This description is going to be a bit vague, because I didn't want to deal with any equations, and it also requires a certain familiarity with some standard semiconductor terminology. I suggest reading this description posted by a professor in Thailand (I kid you not -- ain't the web grand?)
I was interested in understanding a few things in particular. Specifically, why it is a single non-illuminated photodiode kills the entire panel, and why photodiodes are wired sort of backwards -- by which I mean a normal diode wouldn't conduct in the direction they are wired. These are very bizarre properties in my mind, so what's up with that? (My discussion is no better than any others you can find on the web -- just different. That's because I (apparently) have to think about everything differently than everyone else.)
It's all about the N-P junction (obviously). At that point there are a couple things going on. Because of the doping there is a large concentration of electrons in the conduction band on the N side, and a large concentration of holes in the valence band on the P side. Plain old thermodynamic diffusion causes these "majority" carriers to diffuse to the other side (where there are very few -- no electrons in the valence band of the P side, and no holes in the valence band of the N, respectively). This causes a local imbalance of charge in the vicinity of the junction: net positive on the N side, negative on the P side. This creates an electric field across the junction, which in turn causes a potential barrier to form:
___
/ \
____/ \ P ____
N \ /
\___/
I'm sure I got the sign convention wrong, but the point is that an electron trying to get from one side to the other has to cross this barrier (either direction). If you reverse bias it you increase the barrier, making it even more difficult -- thus no current. If you forward bias it you lower the barrier until you reach the band-gap voltage when the barrier disappears altogether -- pure conduction:
Reverse bias: Forward bias:
__
/ \ ____
____/ \ ____ P /
\ / \____/
N \ P ____/ N
\ ____
\ / <-- electron flow <--
\__/
Notice that there is a voltage drop across the diode even when fully forward biased -- ie. a conducting diode is not a perfect conductor. The voltage drop is the band-gap of the semiconductor. According to Dr.Thailand it's on the order of 1.1V in a standard silicon semiconductor. If you continue to forward bias it even more, then resistance in connections or the semiconductor itself will have to make up for the rest of the voltage applied.
Now, there's also these "minority" carriers which are electron-hole pairs created by either plain thermal excitation (these normally rapidly recombine and have very little effect -- just a small "leakage" current in an unbiased diode), or by photons bopping some valence electron up to the conduction band. Any such electron-hole pairs created near the junction are quickly split up by the steep local potential gradient, so they don't recombine right away. For some reason (here's where you gotta really work over the sign conventions) they are separated in the opposite direction that forward-biased current generally flows -- ie. the minority current caused by sunlight flows backwards.
All this theoretically explains why a single photocell in the dark causes an entire chain of cells strung in series to fail completely. Here's a simple case. Short circuit three cells in series: (The diagram is trying to show the potential variation across the cells -- those aren't resistor symbols! Arrows show electron flow.)
1 -> 2 -> 3
____/\ _____/\ _____/\ ____
| \/ \/ \/ |
| |
|______________________________|
<- <- <-
In full illumination the light causes a bunch of electrons to flow across the junction of #1 to the right. Then light has to force those same electrons across the junction of #2 and #3 as well. If any of those cells is dark, then they can't make it across one of the barriers, and it breaks the whole circuit. Because the series is shorted out there's no net gain in voltage across 1-2-3 -- just current.
Now put a load resistor on the bottom:
-> 3
2 _____ _
1 -> ________/ | \
________/ | ) potential created
______/ | _/
| |
| |
|___________/\/\/\/\/\_________|
<- (load) <-
Putting the load resistor in there with the current flowing "backwards" (in the diode's point of view at least) is the same as putting a battery in there in the forward-bias direction. (That is, a voltage drop across a resistor is the same as a voltage increase in the opposite direction due to a battery.) This removes the potential barriers at the NP junctions, but instead creates a potential difference over the entire series of diodes, so each stage is still required to pump the electrons across each diode -- which means that any cell in the dark kills the whole thing again.
Yet increasing the light intensity with a billion-watt light-bulb can't increase the total voltage boost across each cell past the band-gap potential. Any single electron is getting a very specific fixed amount of energy from the incident photons, which is related to the frequency of the light, which in turn is tuned to the band-gap. Increasing intensity just increases the number of photons, not their frequency or energy. However, increasing the intensity of light does increase the rate of electron-hole production, thereby increasing the current without limit. (Theoretically -- obviously there are limiting effects due to resistance of the semiconductor and so on.)
I tried to draw the I-V curve but ASCII doesn't do it justice -- Dr. Thailand's got a nice diagram, albeit with a bizarre choice of sign on the I axis. His shows the leakage current and everything. (Incidentally, his choice of sign is the standard choice -- it just doesn't make any sense to me. What can I say? That's just me.)