Chapter 4: Semiconductor Physics
Crystal structures of solids
Energy band structures of solids
Charge carriers in semiconductors
Carrier transport
PHY4320 Chapter Four (II)
1
Equilibrium Distribution of Electrons and Holes
The densities of electrons and holes are related to the density-of-states
function and the Fermi distribution function.
n(E ) = g c ( E ) f F (E )
p(E ) = g v (E )[1 − f F (E )]
The total electron/hole concentration per unit volume is found by integrating
the corresponding function over the entire conduction/valence band energy.
We need to determine the Fermi energy in order to find the thermalequilibrium electron and hole concentrations.
We first consider an intrinsic semiconductor. An ideal intrinsic
semiconductor is pure semiconductor without impurities or lattice defects. At
T = 0 K, all the energy states in the valence band are filled with electrons and
all the energy states in the conduction band are empty. The Fermi energy
must be somewhere between Ec and Ev. At T > 0 K, the number of electrons
in the conduction band is equal to that of holes in the valence band.
PHY4320 Chapter Four (II)
2
Equilibrium Distribution of Electrons and Holes
(
4π 2m
g c (E ) =
3
h
g v (E ) =
)
* 3/ 2
n
(
4π 2m
h
3
)
* 3/ 2
p
E −E c
Ev − E
If mn* = mp*, the Fermi energy is at
the midgap energy. If mn* ≠ mp*, the
Fermi level for an intrinsic
semiconductor will slightly shift
away from the midgap energy.
PHY4320 Chapter Four (II)
3
The n0 and p0 Equations
n0 = ∫ g c (E ) f F (E )dE
Because the Fermi distribution rapidly approaches zero with increasing
energy, therefore
∞
n0 = ∫ g c (E ) f F (E )dE
Ec
If (Ec−EF) >> kT, then (E−EF) >> kT, so that
f F (E ) =
n0 = ∫
∞
Ec
If we let
(
[− (E − EF )]
1
≅ exp
(
E − EF )
kT
1 + exp
kT
4π 2m
h3
)
⎡ − (E − E F ) ⎤
E − Ec exp ⎢
dE
⎥
kT
⎣
⎦
E − Ec
η=
kT
* 3/ 2
n
PHY4320 Chapter Four (II)
4
Then
(
4π 2m kT
n0 =
h
*
n
3
)
3/ 2
⎡ − ( Ec − E F ) ⎤ ∞ 1 / 2
exp ⎢
η exp(− η )dη
∫
⎥
kT
⎣
⎦0
The integral is a gamma function.
∫
∞
0
So
η
1/ 2
1
π
exp(− η )dη =
2
⎛ 2πm kT ⎞
⎟⎟
n0 = 2⎜⎜
⎠
⎝ h
*
n
2
We define a parameter Nc as
3/ 2
⎡ − ( Ec − E F ) ⎤
exp ⎢
⎥
kT
⎣
⎦
⎛ 2πm kT ⎞
⎟⎟
N c = 2⎜⎜
⎠
⎝ h
Then
*
n
2
3/ 2
Nc is called the effective
density-of-states function
in the conduction band.
⎡ − ( Ec − E F ) ⎤
n0 = N c exp ⎢
⎥
kT
⎣
⎦
PHY4320 Chapter Four (II)
5
Nc is called the effective density-of-states function in the conduction
band. If we were to assume that mn* = m0, then at T = 300 K,
(
)(
⎡ 2π 9.109 ×10 1.381×10
N c = 2⎢
−34 2
6.626 ×10
⎢⎣
For holes:
−31
(
)
− 23
)(300)⎤⎥
⎥⎦
3/ 2
= 2.5 ×1019 cm −3
p0 = ∫ g v (E )[1 − f F (E )]dE
⎛ E − EF ⎞
exp⎜
⎟
1
1
kT ⎠
⎝
1 − f F (E ) = 1 −
=
=
⎛ EF − E ⎞
⎛ E − EF ⎞
⎛ E − EF ⎞
1 + exp⎜
⎟
⎟ 1 + exp⎜
⎟ 1 + exp⎜
⎝ kT ⎠
⎝ kT ⎠
⎝ kT ⎠
If (EF−Ev) >> kT, then (EF−E) >> kT.
