1
Chapter 2
Semiconductor Materials and Their Properties
In this chapter, we will cover the following topics
(1)
Elemental and compound semiconductors
(2)
The valence bond model
(3)
The energy band theory
(4)
Concentration of electrons and holes including Fermi levels, en
ergy distribution,
and temperature
dependence
2.1 Elemental and Compound Semiconductors
Elemental semiconductors:
two important ones:
Si
and
Ge
, both belong to group

IV
with 4 valence electrons in their outermost shell.
They crystallize in a diamond st
ructure. Neighboring atoms are bound
by
covalent bond
s
.
Si is by far the widely used semiconductor for various device applications
Compound semiconductors:
III

V and II

VI compounds.
III

Vs
: GaN, GaP, GaAs, GaSb, InP, InAs, InSb
They crystallize in zin
c blende structure
. 8 valence electrons are shared between a pair of
nearest atoms. Therefore, the bonding has a covalent character. On the other hand, since
the group III elements are more electropositive and group V elements are more
electronegative. Hen
ce, the bonding has a partial ionic character as well. But the covalent
nature is predominant.
II

VIs
: ZnS, ZnSe, ZnTe, CdS, CdSe, CdTe
Crystal structures:
CdS and CdSe: wurtzite (two interpenetrating hexagonal close

packed lattices)
ZnTe and CdTe: zin
c blende
ZnS and ZnSe: can be both (depeding on the substrate on which it is grown)
Bonding: mixture of covalent and ionic types. Stronger ionicity than III

Vs since group

VI elements are considerably more electronegative than group II elements.
The ioni
c character has the effect of binding the valence electrons rather tightly to the
lattice atoms. Thus, the band gaps of these compounds are larger than those of the
covalent semiconductors of comparable atomic weights.
Ternary and quaternary compounds
:
2
G
a
1

x
Al
x
As, ZnS
x
Se
1

x
, Zn
1

x
Mg
x
S
y
Se
1

y
, where
0
≤
≤
1
,
0
≤
≤
1
.
The properties of the resulting compound vary gradually with the fraction
.
Advantages of compound semiconductors
:
Wider choice of bandgap than elemental semiconductors: IR

visible

UV
Direct bandgap materials: optoel
ectronic applications, LEDs, lasers, sensors.
A major difficulty with compounds is that their preparation in single crystal form is more
difficult.
Two models can be used to study semiconductors
(1)
Valence bond model
which describes properties in domain of
space and time.
(2)
Energy band model
which describes properties in energy and momentum.
Energy band model is far more useful.
2.2
The Valence Bond Model
Use elemental semiconductors as example, Si and Se, they form diamond structure where
each atom bound t
o its four nearest neighbors by covalent bonds. These neighbors are all
equidistance from the central atom and lie at the four corners of the tetrahedron.
Fig.
2.1
I
llustration of covalent bond in tetrahedron.
Each bond has two electrons with opposite
spin so that the central atom appears to have
eight electrons with opposite spins.
3
Fig
.
2.2
An 2

D illustration of the diamond structure.
At
=
0
K
, all electrons are tightly bound in the bonds
–
a perfect insulator
At higher temperatures, lattice vibration occasionally shake loose some electrons so they
can move freely inside the crystal. The vacant side created by the broken bond has a net
po
sitive charge known as hole (a particle of positive charge). Both electrons and holes are
responsible for semiconductor conductivity.
One way to visualize the movement of a hole is to consider the neighboring bond
electrons jumps over to the vacant site t
o create another vacant at the position from which
the electron came from. This is equivalent to hole moving from one site to another.
This picture has its limitation. It fails to explain the wave nature of the hole and the Hall
effect.
The above describ
ed generation of electrons and holes
(both are called carriers)
is due to
the thermal excitation. This is the case for
intrinsic semiconductors
where number of
electrons is equal to that of holes. The carrier concentration depends on temperature.
There is
another way of introducing conduction carriers into the semiconductors
–
dopping of impurity atoms intentionally.
For Si and Ge, elementals from group

III and V are commonly used as impurities.
The doped semiconductors are called
extrinsic semiconduc
tors
.
n

type semiconductor
s
4
A small amount of group V elements (As, P, Sb) is added into Si. These impurities
occupy lattice sites that are normally occupied by Si atoms
–
substitutional impurities
Fig
.
2.3
Illustration of
group

V
substitutional impu
rity in Si.
The 5
th
electron is bound to the impurity atom only by weak electrostatic force.
Bohr theory of the hydrogen atom can be used to calculate the radius and the ionization
energy of its ground state
0
=
0
.
53
0
0
∗
Å
=
13
.
6
0
∗
0
2
eV
Two modifications to the hydrogen formula:
(1)
effective mass:
∗
<
0
(free electron mass)
(2)
permittivity of the semiconductor:
>
0
(
permittivity of the free space)
The ionization energy is typically small, therefore, at room temperature all of them
should be ionized to become conduction electrons
–
donors.
n

type semiconductor: majority carriers are electrons, minority carriers are h
oles.
p

type semiconductor
s
Group III
elements occupy a substitutional position in the Si lattice (B, Al, In, Ga)
5
Fig
.
2.4
Illustration of group

