What are peak sun hours?

Peak sun hours are the equivalent number of hours per day at 1000 W/m2 of sunlight. A location with 5 peak sun hours receives the same total energy as 5 hours of full-strength noon sun, even though actual daylight is much longer at varying intensity.

Why apply a derate factor?

Real systems lose energy to inverter conversion, wiring resistance, panel temperature, soiling, and shading. A derate factor of about 0.8 means the array must be sized roughly 25% larger than the raw load to deliver the energy you actually need.

How is the system size calculated?

Daily kWh need is divided by the derate factor to find the generation required, then divided by peak sun hours to get the DC system size in kW. The panel count is the system watts divided by the panel rating, rounded up to a whole panel.

Should I size to 100% of my usage?

Many homeowners size to offset 80–100% of annual use, balancing roof space, budget, and net-metering rules. Sizing above 100% can hit utility export caps. Adjust the monthly figure to target the offset you want.

Is this a final design?

No. This is a first-pass estimate. A real design accounts for roof orientation and tilt, month-by-month sun data, shading analysis, inverter clipping, and local interconnection limits. Always confirm with a site assessment.

Solar PV System Sizing Calculator

The first question in any solar project is “how many panels do I need?” This calculator answers it from your electricity bill, your location’s sunlight, and the panels you plan to use — applying a real-world derate so the array actually covers your load.

How it works

Three steps turn consumption into a panel count.

Daily need. Convert your monthly use to a daily average:

daily kWh = monthly kWh / 30.4

Required generation. The array must produce more than the load to cover system losses, so inflate by the derate factor:

generation needed = daily kWh / derate

System size and panel count. Divide by peak sun hours to get the DC system size, then divide by the panel rating:

system kW   = generation needed / peak sun hours
panel count = ceil(system watts / panel watts)

The bundled table gives average annual peak sun hours for major US cities, from about 6.5 in Phoenix down to 3.7 in Seattle.

Worked example

A home uses 900 kWh/month in Austin (5.3 peak sun hours), with 400 W panels and a 0.8 derate.

Daily need: 900 / 30.4 = 29.6 kWh.
Generation needed: 29.6 / 0.8 = 37.0 kWh.
System size: 37.0 / 5.3 = 6.98 kW → about 7 kW DC.
Panels: 7000 / 400 = 17.5 → 18 panels.

Notes and tips

Use an annual-average month, not a peak summer bill, or you will oversize the array for most of the year.
Peak sun hours vary by season — winter can be half of summer — so a system sized to annual averages will overproduce in summer and underproduce in winter.
The derate factor is where optimism creeps in. Shaded or hot roofs justify 0.75; a clean, well-ventilated, south-facing install can reach 0.85.

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