⎡ − (E F − E ) ⎤
1 − f F (E ) ≈ exp ⎢
⎥⎦
kT
⎣
PHY4320 Chapter Four (II)
6
p0 = ∫
(
4π 2m
Ev
h3
−∞
If we let
Then
p0 =
(
− 4π 2m kT
h
*
p
3
)
3/ 2
)
* 3/ 2
p
⎡ − (E F − E ) ⎤
Ev − E exp ⎢
dE
⎥
kT
⎦
⎣
Ev − E
η' =
kT
⎡ − ( E F − Ev ) ⎤ 0
1/ 2
(
exp ⎢
η' ) exp(− η' )dη'
∫
⎥
kT
⎣
⎦ +∞
⎛ 2πm kT ⎞
⎟
p0 = 2⎜
⎜ h
⎟
⎝
⎠
*
p
2
If we let
⎛ 2πm kT ⎞
⎟
N v = 2⎜
⎜ h
⎟
⎝
⎠
*
p
2
3/ 2
3/ 2
⎡ − ( E F − Ev ) ⎤
exp ⎢
⎥
kT
⎣
⎦
then
⎡ − ( E F − Ev ) ⎤
p0 = N v exp ⎢
⎥
kT
⎣
⎦
Nv is called the effective density-of-states function in the valence band.
PHY4320 Chapter Four (II)
7
The Intrinsic Carrier Concentration
We use ni and pi to denote electron and hole concentrations in an intrinsic
semiconductor. Since ni = pi, we usually use ni to denote either the intrinsic
electron or hole concentration. The Fermi level for an intrinsic semiconductor
is called the intrinsic Fermi energy, EFi.
⎡ − (Ec − E Fi ) ⎤
n0 = ni = N c exp ⎢
⎥
kT
⎣
⎦
⎡ − (EFi − Ev )⎤
p0 = pi = ni = N vexp ⎢
⎥
kT
⎣
⎦
⎡ − (EFi − Ev ) ⎤
⎡ − (Ec − EFi ) ⎤
n = N c N v exp ⎢
exp ⎢
⎥
⎥
kT
kT
⎣
⎦
⎣
⎦
2
i
⎡ − Eg ⎤
⎡ − ( Ec − E v ) ⎤
n = N c N v exp ⎢
= N c N v exp ⎢
⎥
⎥
kT
kT
⎣
⎦
⎣
⎦
2
i
PHY4320 Chapter Four (II)
8
At T = 300 K
Nc, Nv, and ni are constant for a given semiconductor material at
a fixed temperature.
PHY4320 Chapter Four (II)
9
⎛ 2πm kT ⎞
⎟⎟
N c = 2⎜⎜
⎝ h
⎠
*
n
2
3/ 2
⎛ 2πm kT ⎞
⎟
N v = 2⎜
⎜ h
⎟
⎝
⎠
*
p
2
3/ 2
⎡ − Eg ⎤
n = N c N v exp ⎢
⎥
⎣ kT ⎦
2
i
PHY4320 Chapter Four (II)
10
The Intrinsic Fermi Level Position
We can calculate the Fermi level position since the electron and hole
concentrations are equal for an intrinsic semiconductor:
⎡ − (EFi − Ev ) ⎤
⎡ − (Ec − EFi ) ⎤
n0 = N c exp ⎢
= p0 = N v exp ⎢
⎥
⎥
kT
kT
⎦
⎣
⎣
⎦
⎛ Nv ⎞
⎟⎟
− Ec + E Fi + EFi − Ev = kT ln⎜⎜
⎝ N c ⎠ If m * = m *, the intrinsic Fermi
p
n
level will be in the center of the
⎛ Nv ⎞
1
1
⎟⎟
bandgap. If mp* > mn*, the
EFi = (Ec + Ev ) + kT ln⎜⎜
2
2
⎝ Nc ⎠
intrinsic Fermi level will be
*
⎛ mp ⎞
*
3
slightly
above
the
center.