III substitutional impurity in Si.
Radius and ionization energy of the bound hole state can be calcul
ated similarly.
These impurities cn contribute holes by accepting electrons
–
acceptors.
p

type semiconductor: majority carriers are holes, minority carriers are electrons.
If a semiconductor is doped with both donors and acceptors, the extra electrons
attached
the donors fall into the incomplete bonds of the acceptors so that neither electrons nor
holes will be produced
–
compensation.
However, if
𝑁
>
𝑁
, we get n

type; if
𝑁
<
𝑁
, we get p

type.
Ambipolar semiconductors
: can be doped
to become both
n

type
and
p

type
. Elemental
and most of III

V compound semiconductors are ambipolar.
Unipolar semiconductors
: can be doped either n

t
ype or p

type, but not both. Many II

VI compounds are such.
The unipolar behavior is due to the mechanism of self

compensation.
Consider ZnTe doped with iodine (I in group

VII):
Intention: I atoms replace Te atoms to make it n

type
Practice: For I atoms
to replace Te atoms, temperature has to be raised, it will cause Zn
vacancies due to the evaporation of Zn atoms. Each Zn vacancy acts as a double acceptor.
6
∆
Zn
:
energy required to create a Zn vacancy
2
: energy released when two donor electrons drop into the Zn vacancy.
If
2
>
∆
Zn
,
the system energy is lowered, thus ZnTe will remain p

type.
2.3
Energy Band Model
The electronic states in
a crystal is obviously different from that in an isolated atom. But
they are also related because a crystal is formed by binding atoms in a regular order.
Consider we have many isolated atoms (well separated), there are no interactions
between the atoms.
A system of these atoms will have discrete energy levels and each
level is degenerate.
If we bring these atoms close to each other, the interaction between atoms will become
stronger. As a result, the wave functions of electrons in the outermost shell wi
ll begin to
overlap. The degeneracy of each energy level will be
removed. The initially discrete level
will now split into many energy levels. The separation of these split energy levels
depends on the distance between atoms. The stronger the wave function
s overlap, the
larger the energy splitting.
If we have
𝑁
identical atoms bound to form a crystal, each energy level will split into
𝑁
energy levels. Since
𝑁
is usually very large, the density of these energy levels is high. It
can be treated as they form a quasi

continuous energy band
–
allowed bands separa
ted by
forbidden band
–
bandgap.
Fig
.
2.5
Illustration of energy band formation.
7
The overlap of electron wave functions in inner shells is small, the interaction is weak,
therefore, the energy splitting can be neglected. The different properties of cryst
al and
constituent atoms are due to the different states of valence electrons. For example,
isolated neutral atoms (atom gas) do not conduct current. As they are bound together to
form a crystal, very different electronic properties can be determined (cond
uctors,
semiconductors, insulators).
Another example, the optical spectrum associated with the transitions between different
levels in an isolated atom is discrete. However, the spectrum in a solid is continuous.
When
𝑁
atoms form a crystal, one degener
ate level splits into
𝑁
energy levels which
allow totally
2
𝑁
electrons states according to Pauli principle that each energy level can
be occupied by two electrons with opposite spin.
For atoms with one valence electron (Na, sodium and K, pottasium), the
solids formed by
them have their energy band half filled, therefore, they are good conductors (metals). The
reason that
only partially filled bands conduct current
will be explained when we talk
about
the
energy band theory
.
For atoms with two valence el
ectrons (Mg), the outermost electrons are
s
2
. Intuitively,
one would think that the
2
𝑁
states will be completely filled therefore they are insulators.
But the fact is these bands are overlapping each other with higher energy bands. As a
result both bands are partially filled, which leads to a good conductivity (metals).
For C(diamond),
Si and Ge, the situation is more complicated. They all have 4 valence
electrons, s
2
p
2
. When they form a crystal, two energy bands should be produced. One
corresponds to s

state with
2
𝑁
states. The other p

states with
6
𝑁
states. It seems that the
6
𝑁
ba
nd should be partially filled, therefore diamond, Si, and Ge are all good conductors.
But they are not. Actually the orbit mixing between the s

state and p

state has led to a
new combination which result in two energy bands, each having
4
𝑁
states. The low
er
one, called
valence band
, is then completely filled, leaving the upper one called
conduction band
completely empty at low temperatures. Therefore, at low temperatures,
they act like insulators.
Electronic States in a Crystal
A detailed understanding o
f electronic states and the behavior of electrons in a crystal
requires calculations of quantum mechanics. The number of electrons involved in the
system is on the order of 10
23
. This is a complex many body problem, an exact solution
of such a system is im
possible.
The energy band theory
is actually based on the single

eletron approximation. This
approximation takes into account the electron interaction by adding them into the
periodic potential field of the atoms. As a result, electrons can be treated ind
ependently,
while the interactions with other electrons are included in the potential field as a fixed
charge distribution.
8
The Schrodinger equation becomes
−
ћ
2
2
∗
∇
2
+
𝒓
𝛷
𝒓
=
𝛷
𝒓
Where
∗
is the effective mass
which is different from the free electron mass due to the
interaction with other electrons
and
𝒓
is the periodic potential which includes the
electron interaction with lat
tice atoms and other electrons.
Bloch Functions
F. Bloch proved that the solutions of the Schrodinger equation for a periodic potential
must be of a special form
𝛷
𝒌
𝒓
=
𝒌
𝒓
𝒌
∙
𝒓
w
here
𝒌
𝒓
has the period of the crystal lattice
with
𝒌
𝒓
+
𝑻
=
𝒌
𝒓
This is a result of the Bloch theorem which states that the eigen functions of the wave
equation for a periodic potential are of the form of the product of a plane wave
𝒌
∙
𝒓
and a
function
𝒌
𝒓
with the perio
dicity of the crystal lattice. The proof of this theorem can be
found in Solid State Physics by Kittel.
Compared to free electron wave function
𝒌
∙
𝒓
, Bloch function indicates that the
probability of finding an electron in a lattice space is differ
ent from point to point within
one primitive unit cell, but the same between the corresponding points of different unit
cells.
In fact,
𝒌
𝒓
describes the behavior of an electron around a lattice atom.
𝒌
∙
𝒓
demonstrates that the electron
is
not localized, but can propa
gate throughout crystal.
For a crystal with infinitely large volume, the value of
𝒌
can vary continuously. Given a
𝒌
, there exists a set of eigen energies
(
𝒌
)
and corresponding
𝛷
𝒏
𝒌
,
𝒓
. The quantum
number
indicates
different energy band, intuitively can be considered as they are
originated from different atomic energy levels.
(
𝒌
)
is therefore a continuous function
of
𝒌
. Due to the periodicity of a crystal, an electronic does not have a unique val
ue of
𝒌
.
In fact,
𝒌
′
=
𝒌
+
𝑲
can represent the same state as
𝒌
does where
𝑲
is the so

