What happens when supplying an LED with too low voltage? Will it just emit less light? I'm making an octocopter, and need to cut away as much weight as possible. I need to run the LED without cooling if possible.

  • \$\begingroup\$ Which LED? Where is the data spec and how much voltage are you going/wanting to supply? \$\endgroup\$ – deed02392 Mar 25 '14 at 15:45
  • \$\begingroup\$ There is no data spec other than 50W at 33 to 36 V. \$\endgroup\$ – Friend of Kim Mar 25 '14 at 16:14
  • \$\begingroup\$ You mean the one supplier you've found doesn't say what it is publicly. There is always a data spec. \$\endgroup\$ – deed02392 Mar 25 '14 at 17:00
  • \$\begingroup\$ @deed02392 Exactly! Do you mean that the data spec should describe everything in my question? \$\endgroup\$ – Friend of Kim Mar 25 '14 at 17:17
  • \$\begingroup\$ The manufacturer will certainly have written and published a data spec., although you might need to contact the supplier for a copy. But this way you'll know exactly what voltage you can get away with and probably even the relative luminosity (as a % of max given full voltage) with your chosen voltage. This could help you decide on an LED that's designed for lower voltage in the first place that can be just as bright. \$\endgroup\$ – deed02392 Mar 26 '14 at 9:43

Look at a typical LED forward bias graph: -

enter image description here

Y axis is current and x axis is voltage. At about 2V the LED takes 20mA and illuminates nicely. At 2.2V the current has shot-up to nearly 40mA and the LED is bright.

At 1.8V the current is about 5mA and the LED is a bit dim. At 1.6V or below the LED hardly takes any current and probably will barely be visible in a dark room.

Below this there is no business until you start going negative then at about -5V applied the LED usually dies, never to recover.

This is a typical old-fashioned red LED and, may not 100% apply to more modern LEDs exactly/verbatim but the general idea is the same.


Since you have mentioned the word cooling I assume you must be talking about a power LED. Standard LEDs generally do not require cooling. A power LED dissipates roughly 1/3 of the energy supplied to it as heat. Therefore it is possible to reduce the heat dissipation of the LED and hence the need for a heatsink by reducing the power supplied to the LED.

  • \$\begingroup\$ I don't know what the LED is called but it's a 50W LED... \$\endgroup\$ – Friend of Kim Mar 25 '14 at 13:52

As others say, it will depend on that LED datasheet. 50W is pretty big, I doubt you'd be able to run that LED without cooling. Also, you'll want limit the brightness (and power usage / heating) of the power LED with current limiter (eg. driving it with constant current source) or with PWM, not by lowering voltage (LEDs do not behave the same as your typical lightbulb!)

And/or you could always use lower specd LED (with less power drawn, less heating, and of course less light produced).

Alternatively, as it is for octocopter, I'd guess you have 8 pretty good active cooling parts already, maybe you can use them to cool the LED instead of the big passive heatsink on it (just a small one + wind generated by propellers through it).

  • \$\begingroup\$ Thank you for your answer! I ended up having four LEDs, placed underneath the propellers as you suggested. It works very well. \$\endgroup\$ – Friend of Kim Jan 12 '15 at 11:46

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