Formula
Daily energy is the sum of watts × hours. Array watts divide that by peak sun hours and by one minus the system loss. Panel count rounds up. Battery amp-hours are daily watt-hours × days of autonomy, divided by voltage × depth of discharge × inverter efficiency. Peukert’s exponent is left out of the size on purpose. It answers “how long,” which is the runtime page, not “how big.”
array W = daily Wh ÷ (sun hours × (1 − loss))
Ah = daily Wh × days ÷ (V × DoD × inverter efficiency)
controller A = installed W ÷ V × 1.25
surge = simultaneous W × surge factor
The 14% starting loss is the PVWatts historical default derate, about 0.86 remaining. Edit it. Do not also hide battery round-trip loss inside it, or you will count that loss twice. Controller current uses the 125% continuous factor from NEC 690.8 and nothing else from Article 690. Inverter watts default to the sum of the nameplate loads, which is conservative if they never run together. Peak sun hours are whatever you type. The city figures are order-of-magnitude annual averages, not a queried NSRDB station.
Worked examples
100 W for 5 hours and 50 W for 2 hours. 600 Wh a day. At 5 peak sun hours and 14% loss the array is 139.5 W, so two 100 W panels. At 12 V, 80% discharge, 2 days of autonomy, and 85% inverter efficiency the battery is 147 Ah. The controller is 200 ÷ 12 × 1.25 = 20.8 A. The inverter is 150 W continuous and 300 W surge.
The same loads with flooded lead-acid at 50% discharge. Change depth of discharge to 50%. The battery amp-hours rise because less of each amp-hour is treated as usable. The array does not change.
Reference
Starting points only. Edit them.
| Chemistry | DoD | Peukert k |
|---|---|---|
| Flooded lead-acid | 50% | 1.3 |
| AGM | 50% | 1.1 |
| LiFePO4 | 80% | 1 |
Questions
How many watts of panel cover 600 watt-hours at 5 sun hours?
With a 14% system loss, 600 ÷ (5 × 0.86) = 140 W of array. Two 100 W panels cover it. The 14% is the PVWatts historical default. Battery round-trip loss is not inside that 14%. It is applied on the battery.
Why is the battery larger than the daily amp-hours?
Amp-hours are daily watt-hours × days of autonomy, divided by voltage, depth of discharge, and inverter efficiency. Two days, 80% discharge, and 85% inverter efficiency turn 600 Wh into about 147 Ah at 12 V. Peukert’s exponent is not applied here. It is on the runtime page.
What does the 1.25 on the controller mean?
NEC 690.8 sizes continuous current at 125% of the calculated current. Controller amps here are installed array watts ÷ battery voltage × 1.25. That is the continuous-current factor only. It is not a full Article 690 design.
Sources
- NREL PVWatts, historical default system loss near 14% (derate 0.86). Battery round-trip efficiency is not included in that factor here.
- NEC 690.8, the 125% continuous-current factor, applied as installed watts ÷ battery voltage × 1.25. Not a complete Article 690 design.
- Depth of discharge and Peukert exponents are typical manufacturer practice: flooded 50% and k ≈ 1.3, AGM 50% and k ≈ 1.1, LiFePO4 80% and k ≈ 1. They are not a published standard. City sun hours are approximate annual averages and were not re-fetched from a named NSRDB station.