called
reciprocal lattice vector
𝑲
=
1
1
+
2
2
+
3
3
Where
1
,
2
,
3
are integers and
1
,
2
,
3
are primitive vectors
of the reciprocal lattice
which can be constructed by the
primitive vectors of the crystal lattice (
1
,
2
,
3
)
1
=
2
𝜋
2
×
3
1
∙
2
×
3
,
2
=
2
𝜋
3
×
1
1
∙
2
×
3
,
3
=
2
𝜋
1
×
2
1
∙
2
×
3
It is easy to prove that
∙
=
2
𝜋
Actually, the Bloch wave function can be written as
𝛷
𝒏
𝒌
,
𝒓
=
,
𝒌
𝒓
𝒌
∙
𝒓
=
,
𝒌
𝒓
−
𝑲
∙
𝒓
𝒌
′
∙
𝒓
=
,
𝒌
′
𝒓
𝒌
′
∙
𝒓
where
,
𝒌
′
𝒓
=
,
𝒌
𝒓
−
𝑲
∙
𝒓
has the same periodicity
as the lattice since
9
−
𝑲
∙
𝒓
+
𝑻
=
−
𝑲
∙
𝒓
,
𝑲
∙
𝑻
=
2
𝜋
Thus,
𝒌
′
represents the same state as
𝒌
does.
Unlike the situation for free electrons, the nonuniqueness of
𝒌
sug
gests that strictly
speaking,
ћ
𝒌
will not carry the meaning of momentum. However, we will see later
ћ
𝒌
can still be treated as if it is the momentum of an electron in a crystal.
Obviously,
(
𝒌
)
various with
𝒌
periodically and for different n
,
(
𝒌
)
varies within
different energy intervals separated by energy gaps where electron states are forbidden.
Fig
.
2.6
(
𝒌
)
as a function of
𝒌
.
If we take the complex conjugate of the Schrodinger equation, H remains unchanged,
therefore
(
𝒌
)
should have even symmetry with respect to
𝒌
, i.e.
𝒌
=
(
−
𝒌
)
since
𝛷
𝒌
∗
𝒓
=
𝒌
∗
𝒓
−
𝒌
∙
𝒓
is the eigen function of
(
𝒌
)
also.
Brillouin Zone
Since
𝒌
′
=
𝒌
+
𝑲
represents the same electron state as
𝒌
does with
𝑲
being the
reciprocal lattice vector
. We can actually limit
𝒌
within a primitive unit cell in the
reciprocal lattice, because
𝒌
and
𝒌
′
point to the equivalent points within different
primitive unit cell in the reciprocal lattice. Now f
or an fixed electron state, there exists a
unique corresponding
𝒌
.
For example in 1

D, we can limit
–
𝜋
/
2
<
<
𝜋
/
.
A Brillouin zone is a special kind of primitive unit cell in a reciprocal lattice. It is defined
as a Wigner

Seitz cell in the reciprocal
lattice, which is constructed by drawing
10
perpendicular bisector planes in the reciprocal lattice from the chosen center to the
nearest equivalent reciprocal lattice sites.
The Wigner

Seitz cell is a primitive cell which maximally demonstrates the symmetr
y of
the crystal. For discussions on how to construct the Wigner

Seitz cell, one can refer to
Solid State Physics by Kittel.
State Density in

Space
Consider a crystal with a dimension of
1
×
2
×
3
with the dimension of a primitive
cell of
1
×
2
×
3
.
For any state
𝒌
=
1
+
2
+
3
, t
he periodic condition requires
that
𝛷
𝒌
0
,
,
=
𝛷
𝒌
1
,
,
,
𝛷
𝒌
,
0
,
=
𝛷
𝒌
,
2
,
,
𝛷
𝒌
,
,
0
=
𝛷
𝒌
,
,
3
Thus
1
1
=
1
2
𝜋
,
2
2
=
2
2
𝜋
,
3
3
=
3
2
𝜋
.
Since,
1
=
𝑁
1
1
,
2
=
𝑁
2
2
,
3
=
𝑁
3
3
, then
=
𝑁
2
𝜋
,
=
1
,
2
,
3
with
≤
𝑁
2
since
–
𝜋
/
2
<
<
𝜋
/
.
Therefore
,
∆
=
2
𝜋
, then one state in
𝒌

space
takes a volume of
∆
1
∆
2
∆
3
=
2
𝜋
1
2
𝜋
2
2
𝜋
3
=
2
𝜋
3
where
is the volume of crystal.
Ta
king into account that each state can accommodate two electrons with opposite spin
,
the density of states in
𝒌

