[TriEmbed] A little help
dragojdw at gmail.com
Wed Jul 29 10:11:18 CDT 2020
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
On Wed, Jul 29, 2020 at 10:12 AM Brian via TriEmbed <triembed at triembed.org>
> 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
> 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?
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