If
m
p
EFi = Emidgap + kT ln⎜ * ⎟
*, the intrinsic Fermi level
⎜m ⎟
<
m
4
n
n
⎠
⎝
*
will be slightly below the center
⎛ mp ⎞
3
EFi − Emidgap = kT ln⎜ * ⎟
of the bandgap.
⎜m ⎟
4
⎝ n⎠
PHY4320 Chapter Four (II)
11
Donor Atoms and Energy Levels
Real power of semiconductors is realized by adding controlled amounts of
specific dopant, or impurity atoms. The doped semiconductor is called an
extrinsic material. Doping is the primary reason that we can fabricate various
semiconductor devices.
Add a group V element, such as phosphorus, to silicon as a substitutional
impurity. The group V element has five valence electrons. Four of these will
contribute to the covalent bonding with the silicon atoms, leaving the fifth
more loosely bound to the phosphorus atom.
PHY4320 Chapter Four (II)
12
Donor Atoms and Energy Levels
At very low temperatures, the extra electron
is bound to the phosphorus atom. However, it
should be clear that the energy required to
elevate the extra electron into the conduction
band is considerably smaller than that for the
electrons involved in the covalent bonding.
The electron elevated into the conduction
band can move through the crystal to
generate a current, while the positively
charged phosphorous atoms are fixed in the
crystal. This type of atom is called a donor
impurity atom. The donor atoms add
electrons to the conduction band without
creating holes in the valence band. The
resulting material is referred to as an n-type
semiconductor.
PHY4320 Chapter Four (II)
13
Acceptor Atoms and Energy Levels
Consider adding a group III element, such as boron, which has three valence
electrons. One covalent bonding position is empty. If an electron were to
occupy this “empty” position, its energy would have to be greater than that of
the valence electrons, since the net charge of the B atom would become
negative. However, the electron occupying this “empty” position does not
have sufficient energy to be in the conduction band, so its energy is far
smaller than the conduction band energy. The “empty” position associated
with the B atom can be occupied and other valence electron positions become
vacated. These other vacated electron positions can be thought of as holes.
PHY4320 Chapter Four (II)
14
Acceptor Atoms and Energy Levels
The hole can move through the crystal to
generate a current, while the negatively
charged boron atoms are fixed in the crystal.
The group III atom accepts an electron from
the valence band and so is referred to as an
acceptor impurity atom. The acceptor atom
can generate holes in the valence band
without generating electrons in the
conduction band. This type of semiconductor
material is referred to as a p-type
semiconductor.
An extrinsic semiconductor will have either
a preponderance of electrons (n-type) or a
preponderance of holes (p-type).
PHY4320 Chapter Four (II)
15
Ionization Energy
We can calculate the approximate distance of the donor electron from the
donor impurity ion, and also the approximate energy required to elevate the
donor electron into the conduction band. This energy is referred to as the
ionization energy. We will use the Bohr model of the atom for these
calculations.
The permittivity of the semiconductor material instead of the permittivity of
free space will be used, and the effective mass of the electron will be used.
From the Coulomb force of attraction being equal to the centripetal
force of the orbiting electron:
e
2
* 2
mv
=
2
4πεrn
rn
Assume that the angular momentum is quantized,
PHY4320 Chapter Four (II)
m rn v = nh
*
16
nh
v= *
m rn
The Bohr radius is
n = 1,2,3,…
n 2 h 2 4πε
rn =
m *e 2
4πε 0 h 2
a0 =
= 0.053nm
2
m0 e
rn
⎛ m0 ⎞
2
= n εr ⎜ * ⎟
a0
⎝m ⎠
Then
If we consider n = 1 state, and if we consider silicon, for which εr =
11.7, m*/m0 = 0.26, then we have
r1
= 45
a0
r1 = 2.39nm
PHY4320 Chapter Four (II)
For silicon, a = 0.543 nm.
