The calculation
daily kWh = monthly kWh ÷ 30
system kW = daily kWh ÷ (peak sun hours × 0.8)
panels = system watts ÷ panel watts
system kW = daily kWh ÷ (peak sun hours × 0.8)
panels = system watts ÷ panel watts
That 0.8 is the system loss factor, and leaving it out is the single most common way estimates come in too small.
Worked example: 900 kWh a month
Roughly the US average, in a location with 4.5 peak sun hours, using 400 W panels:
| Step | Working | Result |
|---|---|---|
| Daily consumption | 900 ÷ 30 | 30 kWh/day |
| Effective sun hours | 4.5 × 0.8 | 3.6 h |
| System size | 30 ÷ 3.6 | 8.33 kW |
| Panels at 400 W | 8,330 ÷ 400 | 21 panels |
| Roof area | 21 × 21.5 ft² | ≈ 450 ft² |
8.3 kW
system size
21
panels at 400 W
450 ft²
roof needed
Size a system from your own kWh or bill.Solar panel calculator →
By consumption level
All at 4.5 peak sun hours, 400 W panels, 80% system efficiency:
| Monthly use | System size | Panels | Roof area |
|---|---|---|---|
| 600 kWh | 5.6 kW | 14 | 300 ft² |
| 900 kWh | 8.3 kW | 21 | 450 ft² |
| 1,200 kWh | 11.1 kW | 28 | 600 ft² |
| 1,500 kWh | 13.9 kW | 35 | 750 ft² |
Where the losses come from
Typical system losses
Inverter2–4%
Heat / temp5–8%
Wiring2–3%
Soiling2–3%
Mismatch etc.3–5%
Panels are rated at 25 °C in a laboratory. A roof in July is far hotter, and output falls as temperature rises — which is why a hot, sunny climate does not always beat a mild one by as much as expected.
What changes the answer most
- Peak sun hours. The biggest regional variable. Ranges from roughly 3.5 in the cloudy north to 6+ in the desert southwest — enough to change panel count by 40%.
- Roof orientation. South-facing at 30° is ideal in the northern hemisphere. East or west facing typically costs 15% to 20% of annual output.
- Shading. Partial shade on one panel can drag down a whole string unless microinverters or optimisers are used.
- Future load. An EV adds roughly 250 to 400 kWh a month. Size for the house you will have, not the one you have now.
Sizing to 100% of consumption is not always optimal. Where net metering pays less for exports than you pay for imports, a slightly smaller system that offsets daytime use has a better return than one that exports surplus cheaply.
Two related pieces: long DC runs from array to inverter need a voltage drop check, and the payback question is really an annualized return question — a 20-year payback is a very different investment from a 7-year one.
Run your own numbers
FAQ
How many solar panels does an average home need?
A home using around 900 kWh a month in an area with 4.5 peak sun hours typically needs a 7.5 to 8.5 kW system — about 19 to 22 panels at 400 W each. Cloudier locations or higher consumption push that up; sunnier regions pull it down.
What are peak sun hours?
The number of hours per day equivalent to full-strength sunlight of 1,000 watts per square metre. A location with 4.5 peak sun hours may see 10 hours of daylight, but the weaker morning and evening light condenses into the equivalent of 4.5 full-strength hours.
Why apply a system loss factor?
Because panels never deliver their laboratory rating in the field. Inverter conversion, heat, wiring resistance, dust, shading and panel mismatch typically cost 15% to 25% combined. Using 80% of nameplate output is a realistic planning assumption.
How much roof space do I need?
About 21.5 square feet per 400 W panel, so a 21-panel array needs roughly 450 square feet of usable roof — excluding vents, chimneys, setbacks and shaded areas. Usable area is normally well below total roof area.
Do I need batteries?
Not for a grid-tied system. Where net metering exists, the grid acts as the storage. Batteries matter for outage protection or where export rates are much lower than retail rates, and they add substantially to the cost.
Sources
Primary references used for the figures and rules on this page.
- Homeowner's Guide to Going Solar — U.S. Department of Energy
- Electric Power Monthly — U.S. Energy Information Administration