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I've got a set of LPD6803 LED lights controlled by an Arduino that I'd like to connect to a 12v system powered off an auto battery. The lights can run off 5-12V but this particular Arduino board is spec'd for only 4-8V Vin.

The total draw of the lights will be something under 30W. My guess is significantly lower, but I'm basing this on the spec sheet. (In testing I can run them off a 5V/2A AC adapter and it works fine)

Any ideas for the simplest way to power both from the single 12v source? It seems like the lights can run off a 12V circuit while the Arduino runs off a 5V circuit. Or I can run them both off 5V, if the 5V source can provide this power. I'm pretty sure a 12v->USB plug probably wouldn't. Does a 7805 apply here?

(I also have an inverter + AC adapter solution I could use but it's quite a hack!)

Thanks!

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    \$\begingroup\$ A 7805 can only provide 1A max. You'd ideally want a switching regulator to drop the 12-13.5V to 5V while increasing the current and not producing the heat a linear regulator would generate. \$\endgroup\$
    – Majenko
    Commented Aug 25, 2011 at 12:26

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You can use a 7805, but only to power the Arduino; the LEDs should then be powered directly from the 12V supply, since the 7805 can only supply 1A. A higher current, if possible, would also require a considerable heat sink.
And that's where a switcher comes in. Since the lights can run on 12V, but also 5V, there will be huge power losses when running off a 12V supply. 30W at 5V is 6A, and 6A over (12V - 5V) is 42W wasted. The best solution would be to have a switcher that gets you the 5V for both Arduino and LEDs, and you won't get this kind of power loss. Switchers for these voltages can reach efficiencies of more than 85%, so consuming 30W will result in a loss of 5W, down from 42W.

National's Webench designer suggests a design around an LM2743 which is even 94% efficient (only 2W loss in regulator).
Linear Technology also lists several devices that can handle the job.

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  • \$\begingroup\$ Why not connect the LPD6803 to 12V and the Arduino to a linear regulator? Another idea: A datasheet for the LPD6803 suggests that it has a 5V output (if its Vdd is 5V or greater). I'm not sure which current you can draw from that pin or it it's accessible for Steve (he talks about a LPD6803 LED, not a single chip). \$\endgroup\$
    – 0x6d64
    Commented Aug 25, 2011 at 13:18
  • \$\begingroup\$ @0x6d64 - As far as I can tell the LPD6803 is not a switcher, so even when controlling the lights with PWM the voltage difference between 12V and 5V has to go somewhere, in the form of heat. \$\endgroup\$
    – stevenvh
    Commented Aug 25, 2011 at 13:29
  • \$\begingroup\$ "The lights can run off 5-12V", so just run them off 12 V directly. No switchers. \$\endgroup\$
    – endolith
    Commented Aug 25, 2011 at 14:37
  • \$\begingroup\$ Thanks for the great suggestions guys! stevenh - I assume you're right about heat but these are industrial/building signage lights so I assume the heat tolerance is OK for our use which is only occasional (art car). I might try the 7805 for Arduino only first, but I'm concerned about the variable voltage from a car battery. Your Linear Technology search link shows you know what I mean... I don't know how the lights might actually deal with a bump above 12v. \$\endgroup\$
    – Steve Eisner
    Commented Aug 25, 2011 at 20:14
  • \$\begingroup\$ 0x6d64 - thanks, clever idea, unfortunately the LPD6803 chips are encased along with each RGB triad, so I can't easily hack into a 5V out. \$\endgroup\$
    – Steve Eisner
    Commented Aug 25, 2011 at 20:14
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In general the LEDs should be stacked 'higher' to take advantage of the higher voltage. So for example, say you can run 4 leds in series with 5V power you can probably run 9 LEDs in series with 12V at the same current. The extra LEDs in series will have added voltage drop to compensate for the higher rail voltage.

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  • \$\begingroup\$ Thanks bjb .. These come pre-packaged, but you're right that would be best for a known 12v input (ie not 5-12v) \$\endgroup\$
    – Steve Eisner
    Commented Aug 27, 2011 at 7:32
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You don't need to build your own power supply. 5V @ 2A is readily available as a "cigarette lighter" adapter, an aftermarket solution for charging an iPad etc. in a vehicle. They're cheap enough that you can cut the plug off the wire or take the circuit board out of the case if you don't care about using it in a car.

Here are a few examples:
http://www.amazon.com/Micca-2000mA-Power-Adapter-Slim-HD/dp/B003G8OVNA
http://www.amazon.com/Kensington-K33497US-PowerBolt-Charger-Compatible/dp/B003PU01M4

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  • \$\begingroup\$ Steve indeed says it works fine on a 5V/2A adapter. But if the spec really says 30W I presume his adapter will sooner or later end up in smoke (doesn't always happen immediately). I suspect the same may happen with the Micca and Kensington devices. \$\endgroup\$
    – stevenvh
    Commented Aug 27, 2011 at 17:32

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