In solid materials, the behavior of charges depends on their energy.
To understand the electronic characteristics of a material, we need to
know how charges are distributed among the available energy levels.
This distribution is described by two components:
- The Fermi function gives the probability that an electron acquires a given energy, as a function
of temperature.
- The Density of States gives the distribution of energy levels that electrons are allowed to have.
The Fermi function is completely general, and applies to any solid material
in thermal equilibrium.
The Density of States is specific to a particular type of material.
By combining these functions, we can determine the total density of charges
that are available for electrical activity at a given temperature for a
particular piece of material.
The Fermi function.
The
Fermi function ,
FEFE
, gives the probability that a
state
SS
at energy
EE
is
occupied by an electron, given that
EE
is an allowed energy level.
The Fermi function has units of
electrons per state.
The
Fermi energy level ,
E F E F
, is the energy at which the probability of occupancy is exactly
1/21/2
for temperatures greater than zero.
The Fermi function is given by
FE=1 e E-E F /kT +1,FE=1 e E-E F /kT +1,
(1)
where
kk
is the Boltzmann constant and
TT
is the temperature in Kelvin.
Throughout nature, particles seek to occupy the lowest energy state possible.
Therefore electrons in a solid will tend to fill the lowest energy states
first.
Electrons fill up the available states like water filling a bucket, from
the bottom up.
At
T=0T=0
, every low-energy state is occupied, right up to the Fermi level, but no
states are filled at energies greater than
E F E F
.
For
T>0T>0
, some electrons can be excited into higher-energy states.
This is similar to a bucket of hot water.
Most of the water molecules stick around the bottom of the bucket.
The Fermi level is like the water line.
A fraction of water molecules are excited and drift above the water line
as vapor , just as electrons can sometimes drift above the Fermi level.
Recall that an energy level may contain several sublevels, all with the
same energy.
Each sublevel is called a "state," and can be occupied by exactly one electron.
Suppose there are
NN
allowed states at energy
EE
.
Then the probability of finding an occupied state at energy
EE
is
N×FEN×FE
.
As discussed in the next subsection, in continuous-band theory we represent
NN
as a density of states.
The density of states reveals how the allowed sublevels are spread out
across energy bands in a specific material.
Density of states.
In a semiconductor, not every energy level is allowed.
For example, there are no allowed states within the forbidden gap, as illustrat
ed in
Figure 3.
To make use of the Fermi function, we need another function that has units
of
states per energy level per volume.
In a solid with numerous atoms, a large number of states appear at energy
levels very close to each other.
We approximate these states as a continuous "band" and imagine that an "energy level" is a vanishingly small energy interval of width
dEdE
.
The
density of states ,
NENE
, is the fraction of all allowed states that lie within
EE
and
E+dEE+dE
.
This is a density function, meaning
∫ 0 ∞ NE=1∫ 0 ∞ NE=1
.
The Fermi function tells us the probability that a state is occupied.
The density of states complements the Fermi function by telling us how
many states actually exist in a particular material.
We can multiply
NENE
and
FEFE
together, resulting in units of electrons per energy level per unit volume:
e - s t a t e × s t a t e s l e v e l × v o l u m e =e - l e v e l × v o l u m e .e - s t a t e × s t a t e s l e v e l × v o l u m e =e - l e v e l × v o l u m e .
(2)
By integrating over all energy levels, we obtain the total number of electrons.
By integrating over the
conduction band only , we obtain the total number of
mobile electrons (i.e.
electrons that can participate in electric current).
Let
nn
be the total concentration (per volume) of mobile carriers in the conduction
band.
Then
nn
is given by
n=∫ E C E 0 FENEdE,n=∫ E C E 0 FENEdE,
(3)
where
E 0 E 0
is the top of the conduction band, called the vacuum level.
An electron with energy greater than
E 0 E 0
is no longer confined to the solid, and can fly off into space.
Under normal circumstances,
E 0 →∞E 0 →∞
is a sufficient approximation when integrating carrier densities.
In the conduction band, the density of available states has the form
NE=2 πN C E-E C kT 3/2 ,NE=2 πN C E-E C kT 3/2 ,
(4)
where
N C =η2πm n kT/h 2 . 3/2 N C =η2πm n kT/h 2 . 3/2
(5)
For Si,
η=12η=12
and for GaAs,
η=2η=2
.
Electrons vs holes.
In a solid semiconductor at thermal equilibrium, every mobile electron leaves
behind a hole in the valence band.
Since holes are also mobile, we need to account for the density of "hole states" that remain in the valence band.
Because a hole is an unoccupied state, the probability of a mobile hole
existing at energy
EE
is
1-FE1-FE
, as indicated in
Figure 4.
Because holes can move, we should count them when considering the total
concentration of mobile charges.
Let
pp
be the total concentration of mobile charges in the valence band.
Then
p=∫ 0 E V 1-FENEdE.p=∫ 0 E V 1-FENEdE.
(6)
In the conduction band, the density of states for holes is
NE=2 πN V E V -E kT 3/2 ,NE=2 πN V E V -E kT 3/2 ,
(7)
where
N V =22πm p kT/h 2 . 3/2 N V =22πm p kT/h 2 . 3/2
(8)
Mobile carrier density.
Combining the results for
nn
and
pp
, we can determine the total density of mobile carriers in a semiconductor
at equilibrium.
In the conduction band, where
E-E F >3kTE-E F >3kT
, we use the approximation
FE≈e -E-E F /kT .FE≈e -E-E F /kT .
(9)
With a change of variables
x=E-E C /kTx=E-E C /kT
, the integral in
Equation 3 takes the form
n=2 πN C exp-E C -E F /kT∫ 0 ∞ x 1/2 e -x dx.n=2 πN C exp-E C -E F /kT∫ 0 ∞ x 1/2 e -x dx.
(10)
This integral is in standard form and can be found on many integral tables.
The integral evaluates to
π/2π/2
, yielding
n=N C exp-E C -E F /kT.n=N C exp-E C -E F /kT.
(11)
For holes, we use the valence-band approximation
1-FE≈e -E F -E/kT .1-FE≈e -E F -E/kT .
(12)
Integrating
Equation 6 with
Equation 7 and
Equation 12 yields
p=N V exp-E F -E V /kT.p=N V exp-E F -E V /kT.
(13)