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.

I am reading a book and I'm trying to understand the concept of quasi Fermi levels.

For example,

A steady state of Electron Hole pairs are created at the rate of $10^{13}\ \mathrm{cm}^{-3}$ per $\mu$s in a sample of silicon.

The equilibrium concentration of electrons in the sample is $n_0 = 10^{14}\ \mathrm{cm}^{-3}$.

Also, it gives $\tau_n = \tau_p = 2\ \mu\mathrm{s}$. I am not sure what this is but I think this is the average recombination time.

The result is that the new levels of carrier concentrations (under the described steady state) are

$n = 2.0 \times 10^{14}$ ($n_0 = 1.0 \times 10^{14}$)

$p = 2.0 \times 10^{14}$ ($p_0 = 2.25 \times 10^6$)

I follow until here but I get a bit confused after this.

The book goes onto say that this results in two different virtual Fermi levels which are at:

$F_n-E_i = 0.233\ \mathrm{eV}$

$E_i-F_p = 0.186\ \mathrm{eV}$

The equilibrium fermi level ($E_F$) being at $E_F-E_i=0.228\ \mathrm{eV}$

My question:

  1. Why are there two different quasi fermi levels now created?
  2. Why do we not consider two different ones at equilibrium conditions?
  3. Why is it that due to a steady state input of electron hole pairs that we now consider two quasi Fermi levels?
  4. What is the relevance of these new quasi fermi levels?
share|improve this question
    
Can you please provide the name of the book and the page/section? –  NanoPhys Jun 19 '13 at 15:00
    
@NanoPhys "Solid State Electronic Devices" by Ben G Streetman and Sanjay Banerjee. p130. –  midnightBlue Jun 20 '13 at 19:51

2 Answers 2

I believe that the authors, of the reference you provided, explain the reason behind introducing these so-called quasi-Fermi levels at the end of section 4.3.3. For simplicity, let me just repeat it here; perhaps explaining it in different words, with a little more elaboration, would help. I’m sure you're aware of the fact that in $n$- ($p$-) doped semiconductors the Fermi level is closer to the conduction (valence) band as opposed to being close to the middle of the band gap. To be more mathematically precise, the Fermi level of $n$- ($E_{F,e}$) and $p$-doped ($E_{F,h}$) semiconductors is given by $$E_{F,e}=E_{F,i}+k_{B}T\ln\left(\frac{n}{n_{i}}\right)$$ and $$E_{F,h}=E_{F,i}-k_{B}T\ln\left(\frac{p}{n_{i}}\right)$$ respectively. The quantities $E_{F,i}$, $k_B$, $T$, and $n_i$ are the intrinsic Fermi level, Boltzmann constant, temperature, and intrinsic carrier concentration respectively. The quantities $n\approx N_{D}-N_{A}$ and $p\approx N_{A}-N_{D}$ where $N_A$ and $N_D$ are the acceptor and donor concentrations.

Now, when you have a steady light source irradiating your semiconductor sample, you create a large number (compared to thermal excitations) of Electron-Hole Pairs (EHPs). You could think of it as doping of the sample with equal number of electrons and holes; let us call this “photo-doping.” In this photo-doping picture it is much easier to think of two separate quasi-Fermi levels for electrons and holes. You can notice that the above two equations are nothing but a rearrangement of equation (4-15) from the reference you provided. The reason they are called “quasi”-Fermi levels is because the Fermi level is only defined in equilibrium. But in this particular case, steady-state could just be treated as a different type of equilibrium.

To answer your question as to why we don’t have two quasi-Fermi levels in equilibrium I would say: in principle you could define quasi-Fermi levels even when we have EHPs due to thermal excitations. However, the typical concentrations of thermally excited EHPs are so small that it can be incorporated into the broadening of the Fermi-Dirac distribution function. In the mathematical sense, you could also incorporate the case of photo-excitation (or photo-doping) into the broadening of the Fermi-Dirac distribution as well. But then you run into the physically counterintuitive problem of defining what temperature means. First of all, a sample irradiated with light is not under thermal equilibrium. Secondly, since temperature, in the conventional sense, is associated with thermal energy of the system, it makes sense to only incorporate thermally generated EHPs into the broadening of the Fermi-Dirac distribution.

