I'm using a new set-up of a 3.6 kW inverter with two 150 Ah gel batteries in series.

I don't understand why the inverter reports a much larger VA than the watts that are consumed. As I understand it, VA is the actual energy draw from the battery.

As an example, when the reported load is 180 W, the reported VA is 280. That's only around 60% efficiency. Is that normal?

Edit: Attaching some examples (lights are off on purpose, as they would draw +60W):

Case 1

case 1: VA case 1: W case 1: battery A

Case 2

case 2: VA case 2: W case 2: battery A

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    \$\begingroup\$ What load are you driving when you are taking these measurements? \$\endgroup\$
    – vir
    Dec 27, 2022 at 22:58
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    \$\begingroup\$ @GreenScape - The question is what was the load you were using. A load with significant capacitance or inductance or a non-linear load where the AC is rectified to DC will have different characteristics from a simple resistive load such as an incandescent lamp. \$\endgroup\$ Dec 27, 2022 at 23:06
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    \$\begingroup\$ It's not normal to calculate power efficiency that way. It has very little to do with VA. Efficiency is power out divided by power in and, VA hardly ever represents the power inputted to a circuit. \$\endgroup\$
    – Andy aka
    Dec 27, 2022 at 23:49
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    \$\begingroup\$ @GreenScape - all of those items will rectify the AC input to create DC to actually power the device. They will almost certainly have non-linear characteristics and can have a very bad power-factor causing the difference between the true power and VA. As stated by Andy the efficiency is not calculated by the ratio of the two. \$\endgroup\$ Dec 27, 2022 at 23:57
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    \$\begingroup\$ @GreenScape Which then is just DC watts. VA is almost always used in the context of AC and, frankly, I cannot recall it ever being used as just another way to represent DC watts. Except maybe in a textbook saying W=VA. That would be about the only case. \$\endgroup\$
    – jonk
    Dec 28, 2022 at 21:23

4 Answers 4


I think I understand your confusion.

You are trying to power a fridge, don't you?

The VA measure ("apparent" power) has only indirect relation (see the p.s.) to the efficiency. It is a property specific to the AC power setups (not only inverters at any rate).

At any given moment, the power transferred is a product of the current and the voltage in the circuit.

The AC voltage switches direction many times per second (depending on where you live, it is either 50 or 60 cycles per second, this is what the "Hz" measure is for). The AC current does the same, but depending on the load type, it may lag or advance in relation to the voltage. This amounts to the load returning some part of the energy back to the source in each cycle.

If the load misbehaves like this, the "real" (averaged) power transferred from the AC source to the load will be less than the "apparent" power (the product of the averaged current and the averaged voltage).

The ratio between these two "powers" is called a power factor. Good AC loads have a high power factor (near 1.00 or 100%), bad loads have less.

Examples of "good" loads in regard to power factors are incandescent bulbs, heating appliances and modern electronics.

Examples of "bad" loads are most types of electric motors, cheap LED lightbulbs and older electronics.

A sane and efficient inverter is expected to consume input power related to the "real" output power (W) and not to the "apparent" output power (VA). In your case, it could be something like 200W (allowing for ~90% inverter efficiency, normal for a modern inverter).

On the other hand, the inverter output stages need to be engineered for the "apparent" power that may be higher than the "real" power of the load.

This is why inverters have both "real" power (W) and "apparent" power (VA) ratings and this is why your inverter reports both values.

If you keep adding load to your inverter, you could overload it both by the "real" power consumed by the load and by the "apparent" power, independently from each other.

p.s. A minor (in your case) consideration could be that a load with a bad power factor somewhat lowers the inverter efficiency. But this effect is quite minor for a modern inverter and in your case maybe amounts to additional 5W at the inverter input.

p.s. #2 If you want to estimate the real efficiency of your inverter, you need an additional value - the power consumed at the inverter input. If the inverter does not report it (most of them don't), you need a device that measures the battery voltage and the battery discharge current. Devices like this do exist, but they don't add much value to setup like yours so they are rarely used.

p.s. #3 in regard to your comment about the awful efficiency, ~60% is quite low and neither to be expected from a brand new inverter nor particularly safe for it. This would mean that those missing 100W become heat at the inverter. This is a lot and the fan would run almost constantly. One would expect like 2kW load in order to get 100W loss. At least this is how a similar inverter of mine behaves (24V input, 2kW/4kW peak). It has label efficiency of 92-97% and is almost the cheapest I found.