17
e2
v=
4πεnh
* 4
1
m
e
* 2
The kinetic energy is
T= mv =
2
2
2
2(nh ) (4πε )
2
* 4
−
−
e
m
e
The potential energy is
=
V=
4πεrn (nh )2 (4πε )2
− m *e 4
The total energy is
E = T +V =
2
2
2(nh ) (4πε )
The lowest energy state of the
hydrogen atom is E = −13.6
eV. For silicon, the lowest
energy is E = −25.8 meV (Eg =
1.12 eV at T = 300 K).
PHY4320 Chapter Four (II)
18
Group III-V Semiconductors
The donor and acceptor impurities in III-V compound semiconductors is
more complicated than that in Si. When we talk about donors or acceptors in
III-V semiconductors, we need to know for which atoms (III or V) impurity
atoms are substituted. For example, for Si atoms in gallium arsenide
semiconductor, if Si atoms replace gallium atoms, Si impurities will act as
donors. But if Si atoms replace arsenic atoms, they will act as acceptors.
For gallium arsenide
PHY4320 Chapter Four (II)
19
Equilibrium Distribution of Electrons and Holes in
Extrinsic Semiconductors
Adding donor or acceptor impurity atoms
to a semiconductor will change the
distribution of electrons and holes in the
material. Since the Fermi energy is
related to the distribution function, the
Fermi energy will change as dopant
atoms are added.
In general, when EF > Emidgap, the density
of electrons is larger than that of holes,
and the semiconductor is n-type.
PHY4320 Chapter Four (II)
20
Equilibrium Distribution of Electrons and Holes in
the Extrinsic Semiconductor
In general, when EF < Emidgap, the density
of electrons is smaller than that of holes,
and the semiconductor is p-type.
⎡ − ( Ec − E F ) ⎤
n0 = N c exp ⎢
⎥
kT
⎣
⎦
⎡ − ( E F − Ev ) ⎤
p0 = N v exp ⎢
⎥
kT
⎣
⎦
The above are general equations for n0
and p0 in terms of the Fermi energy.
The values of n0 and p0 will change with
the Fermi energy, EF.
PHY4320 Chapter Four (II)
21
Example: consider silicon at T = 300 K so that Nc = 2.8 × 1019 cm−3 and Nv =
1.04 × 1019 cm−3. If we assume the Fermi energy is 0.25 eV below the
conduction band, calculate the thermal equilibrium concentrations of
electrons and holes. The bandgap energy of silicon is 1.12 eV.
Solution: Ec − E F = 0.25eV
E F − Ev = 0.87eV
(
)
⎡ − (0.25) 1.602 ×10 −19 ⎤
15
−3
1
.
8
10
cm
=
×
n0 = 2.8 × 10 exp ⎢
⎥
− 23
(
)
1
.
381
10
300
×
⎦
⎣
⎡ − (0.87 ) 1.602 ×10 −19 ⎤
4
−3
19
=
2
.
6
×
10
cm
p0 = 1.04 × 10 exp ⎢
⎥
− 23
(
)
1
.
381
10
300
×
⎣
⎦
(
(
19
)
)
(
(
(
)
)
)
Comment: electron and hole concentrations change by orders of magnitude
from the intrinsic carrier concentrations (at 300 K, ni = 1.5 × 1010 cm-3) as
the Fermi energy changes by a few tenths of an eV.
In an n-type semiconductor, n0 > p0, electrons are referred to as majority
carriers and holes as minority carriers. In an p-type semiconductor, p0 >
n0, holes are referred to as majority carriers and electrons as minority
carriers.