space is
then
=
2
2
𝜋
3
,
and within the Brillouin zone there are totally
𝑁
=
𝑁
1
𝑁
2
𝑁
3
allowed
𝒌
states, which
equals th
e number of primitive cells.
It can be shown that if the volume of a primitive cell in lattice is
𝛺
, the volume of a
reciprocal primitive cell is
2
𝜋
3
/
𝛺
.
Motion of Electrons in a Crystal
The time

dependent solution to the Schrodinger equation is
𝛷
𝒌
𝒓
,
=
𝒌
𝒓
𝒌
∙
𝒓
−
w
here
=
/
ћ
. Due to the linearity of the time

dependent Schrodinger equation, the
superposition of the above wave function with different
𝒌
𝛷
𝒓
,
=
𝐶
𝒌
𝒓
𝒌
∙
𝒓
−
11
will still be a solution to the Schrodinger equation
.
Since
𝒌
’s are closely packed
in
𝒌

space and can be treated as continuous
, the superposition becomes an integral
𝛷
𝒓
,
=
𝐶
𝒌
𝜋
/
2
−
𝜋
/
2
𝒌
𝒓
𝒌
∙
𝒓
−
𝒌
3
𝒌
Now at s
ome instant of time, one tries to look at a wave packet that consists of Bloch
functions in the neighborhood of
𝒌
,
then the above integral can be rewritten as
𝛷
𝒓
,
=
𝐶
𝒌
+
𝜼
∆
𝒌
−
∆
𝒌
𝒌
+
𝜼
𝒓
𝒌
+
𝜼
∙
𝒓
−
𝒌
+
𝜼
3
𝜼
Since
𝜼
<
∆
𝒌
are small, then
𝒌
+
𝜼
𝒓
≈
𝒌
𝒓
and
𝒌
+
𝜼
=
𝒌
+
∇
𝒌
𝒌
∙
𝜼
+
⋯
which is related to energy by
=
/
ћ
.
Therefore
𝛷
𝒓
,
=
𝒌
𝒓
𝒌
∙
𝒓
−
𝒌
𝐶
𝒌
+
𝜼
∆
𝒌
−
∆
𝒌
𝜼
∙
𝒓
−
∇
𝒌
3
𝜼
The factor in f
ront of the integral is a Bloch function. The integral represents a wave
packet with a center position
𝒓
=
∇
𝒌
which moves with the velocity
𝒗
=
𝒓
=
∇
𝒌
𝒌
=
1
ћ
∇
𝒌
𝒌
which is the group velocity of electron in a crystal.
Now let
’s
consider a 1

D situation, with
𝐶
+
𝜂
=
𝐶
,
𝜂
<
∆
0
,
𝜂
>
∆
Then
𝛷
,
=
𝐶
∙
−
𝜂
∙
−
∇
𝜂
∆
−
∆
=
𝐶
∙
−
sin
∆
−
/
∆
−
/
2
∆
Hence, the wave packet probability
𝛷
,
2
∝
2
sin
∆
−
/
∆
−
/
2
12
Fig.
2.7
1

D electron distribution described by a wave packet with a wave vector
distribution within
+
∆
.
Th
e center peak is bound by
−
/
=
±
𝜋
/
∆
, i.e. the half width
∆
is related to
∆
by
∆
∆
=
𝜋

exactly what is required by the uncertainty principle. For a wave
packet with a space expansion of lattice constant
, we have
∆
=
𝜋
/
covers t
he entire
Brillouin zone, i.e.
is completely uncertain, and vice versa. However, when we are
dealing with the ranges of
𝒌
and
𝒓
that are much greater than the uncertainty
∆
𝒌
and
∆
𝒓
,
we can consider both
𝒌
and
𝒓
have precise values
–
quasi classic
.
Effective Mass
Suppose that an external field is applied, and the electron moves a distance
𝒓
, the work
done on the electron is
=
∙
𝒓
=
∙
𝒗
=
∙
∇
𝒌
=
∙
∇
𝒌
/
ћ
The work done
on
the electron causes its energy to change
𝒌
=
∇
𝒌
∙
𝒌
=
∇
𝒌
∙
𝒌
/
Since
=
, therefore
=
ћ
𝒌
/
=
𝒑
/
which is analogue to the Newton’s
law with
𝒑
=
ћ
𝒌
representing the momentum of the electron, similar to that of free
electron.
The acceleration of an electron
in a crystal is
𝒗
=
1
ћ
∇
𝒌
=
1
ћ
𝒌
∙
∇
𝒌
∇
𝒌
=
1
ћ
2
∙
∇
𝒌
∇
𝒌
In tensor form, we can write
𝒗
=
1
∗
where the inverse effective mass tensor
13
1
∗
=
1
ћ
2
2
/
2
2
/
2
/
2
/
2
/
2
2
/
2
/
2
/
2
/
2
is a function of
𝒌
. The tensor is symmet
ric,
we can choose a proper coordinate system so
that the tensor is diagonalized
. Therefore, we can have for each direction
𝛼
∗
𝛼
=
𝛼
,
𝛼
=
,
,
which
is analogue of Newton’s 2
nd
law.
Usually in semiconductors, we only deal w
ith those states that are close to band edges (
and
) because electrons are distributed a few
around the band edges.
For conduction band near
with
𝒌
=
0
, we can have
𝒌
=
0
+
1
2
2
2
2
+
2
2
2
+
2
2
2
+
⋯
=
0
+
ћ
2
2
2
,
∗
+
2
,
∗
+
2
,
∗
+
⋯
Since
0
is a minimum,
𝛼
=
0
, and
2
2
>
0
, therefore,
,
𝛼
∗
>
0
. Near the band
edge,
𝒌
has a parabolic relation with
𝒌
.
For crystals with cubic symmetry, we have
,
∗
=
,
∗
=
,
∗
=
∗
at
𝒌
=
0
. Then
𝒌
=
0
+
ћ
2
2
2
∗
a
nd
∗
𝒗
=
, the same expressions as free electrons except that we have to use the
effective mass
∗
to replace the free electron mass.
For valence band near the top
with
𝒌
=
0
, we can have
𝒌
=
0
+
1
2
2
2
2
+
2
2
2
+
2
2
2
+
⋯
=
0
+
ћ
2
2
2
,
∗
+
2
,
∗
+
2
,
∗
+
⋯
=
0
+
ћ
2
2
2
∗
+
⋯
for cubic crystal
. Since
0
is a maximum,
2
2
<
0
which will result in negative
electron effective mass
(
∗
<
0
)
near the top of the valence band
.
However, when we
look
at the Newton
’s
law expression of these electrons under an electric field
,
∗
𝒗
=
=
−
14
the holes are introduced as having positive charge, therefore, the above Newton’s law
should be modified
−
∗
𝒗
=
∗
𝒗
=
where the hole effective mass
∗
=
−
∗
is actually positive
at the top of the valence
band
.
The effective mass is directly related with the curvature of the energy band, and therefore
related with the energy band width.
Thus, near the t
op of the valence band we can write
𝒌
=
0
−
ћ
2
2
2
∗
Fig
.
2.8
Illustration of the band curvature with the effective mass
.
Energy Band Structures
The energy band structure is normally described as the relationship between
energy
and
𝒌
=
,
,
which is difficult to plot in 3

