# How long would a fully charged 230AH 12V battery last when using it with a 12V->230V DC>AC inverter, drawing an average of 1500 watts power?

• This sort of battery.
• This sort of transformer.

Powering a 1500Watt PA for how long, and how do you work out the AH when using a transformer?

As best as I can figure out it would be something like:

230AH * 0.85 (roughly 15% lost through the transformer) = 195.5AH

Using 1500 watts with 230 volts means the current required would be (I=P/V) = 1500/195.5 = 7.7Amps

This would mean the battery would would last for 195.5AH / 7.7 amperes = 25.4.hours, which seems very unlikely.

I changed the formula to use the voltage of the battery instead of the transformer:

Using 1500 watts with 12Volts means the current required would be (I=P/V) = 1500/12 = 125 amperes

This would mean the battery would would last for 195.5AH / 125.amperes = 1.6 hours, which seems more likely.but is it correct?

• Assuming "PA" means "public address" aka "amplifier for those huge ass speakers used in concerts," then it gets more complicated. They don't draw full power constantly. The run time you calculate will be "worst case" - non-stop heavy metal at full volume. You'll probably have a longer run time than you calculate, and you can extend it some extent by turning down the volume.
– JRE
Jul 3, 2020 at 11:07
• ... and if you can find a PA with a class D output rather than a class B/AB, you'll increase your run time significantly again. Jul 3, 2020 at 12:28
• @JRE Having worked with PA amplifier design, we always used 1/8 of the rated RMS power (a bit of a misnomer in itself) as "music average". If your music is dubstep or ironically Enya, this value is much higher. Radio chatter with some music and it's much lower. There are AES normes for creating pulsed pink noise with said avaerge and peaks to emulate average music with various crest factors. Jul 3, 2020 at 12:46

Say you draw 1700w due to inverter efficiency, which is 142Amps. 230Ah capacity is usually at 0.01C which means this is the capacity at (Battery Ah)*0.1 Amps Draw - which is 23Amps in your case. If you go to 0.5C (115A) for example, your capacity is around 70%. After understanding this, there are a few more things to be aware of:

• On such amperage you are going to create a lot of heat, I don't think you want to run on such amperage for a long time. (Rule of thumb - don't let your battery get to 50C). Batteries that get warm while operating/in storage die very early
• Lead Acid batteries should not be discharged to more than 50-70% (depending of their technology and quality) of their full capacity - on your case it means DO NOT discharge more than 115Ah! (This is called DoD, discharging more than this will shorten the life of your battery drastically)
• This amperage requirs very thick cables, consider going to higher voltage batteries (in series) and inverter to lower the amperage

The converter efficiency might mean that a load of 1500 watts might cause a power input draw from the 12 volt battery of 1700 watts. That's a battery current of 1700/12 = 142 amps. Given also that you won't be able to "use" all the ampere hours of the battery, I reckon you'll be looking at 180 Ah at best so, the duration becomes about 1 hour 15 minutes.

• thanks, this is really useful! Jul 3, 2020 at 11:08
• Plus you don't want to drain a lead-acid battery below 50% charged very often, so unless it's a LiFePO4 battery, the practical duration becomes 40 minutes. Jul 3, 2020 at 11:44
• @pampthehorn Also, the inverter is only rated for 1000W, so it won't power a 1200W constat load. Fortunately, the PA system does not constantly draw 1200W. So time time depends on the actual average power, which is smaller. Jul 3, 2020 at 12:32

Your mistake is treating amp-hours (Ah) as energy. They are not. Amp-hours are units of charge. The efficiency of the inverter applies to the power, not the current:

$$\rm output\ power = input\ power \times efficiency$$

So if you have a 1500 watt load, your average input power would be:

$$\frac{1500 \mathrm W}{85\%} \approx 1765 \mathrm W$$

Which means your average input current would be:

$$\frac{1765 \mathrm W}{12 \mathrm V} \approx 147 \mathrm A$$

Which gives a run time of:

$$\frac{230 \mathrm{Ah}} {147 \mathrm A} = 1.56\ \mathrm {hours}$$

Note that 147 amps is a dangerously large current, so be sure your battery is wired to the inverter properly.