EDIT: on hold while I check conditions

@Harper pointed to the fact that my battery is rated to draw 165A continuously for an hour, which should cover the needs of my inverter when my vacuum starts up. I do sometime have to switch over to a backup battery, and I need to verify this hasn't somehow interfered with the measurements. In the worst case, it might mean that the measurements were made using a 136Ah AGM battery from Sunwind with a rating of 67Ah with C=1 - about half that of the Concorde.

Still, the points on wire thickness might be relevant ... I usually employ really thick (35mm2 - 70mm2) jumper cable wires and the length between the inverter and battery is less than a metre, but in this case I have just relied on the cables supplied with the inverter, which are far from thick: maybe 2,5mm2.

I have a single 12V battery (a 305Ah Concorde Sun Xtender 2580L) feeding the power needs of my off-grid cabin. This has worked well, but I recently got a 2000W inverter which introduced a new problem: power surge and continuous high-power delivery.

I had figured out that my setup would be fine, since I only needed this inverter for ten minutes or so during a weekend (powering a 220V 1400 Watt vacuum cleaner), but this failed to work: as soon as the vacuum powers up, the shown voltage on the inverter drops from 12.3V to 9.3V. I take this as a sign that a single battery is unable to sustain such a high load. What I'd like to know is this:

How can I know how "big" a battery bank I need to sustain X Watts in continuous power draw? Is there some indication in the battery specs that will help me in finding the numbers, or are people just using some heuristic, like "max power of a battery bank is approx 2x its numeric rating in Ah"?

P.S. Feel free to talk to me in Volts, Ampere hours and Watts. I can do the math :)

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    How long are the wires between battery and inverter? What size are they? What is the nameplate power draw of the vacuum cleaner, in either watts, VA or amps+volts (120 v 230)? Sep 12 '19 at 16:58
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    Consider using a sweeper that runs off its own, built-in rechargeable battery. Sep 12 '19 at 17:15
  • I put the question "on hold" after seeing the comments. I need to verify that everything works as I said, because it might be that I was measuring using a battery that could only deliver 67A (C=1). In any case, the cable length is less than 1m, the power draw is 1400W (at 220V, which shouldn't matter), but the cable thickness is as delivered with the inverter: thin.
    – oligofren
    Sep 13 '19 at 13:30

I spent way too long thinking I was going off-grid and then didn't, as the system cost to do what I needed and the grid connection cost crossed paths when the power company had a slow-down in new connections and got suddenly reasonable...

305 Ah battery is going to have a C/20 rate of 15 amps, for roughly 183 watts on a 12V system. C/20 is the discharge rate over a 20 hour period and is typically how the "305 Ah" is rated in the first place (you get more Ah capacity discharging slower and less discharging faster - and the Ah tends to drop from rated capacity as the system ages.)

A good "rule of thumb" on (lead/acid) battery bank size to support a certain draw is to stick at or below the C/20 rate. If your inverter supports multiple input voltages, I would strongly suggest going to a higher voltage (24, 36 or 48V) since maintaining a battery bank charged in parallel is much more prone to problems than a series string.

Or, get a different vacuum, or a non-electric sweeper. You don't provide details about your vacuum cleaner, but many are quite high Amperage draw at 120V (or 240V), which will be more than 10 times (or 20 times) as much draw at 12V (inverters are not 100% efficient.) That will result in a very significant increase in the size of the battery bank you need. Alternatively, you fire up a generator for 15 minutes once a week to meet your 10 minutes once a week of high power draw, and haul fuel for it, and maintain it.

  • The battery does not publish a C/20 rate, but does publish a C/24 rate of 258AH or 11.5A at that rate. That is 138W. Sep 12 '19 at 20:27

The battery is fine. It's so totally your wires!

TLDR: you guess 2.5mm2; at 165A that right there would for 2.3V of your drop, assuming the wire is legit and the wire doesn't have even more resistance from glowing cherry red at these currents. You need more like 100mm Al.

Nah, I was guessing you had thermally legal wire size but a long distance. Sizing issues. Getcha everytime.

By the way, that thing that happened, where everyone jumped on the wrong bandwagon, that sometimes happens when the asker leads with a theory or presumption.

Forget the battery; it could start a truck

Literally. It's not optimized for engine starting, but it could do it. Lead-acid batteries' redeeming quality is they can surge crazy amounts of current. That's why we love them for engine starting. Indeed, this battery is rated 165AH at a 1-hour rate, meaning it's rated to pull 165A or about 2000 watts continuously for an hour - far more than a vacuum cleaner needs to start and run. And sure enough, the vacuum does start; that's not your complaint. Your complaint is the voltage drop at the inverter.

Well, we know all about voltage drop. We keep having this "3%" conversation.

Voltage drop is a different kettle of fish

First, I need to say four words: "Aluminum Is My Friend". It was not the friend of people wiring 15/20A 120V branch circuits with AA-1350 alloy in the 1970s, but it is your friend for a feeder if you are going any significant distance.

