If you attempt to apply a forward bias voltage larger than the diode's "natural" forward voltage, the diode will draw more current in an attempt to bring the applied voltage down to its "natural" voltage. If the power supply is sufficiently "stiff" (maintains its output voltage regardless of current), the diode will release its magic smoke, and become an open circuit.
Now, if you have a higher concentration of dopants, you end up with more diffusion pressure which means the electric field must be bigger in order to oppose it, meaning that the depletion/space charge region must be wider. However, what I just said is incorrect, but i don't understand where my reasoning is wrong.
If you understand that there's a built-in voltage, then you should also know that distance and path is unimportant to a voltage. Voltages are scalar. 5V over 1 billion miles is the same as 5V over 2um. Now the electric fields associated with those two voltage are entirely different. Now the reason that the depletion region gets smaller with higher dopants is that there is more static charges exposed in the depletion region. Think of it like this, with a low dopant level I'd have to take 3mm of space to free "20" charges, but if they were a higher density, I'd only have to take 3um to free "20" charges. Remember it's the exposed charges in the depletion region that create the field, not the charges in the drift region.
Now again, if you have a higher electric field applied you should get proportionally less diffusion current meaning that the width of the depletion region can be smaller and still fully oppose the diffusion current. This is also incorrect reasoning, but where did i go wrong?
You went wrong when you thought about the field inside the depletion region effecting things outside the depletion region directly. The field is stuck there. It can't directly effect anything. It does have some secondary effects though. For instance the wider the depletion region, the easier it is to diffuse across the non-depletion areas. This is because there is less distance to the end of the device when there's more length in the field. We use this all the time in BJT's. The other things that a field will do is create a build up of charges on it's edges (You know, all those ones you ripped out to create the depletion region, they want back to where they were). These "free charge" (free because they are able to move, unlike in the depletion region where the charges are held in the atoms of the crystal lattice) areas are actually what is driving the diffusion in the other areas. The free charges are more concentrated by the depletion region than in other places. Again, they want back, but the built in field removes them every time they try so they're stuck at the edge. This creates the "High to Low pressure" that drives diffusion. The higher the field, the more charges you ripped away that want to go back that the field at the boarder resists them, the more concentrated the free charges are, the more diffusion there is. All linked. I'll Try to get picture in here. It really is difficult to see without picture.
Edit: I've found a good Youtube video with all the particles present and a very decent explanation. I hope this helps.
Best Answer
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.