One of the noted disadvantages, of avalanche photo-diodes being used a photo-receptors within the context of fiber optics systems is that they have a large dark current. Why is this considered a disadvantage? Is it just seen as another source of noise?
Electronic – Why is dark-current considered a disadvantage in photo-detectors
diodesphotodiode
Related Solutions
There are, no doubt, many reasons why the dark current increased, but to answer the question about whether the diode is broken - badly or otherwise - it's just necessary to measure the current with the diode illuminated. If you're using the diode in reverse-biased mode, measuring the reverse current with various levels of illumination will give you some idea of how its response conforms to the data sheet specifications.
If you're using it in photovoltaic mode, then subjecting it to various levels of illumination and load and plotting its response as a load line will also yield salient data.
In the real world, though, if you've exceeded the manufacturer's maximum temperature spec by what... six or seven times? I'd say you've definitely let the magic smoke out.
I'm assuming your circuit looks like
simulate this circuit – Schematic created using CircuitLab
and your expectation that "it should saturate" is exactly correct. It IS saturating. The op amp is trying to drive the output below zero, but it cannot since it's a single-supply amplifier.
If you're going to use a TIA you have to use a negative bias to get a positive output.
ETA: Oh yes, and just to make your day - you may well have killed your op amps. Maybe not, since designers have gotten pretty good at protecting chips against overload, but this is pretty extreme. Not as extreme as getting hit by lightning, but that kills most chips, too. You need to check them in a less stressful circuit if you want to reuse them.
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Best Answer
The dark current can be seen as interference source in 2 aspects:
Both causes come from the same source actually (flicker noise is created by "quick" breakdown current fluctuations which have frequency comparable to your useful signal, and drift represents the slower ones that are easy to filter out if your signal has no constant part). Anyway, the more dark current is, the more interference it creates.