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I'm building a lighting project for Burning Man and I have both 12V inputs and 5V inputs that need to be powered off of batteries. Here are the max loads:

  • ATmega328p - 25mA at 5V
  • Addressable LEDs - 100mA at 5V
  • Ultra-bright LED - 1.1A at 5V
  • EL Sequencer/wires, 15 meters - 300mA at 12V

Looks like peak load of 1.5A, although it'll average closer to ~500mA at rest. The LEDs will be dim/off until somebody engages with it.

What would be the best way to power the whole setup? I'm going to have 20 of these, so multiply everything by 20, which may change which power source I use. I also have to run this for 7 nights (12 hours * 7 nights). I want to avoid using a generator due to noise considerations.

Would a couple car batteries be enough? Should I run the addressable LEDs and ATmega on individual 4⨉AAs and the rest on a couple shared car batteries, or would the conversion from 12V to 5V be negligible compared to the ultra-bright LED and EL wires?

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  • \$\begingroup\$ With the numbers you provided, you're talking about a capacity of 7 * 12 * .5 * 20 = 840 AH -- 10 amps continuous, without recharging for a week. Not going to happen, unless you're willing to have a couple of huge batteries together weighing 250 lbs and costing close to $2000. You need to bring your expectations down by a factor of 5 or preferably 10 (resulting in 80AH, which is doable). \$\endgroup\$
    – tcrosley
    Jun 20 '14 at 0:56
  • \$\begingroup\$ Actually $2k and 250 lbs sounds about right. We have a budget of $10k, of which about $6-$8k is for electronics. And 250 lbs is easy, since it'll sit on the ground. I realize that being a battery and totally not hand-portable makes for an odd requirement. \$\endgroup\$ Jun 20 '14 at 0:59
  • \$\begingroup\$ In case you'd like to see the budget, I made it public and commentable on Google Docs/Sheets: docs.google.com/spreadsheets/d/… \$\endgroup\$ Jun 20 '14 at 1:02
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First of all, you don't want to use a car battery, since they are designed to deliver brief high-current bursts ("cranking amps") and then be pretty much constantly charged by an alternator. Instead, you want to use a deep-cycle marine battery, which are designed to be charged infrequently and then discharged down to 25% or so of their capacity.

EDIT: I modified the calculation so the batteries aren't discharged below 20% of their capacity.

You need a capacity of 7 * 12 * .5 * 20 = 840 AH. The first battery I looked is a 6v one:

GPL-6CT: 300 AH 90 lbs $380

It will deliver 8 amps for 2205 minutes, so it could deliver 10 amps for 29.4 hr; derating by 20% (since you don't want to completely drain it) gives 23.5 hr, a little over one quarter what is needed. So you would need eight of these since they are 6v. You would put two of the 6v batteries in series, and power a quarter of your display with that.

The second one is 12v:

GPL-4DL: 210 AH 124 lbs $550

It will deliver 8 amps for 1375 minutes, so it could deliver 10 amps for 18.3 hr; derating by 20% (since you don't want to completely drain it) gives 14.6 hr, one sixth what is needed. So you would also need six of these. However 20 installations doesn't divide by 6 evenly, so this is not a good option to consider.

It might seem to be tempting to run your 5v off the "bottom" 6v battery in each string using a LDO (low-drop out) regulator, but since the majority of your amps goes to the 5v rail, this won't make good use of the "top" battery. So use a high-efficiency switching regulator (like the PYB30-Q24-S5-U) to drop the 12v down to 5v @ 6A. You would use four of these, one for each set of 6v series batteries.

The above calculations and recommendations assume you will not be charging the batteries at all during the seven days. If you do have access to a charger, then the AH requirements are cut by a factor of seven: 12 * .5 * 20 = 120 AH. This can easily be handled by this 12v battery:

GPL-4DL 210 AH 124 lbs $550

It will deliver 8 amps for 1375 minutes, so it could deliver 10 amps for 18.3 hr; derating by 20% (since you don't want to completely drain it) gives 14.66 hr -- so you would only need one of these assuming you charged it each day. (I though of another option where you would use, for example, 10 smaller batteries, one for every two installations, but then you would have to deal with 10 chargers which is not a good idea.)

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  • \$\begingroup\$ I should add that the battery can be charged during the day from other sources. Would that change your answer, in that I can get away with lower spec'ed 6V batteries and recharge them daily. \$\endgroup\$ Jun 20 '14 at 3:37
  • \$\begingroup\$ That changes things considerably -- I'll take another look at lower capacity marine batteries, and let you decide whether you want to deal with chargers or not. \$\endgroup\$
    – tcrosley
    Jun 20 '14 at 3:40
  • \$\begingroup\$ Just to clarify on the high-efficiency switching regulator, I have two options: 1) give all 20 of the installations each their own regulator and run half as many power/ground wires from the marine battery all carrying 12V, or 2) use a single regulator and then run double the wires for both 5V and 12V. Would a buck converter like the LM2596 (12V to 5V with 3A max) work? \$\endgroup\$ Jun 20 '14 at 3:51
  • \$\begingroup\$ Yes, that would work. I went the other way, and chose a 15A one so you would only need two, one for each pair of 12v battery strings. See my revised answer. \$\endgroup\$
    – tcrosley
    Jun 20 '14 at 4:00

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