystal, the discrete energy levels broaden. The core electrons are tightly bound and not effected.
Hence, it is the outermost electrons forming the allowed and forbidden bands, and the properties of the semiconductors are essentially determined by these valance electrons.
The wavelengths that are allowed in a crystal is governed by the k-vector such that;
k = 2 / ------2.31
Hence, the shortest wavelength that is allowed in a crystal depends on spacing between the two adjacent lattice points.
3-THE HYDROGEN ATOMAND QUANTUM NUMBERS
3.1 THE HYDROGEN ATOM
Hydrogen atom forms a relatively simple system with one electron and one proton.
Electron can stay in any stable energy level and can move to a higher or a lower energy state by absorbing or radiating a well defined quantum of energy, as shown in Fig 3.1 below.
------
electron in orbit at certain
energy level
-a-
Free electrons
Vacuum level
Bound
electrons
300 (310) (320) or 3s (3p) (3d)
13.6 eV
-b- 200 (210)or 2s (2p)
100 or 1s
Fig. 3.1 a- Hydrogen atom b- Energy levels.
Mathematical model of Hydrogen atom can be used, as a simplification, while studying the doped semiconductors. Here an electron is bound to an ionised impurity.
The Schrodinger Equation for the electron-proton system is;
(-h'2/2mo e2 )-( h'2/2mph2 )- e2/(4ore-rp2)(r) = Etot (r) -3.1
where the first and second terms on the left-hand side of the equation represent the kinetic energy of an electron and a proton respectively, and the third term represent the potential energy of the system (e.g. coulomb energy between electron and proton at points re and rp respectively.)
If we now consider the relative motion of the electron and proton we have;
(-h'2/2mr2 ) - e2 / (4o / r)(r) = E(r) ------3.2
where r = re – rp , is the relative coordinate and mr is the reduced mass such that;
1 / mr = 1 / mo + 1 / mp 1/ mo , since mp mo .
The energy values are;
En = -mo e4 / (4o )2 h'2)1 / n2 = 13.6 / n2 eV------3.3
Where “n” is an integer.
The general wavefunction has the form,
nml ( r,,), where “ n,m,l “ are quantum numbers.
The ground state wavefunction has the form;
100 ( r,,) = (1 / a3 )exp (-r / a )------3.4
where “a” is called Bohr radius, given as;
a = h'2 / mo2 = 0.53 A------3.5
Some energy levels with corresponding quantum numbers are
shown in Fig 3.1b. For quantum number l = 0, 1, 2, 3,… the
representation used is s, p, d, f,….respectively.
Hence;
100 is called 1s, and
210 is called 2p states etc…
3.2 QUANTUM NUMBERS
Symbol “n” Principal quantum number. It determines the
total energy and can have integers, 1,2,3,4, ...
Symbol “l” Azimuthal quantum number. It determines the
Orbital momentum and can have integers, 0,1,2,3
4, ……(n-1).
Symbol “m” Magnetic quantum number. It determines
magnetic moments. It can have values in the
range -l,……..0………..,+l
Spectrographic designation of Azimuthal quantum numbers are;
Value of “l” Designation
0s (sharp)
1p (principal)
2d (diffuse)
3f (fundamental)
nlmSpectrographic designations
100 1s2 electrons
------
200 2s2 electrons
1 -1,0,+1 2p6 electrons
------
300 3s2 electrons
1 -1,0,+1 3p6 electrons
2-2,-1,0,+1,+2 3d10 electrons
------
400 4s2 electrons
1 -1,0,+1 4p6 electrons
2-2,-1,0,+1,+2 4d10 electrons
3 -3,-2,-1,0,+1,+2,+3 4f14 electrons
Subscripts or superscripts on s,p,d,…. Levels ( eg. s2 or s2 means 2 electrons) represent the number of electrons in that level. Hence;
Group 4 Semiconductors
C (Carbon)1s22s22p2
Si (Silicone)1s22s22p63s23p2
Ge (Germanium)1s22s22p63s23p63d104s24p2
Group 3-5 Semiconductors
Ga (Gallium)1s22s22p63s23p63d104s24p1
As (Arsenic)1s22s22p63s23p63d104s24p3
Now, as the “p” level have ( -1, 0, +1 ) but all have the same energy, they are called 3-fold degenerate, eg. in a magnetic field they split into 3 levels.
Similarly, “d” levels are 5-fold degenerate.
Example 3.1
Calculate the wavelength associated with a 1 eV a) photon, b) electron, c) neutron. (h = 6.6 10 –34 Js; c = 3 10 8 m/s;
e = 1.6 10 –19 J; mo = 0.91 10 –30 kg; mn = 1824 mo) .
Solution
a)Using the equation; E = hv or E = hc / or ph = hc / E
Hence, ph = (6.6 10 –34 Js)(3 10 8 m/s) / 1.6 10-19 J = 1.24 m.
b)From Eq.2.31;we have k = 2 / and for an electron the energy is given by;
E = (h'2 k2 )/2mo which gives for “k”, k = (2moE) / h'21/2 . Since we have, = 2 / k, now substitute for “k” and h'= h / 2 to get;
= (h / 2) 2 / (2moE)1/2 = h / (2moE)1/2 = (6.6 10 –34 Js) / 2(0.91 10 –30 kg )( 1.6 10-19 J )1/2 = 12.3 A
c)As we see in part b) we have a relationship 1 / (m)1/2 and we are given that, mn (neutron mass) = 1824 mo , then,
n /e = (mo)1/2 /(mn)1/2 = 1 /(1824)1/2 and n = 12.3 A(1 / 1824)1/2
n = 0.28 A
Example 3.2
Calculate the wavelength associated with an electron having 1eV energy, if the electron is inside GaAs crystal where its effective mass is 0.067 mo. ( mo = 0.91 10 –30 kg)