Some tips and educated guesses:
This would be a great job for an adjustable benchtop power supply. Set
the current limit to something relatively small, say, 100 mA, and start
bringing up the voltage until the current limit is hit. Many high-power
LEDs like to run at 750 mA, so 100 mA won't be nearly enough power
dissipation to worry about overheating them just in the process of
finding out the voltage and polarity. You should find a point where
current starts increasing rapidly with small increases in voltage, and
that'll get you in the ballpark. Then you can set your current limit to
750 mA and turn the voltage all the way up; whatever voltage ends up
across the LED at 750 mA is the rated voltage. As others have said, do
NOT run these at full power for more than an instant without proper heat
sinking.
Note that LEDs, being diodes, won't conduct at all until the potential
across them exceeds the junction voltage, and that these modules are
often individual chips wired in series. If you see 9 chips in there, my
guess is that the operational voltage is probably somewhere around 12 V
(1.something volts for each junction, which is very reasonable).
If you wanted to take a 12-V power supply and just quickly tap the leads
one way and then the other, you probably wouldn't toast the module (but
you do have more than one, right? ;-) ). You could still put a few
hundred ohms in series as ballast if you wanted to be careful.
Finally, the "tray" that the silicon sits in is always the cathode, but
it looks like you might not be able to see through the cover. It might
be a reasonable guess to assume the chassis is the cathode (one lead
would be clearly isolated from the rest of the chassis, while the other
is not).
I take it there are no identifying markings on the device at all? No
numbers, codes, anything?
HTH,
-B
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