July 5, 2011 7:02 PM Subscribe

I'm setting up my "off the grid" solar lighting system in my home, and I'd like to connect the battery to some LED lights. It's a 12 volt DC battery - Do my lights have to be 12 volt also? If they're LED bulbs, they'll have a transformer inside; will that allow me to use them, or do I have to have a voltage converter or something? I've got some edison-style LED bulbs around the house for 85-265 volts.
Thanks in advance for any help. pictures and web links that can explain things simply are helpful. (An earlier, relavent post:
http://ask.metafilter.com/182209/How-to-run-some-lights-from-a-solar-panel)

posted by eggrollover to Technology (4 answers total) 5 users marked this as a favorite

posted by eggrollover to Technology (4 answers total) 5 users marked this as a favorite

It's much more efficient to avoid the double conversion; solar-off-the-grid people often have a whole house's worth of 12v or 24v wiring and specialized or modified appliances. The efficiency translates directly to how long you can keep your lights on in the winter, not just a change in your electric bill. You might try browsing some issues of homepower or companies like backwoods solar or a class from SEI.

posted by hattifattener at 10:09 PM on July 5, 2011

posted by hattifattener at 10:09 PM on July 5, 2011

An inverter can be a real power hog in itself. I would find some 12 volt leds.

posted by psycho-alchemy at 1:23 AM on July 6, 2011

posted by psycho-alchemy at 1:23 AM on July 6, 2011

Losses from double conversion are real, but depending on your wiring configuration, might well end up being the least available evil.

The formula relating voltage (V) in volts, current (I) in amps, and power (P) in watts, is P = I * V; this implies that a 120V appliance will draw*one tenth* the current drawn by a 12V appliance of equivalent power.

The formula relating resistance (R) in ohms, current (I) in amps, and power (P) in watts, is P = I^{2} * R. So, for a wiring run of any given resistance R, cutting the current by a factor of 10 will *reduce the wiring loss by a factor of 100.*

Another way of looking at this is that for any given 120V power distribution network, you*can* design a 12V network that wastes no more power; but you need to use wires with a hundred times the cross-sectional area (ten times the diameter) which apart from being wasteful is very very costly. And this is exactly why mains electricity is supplied at voltages high enough to be dangerous.

Just for the sake of getting a feel, let's say you want to power 100W of 12V lighting 50m from your battery, and let's run some numbers for two kinds of cable: two core twin sheath 110A rated 6mm^{2} at $12.95/metre, and two core twin sheath 20A rated 1.84mm^{2} (which is about the same as ordinary 110V lighting cable) at $1.95/metre.

First thing is cable cost: a 50m run of the thick one would cost you $647, vs. $97.50 for the thin.

Now for power loss. The resistance of 100m (there and back) of the thick cable would be 0.28 ohm; thin cable: 0.91 ohm. 100W at 12V is 8.3 amps, so wiring losses would be (8.3 amps)^{2} * 0.28 ohm = 19W for thick cable; (8.3 amps)^{2} * 0.91 ohm = 63W. That wiring loss is so high compared to the actual lighting load that you wouldn't actually get anywhere near 12V at the lights by feeding the wiring from a 12V battery.

Now let's look at the 120V case: 100W at 120V is 0.83A. Thick cable loss is therefore (0.83 amps)^{2} * 0.28 ohm = 0.19W; thin cable: (0.83 amps)^{2} * 0.91 ohm = 0.63W.* In both cases, wiring losses are now negligible.*

So if you're going to run anything more than a trivial amount of lighting in anything bigger than a small house, you might well be better off putting a decent inverter close to your battery bank, and using conventional 110V wiring and 110V-compatible LED fixtures throughout the house. Inverters can be had with better than 90% efficiency, which would keep your wastage down around 10W for a 100W load.

Those 85-260V LED lamps you have will have a switching power supply inside, designed to convert whatever mains voltage they see into a constant current for the actual LEDs. I would not be at all surprised to find better than 90% efficiency inside that supply as well, which is probably about as good as you'd get from*any* circuit designed to match the constant-current requirement of a high-power LED to *any* constant-voltage supply.

Wiring losses between panels and battery bank, and between battery bank and inverter, will also be lower as the battery bank's design voltage increases. You'd only ever use a 12V battery bank for a really small, almost toy installation; anything serious and you're looking at a 48V bank almost straight away.

posted by flabdablet at 1:46 AM on July 6, 2011 [5 favorites]

The formula relating voltage (V) in volts, current (I) in amps, and power (P) in watts, is P = I * V; this implies that a 120V appliance will draw

The formula relating resistance (R) in ohms, current (I) in amps, and power (P) in watts, is P = I

Another way of looking at this is that for any given 120V power distribution network, you

Just for the sake of getting a feel, let's say you want to power 100W of 12V lighting 50m from your battery, and let's run some numbers for two kinds of cable: two core twin sheath 110A rated 6mm

First thing is cable cost: a 50m run of the thick one would cost you $647, vs. $97.50 for the thin.

Now for power loss. The resistance of 100m (there and back) of the thick cable would be 0.28 ohm; thin cable: 0.91 ohm. 100W at 12V is 8.3 amps, so wiring losses would be (8.3 amps)

Now let's look at the 120V case: 100W at 120V is 0.83A. Thick cable loss is therefore (0.83 amps)

So if you're going to run anything more than a trivial amount of lighting in anything bigger than a small house, you might well be better off putting a decent inverter close to your battery bank, and using conventional 110V wiring and 110V-compatible LED fixtures throughout the house. Inverters can be had with better than 90% efficiency, which would keep your wastage down around 10W for a 100W load.

Those 85-260V LED lamps you have will have a switching power supply inside, designed to convert whatever mains voltage they see into a constant current for the actual LEDs. I would not be at all surprised to find better than 90% efficiency inside that supply as well, which is probably about as good as you'd get from

Wiring losses between panels and battery bank, and between battery bank and inverter, will also be lower as the battery bank's design voltage increases. You'd only ever use a 12V battery bank for a really small, almost toy installation; anything serious and you're looking at a 48V bank almost straight away.

posted by flabdablet at 1:46 AM on July 6, 2011 [5 favorites]

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You can easily convert from 12V DC to 110V AC with a common power inverter. They can be had for about $30. You connect the battery terminals on one side, and on the other side they have a regular wall outlet so you can plug in your lights. Here is an example. They can get more expensive if you need one to run high-wattage gear like amplifiers and outdoor spotlights; the one I linked is 400W which could handle a few dozen LED bulbs no problem. If you just have screw-in 110V bulbs and need to connect them to something, just buy an ordinary fixture or lamp with a wall plug and plug it into the inverter.

Alternatively you can get 12V DC bulbs and connect them directly to the battery. This would be more efficient because you will avoid power losses in the inverter, which might be something to think about if you are relying on solar power and the power budget is tight. AC LED bulbs are actually converting 110V AC back into DC internally so there is a double conversion going on which means you pay the penalty twice.

Burning Man attendees often grapple with these issues which is where I got my chops. I found this page to be a nice summary.

posted by PercussivePaul at 7:28 PM on July 5, 2011