Understand, voltage drop on 12V systems is greatly amplified by the severity of voltage drop when you only have 12V to begin with. On a 120V circuit we are content with a 4 volt drop (3.3%) but on a 12V system, 4V is 33% (10 times worse).

And to add to the tyranny, to pull the same usable wattage/power, current/amperage must normally be 10 times more; that makes the voltage drop problem 10 times worse again. So, 100 times worse. Seriously.

Once you're up to 120V, the voltage drop issue is only 1/100 as bad. Pardon me, 240V, only 1/400 as bad.

Your voltage drop

So you say your setup is "wire that looks like 2.5mm" from the battery to the inverter, and you acknowledge that's a bit small.

Yeah. 2.5mm wire has 0.007 ohms per metre. E=IR, 0.007*165 = 1.15 volts. Normally nothing to worry about, but this is 12V so we're down 10% already. And the wire is a round trip, so 2.3 volts.

  • But the inverter is dynamic; presented with 20% voltage drop it will draw 20% more current for the same wattage output. So recalc with 199 amps, rinse wash repeat.
  • Further, the 2-wire cable would be emitting 10 watts per inch which is enough to anneal the wire, changing its resistance.
  • And since UL, BSI, TUV etc. would never list an inverter that sold with 2.5mm wire on the primary, we could be dealing with cheap Cheese, and they love using "fake" wires that are "all insulation" and very little metal.
  • In the US, a 165A circuit would require #2/0 Cu (67mm2) or #4/0 Al (107mm2) simply to not exceed thermal limits, nevermind voltage drop. So you're off by a factor of 25 here.

Even if some of the above facts are a bit off, the overall arc is that this setup is totally unworkable by an order of magnitude, and proper cables should solve the problem.

One more thing; off-grid power is 10% generation and 90% conservation. It's cheaper to buy efficient appliances than build out capacity to support inefficient ones. You need to replace those wires regardless, but I would look at a lower amperage vacuum cleaner before I'd look at increasing inverter/battery size.

  • The wire size may well be an issue, but I do want to note that a battery rated to deliver 165A may not be rated to deliver that without any voltage drop -- batteries have internal resistance, and dropping to 9 or 10V when 165A are pulled may be normal for this battery. If that's the case, since the battery is rated for 165A draws, if the inverter is okay with that input voltage, the correct answer may be "don't worry about it".
    – Nate S.
    Sep 12 '19 at 19:33
  • @NateStrickland a battery that large having such high internal resistance doesn't match my experience with large lead-acids. That would suggest a defective battery. Sep 12 '19 at 20:28
  • 1
    "Lead acid has a very low internal resistance and the battery responds well to high current bursts that last for a few seconds. Due to inherent sluggishness, however, lead acid does not perform well on a sustained high current discharge; the battery soon gets tired and needs a rest to recover. Some sluggishness is apparent in all batteries at different degrees but it is especially pronounced with lead acid. This hints that power delivery is not based on internal resistance alone but also on the responsiveness of the chemistry, as well as temperature."
    – Nate S.
    Sep 12 '19 at 22:48
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    To verify that it's voltage drop from your wires being too long and/or too small of gauge, measure the voltage of the battery with a voltmeter when the inverter input voltage is 9.3V. If your battery is at a significantly higher voltage at the same time, then your wires are definitely the problem. Shorten the length or increase the size of your 12V wires.
    – Dotes
    Sep 13 '19 at 13:12

The battery is only 305 ah I would be looking for a 1000 ah deep cycle starting battery. The battery voltage diving is a sure fire indication that the battery can not supply the current the inverter is trying to draw. If the inverter was over heating that would indicate it was an inverter problem, but with the significant voltage drop the battery just can not provide that much. You can also use multiple batteries depending on your inverter and charge controller they might be able to be put in series I like this method if possible because a 2nd battery in parallel sometimes 1 battery has a bad connection so it doesn’t fully charge. And you don’t know. parallel is more common is small systems. Both batteries should be the same size, I normally swap them in pairs if in parallel but it would be less expensive for you to get another battery the same as your first and put it in parallel, this would increase the available current and reduce the drop in voltage to a point it will probably work.

  • That doesn't really answer the question, though. I was wondering how to compute/find the max power rating. It seems other have pointed to the tech specs of using the discharge values for various time lengths to get the number. The 305Ah was at C=100, but at C=1 it was 165Ah, meaning it should sustain 165A over one hour. That's 165*12VA=1980W.
    – oligofren
    Sep 13 '19 at 13:34

Your battery is struggling - 2000W on a 12V battery needs 166A and that does not include any efficiency loss.

Also, when some items start (big with motors etc), they need to draw more for a small amount of time, a rule of thumb is to multiply the load by 4... So, a 1500W vacuum cleaner may well need a 6000W inverter to start easily...

  • I know my battery is struggling. That was the original issue. My question was how to calculate in advance what kind of setup I would need to drive it. I know a 100Ah LiFePo4 battery can deliver a peak of 1280W, but AGM batteries have much higher internal resistance.
    – oligofren
    Sep 13 '19 at 12:37

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