Why is the quasi-Fermi level flat across the depletion region in a pn diode under forward bias?
I get that EFn decreases on the p-type side and EFp decreases on the n-type side due to recombination, but why are they flat in the depletion region? What kind of approximation is used?
The model governing the (simplified) behavior of the quasi Fermi levels in the space charge region (and which name you are seeking to know) is the law of mass action for semiconductor physics. The constant splitting value of quasi Fermi levels can be derived from this law.
TL;DR)
The Fermi level is a parameter in the equations used to express equilibrium values, for example, of carrier concentrations. Obviously, forward biasing causes the current flow through an pn junction and therefore disturbs the junction's thermal equilibrium.
The energy state with the energy lying in the conduction band is an electron quasiparticle and the state with energy in the valence band is a hole quasiparticle. When all the states are in thermal equilibrium, there is a Fermi level parameter common for all states: the energy level which is occupied with an 1/2 probability. This is true for both electron quasiparticles and hole quasiparticles. These quasiparticles, in the context of semiconductor device physics, are called simply electrons and holes.
When the pn junction is put out of equilibrium, either by forward biasing or by light irradiation, it relaxes to a new global state of equilibrium through interactions between quasiparticles. (The interaction with the crystal lattice is accounted for by the quasiparticle concept).
The equilibrium within the ensemble of electron quasiparticles, as well as within the ensemble of holes, is restored faster than the equilibrium between the quasi-electron ensemble and the hole ensemble: the equilibrium within the ensembles is restored through collisions, and the equilibrium between the ensembles is restored through recombination, with the collision times being much shorter than the recombination time.
For time intervals lying between these characteristic (recombination and collision) times, the ensembles are in quasi-equilibrium. For each of the two ensembles (electron and hole ones), we can define individual quasi Fermi levels, which are used with the Fermi-Dirac distribution for particles in each individual ensemble. As concerns your question, there is another view of these quasi Fermi levels from the calculation perspective.
Actually, the law of mass action enable us to only confirm equidistance of the quasi Fermi levels in the space charge region. In thermal equilibrium, the product of electron and hole concentrations is constant \$n_i^2\$, the square of intrinsic concentration. The injection of carriers either through electrodes (when forward biasing) or through light irradiation increases this product: \$p_{non-eq}n_{non-eq} > n_i^2\$. The model declares that this product value increases when thermal equilibrium is disturbed, but it is still constant for uniform carrier concentrations injected with a constant given bias or light irradiation. We extend the applicability of the law of mass action by introducing two quasi Fermi levels \$E_{Fp}\$ and \$E_{Fn}\$, splitting the "intrinsic" \$E_{i}\$ into \$E_{Fp}\$ for holes and \$E_{Fn}\$ for electrons.
Now, the electron concentration is \$n = n_i \exp((E_{Fn}-E_i)/kT)\$ and the hole concentration is \$p = n_i \exp((E_i-E_{Fp})/kT)\$, so their product is \$pn = n_i^2 \exp((E_{Fn}-E_{Fp})/kT)\$. The law of mass action states that this product is constant, therefore, the difference of the two quasi Fermi levels is also constant throughout the entire space charge region, which is the depletion region in the depletion approximation.
The depletion region consists of immobile charge carriers. So on the p side, these are the electrons that are loosely bound to the crystal lattice. On the n side, these are the holes that are loosely bound to the crystal lattice, i.e. the positively charged ionized dopants left behind from those electrons injected into the p side during the formation of the depletion region.
The electrons that comprise the p side's depletion region prevent further diffused (mobile) electrons from recombining in that region; they're already occupying states there. Farther into the p side, diffused electrons can recombine, since there are unoccupied states, i.e. holes. So the concentration of mobile electrons drops off very rapidly right after the depletion region ends. This is why we see the steep drop-off of \$E_{F_{n}}\$ on the p side.
Similarly, the holes that comprise the n side's depletion region prevent further diffused holes from recombining in that region. Farther into the n side, diffused holes can recombine there. This coincides with the sharp increase of \$E_{F_{p}}\$, as there is a sharp increase in the concentration of mobile electrons.
That's for the forward biased case. For the reverse-biased case, we see something like this:
We can make similar arguments in this case. The depletion region of each side is formed in the same way as in the forward bias case, though each gets a little wider in reverse bias. This widening happens on the n side because the reverse biasing pulls the mobile electrons on the n side away from the junction (i.e. closer toward that side's terminal), leaving behind more positively ionized donor atoms towards the junction. Conversely, on the p side, the reverse biasing pulls the mobile holes on the p side away from the junction, leaving behind more negatively ionized acceptor atoms towards the junction.
Since we assume low-level injection, the majority carriers of each side are also assumed to be distributed evenly/uniformly; i.e. any gradient/change in majority carrier concentration for either side is negligible because the majority carrier concentration dwarfs the minority carrier concentration. That is, \$p_{p} >> n_{p}\$ such that \$p_{p} + \Delta p_{p} \approx p_{p}\$, and \$n_{n} >> p_{n}\$ such that \$n_{n} + \Delta n_{n} \approx n_{n}\$. Thus, for the p side, the majority holes are assumed to be evenly distributed across the p region. This coincides with \$ E_{F_{p}} \$ in the p side's quasi-neutral region (QNR) (the region of the p side that is not the depletion region) being constant/flat in the diagram above. The mobile minority electrons of the p side's QNR, however, are more concentrated away from the junction. This is because the p side's depletion region is negative, so it repels the minority electrons on that side away from there. This is why we see a dip in \$ E_{F_{n}} \$ on the p side's QNR. (Conversely, the n side's depletion region is positive, so it repels the minority holes on that side, in the same way, away from the junction.)
The result is that we end up with a minority carrier distribution that looks like the one shown below:
Note that this figure/example above assumes a p region that's doped more heavily than the n region. This is why we see such a small/nearly negligible change in the minority electron concentration of the p region. Also note that the space between the vertical axes \$ n(x_{p}) \$ and \$ p(x_{n}) \$ is the depletion/space charge region of the pn junction.
This is why \$E_{F_{p}}\$ and \$E_{F_{n}}\$ "dip into" the conduction and valence bands (respectively), in the first figure above. On the p side, the "dipping" of \$E_{F_{n}}\$ into the valence band occurs near the beginning of its depletion region and indicates the lack of mobile (minority) electrons there (as described above). Only \$ E_{F_{n}} \$ dips down here because, again, we're assuming that the majority carriers are distributed uniformly, i.e. \$E_{F_{p}}\$ is constant across the p side, due to low-level injection.
Parallel arguments can be made to the n side and majority electrons/minority holes, but this answer has gone long enough. I'll leave that part to the reader as an exercise :)
