Can an inverter be supplied by a battery with lower AH than required for designed VA?

Question

I'm looking to provide backup power to a home CCTV system.

The PoE switch has a maximum rated dissipation of 78W. Each camera draws a peak of 7W. Since each camera is powered by the PoE, I assume the dissipation of each camera does not need to be included separately.

The video recorder has a rated dissipation of '< 2W (without HDD)'.

The HDD specification sheet itself states a R/W demand of 4W.

The sum value then is 78+2+4 = 84W.

Allowing for around 60% efficiency, 84/0.6 = 140W.

To be able to provide 6 hours of backup using a 24V system, the batteries need to provide 35AH. The only purpose of this backup system - at present - is to provide juice to the CCTV System. I may in the future add the perimeter lights, and well-pump. Can the inverter be a high 900VA device supplied by low AH batteries E.g. 2 Car batteries rated at 45AH each?

Answer 1

The PoE switch has a maximum rated dissipation of 78W. Each camera draws a peak of 7W. Since each camera is powered by the PoE, I assume the dissipation of each camera does not need to be included separately.

I would expect that the PoE would be ~75% efficient so 14W load means ~20W requirement.

Can the inverter be a high 900VA device supplied by low AH batteries E.g. 2 Car batteries rated at 45AH each?

It can, as it will only take what power it needs to supply the load. But it will be supplying about 26W which is well under 10% of its maximum, so will likely not be efficient at that level of load. A smaller inverter will probably be a better choice.

Answer 2

Amp*Hours is a unit of capacity. It is the product of current (in A) and time (in hours). A battery rated for 45Ah has enough capacity to output 1A for 45h, or 10A for 4.5h, etc. In theory.

Thanks for Ecnerwal's comment: always read the fine print. Batteries run out of steam at high current due to internal resistance, so the manufacturer may quote capacity at a very low current to get a bigger number. At higher current, it will be lower. Always read the fine print...

Amp*Hours has an indirect relationship to the maximum current and maximum power the battery can provide, which is really what you're asking. A bigger battery with a higher capacity in Ah will usually also offer higher sustained current (Amps) and power (Watts). This is simply because it's bigger, so it has more metal, more area for chemical reactions, lower resistance, etc.

However, the number of Ah is not directly related to the number of Amps. To use a car analogy, the capacity in Ah is the size of your fuel tank, but that's not directly related to the peak power of the engine.

Note if your inverter is too large, it could draw more power than a smaller one even if it is idling.

Anyway. Lead-acid batteries intended for cars are designed to deliver huge current to crank the engine, but they are not designed for high capacity. The car has an alternator, so once it is started, it'll charge the battery, making its capacity irrelevant. This type of battery construction has a huge disadvantage in your application: if you discharge the battery too deep, even once, it will die.

If you want to use lead-acid, you need a deep-cycle battery, which is designed for this type of use. It will survive charge/discharge cycles much better, but you still need the inverter to monitor battery voltage and shutdown before the battery is discharged too deep, otherwise it will die.

These are more expensive according to the Ah number on the label, but contrary to the car batteries, you can actually use (most of) the rated capacity without killing them.

It's not really a problem to use a 900VA inverter on a small battery as long as the inverter does not draw more amps than what the battery can provide. If you get deep-cycle batteries intended for storing energy and providing back-up power, the maximum current they can provide should be specified on the label, along with maximum charge current and other information that is useful to setup the backup system.

Answer 3

You're getting your units mixed up. VA (Volt-Amps) is loosely related to W (Watts) and is the instantaneous amount of power used buy a device or delivered by a source.
Ah (Amp-Hours) is the energy storage capacity of a battery, and does not by itself tell you what the instantaneous power delivery capability of the battery is - it tells you how much current can be delivered for a period of hour, leaving the battery completely dead flat (which isn't something you should do to a car battery if you want it to survive for long).

So to answer your question: Yes a 'high' 900VA inverter can be quite happily powered by a pair of 'low' 45Ah batteries - there's no conflict here because those are 2 different units of measurement and they're measuring 2 different things.

