8 Calculation of Density of States (DOS)
3.3.1 DOS in 3D
3.3.2 DOS in 2D
3.3.3 DOS in 1D
3.3.4 DOS in 0D
CALCULATION OF DENSITY OF STATES (DOS): Quantum Wells, Wires and Dots
3.3.1 DOS for 3 dimensions (Bulk)
Consider the volume in “k” space
where for a particle in this space
Figure . Electron state is defined by a point in k-space.
Note that the 2 arises from the constraints of periodic boundary conditions as proposed to the more general where n=0, 1, 2, 3… The volume of a given mode is then= . The number of modes (N) in the sphere is,
Say the particle in an electron and we consider spin (up and down), then we multiply N by 2.
is the total number of states in sphere. Now consider the density
Figure . Density of states in 3 dimension (Eq.3.48)
3.3.2 DOS in Two Dimensions (well)
Here we have 1D that is quantized. Let’s us assume it is the z-direction. The total energy of this system is a sum of the energy along the quantized direction plus the energy along the other 2 free directions. It is expressed as
where for the particle
The area of a given mode is then ???? with the total number of modes (N) in the area being
Again if the particle is an electron and we consider spin, multiply by 2 to get
This is the energy density of the sub-band for a given or (En). For each successive there will be an additional ћ2 and hence another subband. Therefore the density of the states is written
Where Ѳ is the heavy side function.
Figure Density of states in 2 dimension. Shaded area presents occupied states.
3.3.3 DOS in One Dimensions (Wire)
Consider now the situation where there are two dimensions confined and only 1 degree of freedom (say the x-direction). The total energy of the system can be written as
Figure Density of states in 1 dimension. Shaded area presents occupied states.
3.3.4 Zero dimensions (Quantum Dot)
Here since all three dimensions are confined. The density of states is basically a series of delta functions. The total energy of the system is
3.3.5 More density of states
Density of states in the conduction band
For this we need to know the probability that an electron will occupy a given stats of energy E. The Probability, P(E), is referred as the Fermi Dirac distribution. In addition we need to know the density of states ( ′). The density of states has units of number of unit volume per unit energy. Therefore ′ is the number of states per unit volume. The number of occupied states at a given energy per unit volume is therefore
Here is the Fermi energy.
the total concentration of electrons in the conduction band is therefore the integral over all available energies
where?c is the energy where conduction band starts. For the case of three dimensional material
Taking account into conduction band begins, the density of states can be written as
The integral is called the Fermi integral or Fermi Dirac integral.
Consider the case where, E − Er ≫ KTand the Fermi Dirac distribution function becomes
This is the expression for the effective density of states of the conduction band.
Density of states in the valance band
The number of holes at a given energy per unit volume is given as
Where Ev is the energy where valance band starts. The total concentration of holes in the valance band is the integral over all energies.
This is the effective density of the states in the valance band.
Summary
Fermi level of an intrinsic semiconductor If the bulk semiconductor is intrinsic, there has been no doping of the material and hence no extra electrons or holes anywhere. in this situation
One can therefore see that at T=0 the Fermi energy of an intrinsic semiconductor is at the halfway point between the top of the valance band and the bottom of the conduction band.
Density of states in the conduction band
We start with the Fermi Dirac distribution for electrons and also the density of states
Consider only one of the subband. In this case the density of states simplifies to
Now recall from the previous section that the number of states at a given energy per unit volume
the total concentration of electrons in this first subband is the integral over all available energies. Rather than use ntot as before let’s just stick to nc from the start
Since the band really begins at en as opposed to Ec like in the bulk the integral change from
Density of states in the valance band
As with the conduction band case we need the probability of occupying a given state in the valance band. This denoted ℎ( ) and is evaluated from
Fermi level position :2D
The procedure for finding the Fermi level position is the same as in the 3D Consider a spherical volume of
Now the density is
is the number of states per unit volume and the energy density is given as
Divide by 2 to go back to only 1 spin orientation since in an optical transition spin slips are generally forbidden
The expression applies to either conduction band or valance band. Applying the following equivalence ( )
where is the desired joint density of the states. Now from the conservation of momentum, transition in k are vertical such that the initial k value in the valance band is the same k value as in the conduction band (ka=kb=k) where ka is the k value in the valence band and kb is the value in the conduction band. The energy of the initial state in the valance band is
Likewise the energy of the final state in the conduction band is
Where for notational simplicity we have used the reduced mass
2D Well
Area in k-space
= ?? = 4??
Where the area occupied by a given mode or state is .Here we assume that represents the confined direction
Together, the number of modes in the area is
Now consider the density
With the energy density given by
Starting with the energy density
Divide by 2 to get rid of the spin since formally speaking, spin flip optical transitions are forbidden
wheree??(?) is the desired joint density of states. As before in the 3D case, the conservation of momentum means that transition in k-space are vertical. That is the initial k value in the valance band is the same as the final k value in the conduction band ( = = ) where ( ) is the valance (conduction) band values.
The energy of the initial state in the valance band is
Likewise the energy of the final state in the conduction band is
Such that when replaced into our main expression the desired expression for the joint density of states is
1D wire
Consider the length in k-space
Lk=2k
The length occupied by a given mode or state is where
The number of states in the given length is
Multiply this by 2 to account for spin, we get total number of states as
Consider the density ie number of states per unit length
And the energy density is given by
Or alternately
Starting with the energy density
Divide by 2 to consider only one spin orientation since spin flip transition are generally forbidden
Now apply the following equivalence
wherepj (E) is the desired joint density of states. As before in the 3D and 2D case, the conservation of momentum means that transition in k-space are vertical so that Ka=Kb=K ) where ka(kb) is the valance (conduction) band values.
The energy of the initial state in the valance band is
Likewise the energy of the final state in the conduction band is
Such that when replaced into our main expression the desired expression for the joint density of states is
Now to continue towards our final expression we express k fully. Since
This leads to the final expression for the joint density of states
