Chapter 2: Semiconductor Fundamentals |
Before we can calculate the density of carriers in a semiconductor, we have to find the number of available states at each energy. The number of electrons at each energy is then obtained by multiplying the number of states with the probability that a state is occupied by an electron. Since the number of energy levels is very large and dependent on the size of the semiconductor, we will calculate the number of states per unit energy and per unit volume. |
2.4.1 Calculation of the density of states |
The density of states in a semiconductor equals the density per unit volume and energy of the number of solutions to Schrödinger's equation. We will assume that the semiconductor can be modeled as an infinite quantum well in which electrons with effective mass, m^{*}, are free to move. The energy in the well is set to zero. The semiconductor is assumed a cube with side L. This assumption does not affect the result since the density of states per unit volume should not depend on the actual size or shape of the semiconductor. |
The solutions to the wave equation (equation 1.2.14) where V(x) = 0 are sine and cosine functions: |
(2.4.1) |
Where A and B are to be determined. The wavefunction must be zero at the infinite barriers of the well. At x = 0 the wavefunction must be zero so that only sine functions can be valid solutions or B must equal zero. At x = L, the wavefunction must also be zero yielding the following possible values for the wavenumber, k_{x}. |
(2.4.2) |
This analysis can now be repeated in the y and z direction. Each possible solution then corresponds to a cube in k-space with size np/L as indicated on Figure 2.4.1. |
Figure 2.4.1: | Calculation of the number of states with wavenumber less than k. |
The total number of solutions with a different value for k_{x}, k_{y} and k_{z} and with a magnitude of the wavevector less than k is obtained by calculating the volume of one eighth of a sphere with radius k and dividing it by the volume corresponding to a single solution, , yielding: |
(2.4.3) |
A factor of two is added to account for the two possible spins of each solution. The density per unit energy is then obtained using the chain rule: |
(2.4.4) |
The kinetic energy E of a particle with mass m^{*} is related to the wavenumber, k, by: |
(2.4.5) |
And the density of states per unit volume and per unit energy, g(E), becomes: |
(2.4.6) |
The density of states is zero at the bottom of the well as well as for negative energies. |
The same analysis also applies to electrons in a semiconductor. The effective mass takes into account the effect of the periodic potential on the electron. The minimum energy of the electron is the energy at the bottom of the conduction band, E_{c}, so that the density of states for electrons in the conduction band is given by: |
(2.4.7) |
Example 2.3 | Calculate the number of states per unit energy in a 100 by 100 by 10 nm piece of silicon (m^{*} = 1.08 m_{0}) 100 meV above the conduction band edge. Write the result in units of eV^{-1}. |
Solution | The density of states equals:
So that the total number of states per unit energy equals: |
2.4.2 Calculation of the density of states in 1, 2 and 3 dimensions |
We will here postulate that the density of electrons in k–space is constant and equals the physical length of the sample divided by 2p and that for each dimension. The number of states between k and k + dk in 3, 2 and 1 dimension then equals: |
(2.4.8) |
We now assume that the electrons in a semiconductor are close to a band minimum, E_{min} and can be described as free particles with a constant effective mass, or: |
(2.4.9) |
Elimination of k using the E(k) relation above then yields the desired density of states functions, namely: |
(2.4.10) |
for a three-dimensional semiconductor, |
(2.4.11) |
For a two-dimensional semiconductor such as a quantum well in which particles are confined to a plane, and |
(2.4.12) |
For a one-dimensional semiconductor such as a quantum wire in which particles are confined along a line. |
An example of the density of states in 3, 2 and 1 dimension is shown in the figure below: |
Figure 2.4.2: | Density of states per unit volume and energy for a 3-D semiconductor (blue curve), a 10 nm quantum well with infinite barriers (red curve) and a 10 nm by 10 nm quantum wire with infinite barriers (green curve). m^{*}/m_{0} = 0.8. |
The above figure illustrates the added complexity of the quantum well and quantum wire: Even though the density in two dimensions is constant, the density of states for a quantum well is a step function with steps occurring at the energy of each quantized level. The case for the quantum wire is further complicated by the degeneracy of the energy levels: for instance a two-fold degeneracy increases the density of states associated with that energy level by a factor of two. A list of the degeneracy (not including spin) for the 10 lowest energies in a quantum well, a quantum wire and a quantum box, all with infinite barriers, is provided in the table below: |
Figure 2.4.3: | Degeneracy (not including spin) of the lowest 10 energy levels in a quantum well, a quantum wire with square cross-section and a quantum cube with infinite barriers. The energy E_{0} equals the lowest energy in a quantum well, which has the same size |
Next, we compare the actual density of states in three dimensions with equation (2.4.10). While somewhat tedious, the exact number of states can be calculated as well as the maximum energy. The result is shown in Figure 2.4.4. The number of states in an energy range of 20 E_{0} are plotted as a function of the normalized energy E/E_{0}. A dotted line is added to guide the eye. The solid line is calculated using equation (2.4.10). A clear difference can be observed between the two, while they are expected to merge for large values of E/E_{0}. |
Figure 2.4.4: | Number of states within a range DE = 20 E_{0} as a function of the normalized energy E/E_{0}. (E_{0} is the lowest energy in a 1-dimensional quantum well). See text for more detail. |
A comparison of the total number of states illustrates the same trend as shown in Figure 2.4.5. Here the solid line indicates the actual number of states, while the dotted line is obtained by integrating equation (2.4.10). |
Figure 2.4.5: | Number of states with energy less than or equal to E as a function of E_{0} (E_{0} is the lowest energy in an 1-dimensional quantum well). Actual number (solid line) is compared with the integral of equation (2.4.10) (dotted line). |
ece.colorado.edu/~bart/book | Boulder, August 2007 |