  • \$\begingroup\$ Thanks a lot, great insights! In my case the load is: a PC, two routers and a GPON terminal, plus a led light bulb, occasionally. Hopefully it's all like you're describing and the inverter is showing me VA just so that I can see both 'dimensions' of its limitation: the VA and W. Unfortunately, I don't have any devices to monitor power consumption at the battery side. But I'm going to monitor the readings of inverter itself (it can show a battery draw current), for some time. An maybe return with some photos of its readings. \$\endgroup\$
    – GreenScape
    Dec 28, 2022 at 16:04
  • \$\begingroup\$ If the inverter reports both the battery voltage and the draw current, then you can get the battery-side consumption just by multiplying these values (unlike AC, no caveats there). \$\endgroup\$
    – fraxinus
    Dec 28, 2022 at 16:24
  • \$\begingroup\$ If an inverter is only intended to deal with loads that will never push much power back at it, having it dissipate any such power that is pushed back may be cheaper and easier than having it either store such power until the next cycle or return it to the source. Such an inverter may be "sane and efficient" when used for such loads, but be very inefficient with some kinds of small reactive loads and be quickly forced into thermal shutdown by moderate-sized reactive loads whose apparent power would be well within spec. \$\endgroup\$
    – supercat
    Dec 28, 2022 at 18:50
  • \$\begingroup\$ @supercat pretty much right, but I only hope that inverters like these are not mass-produced anymore. They will even not be cheaper than the proper ones, at least not with the modern element base. \$\endgroup\$
    – fraxinus
    Dec 28, 2022 at 19:05
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    \$\begingroup\$ And the most important - the inverter itself is not a measurement tool. It has functional indicators that can be off by e.g. 15%. \$\endgroup\$
    – fraxinus
    Dec 28, 2022 at 23:55

The ratio of watts to VA is the power factor. Ideally, it should be 1.0 or 100%. But electronic appliances can often have very poor power factors. 0.6 or 60% is believable.

The problem here is in the loads connected, not the inverter.


Watts is the amount of energy going out, VA is the RMS current multiplied by RMS voltage, which may be a little more or perhaps significantly more than the power figure. In practice an inverter will have a maximum power capability which is limited by the magnetics and a maximum current capability that’s limited by the silicon (or perhaps copper). A designer would choose values that are appropriate for the expected load characteristics.

  • 1
    \$\begingroup\$ Unfortunately, it's not about the power capacity. I'm bit worried about the efficiency. It's a gel battery, and it it's recommended to discharge it below 50%. Add to that a supposed low battery-to-consumers energy efficiency and the usable energy of a battery becomes a very sad picture. \$\endgroup\$
    – GreenScape
    Dec 28, 2022 at 7:04
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    \$\begingroup\$ @GreenScape in that case don’t worry about the VA figure; an awful power factor will reduce efficiency but not as badly as it might seem. A power output of 180W might mean an input of maybe 200W but it’s most unlikely to be 280W. If the load is highly inductive or capacitive then the VA figure might represent energy essentially bouncing back and forth between inverter and load, or if the load is complex (eg switch mode supply) then it could be drawing current in a non-sinusoidal way, which isn’t necessarily bad for efficiency. \$\endgroup\$
    – Frog
    Dec 28, 2022 at 9:22
  • \$\begingroup\$ floppydisk, in his question to my OP, has pointed me to look at the battery current. So VA seems to be exactly the battery draw current * battery voltage. In my case it's 11 amps at ~25v. Unfortunately, it's seems to be more likely an awful efficiency :( Can't really understand why is that. BTW, when the load jumps to 2kW (+1.8kW) the VA matches W. Those +1.8kW is a well pump. \$\endgroup\$
    – GreenScape
    Dec 28, 2022 at 15:56
  • \$\begingroup\$ @GreenScape VA measurements for inverters are typically based on the output voltage and current, since VA is a measurement specifically designed for AC systems. It is possible to make a VA measurement for a DC system, just as you could express the battery’s output in dBmW. I doubt very much that this is what the inverter is reporting, have you measured the battery current? \$\endgroup\$
    – Frog
    Dec 28, 2022 at 18:47
  • \$\begingroup\$ please see my update. I've added pics of inverter's readings \$\endgroup\$
    – GreenScape
    Dec 28, 2022 at 19:50

VA is the entire AC sinewave which the generator must deliver.

