Intuitive explanation: how does a LED work? I'm looking for an intuitive explanation of how a LED works. I have a bachelor level understanding of math (I did take some physics courses as well, but nothing too advanced). I'll explain what I know and what I've figured out (hoping it's correct), and point out where my doubts are.
A LED is built putting side by side two pieces of differently doped semiconductor material, p-type and n-type. On one side, there are electron holes, which basically means that atoms are missing one electron to have their outer shell complete. On the other side, there are free electrons. The idea is that if you have a group of atoms in which each accepts 4 bonds, and you put there one atom that only accepts 3, you get one free electron. The same thing the other way around gives the electron holes.
Now, when a voltage is applied in the right direction, free electrons flow to the part where the holes are. Every time an electron meets a hole, it basically "falls into it" (i.e. it completes the outer shell of the atom having a missing bond), and loses some energy. For this reason, when an electron meets a hole, it emits a photon.
Now to my question. What keeps a LED on? Why doesn't it make a super-short burst of light and then stays off? When an electron meets a hole and falls into a lower energy state, it should stay there, I guess. So it should come a time when all holes are filled up and there's nowhere for the electrons to go. Instead, there must be something that empties the holes for new electrons to come. How does this happen?
 A: A diode in its grounded state is an insulator. Start with that.
The way we build the diode is, we take two insulators that are very close to being conductors but not quite there, and we make them conductors, in one case by inserting atoms which have extra electrons to give (N-doping) and in the other case by inserting atoms which have a tendency to instead suck up the electrons, so that in a sense it has electron “holes” to give (P-doping).
When we fuse these two now-conductors together, an interesting thing happens: the electrons from one side fall into the holes from the other side at equilibrium. So the boundary between the two, because of the diffusion of electrons across the gap, is a standard insulator like the one we started out with.
Once you have started there, the question is what a voltage does. Keep in mind that a voltage is a sort of pressure for electrons, they want to move from low voltage to high voltage. So if you put this pressure on in one way, you pull the holes further away from the insulating junction barrier, and you pull the electrons further away in the other direction, and therefore the depletion region between the two, the region where the thing is an insulator, grows. This is called reverse bias, the diode cannot conduct electricity this way until it breaks, some diodes are manufactured to break in interesting and predictable ways, but for the most part we just imagine that the diode cannot conduct electricity that way.
If you raise the voltage the opposite direction, the thing remains an insulator but the depletion region shrinks. At a certain voltage called the device’s “diode drop”, typically 0.7 V for silicon diodes, the depletion region has shrunk to zero and current flows through the junction now, it has become a conductor. This “drop” in voltage is the only major cost, if you try to put more voltage across the device it will do what conductors do and force more current through the junction until it explodes, so you have to design these circuits with resistors to regulate the currents involved, it's not like an incandescent light bulb where the bulb filament itself provides the resistance; you turned it into a conductor and now small voltage changes imply large current changes.
So to answer your question, the reason they are flowing is because you provided an electric field to push them one way, it had to be strong enough to push them through this space that “wants to be” an insulator. A diode is kind of like a one way gap for sparks to flow, you are creating a breakdown in an insulator so that it becomes a conductor, just like a spark is insulating air breaking down to become a conductor. If you don't push the electron hard enough to get through this region, it doesn't.
A: 
Now, when a voltage is applied in the right direction, free electrons flow to the part where the holes are.

This happens independently of an external voltage, because the electrons in the n-doped part feel the urge to fill the holes in the p-doped part. But because electrons move from the n-doped part to the p-doped part, the n-doped part becomes positively charged and the p-doped part becomes negatively charged. Thus, the flow of electrons leads to increasing electric field. When an equilibrium of forces is reached, the flow terminates and we are left with the depletion region.

Instead, there must be something that empties the holes for new electrons to come. How does this happen?

By applying a voltage.