PHY4320 Chapter Four (II)
22
We can derive another form of the equations for the thermalequilibrium concentrations of electrons and holes:
⎡ − ( Ec − E F ) ⎤
⎡ − (Ec − EFi ) + (EF − E Fi ) ⎤
n0 = N c exp ⎢
= N c exp ⎢
⎥
⎥
kT
kT
⎣
⎦
⎣
⎦
⎛ EF − EFi ⎞
n0 = ni exp⎜
⎟
⎝ kT ⎠
⎡ − (EFi − Ev ) + (EFi − E F ) ⎤
⎡ − ( E F − Ev ) ⎤
p0 = N v exp ⎢
= N v exp ⎢
⎥
⎥
kT
kT
⎣
⎦
⎣
⎦
⎡ EFi − EF ⎤
⎡ − (EF − E Fi ) ⎤
p0 = ni exp ⎢
= ni exp ⎢
⎥
⎥
kT
⎣ kT ⎦
⎣
⎦
PHY4320 Chapter Four (II)
23
The n0p0 Product
⎡ − ( E F − Ev ) ⎤
⎡ − ( Ec − E F ) ⎤
exp ⎢
n0 p0 = N c N v exp ⎢
⎥
⎥
kT
kT
⎦
⎣
⎦
⎣
⎡ − Eg ⎤
2
n0 p0 = N c N v exp ⎢
⎥ = ni
⎣ kT ⎦
The product of n0 and p0 is always a constant for a given semiconductor
material at a given temperature. It is one of the fundamental principles of
semiconductors in thermal equilibrium.
It is important to keep in mind that the above equation is derived using the
Boltzmann approximation.
We may think of the intrinsic concentration ni simply as a parameter of the
semiconductor material.
PHY4320 Chapter Four (II)
24
The Fermi-Dirac Integral
∞
n0 = ∫ g c (E ) f F (E )dE
Ec
4π
n0 = 3 2mn*
h
(
)
3/ 2
E − Ec
η=
kT
If we define
E − Ec dE
∫Ec
⎛ E − EF ⎞
1 + exp⎜
⎟
⎝ kT ⎠
EF − E c
η
=
and
F
kT
∞
⎛ 2m kT ⎞
⎟⎟
n0 = 4π ⎜⎜
⎝ h ⎠
*
n
2
F1/ 2 (η F ) = ∫
∞
0
3/ 2
∫
∞
0
η 1/ 2 dη
1 + exp(η − η F )
η 1 / 2 dη
1 + exp(η − η F )
PHY4320 Chapter Four (II)
Fermi-Dirac integral
25
For holes:
p0 = ∫ g v (E )[1 − f F (E )]dE
Ev
−∞
4π
p0 = 3 2m*p
h
(
)
3/ 2
Ev − E dE
∫−∞
⎛ EF − E ⎞
1 + exp⎜
⎟
⎝ kT ⎠
Ev
If we define
Ev − E
η' =
kT
We need to use the FermiDirac integral when EF is
above Ec or below Ev.
⎛ 2m kT ⎞
⎟
p0 = 4π ⎜
⎟
⎜ h
⎠
⎝
PHY4320 Chapter Four (II)
*
p
2
and η' F =
3/ 2
Ev − E F
kT
(η' )1/ 2 dη'
∫0 1 + exp(η' −η' F )
∞
26
Degenerate and Nondegenerate Semiconductors
When discussing donors and acceptors, the concentration of dopant atoms is
assumed to be small compared to the density of host atoms. The impurities
introduce discrete, non-interacting donor and acceptor energy states in the ntype and p-type semiconductor, respectively. These types of semiconductors
are referred to as nondegenerate semiconductors.