Watts is the portion of the AC sinewave your load actually uses.

This is a mild deal for motors and other inductive loads, but a very big deal for electronic loads which can be choosy about which portion of the sinewave they want.

So it's your loads. If you doubt that, plug in a purely resistive load like a toaster/and you should see VA = W. You can also use a Kill-a-Watt energy monitor on utility power to read out the appliance's normal behavior on utility AC.

Traditional utility equipment like transformers have to deliver the whole sinewave. Generators slightly less so, and inverters muchly less so. If the load isn't drawing it in that microsecond, the inverter doesn't actually have to deliver it. We don't have a transformer core or generator stator winding that acts like a huge inductor to worry about.

So VA seems to be exactly the battery draw current * battery voltage.

VA has no relevance in the DC world. You might have a spiky choppy load, but in DC, stick a sufficiently large capacitor in front of it and it's not spiky anymore.

Why are you bothering with AC anyway?

In my case the load is: a PC, two routers and a GPON terminal, plus a led light bulb, occasionally.

Why are you inverting? All that stuff will run cheerfully on 12 volts DC.

The routers probably have a wall wart that makes 12 volts DC, or you can get routers that do. 12 volt LEDs are really not a problem. A PC power supply, internally they are 5-12 volts and plenty of people sell PC PSUs that input 12 volts. Google kinda started it, as an energy saving measure, so they could run their cheap PC arrays directly off the UPS batteries instead of the inefficiency and stupidity of a double conversion.

Efficiency is job ONE

Nothing annoys me more than off-grid people who rack $4000 of batteries specifically so they can keep using their 1980s piece of crap refrigerator that draws an average of 400 watts. They need to spend $400 on batteries and then $900 on a new fridge that averages 40 watts.

Conservation, specifically replacing inefficient stuff, must come before the battery budget.

Your biggest problem, beyond double conversion, is that ??? PC that was not designed in any way, shape or form for energy efficiency. I don't know if your application is off-grid or just a homebrew UPS for "enough time to close down databases and flush disk caches"... but if you want it to be a runner, your #1 way to save on batteries is an efficient PC, such as the 2018 Intel Mac Mini + BootCamp. (The last Intel Mini). Keep a small partition with MacOS just for maintenance, and run Linux or whatever on the bare metal. The Mac Mini runs natively on 12V and conversion isn't hard. And a 2018 Mac Mini is cheaper than a few hours worth of batteries.

  • 1
    \$\begingroup\$ Harper, how much big is a fridge that averages 400W? A week ago i did throw away a broken 250-liter 1987 freezer that with its profoundly degraded insulating foam ate at most 1.0kWh per day. \$\endgroup\$
    – fraxinus
    Dec 29, 2022 at 0:03
  • \$\begingroup\$ @fraxinus it was a hypothetical example. Though I have an old 80s vodka freezer (I assume that, since it has a glass door) which pulls nearly that. I rarely ever run it. \$\endgroup\$ Dec 29, 2022 at 0:29
  • \$\begingroup\$ Overnight it is obvious - a big (family size, 250-300l) absorption (ammonia bubbling) fridge will be like 500W on average. But these were not made much even in 1980s \$\endgroup\$
    – fraxinus
    Dec 29, 2022 at 6:56
  • \$\begingroup\$ It's a whole-house inverter. So, unfortunately, not way to supply 12V to PC. Moreover, I doubt I'll find a PC's power supply that would accept 12V. Novaday, almost all devices use DC, so they do a double conversion indeed. And my fridge is pretty modern and consumes like <50W while running. My question, mostly was regarding the efficiency. It was hard to believe that efficiency can go this low, so I wanted some thought of the experts. And I've got a lot of the, so I appreciate all. My specific use case is to be able to work from home during blackouts that can last multiple hours. \$\endgroup\$
    – GreenScape
    Dec 29, 2022 at 8:28
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    \$\begingroup\$ @GreenScape Google "12V psu for pc" and you'll get plenty, I had no trouble finding them. I recommend you get a Kill-a-Watt and look at the VA vs watts of the appliances when they are on utility power so you have a baseline. I know your question was about the efficiency but you don't have the topic knowledge to understand the answer. Trying to help with that! Sorry if it annoys. \$\endgroup\$ Dec 29, 2022 at 9:25

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