Question regarding FD distribution and doped SC

[niehls]

Consider an n-doped semiconductor. I am trying to figure out at which temperature half of all the donors will be ionized, given energies of donor level and band gap.

I was thinking that, because the chance is 50% that the donors be populated, the Fermi level must lie exactly on the donor level energy, $$E_d$$. I was thinking that if we integrate the FD from the conduction band edge to infinity and set this integral equal to one half:

$$\int_{E_c}^{\infty} \frac{1}{e^{(\epsilon - \mu)/kT} + 1} d\epsilon = \frac{1}{2}$$

Obviously, this is the wrong way about to solve this problem, since i get the temperature 8570K(Using $$E_d = E_C - 0.025, E_g = 1.12$$ as example values (in eV). I am not very used to working with the FD distribution as you see, and i would need some input.

Another way to go about this problem i guess, is multiplying the FD distribution with the density of states and integrate, setting the integral equal to N_d/2, where N_d is the number of donor atoms.

Thanks

 

[Jhero]

for reaching out for input on this problem. It seems like you have a good understanding of the Fermi-Dirac distribution and its relationship to donor ionization in semiconductors. However, your initial approach does have some flaws.

Firstly, the integration from the conduction band edge to infinity assumes that all states above the donor level are occupied by electrons, which is not necessarily true. Additionally, the integration should be performed with respect to energy, not just the difference between the Fermi level and the donor level.

A more accurate approach would be to use the relationship between the Fermi-Dirac distribution and the density of states:

n = \int_{E_C}^{\infty} g(E) \frac{1}{e^{(E-\mu)/kT} + 1} dE

Where n is the electron concentration and g(E) is the density of states. For an n-doped semiconductor, the donor states are the dominant contributors to g(E), so we can approximate g(E) as a delta function centered at the donor level energy, E_d.

g(E) = \delta(E - E_d)

Substituting this into the above equation and solving for T, we get:

T = \frac{E_d - \mu}{k ln(\frac{n}{N_d - n})}

Where N_d is the total number of donor atoms. This equation gives the temperature at which the donor ionization is 50%. Note that this is an approximation and assumes that the donor states are non-degenerate.

Alternatively, you could use the same approach as your initial attempt, but integrate from the donor level energy to the conduction band edge, and set the integral equal to N_d/2. This will give a slightly different result, but should still be in the same ballpark.

I hope this helps. Let me know if you have any further questions or if you would like me to clarify anything. Good luck with your research!