Size the inverter for the peak load, not the average, and manage the loads so the battery is never run flat. A 250 W compressor surges to 750-1,500 W when it starts, so the inverter must deliver that surge even though the day’s average is a fraction of it; add up the watts that can run together, allow 3 to 6 times running power for motors on starting, and shed the large loads first as the battery falls toward 11.8 V.
Part of Advanced Wind Turbine
An inverter converts the battery’s DC into AC for ordinary appliances; load management is the habit and the hardware that decide which loads run when.
Peak Decides The Inverter
The carrier says to size the inverter for the peak load, not the average consumption. It notes that modified sine wave inverters, common in salvage, are adequate for motors, lights and tools, and that pure sine wave inverters are needed for sensitive electronics. Inverter Basics covers how an inverter works. A motor draws 3 to 6 times its running current on starting, according to Load Control.
Worked example: a village load list of about 2 kWh a day
| Load | Running watts | Hours a day | Energy | Start surge (3 to 6 times) |
|---|---|---|---|---|
| LED lights (10 at 8 W) | 80 W | 5 | 400 Wh | none |
| Phone and radio charging | 40 W | 3 | 120 Wh | none |
| Fridge compressor | 250 W | 4 (running time) | 1,000 Wh | 750-1,500 W |
| Drill or small workshop motor | 500 W | 1 | 500 Wh | 1,500-3,000 W |
| Total | 870 W if all on | 2,020 Wh |
The energy, 2,020 Wh, is the carrier’s 2 kWh a day. The continuous load is 80 + 40 + 250 + 500 = 870 W if everything runs together. The worst start is the drill starting while the others run: 80 + 40 + 250 + (1,500 to 3,000) = 1,870 to 3,370 W. There are two answers:
- Buy surge capacity. An inverter with a continuous rating over 870 W and a surge rating over 3.4 kW. At 48 V that peak is 3,370 ÷ 48 = 70 A from the battery; at 12 V it is 281 A, which needs very heavy cable (Electrical System).
- Manage the starts. A rule that the drill is started only when the fridge compressor is off drops the worst start to 80 + 40 + (1,500 to 3,000) = 1,620 to 3,120 W. A soft-start device or a smaller motor cuts it further.
The inverter’s own losses add to the battery’s burden. If the inverter is 90% efficient (an assumed value; check yours), 2,020 Wh of AC needs 2,020 ÷ 0.90 = 2,244 Wh from the battery. Run the lights and chargers on DC directly and they avoid that loss.
Load Management: Which Loads Run When
Step 1: Make a priority list. Lights, communication and refrigeration first; workshop tools and heating last.
Step 2: Fit a low-voltage disconnect. Drop the large loads when the bank approaches the depleted voltage, 11.8 V on a 12 V system (Charge Controller). Battery life depends on never reaching empty.
Step 3: Run big loads when there is surplus. When the battery is full and the dump load is taking power, that power is being thrown away as heat. Schedule the drill, the pump or the charger of tools for those windy hours, and the same energy costs the battery nothing (Dump Load & Overspeed Protection).
Step 4: Turn the inverter off when no AC load is running. Idling consumes battery power. Recalled, not corpus.
Step 5: Track the battery voltage in a log to see the trend over the week and size the bank (Battery Bank Sizing).
Common Mistakes
- Sizing on watts running only — the surge trips the inverter at the first start.
- All big loads in the evening — the bank is lowest when the demand is highest.
- No low-voltage disconnect — a long calm kills the bank.
Safety
AC from a battery is as dangerous as mains
A 120 V or 230 V inverter output can kill, and the battery behind it has huge short-circuit current. Fuse the inverter’s DC cable at the battery, use properly rated breakers on the AC side, earth the frame, and keep wiring dry (Electrical Safety). An overloaded inverter overheats: do not cover it.
What’s Next
- Learn next: Dump Load & Overspeed Protection for the surplus path
- Combine with: Circuit Breakers