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  − E_r ≫ 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

Multiply by 2 to account for spin

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

L_k=2_k

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

wherep_j (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 K_a=K_b=K ) where k_a(k_b) 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