# How to calculate battery capacity for inverter systems

To find out how many batteries for your inverter. The rule is“*maximize run time, minimize the battery size and cost*.”

The formula is :

Battery Capacity(WH)*Discharge coefficiency*Inverter efficiency=Load wattage(W) * Runtime(H)

## If you know the load watts instead of amps, follow the following procedure

**Step A: Convert watts to amps**

Actually, watts is the fundamental unit of power and watt-hours is the energy stored. The key is to use the watts you know to calculate the amps at the battery voltage .

For example, say you want to run a 250 watt 110VAC light bulb from an inverter for 5 hours.

Watt-hours = watts * hours = 250 watts * 5 hours = 1250 watt hours

Account for the efficiency of the inverter, say 85%

Watt-hours = watts * hours / efficiency = 1250 / 0.85 = 1470 watt-hours

Since watts = amps * volts divide the watt hours by the voltage of the battery to get amp-hours of battery storage

Amp-hours (at 12 volts) = watt-hours / 12 volts = 1470 / 12 = 122.5 amp-hours.

If you are using a different voltage battery the amp-hours will change by dividing it by the battery voltage you are using.

**If you know the amps drawn on battery, follow the following ****procedure**

**Step 1.**

If the current drawn from battery is *x* amps, the time is *T* hours then the capacity *C* in amp-hours is

*C = xT*

For example, if your pump is drawing 120 A and you want it to run for 24 hours

*C = *12 Amps * 24 hours = 2880 amp hours

**Step 2**. Cycle life considerations

It isn’t good to run a battery all the way down to zero during each charge cycle. For example, if you want to use a lead acid battery for many cycles you shouldn’t run it past 80% of its charge, leaving 20% left in the battery. This not only extends the number of cycles you get, but lets the battery degrade by 20% before you start getting less run time than the design calls for

*C’ = C/0.8*

For the example above

*C’ = 2880 AH / 0.8 = 3600 AH*

**Step 3**: Rate of discharge considerations

Some battery chemistries give much fewer amp hours if you discharge them fast. This is called the Peukart effect. This is a big effect in alkaline, carbon zinc, zinc-air and lead acid batteries. For example if you draw at 1C on a lead acid battery you will only get half of the capacity that you would have if you had drawn at 0.05C. It is a small effect in NiCad, Lithium Ion, Lithium Polymer, and NiMH batteries.

For lead acid batteries the rated capacity (i.e. the number of AH stamped on the side of the battery) is typically given for a 20 hour discharge rate. If you are discharging at a slow rate you will get the rated number of amp-hours out of them. However, at high discharge rates the capacity falls steeply. A rule of thumb is that for a 1 hour discharge rate (i.e. drawing 10 amps from a 10 amp hour battery, or 1C) you will only get half of the rated capacity (or 5 amp-hours from a 10 amp-hour battery). Charts that detail this effect for different discharge rate can be used for greater accuracy.

For example, if your portable guitar amplifier is drawing a steady 20 amps and you want it to last 1 hour you would start out with Step 1:

C=20 amps * 1 hour = 20 AH

Then proceed to Step 2

C’ = 20 AH / 0.8 = 25 AH

Then take the high rate into account

C’‘=25 /.5 = 50 AH

Thus you would need a 50 amp hour sealed lead acid battery to run the amplifier for 1 hour at 20 amps average draw.

**Step 4.** What if you don’t have a constant load? The obvious thing to do is the thing to do. Figure out an average power drawn. Consider a repetitive cycle where each cycle is 1 hour. It consists of 20 amps for 1 second followed by 0.1 amps for the rest of the hour. The average current would be calculated as follows.

20*1/3600 + 0.1(3599)/3600 = 0.1044 amps average current.

(3600 is the number of seconds in an hour).

In other words, figure out how many amps is drawn on average and use steps 1 and 2. Step 3 is very difficult to predict in the case where you have small periods of high current. The news is good, a steady draw of 1C will lower the capacity much more than short 1C pulses followed by a rest period. So if the average current drawn is about a 20 hour rate, then you will get closer to the capacity predicted by a 20 hour rate, even though you are drawing it in high current pulses. Actual test data is hard to come by without doing the test yourself.

Thank you very much!

so if I have a small fridge that draws 8.4 amps that I need to run for six hours

I will need a battery load of 63ah/h to do that for me? if I work out your calculation right

Raymond, is it 220Vac? If so, the power for the fridge is 220V*8.4A*6H=11088VA. Considering the DC to AC efficiency, your battery should discharge 11088VA/0.85=13044VA power to support it.