D. Typically, the relationship is
plotted along the principle directions of the crystal.
15
Fig
.
2.9
Band structure along the principle directions of the crystal
(Si)
.
Direct b
and
gap
: the minimum of the conduction band is located at the same
𝒌
as the
maximum of the valence band.
Indirect band
gap
: the minimum of the conduction band and the maximum of the
valence band are located at the different
𝒌
’s in
𝒌

space.
For direct
band gap semiconductor, a photon with energy
ћ
=
=
−
can excite
an electron from the top of the valence band to the bottom of the conduction band. But
for indirect band gap semiconductor, such a transition requires that a photon with a
n
energy
ћ
>
=
−
, because photons have very small momentum and the
absorption of a photon needs to be a vertical transition in
𝒌

space.
Electrons and holes are populated at the bottom of the conduction band and top of the
valence band
, respectively. For direct band gap semiconductors, the probability of
electrons and holes recombine to emit photons is much higher than that of indirect
semiconductors. Therefore, direct band gap semiconductors have important applications
in optical devic
es.
Indirect: Si and Ge
Direct: most III

V and II

VI compounds
Constant energy surface is plotted as an surface area in
𝒌

space where each point on the
surface has the same energy.
For a semiconductor with its conduction band minimum at
𝒌
=
0
and isotropic effective
mass, obviously the constant energy surface is spherical
,
16
=
+
ћ
2
2
2
∗
Fig
.
2.10
C
onstant energy surface for conduction band minimum at
𝒌
=
0
and isotropic
effective mass.
If the tensor of the effective mass is anisotropic, the constant energy surface is an
ellipsoid.
For Ge and Si, the minimum of conduction band is not at
𝒌
=
0
, the ce
nter of the
constant energy surface is not located at
𝒌
=
0
. Due to the crystal symmetry, there should
be more than one constant energy surfaces, e.g. for Si there are 6 constant energy surfaces
along 6 equivalent principle axis
100
, the centers of the 6
ellipsoids are located at about
¾
of the distance from the Brillouin zone center.
𝒌
=
+
ћ
2
2
−
0
2
∗
+
2
+
2
∗
w
here
∗
and
∗
are the longitudinal and transverse effectiv
e mass.
Fig
.
2.11
Constant energy surfaces in Si and Ge.
17
For Ge, the centers of constant energy ellipsoids are along each of the 8
100
directions.
The Brillouin zone boundary goes through the center of each
ellipsoid. There are 8 half
ellipsoids withi
n the 1
st
Brillouin zone, making 4 full ellipsoids within the 1
st
Brillouin
zone.
The above description of energy band structure is detailed and emphasize on the
−
𝒌
relationship. There are times when the detail description is not necessary when we are
only interested in the band gap
and band alignment in real space.
At 0K the valence band is completely filled, the conduction band has no electrons. Unde
r
this condition, the semiconductor behaves like an insulator.
This behavior is due to the fact that completely filled energy band as well as completely
empty band do not conduct electric current.
Fig
.
2.12
Simplified band
diagram
.
Every moving elect
ron contributes to electric current. But current is the total effect of all
electrons in a crystal. Within an energy band, if there is a state with momentum
𝒌
and
energy
𝒌
, there must be another state with momentum
−
𝒌
and
−
𝒌
=
𝒌
due to
the symmetry of the crystal. Obviously,
∇
𝒌
=
−
∇
−
𝒌
which implies that
electron at
𝒌
has same magnitude but opposite velocity as the electron at
−
𝒌
. If the band
is completely filled, then both states are occupied with electrons. The contribution to
current from the two electrons with
𝒌
and
−
𝒌
will be cancelled. Hence, the total current
is zero.
18
Fig
.
2.13
Illustration of the symmetry of
−
relationship
and the corresponding
electron velocity.
Under the condition of applied external field, the distribution of electrons within a
Brillouin zone is unchanged, even all electrons in

