[TriEmbed] A little help

Josh Wyatt dragojdw at gmail.com
Wed Jul 29 10:37:49 CDT 2020


The LED is a diode. It will "clamp" to a specific voltage, until its
dissipation limits are exceeded. Use something in series to limit the
current, and form a voltage divider. Else your high current power supply
will be half of the divider :)
Thanks,
Josh

On Wed, Jul 29, 2020 at 11:23 AM Mauricio Tavares <raubvogel at gmail.com>
wrote:

> On Wed, Jul 29, 2020 at 11:11 AM Josh Wyatt via TriEmbed
> <triembed at triembed.org> 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 <
> triembed at triembed.org> 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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