How many solar panels do I need? The calculation step by step

September 29, 2026 · 5 min read · Danilo Nascimento

Work out the number of panels from monthly consumption, peak sun hours and system efficiency, with a complete example and an interactive calculator.

The calculation in three steps

Working out how many panels a home needs is a short calculation. The starting point is daily consumption, which you get from how to estimate home energy use from your bill.

1
Consumption per day
300 kWh per month ÷ 30 days = 10 kWh per day.
2
Power needed (kWp)
Divide daily consumption by the site’s peak sun hours times the system efficiency.
3
Number of panels
Divide the power needed by the power of one panel and round up.

Peak sun hours (PSH)

One peak sun hour equals 1 kWh of radiation per m², or 1,000 W/m² for 1 hour. The sun does not shine at that intensity all day: adding up a day’s radiation gives the equivalent of a number of “full” hours. That number, the PSH, measures how much sun a site receives.

Peak sun hours per day (order of magnitude)

Northern Europe≈ 2.5 h

Central Europe≈ 3.5 h

Mediterranean and tropics≈ 5 h

Desert regions≈ 6.5 h

Why system efficiency is not 100%

Nameplate power is a reference measured under standard conditions: real power varies with irradiance and temperature. On top of that, heat, dirt, cables and the inverter itself take part of the energy. For a quick calculation use a system efficiency (overall performance factor) of about 80%. The article on losses and the performance ratio shows where each share comes from.

10 kWhdaily consumption÷5 h × 0.8peak sun × efficiency=2.5 kWppower needed

Calculate the number of panels

Adjust the consumption, the site’s sun and the panel. The result rounds up, because there is no such thing as half a panel, and it also shows the estimated generation and the roof area.

A complete example

1
1. Daily consumption
A bill of 300 kWh per month: 300 ÷ 30 = 10 kWh per day.
2
2. Power needed
With 5 peak sun hours and 80% efficiency: 10 ÷ (5 × 0.8) = 2.5 kWp.
3
3. Number of panels
With 550 W panels: 2,500 ÷ 550 = 4.55, rounded up to 5 panels.
4
4. Check
5 × 550 W = 2.75 kWp, which generate about 2.75 × 5 × 0.8 × 30 = 330 kWh per month, 10% above consumption.

5550 W panels

2.75 kWpinstalled power

≈ 13 m²of roof, with 2.6 m² panels

What limits the number of panels

Roof area: each 550 W panel takes about 2.6 m²
Shade: chimneys, trees and buildings reduce generation and may rule out part of the roof
Inverter: the panels’ power has to match the inverter (see how to size the inverter)
Utility rules: there are power limits depending on the connection
Budget: each extra panel generates more, but its return falls once generation already covers consumption

Draw it in Editor Solar

This is Editor Solar itself, already loaded with the diagram from this article. Move the equipment, edit labels, connect new parts and export to PNG, SVG or PDF. It is free and runs right in your browser.

On-grid system with 5 panels of 550 W (2.75 kWp), sized for 300 kWh per month.

Try it: Change the number of panels to the result for your own bill and write the total power in kWp on the diagram. Then note how many square metres of roof they need.

Ready-made templates: see all diagram templates

Open this diagram in the full editor ↗

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