space move
according to
ћ
𝒌
=
−
This is because some electrons will flow out of the Brillouin zone on the right side, more
electrons will flow into
the Brillouin zone from the left side to fill up the empty states.
For partially filled energy band, it is easy to see why current is n
ot zero under an electric
field.
Fig
.
2.14
Occupied
−
states
in partially filled band
with zero and nonzero field.
Holes as empty states in valence band
The above analysis also provides an explanation for how an empty state in a valence band
act
s as a hole
–
a conductive carrier.
19
For completely filled band, the current density
=
−
𝒗
4
𝑁
=
1
=
0
where
is the volume since each band has
4
𝑁
states. Now assume that some state
in
the valence band is empty, we the
n can write the c
urrent density
=
=
−
𝒗
4
𝑁
=
1
,
≠
=
−
𝒗
4
𝑁
=
1
−
𝒗
=
𝒗
One empty state in the valence band acts as if it carries a positive charge
which
conducts current.
Energy levels of impurities
Usually impuri
ty levels lie in forbidden energy region. For intentionally doped donors
and acceptors, they are normally shallow levels in the range of
~
0
.
001
~
0
.
01
. At
room temperature, they are all ionized.
Fig
.
2.15
Impurity levels of donors and acceptors
in forbidden band.
But some impurities tend to give deeper levels in the forbidden band, e.g. Au in Si; O in
Ge. They serve as recombination centers.
Compensation
Fig
.
2.16
Semiconductors doped with both donors and acceptors.
20
Electrons from the level
instead of going to the conduction band will drop into the
acceptor states at
. Each such transition eliminates one electron

hole pair that would
have been there if the transition could be prevented. Obviously, if
𝑁
>
𝑁
, p

type; a
nd
if
𝑁
<
𝑁
, n

type.
2.4
Equilibrium Carrier Concentration
In order to calculate the electron and hole concentrations in the conduction and valence
bands, we need to know the density of states and probability of occupancy of these states.
Dens
ity of States
𝑁
As we have learned the density of states in
𝒌

space is
=
2
2
𝜋
3
, we need to convert this
into the density of states
in energy
𝑁
since the probability of occupancy described by
the Fermi function is given in energ
y.
Assume that the constant energy surface is spherical so that the effective mass is a scalar
=
+
ћ
2
2
2
∗
A constant energy surface of
−
, the radius of the spherical surface is
=
2
∗
−
/
ћ
. The vol
ume encircled by this energy surface
4
𝜋
3
𝜋
3
=
4
𝜋
3
𝜋
2
∗
−
3
/
2
ћ
3
The total number of states
2
2
𝜋
3
4
𝜋
3
𝜋
3
=
2
2
𝜋
3
4
𝜋
3
𝜋
2
∗
−
3
/
2
ћ
3
=
8
𝜋
3
2
∗
3
/
2
3
−
3
/
2
=
𝑁
−
0
The density of state in energy
𝑁
=
4
𝜋
2
∗
3
/
2
3
−
1
/
2
for conduction band
>
.
Similarly
𝑁
=
4
𝜋
2
∗
3
/
2
3
−
1
/
2
for valence band
<
.
21
Fermi

Dirac Distribution
The probability of occupancy of a state with energy
by an electron
=
1
1
+
exp
−
/
where
is the Fermi energy
Fig.
2.17
Fermi

Dirac distri
bution
At
=
0
,
=
0
,
>
1
,
<
,
in general,
=
>
1
2
,
<
=
1
2
,
=
<
1
2
,
>
In extreme cases when
−
≫
1
,
=
exp
−
−
/

Maxwell

Boltzmann
D
istribution
, and when
−
≪
1
,
=
1
.
Electron concentration
The number of electrons in an energy interval
dE
within the conduction band
=
𝑁
=
4
𝜋
2
∗
3
/
2
3
−
1
/
2
1
+
exp
−
−
1
Total electron concentration distributed within the conduction band
22
=
4
𝜋
2
∗
3
/
2
3
−
1
/
2
top
1
+
exp
−
−
1
This expression can be simplified if we assume that
lies below
by more than
3
so
≈
exp
−
−
/
,
an analytical expression can be obtained
=
𝑁
exp
−
−
/
where
𝑁
=
2
2
𝜋
∗
/
2
3
/
2
representing the density of states required at t
he conduction band edge
which gives the
concentration in the conduction band after multiplying with the probability of occupancy
at band edge.
In reality, the density of states at
is zero as given by
𝑁
∝
−
1
/
2
.
Hole
concentra
tion
The probability of a state not occupied by an electron
=
1
−
=
1
1
+
exp
−
/
which can be reviewed as the probability of a state occupied by a hole. Similarly, the hole
concentration
𝑝
=
4
𝜋
2
∗
3
/
2
3
−
1
/
2
bott
om
1
+
exp
−
−
1
Assuming that
lies above
by more than
3
so
≈
exp
−
−
/
,
an analytical expression can be obtained
𝑝
=
𝑁
exp
−
−
/
where
𝑁
=
2
2
𝜋
∗
/
2
3
/
2
representing the density of states required at the valence band edge
.
For semiconductors with anisotropic effective mass and multiple
equivalent energy
minima
in conduction band
, all expressions concerning
2
∗
3
/
2
should be modified as
8
∗
∗
∗
1
/
2
or
8
∗
∗
2
1
/
2
where
is the number of equivalent energy valleys
(minima).
We thus define a
density

of

states effective mass with
2
∗
3
/
2
=
8
∗
∗
2
1
/
2
such that
∗
=
2
∗
∗
2
1
/
3
.
For intrinsic semiconductors (no doping,
𝑁
=
0
and
𝑁
=
0
), we should have
=
𝑝
=
, and we let
=

the Fermi level of intrinsic semiconductor
, the
n
=
𝑁
exp
−
−
/
=
𝑁
exp
−
−
/
We can write in general
=
exp
−
/
,
𝑝
=
exp
−
/
23
The electron

hole concentration product
𝑝
=
2
=
𝑁
exp
−
−
/
𝑁
exp
−
−
/
=
𝑁
𝑁
exp
−
−
/
=
32
𝜋
2
∗
∗
4
3
/
2
3
exp
−
/
=
2
3
exp
−
/
b
oth
∗
and
∗
are density

of

states effective masses.
The above expressions for electron and hole concentrations are valid for nondegenerate
semiconductors because the carrier concentrations are low so that the Fermi

