Show that product of electron
equilirium does not depends on Fermi
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Answer:The density of electrons in a semiconductor is related to the density of available states and the probability that each of these states is occupied. The density of occupied states per unit volume and energy, n(E), ), is simply the product of the density of states in the conduction band, gc(E) and the Fermi-Dirac probability function, f(E), (also called the Fermi function):
(2.6.1)
Since holes correspond to empty states in the valence band, the probability of having a hole equals the probability that a particular state is not filled, so that the hole density per unit energy, p(E), equals:
(2.6.2)
Where gv(E) is the density of states in the valence band. The density of carriers is then obtained by integrating the density of carriers per unit energy over all possible energies within a band. A general expression is derived as well as an approximate analytic solution, which is valid for non-degenerate semiconductors. In addition, we also present the Joyce-Dixon approximation, an approximate solution useful when describing degenerate semiconductors.
The density of states in a semiconductor was obtained by solving the Schrödinger equation for the particles in the semiconductor. Rather than using the actual and very complex potential in the semiconductor, we use the simple particle-in-a box model, where one assumes that the particle is free to move within the material.
For an electron which behaves as a free particle with effective mass, m*, the density of states was derived in section 2.4, yielding:
(2.6.3)
where Ec is the bottom of the conduction band below which the density of states is zero. The density of states for holes in the valence band is given by:
Explanation:
Answer:
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