On Wed, Jul 29, 2020 at 11:11 AM Josh Wyatt via TriEmbed
<[email protected]> wrote:
>
> Take a little time and characterize the LED parameters a bit...
>
> I would use a high-ish voltage DC source, like a 12 volt supply, and put a 
> 100 ohm or so resistor in series with the LED. The resistor will limit 
> current to 90mA in the worst case (3 volt LED on 12 volt source), and will 
> allow you to measure the voltage drop across the LED. Once you have that, use 
> Ohm's law to calculate the current required for "full" rating in wattage; 
> this might require some guesswork by measuring heat dissipation with a 
> heatsink, etc. But a safe figure might be 1/2 watt or less for basic testing.
>
      What about those variable voltage power supplies? My crappy one
allows me to set the voltage and then shows how many amps are being
sucked by the circuit.

> -j
>
> On Wed, Jul 29, 2020 at 10:12 AM Brian via TriEmbed <[email protected]> 
> wrote:
>>
>> 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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