A battery bank has three capacity figures and only one of them is useful. The nameplate figure is amp hours times voltage: 100 Ah at 12 V is 1200 watt hours. Usable capacity is that multiplied by the depth of discharge you are willing to use, and then again by inverter efficiency, because converting DC to mains AC costs energy before the load sees any of it.
Depth of discharge is a chemistry question, not a preference. Lead-acid batteries lose cycle life quickly below about 50 per cent, so 50 is the conventional limit; lithium iron phosphate is routinely specified at 80 to 90 per cent. Inverter efficiency for a decent pure sine wave unit runs around 85 to 90 per cent under reasonable load, and worse when lightly loaded. Both are editable because both depend on the hardware you own.
So 100 Ah at 12 V with 80 per cent depth of discharge and 88 per cent inverter efficiency gives 845 usable watt hours. Carrying a 300 watt load, that is 2.82 hours. Note that lead-acid capacity ratings assume a slow 20 hour discharge, and drawing hard reduces the effective capacity further under Peukert losses that this calculator does not attempt to model.
The formula
Ah- Amp hour rating of the bank
V- Nominal bank voltage, 12, 24 or 48
DoD- Depth of discharge you are prepared to use
efficiency- Inverter conversion efficiency, typically 85-90%
load- Continuous load in watts
How it works, step by step
- Enter the amp hour rating and nominal voltage of the bank.
- Set depth of discharge to suit the chemistry: 50% lead-acid, 80-90% lithium.
- Set inverter efficiency, or leave it at 88% for a decent sine wave unit.
- Enter the continuous load in watts to read the runtime.
Worked examples
100 Ah at 12 V carrying 300 W
Nameplate is 1200 Wh. At 80 per cent depth of discharge that is 960 Wh, and after an 88 per cent inverter you have 845 Wh. Divided by 300 W that is 2.82 hours. The bank is supplying 28.4 amps while it does so.
Lead-acid versus lithium on the same bank size
The same 100 Ah at 50 per cent depth of discharge yields only 528 Wh and 1.76 hours. Lithium at 90 per cent gives 950 Wh and 3.17 hours — nearly double the runtime from an identical amp hour figure.
How to read your score
Frequently asked questions
Why can I not use the full capacity?
Because discharging a battery to empty damages it. Lead-acid loses a large fraction of its cycle life if regularly taken below 50 per cent, and while lithium iron phosphate tolerates 80 to 90 per cent, its battery management system will usually cut off before zero to protect the cells.
Does inverter efficiency really cost me that much?
Yes, and more than the headline figure suggests. A 90 per cent efficient inverter also draws a standby current of perhaps 10 to 20 watts just to stay switched on, which matters a great deal on a small load overnight. Turn it off when nothing needs it.
Should I choose 12, 24 or 48 volts?
Higher voltage for higher power. The current for a given load falls in proportion to voltage, and cable losses go with the square of current, so a 3 kW load on 12 V needs impractically heavy cable while on 48 V it is straightforward.
Why does my real runtime fall short?
Mostly the Peukert effect: lead-acid capacity is rated over a slow 20 hour discharge, and drawing hard can cut effective capacity by 20 to 40 per cent. Cold also reduces available capacity noticeably. Lithium is far less affected on both counts.
Battery runtime reference
| Bank | 100 W | 300 W | 600 W | 1000 W | 2000 W |
|---|---|---|---|---|---|
| 50 Ah at 12 V = 0.60 kWh | 4.22 h | 1.41 h | 0.70 h | 0.42 h | 0.21 h |
| 100 Ah at 12 V = 1.20 kWh | 8.45 h | 2.82 h | 1.41 h | 0.84 h | 0.42 h |
| 200 Ah at 12 V = 2.40 kWh | 16.90 h | 5.63 h | 2.82 h | 1.69 h | 0.84 h |
| 100 Ah at 24 V = 2.40 kWh | 16.90 h | 5.63 h | 2.82 h | 1.69 h | 0.84 h |
| 200 Ah at 24 V = 4.80 kWh | 33.79 h | 11.26 h | 5.63 h | 3.38 h | 1.69 h |
| 200 Ah at 48 V = 9.60 kWh | 67.58 h | 22.53 h | 11.26 h | 6.76 h | 3.38 h |
Lead-acid at 50% DoD roughly halves every figure here. Cold weather reduces it further.