If you really do want to work out if the batteries are properly matched to the UPS, then you'll need some additional information about the batteries - and for car batteries you should be able to find their CCA (Cold Cranking Amps) spec.
Since a car battery's CCA is only expected to be used for no more than 30 seconds or so and a typical UPS is expected to run for at least 5-10 minutes at full load, you'll want to divide that CCA down to maybe 20% for the next calculations.
A typical 12V 45Ah car battery might have a CCA rating of 250A - so 20% of that is 50A.
If your inverter as 2 batteries connected, they'll either be in series or in parallel, so there's 2 ways to do the next calculation - but the result will be the same.

If they're in series, then you have: (12V + 12V) x 50A = 1200W
If they're in parallel, then you have: 12V x (50A + 50A) = 1200W

We won't be too optimistic about the efficiency of the inverter and estimate it manages to turn 80% of its input power into output power, so 80% of 1200W in is 960W out.
As it happens, this result comes quite close to your UPS's advertised 900VA rating - but this is just coincidence.

Answer 4

Does all that equipment really need to run on 120V?

The #1 waste I see in designing this sort of system is insisting on using inverter to make 120VAC, and then plugging 12 volt DC power supplies into the 120VAC, and the equipment actually runs on 12 volts DC.

The fallacy of that should be obvious. Why not just hot-shot the equipment straight off battery? (perhaps at a different battery voltage).

Because you will have considerable conversion losses both ways, and I see where you are trying to account for that. But what you're overlooking is inverter losses aren't just a percentage, they are also a "flat rate" required to have the inverter "spun up" at all. For low power loads like yours, the parasitic loss of the inverter merely being spun up could be the largest load in the system.

78W is not an accurate number for the switch.

The 78W is a nameplate limit of the PoE switch, you're not allowed to hang more than 78 watts of PoE load on the switch. However the switch's actual draw will be the sum of

• the actual load of the PoE devices which are connected <=78W
• conversion losses for that load
• its parasitic load simply for being powered up and switching

The 78W only accounts for the first line item, but is situational based on your actual loads. Example, suppose you have 5 cameras, that's 35W (well within 78W limit), 40% conversion loss for 14W, and 10W of parasitic "just to be powered up" load. So 59W, that is an example of the switch drawing less than 78W, but it could be more too.

Those other things will need a lot more power

I may in the future add the perimeter lights, and well-pump. Can the inverter be a high 900VA device supplied by low AH batteries E.g. 2 Car batteries rated at 45AH each?

The inverter might suffice but the battery sure won't. It can barely carry the load you are planning. The new loads you want to add are massive by comparison, and will require a much larger battery.

They make DC lights. The well pump might need to be AC but you could give it its own inverter and only turn it on when the pump needs to be running.

But since you do not have the extra loads yet, you'd be spinning up a heavy inverter and eating its larger parasitic ("spun up and ready") load.

Choose appropriate batteries and use within their range.

You see lead-acid batteries with an amp-hour spec, and you think "well, I can use all of that every day for years". Nope. you can use that once.

They don't tell you this, because it's "common knowledge" in the industry. But lead-acid batteries take damage from being discharged. Best practice is to size lead-acid battery banks so they normally do not use more than 25-33% of battery capacity.

Yeah. Ouch.

But the upside of lead-acid is it's cheap. That's the appeal, of course.

Note that all lead-acids are not built the same. "Car batteries" are built for an enormously large energy impulse (starting an engine) and NOT optimized for deep cycling, and would soon fail even if only 25% of the cycle were used daily. For that, you need a different optimization called a "deep cycle battery". Again only 25-35% though.

Probably the ideal lead-acid deep cycle battery is "golf cart batteries", especially since golf cart owners consider them scrap when they don't make it a full 18 holes once. And so used golf cart batteries are cheap and have a lot of usable life left.

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Just the same, lithium and other technologies are not without their limits. For instance some lithium technologies shouldn't be used regularly across more than 80% of their nameplate range without degrading pack life, though the effect is nowhere as severe as lead-acid. (also 80% is certainly more reasonable than 30%).