Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose there is a semiconductor with Fermi energy $E_f$ and that there are $N$ bound electron states.

I'd like to know why the mean number of excited electrons takes the form $$\bar n={N\over \exp\beta(\mu-E_f)+1}$$

where $\mu$ is the chemical potential.


I can see that the Fermi Dirac statistics say that for one fermion the mean occupation number is $$\bar n = {1\over \exp\beta(E-\mu)+1}$$ I am not sure however as to how for the semi-conductor it should assume the above form. The $N$ is clearly due to the $N$ bound $e^-$ states. I am not quite so confident as to why $$\mu\to E_f$$ $$E\to \mu$$

Could someone please explain?


Anybody? :(

share|improve this question
    
By definition? I mean, number of electrons is exactly the number given by FD statistics. –  Misha Mar 29 '13 at 9:15
1  
Are you sure that your first formula is correct? If you change the chemical potential with E, and given that for semiconductors, the chemical potential is taken as Ef(T), you simply get N by the FD distribution. –  neutrino Mar 29 '13 at 19:59
    
@Misha: Thanks, but I don't understand how we get the first form given the FD distribution! What definitions are involved to give the change in variables...? –  Henrietta James Mar 29 '13 at 21:28
    
@neutrino: Thanks, I am quite sure about the accuracy of the 1st equation :/ and I don't think I quite understand what you said...? Would you mind elaborating a little? Thanks again! –  Henrietta James Mar 29 '13 at 21:30

1 Answer 1

The first equation is not correct, the Fermi-level is the chemical potential of the electrons.

In semiconductor the carrier density $n$ (units of $\text{m}^{-3}$) in the conduction band is the integral of the Fermi-Dirac function over the conduction band density of states $g_c(E)$ (units of $\text{m}^{-3}\text{J}^{-1}$),

$$n = \int_0^{\infty} g_c(E) f(E, E_f) dE$$

where $$ f(E, Ef) = \frac{1}{\exp\left(\frac{E - E_f}{k_BT}\right)}$$

In practice you can actually use the Boltzmann distribution provided $E_f << E_c$ where $E_c$ is the conduction band energy. This allows the so called effective density of state to be used which looks like slightly your original equation,

$$n \approx N_c \exp\left(-\frac{E - E_f}{k_BT}\right)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.