Dirac (F
D)
distribution function can be reduced to Maxwell

Boltzmann (MB) distribution.
For degenerate semiconductors, FD cannot be reduced to MB, the integral for
n
and
p
are
to be carried out numerically.
Temperature dependence of intrinsic carrier concentrati
on
=
3
/
2
exp
−
/
2
For most semiconductors, the band gap
decreases with the increase of temperature
,
=
0
−
accurate for
100
<
<
400
but
inaccurate for low temperatures,
where
0
is the extrapolated value of the band gap at
=
0
.
The intrinsic
concentration
=
1
3
/
2
exp
−
0
/
2
w
here
1
=
exp
/
2
.
Carrier concentrations in extrinsic semiconductors
Ionization of impurities
As donors in se
miconductors can lose one electron to become positively charged
e.g.
As
0
−
↔
As
+
Let
𝑁
: total donor concentration
𝑁
0
: neutral donor concentration
𝑁
+
: ionized donors
We obviously should have
𝑁
=
𝑁
0
+
𝑁
+
. The occu
pation probability of impurity level
is different from the regular FD distribution because of the spin degeneracy of the
donor levels. When a donor is ionized, there are two possible quantum states
corresponding to each of the two allowed spins. An
electron can occupy any one of these
states with the condition that as soon as one of them is occupied, the occupancy of the
other is prohibited.
Taking this into consideration, the FD distribution needs to be modified for impusities
=
𝑁
0
𝑁
=
1
+
1
2
exp
−
−
1
24
The concentration of ionized donors
𝑁
+
=
𝑁
−
𝑁
0
=
𝑁
1
+
1
2
exp
−
−
1
Similarly for acceptors with ionization process
Al
0
+
↔
Al
−
we should have
𝑁
=
𝑁
0
+
𝑁
−
, occupation probability of the acceptor level
=
𝑁
−
𝑁
=
1
+
1
2
exp
−
−
1
For n

type semiconductors,
−
is on the order of
at room temperature (
=
300
K
), and for no
ndegenerate semiconductors,
−
>
3
. The ratio of electrons
attached to the impurities to that in the conduction band
𝑁
0
=
𝑁
1
+
1
2
exp
−
−
1
𝑁
exp
−
−
Under normal condition
,
−
>
3
, then
exp
−
≫
1
and
𝑁
0
=
2
𝑁
𝑁
exp
−
~
𝑁
𝑁

on the same order, since
−
~
. For typical doping
𝑁
<
10
17
/
3
, most
impurities a
re ionized at room temperature with less than 1% remain unionized. We
should have
=
𝑁
.
Consider a nondegenerate semiconductor to which impurities have been introduced. To
keep the discussion in general, we assume the
re are donors (
𝑁
) and acce
ptors (
𝑁
).
Based on the condition of
charge
neutrality,
𝑁
+
+
𝑝
=
𝑁
−
+
which can
be
separated according to mobile and immobile charges as
−
𝑝
=
𝑁
+
−
𝑁
−
.
At
>
100
, we should have practically all impurities ionized,
𝑁
+
=
𝑁
and
𝑁
−
=
𝑁
,
thus
−
𝑝
=
𝑁
−
𝑁
.
Since
𝑝
=
2
, we arrive
=
1
2
𝑁
−
𝑁
+
𝑁
−
𝑁
2
+
4
2
𝑝
=
1
2
𝑁
−
𝑁
+
𝑁
−
𝑁
2
+
4
2
For intrinsic situation where
𝑁
=
𝑁
=
0
as well as for completely compensated doping
where
𝑁
=
𝑁
, they reduces to
=
𝑝
=
.
For net n

type doping where
𝑁
−
𝑁
≫
2
, we obtain
25
≈
𝑁
−
𝑁
≈
𝑁
if
𝑁
≫
𝑁
and
𝑝
=
2
/
𝑁
(case of
strongly extrinsic)
For net p

type doping where
𝑁
−
𝑁
≫
2
, we obtain
𝑝
≈
𝑁
−
𝑁
≈
𝑁
if
𝑁
≫
𝑁
and
=
2
/
𝑁
(case of strongly extrinsic)
2.5
The
Fermi level
and Energy Distribution of Carriers
I
ntrinsic
semiconductors
(
=
𝑝
)
, we have
𝑁
exp
−
−
=
𝑁
exp
−
−
Then
=
+
2
+
1
2
ln
𝑁
𝑁
=
+
2
+
3
4
ln
∗
∗
w
here
+
2
is the cent
er of the band gap. Obviously,
f
o
r
∗
=
∗
,
is at the midpoint
exactly;
f
or
∗
>
∗
,
is above the midpoint;
f
or
∗
<
∗
,
is below the midpoint.
As a good approximation,
3
4
ln
∗
∗
is small for most semiconductors,
is roughly at
the midgap
for intrinsic case.
Extrinsic semiconductors
n

