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How residential solar systems are actually sized.

"How many panels do I need?" is the wrong question. The right question is: how much energy do I want to displace from the grid, and what system size is required to produce it on my specific roof?

Three quantities you need first

1. Annual electricity consumption

Pull your most recent 12-month bill history from your utility's online portal (every major US utility now offers this; it is often a CSV download). Sum the kWh column. This is your annual consumption. The national average is about 10,800 kWh per residential customer per year, with wide variation: California averages closer to 6,500 kWh due to mild climate; Louisiana and Texas push 14,000 kWh due to air conditioning load.

If you have made or plan major load changes — installing a heat pump, electrifying a hot water heater, buying an electric vehicle — adjust upward. An EV driven 12,000 miles per year typically adds 3,000–4,000 kWh of household consumption.

2. Peak sun hours at your location

A peak sun hour is one hour at which solar irradiance averages 1,000 W/m². That is the standard test condition used to rate a solar panel's nameplate output. If your location averages 5.5 peak sun hours per day, a 1 kW (nameplate) system will produce approximately 5.5 kWh on an average day — before applying real-world losses.

State-level averages range from about 3.0 in Alaska to 6.5 in Arizona and New Mexico. NREL's National Solar Radiation Database is the canonical source; our calculator pulls population-weighted state averages from it.

3. Performance ratio

A real installed system never produces the nameplate-times-sun-hours figure. The performance ratio (PR) captures everything between the DC nameplate and the AC kWh that lands on your bill:

The industry default PR is 0.80, codified in NREL's PVWatts engine and used throughout this site. Premium installations with cool climates and oversized inverters can achieve 0.83–0.86; hot, dusty, or partially shaded sites may fall to 0.70.

The sizing equation

system_size_kW = annual_kWh / (peak_sun × 365 × performance_ratio)

Worked example: a household in Texas consuming 14,000 kWh/year at 5.3 peak sun hours and PR 0.80 needs:

14,000 / (5.3 × 365 × 0.80) = 14,000 / 1,547.6 = 9.05 kW DC

That is approximately 23 × 400 W panels — a typical 9 kW residential installation in the Texas market.

Should you size to 100% offset?

Not always. There are good reasons to undersize, and several places where oversizing is actively destructive:

Why panel wattage isn't the right unit

Manufacturers compete intensely on per-panel wattage — last decade's premium 300 W panels are this decade's mid-range 400 W panels. But what matters is the system's total nameplate capacity and its area-adjusted efficiency. A 9 kW system made of 22 × 410 W panels at 20.5% efficiency and a 9 kW system made of 25 × 360 W panels at 18.5% efficiency produce essentially identical energy. The difference shows up in roof area required, not in output.

The relevant trade-off:

Site-specific accuracy: use PVWatts

The four-input model on our home page produces a reasonable estimate, but it cannot model your specific roof's geometry. For a free site-specific calculation that does, use NREL's PVWatts calculator. PVWatts takes:

For a homeowner doing due diligence on an installer quote, a 5-minute PVWatts run is the single most valuable cross-check available. If the installer's projected first-year production differs from PVWatts by more than 10%, ask why.

Last reviewed May 2025. PVWatts and NSRDB references verified against the current NREL documentation.

Independent · Source-cited

Try the calculator yourself.

Four inputs. Sixty seconds. Every number traceable to a public source.