As the impurity concentration increases, the distance between the impurity
atoms decreases and the electrons from the impurity atoms will begin to
interact. When this occurs, the discrete donor or acceptor energy level will
split into a band of energies. As the impurity concentration further increases,
the band of donor or acceptor states widens and overlaps with the bottom of
the conduction band or the top of the valence band. When the concentration
of electrons in the conduction band exceeds the density of states Nc, the Fermi
energy lies within the conduction band. This type of semiconductors is called
degenerate n-type semiconductors. When the concentration of holes
exceeds the density of states Nv, the Fermi energy lies in the valence band.
This type of semiconductors is called degenerate p-type semiconductors.
PHY4320 Chapter Four (II)
27
Degenerate and Nondegenerate Semiconductors
Nondegenerate
Degenerate
PHY4320 Chapter Four (II)
28
Statistics of Donors and Acceptors
The probability of the donor energy level being occupied is
1
1
⎛ Ed − E F ⎞
1 + exp⎜
⎟
2
⎝ kT ⎠
Each donor level could be empty, contain one electron of
either spin, or two electrons of opposite spins. However, the
Coulomb repulsion of two localized electrons raises the
energy of the doubly occupied level so high that double
occupation is essentially prohibited. This is the reason for
the factor of ½ appearing in the probability function.
Nd
nd =
1
⎛ Ed − E F ⎞
1 + exp⎜
⎟
2
⎝ kT ⎠
nd = N d − N d+
Na
pa =
1
⎛ E − Ea ⎞
1 + exp⎜ F
⎟
g
⎝ kT ⎠
pa = N a − N a−
nd is the density of electrons occupying the
donor level.
Nd is the concentration of donor atoms.
Nd+ is the concentration of ionized donors.
g is a degeneracy factor. The ground state g is
normally taken as 4 for the acceptor level in Si and
GaAs because of the detailed band structure.
PHY4320 Chapter Four (II)
29
Complete Ionization and Freeze-Out
If (Ed − EF) >> kT, we then have
nd =
Because
Nd
⎡ − (Ed − E F ) ⎤
≈ 2 N d exp ⎢
⎥
1
kT
⎛ Ed − E F ⎞
⎦
⎣
1 + exp⎜
⎟
2
⎝ kT ⎠
⎡ − ( Ec − E F ) ⎤
n0 = N c exp ⎢
⎥
kT
⎦
⎣
We can determine the
⎡ − (Ed − E F ) ⎤
percentage of electrons
2 N d exp ⎢
⎥⎦
n
kT
⎣
d
in the donor state
=
nd + n0
⎡ − (Ed − E F ) ⎤
⎡ − ( Ec − E F ) ⎤
compared with the
N
+
2 N d exp ⎢
exp
c
⎥⎦
⎢⎣
⎥⎦
kT
kT
⎣
total number of
electrons:
nd
=
nd + n0
1
Nc
⎡ − ( Ec − E d ) ⎤
1+
exp ⎢
⎥⎦
kT
2Nd
⎣
PHY4320 Chapter Four (II)
30
For phosphorus-doped silicon at T = 300 K, Nc = 2.8 × 1019 cm-3, Nd = 1016
cm-3, and the ionization energy is 0.045 eV:
nd
=
nd + n0
1
2.8 ×10
⎛ − 0.045 ⎞
exp⎜
1+
⎟
16
2 10
⎝ 0.0259 ⎠
19
= 0.004 = 0.4%
( )
At room temperature, the donor states are almost completely ionized,
which is also true for the acceptor states at room temperature.
Complete ionization
PHY4320 Chapter Four (II)
31
At T = 0 K, all electrons are in their lowest energy state. For an n-type
semiconductor, each donor state must contain an electron, therefore, nd = Nd.
⎧exp[(Ed − EF ) / kT ] = 0
⇒ E F > Ed
⎨
⎩T = 0
The Fermi level is above the donor level for an n-type semiconductor at
T = 0 K. Similarly, the Fermi level will be below the acceptor level for a
p-type semiconductor at T = 0 K.
Freeze-out (T = 0 K)
PHY4320 Chapter Four (II)
32