type:
=
𝑁
+
, then
𝑁
exp
−
−
=
𝑁
1
+
2
exp
−
−
1
Introduce
𝜒
=
𝑁
2
𝑁
1
/
2
exp
−
2
where the ionization energy
=
−
. The
above equation can be
written as
1
1
+
2
exp
−
=
2
𝜒
2
exp
−
Then
=
+
ln
4
+
𝜒
2
−
𝜒
4
𝜒
And
=
𝑁
𝜒
2
4
+
𝜒
2
−
𝜒
Now let us discuss the above expressions under different temperatures corresponding to
:
(1)
Weak ionization
𝜒
≪
1
=
+
1
/
2
𝜒
(
above
)
and
=
𝜒𝑁
(few donors are ionized)
(2)
Complete ionization
𝜒
≫
1
26
=
+
1
/
2
𝜒
=
+
𝑁
/
𝑁
(
below
)
and
=
𝑁
(all donors are ionized)
Since
𝜒
is a function of temperature, given
𝑁
and
(type of donors), we can determine
a temperature wh
ich divides the situations of weak ionization and complete ionization
(transition temperature from weak to complete ionizations:
𝜒
=
1
)
. This temperature can
be solved by
𝜒
=
𝑁
2
𝑁
1
/
2
exp
−
2
=
1
And keep in mind that
𝑁
=
𝑁
.
It can be seen that as T increases,
𝜒
increases, and
moves from above
to below
.
In fact,
decreases approximately linearly with the increase of temperature.
It is obvious that as the temperature continues to in
crease,
approaches
. At this
point, the hole concentration approaches the electron concentration and cannot be
neglected any more
, we must use
=
𝑝
+
𝑁
With
𝑝
=
2
, then
=
1
2
𝑁
1
+
1
+
4
2
𝑁
2
1
/
2
(1)
/
𝑁
≪
1
,
=
𝑁
complete ionization
(2)
/
𝑁
≫
1
,
=
intrinsic
The transition temperature from complete ionization to intrinsic can be determined by
setting
=
𝑁
.
Temperature dependence of carrier concentration
in
n

type semiconductors
(1)
At low temperatures,
>
,
<
𝑁
,
𝑝
≪
, weak ionization
(2)
At moderate temperatures,
<
,
=
𝑁
,
𝑝
≪
, complete ionization
(3)
At high temperatures,
→
,
≈
≫
𝑁
,
𝑝
≈
, intrinsic
27
F
ig
.
2.18
Temperature dependence of carrier concentration
Compensation
Consider semiconductors doped with both donors and acceptors, the temperature
dependence of the carrier concentration is different from that of doped with either donors
or accepters on
ly at low temperatures.
Assume
𝑁
>
𝑁
, there will be
𝑁
electrons make a transition from
to
. The
remaining donors
𝑁
−
𝑁
can be excited into the conduction band.
The neutrality condition is then
+
𝑁
=
𝑝
+
𝑁
+
since all acceptors are ionized (occupied by electron)
𝑁
−
=
𝑁
. Since donors are
partially ionized at least
,
is in the neighborhood of
much closer to
than
, thus,
𝑝
≪
, or
+
𝑁
=
𝑁
+
=
𝑁
1
+
2
exp
−
But
exp
−
=
𝑁
exp
where
=
−
is the ionization energy. Rearrange
the above equation
+
𝑁
𝑁
−
𝑁
−
=
𝑁
2
exp
−
Let’s assume lig
ht compensation, i.e.
𝑁
≫
𝑁
, we can examine two cases
(1)
≪
𝑁
(extremely low temperature)
28
=
𝑁
𝑁
−
𝑁
2
𝑁
exp
−
Since
=
𝑁
exp
−
−
, we have
=
+
ln
𝑁
−
𝑁
2
𝑁
does not depend on the parameters of conduction band, only depends on
,
𝑁
,
𝑁
,
in
this case. This is because the electrons are mainly distributed between the donor and
acceptor levels. The temperature dependence of
is th
en
ln
=
𝑁
𝑁
−
𝑁
2
𝑁
−
1
(2)
≫
𝑁
(slightly higher temperature)
We still have
≪
𝑁
, then
=
𝑁
𝑁
2
1
/
2
exp
−
This is the same expression for electron concentration at low
temperature (weak
ionization) when the semiconductor was doped with donors only. This is because the
electrons are mainly distributed between the donor level and the conduction band, the
existence of small quantity of acceptors does not alter the distribu
tion
=
+
2
+
1
2
𝑁
2
𝑁
The temperature dependence of
is then
ln
=
𝑁
𝑁
2
1
/
2
−
2
1
Apparently, the slopes for
ln
~
1
are
different between the two cases: a factor of ½.
The
slope at low temperature is twice as much as that at slightly higher temperature. At the
transition temperature, we should have
=
𝑁
. Based on this behavior, one can tell
whether the semiconductor is compensated or not.
29
Fig
.
2.19
Temperature depen
dence of
n
for compensated semiconductors
.
As temperature increases beyond the weak ionization, we can have all donors ionized, but
not all electrons will be in the conduction band, actually,
=
𝑁
−
𝑁
(complete
ionization)
.
Further increase the
temperature, the semiconductor reaches the intrinsic
region where
=
𝑝
=
≫
𝑁
−
𝑁
.
The major difference between a compensated and
uncompensated semiconductor is the temperature dependence at the low temperature
region.
Energy Distribution
of Carriers
Both
=
𝑁
exp
−
−
and
𝑝
=
𝑁
exp
−
−
do not tell how electrons and holes
are distributed within the conduction and valence bands. If we need to know the
concentration as a function, we
need
𝑁
where
𝑁
is the density of states and
is the FD function.
Fig
.
2.20
Illustration of
𝑁
,
, and
𝑁
.
Comments 0
Log in to post a comment