# Quasi fermi-level

1. Jul 25, 2004

### marlon

can anyone provide me with a good definition of the quasi-fermi-level?

I know it originates when there is light present in a semiconductor-laser-system, this light is used for the stimulated emission.

thanks

nikolaas van der heyden

2. Jul 25, 2004

### chroot

Staff Emeritus
marlon,

Please do not crosspost.

The fermi energy is the energy of the highest filled state in a system at absolute zero. At absolute zero, every state with energy below the fermi energy is filled, and every state with energy above it is empty. At temperatures above absolute zero, thermal excitation permits some of the states above the fermi energy to be filled, leaving some below it empty.

- Warren

3. Jul 25, 2004

### QuantumCowboy

... and a quasi-fermi level is the fermi level (chemical potential) that a system takes on when an external voltage is applied.

Quantum Cowboy

4. Jul 26, 2004

### marlon

ok, sorry bout the crosspost, won't happen again.

If an external voltage is applied....ok, then how is this new fermi-level generated. What is the underlying process. I don't minde if the explaination is in heavy QM-language, feel free to indulge yourselves, please I insist.

So it has nothing to do with a prior present radiationfield which is then used for stimulated emission. the external voltage is used for population-inversion in semiconductor-lasing-systems, right ???

already thanks for helping me out

grazie mile

5. Sep 12, 2011

### Kasko87

Essentially, this is the plot:

When a system (such as a semiconductor) is in thermal equilibrium and no bias (voltage, EM radiation, ecc) is applied, the distribution function that describes the occupation of the quantum states is the Fermi-Dirac distribution (FDD), given by

$f_0(E,E_F,T) = \frac{1}{e^{(E-E_F)/k_B T} +1}$

where E is the energy of the state, EF is the Fermi energy (also called Fermi level), and T is the temperature. This distribution law is valid for any energy level of the system, independently of the fact that it is above or below the Fermi energy (conduction or valence band state).

When the system is under bias, the FDD doesn't hold anymore. However, if the bias is not to great, or not changing to quickly, it is still possible to describe the occupancy of the quantum states of the different bands using a distribution law of the same form of the FDD, but with different Fermi energy for different bands (One says that there is a situation of quasi-thermal equilibrium). Hence one has

$f_c = f_0(E,E_{F_n},T)$

$f_v = f_0(E,E_{F_p},T)$

where fc and fv is the probability of finding an electron in the conduction and valence band, respectively, and EFn and EFp are called the quasi Fermi levels for the conduction and the valence band.

You can look at the topic by this point of view:

Let us suppose that a semiconductor is in thermal equilibrium (no bias). Its density of free carriers in the conduction band (electrons) and in the valence band (holes) is given by

$n = \int_{E_c}^{\infty} {g_c(E) f_0(E, E_F, T) dE}$

$p = \int_{-\infty}^{E_v} {g_v(E) ( 1 - f_0(E, E_F, T) ) dE}$

where Ec is the CB minimum, Ev is the VB maximum and gc and gv are the density of state of the CB and VB, respectively. Suppose that a photon field now hits the systems. This rise both n and p and it is impossible to describe this new (quasi) equilibrium by means of the last two formulas, unless one replaces the single parameter EF with the two parameters EFn and EFp:

$n_{bias} = \int_{E_c}^{\infty} {g_c(E) f_0(E, E_{F_n}, T) dE}$

$p_{bias} = \int_{-\infty}^{E_v} {g_v(E) ( 1 - f_0(E, E_{F_p}, T) ) dE}$

Physically, the quasi Fermi level relative to a band is the effective Fermi level that brings to the same density of carriers in the same band when the system is not perturbed.

Since, in the example above, $n_{bias} > n$ and $p_{bias} > p$ (photons increase both carrier densities), one has $E_{F_n} > E_F$ and $E_{F_p} < E_F$.

For more details you can read Chapter 3 of The Physics of Solar Cells by J. Nelson or many other books.

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