Calculate This Power · Solar
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Solar
Panel Sizing

Size a solar array from daily energy use, peak sun hours, and battery bank. Outputs panel count, recommended charge controller, and array layout - for off-grid, RV, or backup applications.

See the math behind this calculator
US sun hours range
3.5 – 6.5 h/day

Modern panel typical
400 – 450 W

System losses
~25% real-world
Daily energy needs
Location & sun
Realism
Panel choice
Array size needed - W
Number of panels-
Battery bank (Ah)-
Charge controller-
Inverter min-
Daily generation-
Annual generation-

The math

Solar sizing combines three numbers: how much energy you use, how much sun you get, and how lossy your real-world system is. The third one is what catches first-time off-gridders.

array_W = daily_Wh ÷ (sun_hours × efficiency)

To turn 1 kWh per day into a panel size at 4.5 sun hours and 75% efficiency: 1000 ÷ (4.5 × 0.75) ≈ 296W of solar. Round up to the next stocked panel size.

Frequently asked

How do I know my peak sun hours?

The NREL PVWatts calculator gives location-specific monthly averages. For a rough number, use the regional preset here. For sizing a critical system, look up your worst month (typically December or January) and use that - anything you size based on annual average will fall short in winter.

Why such a big efficiency haircut?

Because panels are rated at ideal conditions. A 400W panel rarely produces 400W. Real output averages 70-80% of nameplate across a year. Cold winter days actually beat the rating; hot rooftop summer days fall well below it.

What size charge controller?

Match it to your array's maximum current. The calculator gives a minimum amp rating - go one size up for safety margin. MPPT controllers are 10-30% more efficient than PWM and are worth the cost for arrays over a few hundred watts.

Can I add panels later?

Yes if your charge controller and wiring have headroom. Plan for it: oversize the controller by 25-50% on initial install, run heavier wire than you think you need. Adding capacity later is much cheaper if the infrastructure supports it.

Show the working

Array sizing is one division. The number that actually decides your answer is the system efficiency, and it is far lower than most people expect. If any of it looks wrong, it might be - tell us what we got wrong.

Daily energy normalised to watt-hours kWh → Wh = value × 1000 Ah at 12 V → Wh = value × 12 Ah at 24 V → Wh = value × 24 Ah at 48 V → Wh = value × 48
Array size required array watts = daily Wh ÷ (sun hours × system efficiency)
Panels needed panels = ceiling(array watts ÷ panel watts) actual array = panels × panel watts
What the array actually generates daily generation Wh = actual array watts × sun hours × system efficiency

Constants used

  • 4.5 sun hours — default peak sun hours per day — this varies enormously by latitude and season and is worth replacing with a figure for your location
  • 75% system efficiency — combined losses from the charge controller, wiring, heat, dust and panel mismatch

Peak sun hours are not hours of daylight. It is the equivalent number of hours at full rated irradiance, so a long overcast winter day can be under two, while the same location in June might be six.