How this calculator works
Your average daily use is divided by the energy one kilowatt of panels produces per day — peak sun hours reduced by system losses (inverter, wiring, dirt, shading). That gives the system size in kW; dividing by the panel rating gives the panel count.
Daily use (kWh) = Monthly kWh × 12 ÷ 365 × Offset%
System kW = Daily use ÷ (Peak sun hours × (1 − Loss%/100))
Panels = round up( System kW × 1000 ÷ Panel watts )
Assumptions and tips
- 14% system losses is the PVWatts default (soiling, shading, wiring, mismatch and more).
- Roof area per panel varies by model; check the panel’s dimensions.
Worked example
Using the default values in the calculator above:
| Input | Value |
|---|---|
| Monthly electricity use | 900 kWh |
| Share of use to cover | 100 % |
| Peak sun hours per day | 4.5 h |
| Panel rating | 400 W |
| System losses | 14 % |
| Result | Value |
|---|---|
| System size | 7.65 kW |
| Panels needed | 20 × 400 W |
| Installed size with whole panels | 8 kW |
| Estimated yearly production | 11,300 kWh |
| Roof area (≈18 sq ft per panel) | 360 sq ft |
Step by step:
- Daily use = 900 × 12 ÷ 365 × 100% = 29.59 kWh
- System = 29.59 ÷ (4.5 × 0.86) = 7.65 kW
- Panels = 7,646 W ÷ 400 → 20
Solar system size by monthly usage
4.5 peak sun hours, 14% losses, 400 W panels.
| Monthly use | System size | Panels needed |
|---|---|---|
| 500 kWh | 4.25 kW | 11 × 400 W |
| 750 kWh | 6.37 kW | 16 × 400 W |
| 900 kWh | 7.65 kW | 20 × 400 W |
| 1200 kWh | 10.19 kW | 26 × 400 W |
| 1500 kWh | 12.74 kW | 32 × 400 W |
Common mistakes to avoid
- Sizing from one summer bill instead of a 12-month average.
Frequently asked questions
How many solar panels do I need for 900 kWh a month?
With 4.5 peak sun hours and 14% losses, about 7.6 kW — roughly 20 panels of 400 W.
What are peak sun hours?
The number of hours per day the sunlight would equal 1,000 W per square meter. It varies by location and season.