In your particular example, we have an $n$-doped system. Consequently, the concentrations $n_0 = 10^{14}$ cm$^{-3}$ and $p_0 = 2.25 \times 10^{6}$ cm$^{-3}$ exist under thermal equilibrium. When (say) you shine light on it, we have $n \approx p = 2 \times 10^{13}$ cm$^{-3}$ due to photo-excitations. In this particular example, the quasi-Fermi level of electrons is not that far away from the Fermi level in equilibrium. The quasi-Fermi level of the holes, however, is significantly away from the Fermi level at equilibrium. This is a direct consequence of the fact that the concentration of minority carriers jumps by a large amount. Now, if we were to define separate Fermi-Dirac distribution functions for electrons and holes, under this steady-state condition, and we computed the respective Fermi-Dirac integrals using the density of states of the conduction and valence bands, then we would obtain the correct electrons and hole concentrations (i.e. $2 \times 10^{13}$ cm$^{-3}$). This is another way of justifying two separate quasi-Fermi levels.

It makes sense why the separation between the quasi-Fermi levels would be proportional to the rate of EHP generation due to photo-excitation. As a result, the separation between the quasi-Fermi levels, just as the authors say, is a measure of how much the system is out of equilibrium. So, in summary, I believe that the best way to get an intuition for quasi-Fermi levels is by considering the photo-doping picture. Although the authors don’t mention this explicitly, I have a hunch that the concept of quasi-Fermi levels was inspired from conventional doping (using donor or acceptor atoms).

share|improve this answer

Fundamentally, electrons are fermions, so a set of electrons in thermal equilibrium are always characterized by a fermi level (a.k.a. chemical potential) and temperature. The probability that a state of energy $E$ is occupied is $$P(E) = 1 / (1 + \exp((E-E_F)/kT)$$ (This is neglecting electron-electron interactions, which is usually OK in a semiconductor.) This general law is true regardless of what the possible values of $E$ are -- there can be a conduction-band and a valence-band, that's fine, it's still one fermi level and one temperature. This fact about thermal-equilibrium fermionic systems is very generally true and you can find proofs and explanations in statistical mechanics books.

When we talk about out-of-equilibrium states, like a semiconductor under bias and/or under illumination, all bets are off. There is no guarantee about what states are or aren't occupied with what probability.

However, there is a relatively common situation which is the following:

  • Conduction-band electrons interact very quickly with each other and with phonons, through random collisions.

  • Valence-band electrons interact very quickly with each other and with phonons, through random collisions.

  • Random (thermal) interactions which increase or decrease the number of conduction-band electrons or valence-band electrons, such as thermal electron-hole-pair generation or electron-hole-pair recombination, occur relatively slowly.

What does "quickly" and "slowly" mean? We imagine some process that is pushing the system away from equilibrium, like light being absorbed; and it is counteracted by random thermal processes which (always) have a strong tendency to push the system towards an equilibrium state. If the random thermal processes occur sufficiently fast, then the system just stays more-or-less in an equilibrium state. The process pushing the system away from equilibrium can't do anything when it's fighting against this much stronger counteracting tendency of the system to stay in equilibrium. That's the definition of a "quick" randomization (a.k.a. "thermalization") process.

A "slow" randomization process is the opposite; the system is only slightly inclined to move gradually towards equilibrium. So in this case you can get a non-equilibrium steady-state where the external process moving the system away from equilibrium (like light being absorbed, or a bias injecting charges) is exactly balanced by the internal random processes moving the system towards equilibrium (like if there's an unusually large number of electron-hole pairs, they will tend to recombine when they randomly collide).

Going back to the three bullet points above: In this situation, the consequence is that the conduction-band electrons have to have a fermi-dirac distribution with temperature $T$ (the material temperature); the valence-band electrons have to have a fermi-dirac distribution with temperature $T$; but because of the third bullet point, the fermi level of the conduction band does not need to be the same as the fermi level of the valence band.

It's useful to think about situations where the three bullet points are not true. In hot-electron and ballistic-electron devices, the first bullet point is not valid. The electrons in these devices can have a different temperature than the material temperature, or they can even have no temperature at all (i.e. there may be no value of $(E_F,T)$ that make the equation $P(E) = (1 + \exp((E-E_F)/kT)^{-1}$ valid.)

Another example is in a material where electrons and holes combine extremely quickly, so the third bullet point is not true. Then it is actually required for the conduction and valence band to have the (approximately) the same quasi-fermi level. The more quickly they recombine, the closer the two quasi-fermi levels must be, other things equal.

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.