Current for 2026As of: July 2026

Solar Panel Calculator calculate solar yield.

Enter system capacity in kWp, location peak sun hours and performance ratio – get the annual yield in kWh instantly

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Solar Panel Calculator

Calculate your annual solar energy yield in kWh – enter system capacity, location and performance ratio.

1 kWp30 kWp
700 h/Jahr1,150 h/Jahr

Performance Ratio (PR)

Default: 80% – accounts for inverter losses, shading and temperature

Annual solar yield

7,600 kWh

760 kWh/kWp · PR 80%

Per day

21 kWh

Per month

633 kWh

Per year

7,600 kWh

Tip: Germany has 800–1,100 peak sun hours depending on the region. Southern Germany (Bavaria, Baden-Württemberg) reaches 950–1,050 h, northern Germany 800–900 h.

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Calculating photovoltaic yield correctly

Formula, location factors and performance ratio at a glance

The annual solar yield results from three factors: system capacity in kWp, location-specific peak sun hours, and the performance ratio. The formula is: yield = kWp × peak sun hours × PR. A 10 kWp system in Bavaria with 950 peak sun hours and a PR of 0.80 produces exactly 7,600 kWh/year – enough for a well-supplied four-person household without a heat pump.

Peak sun hours by region: Germany has very different levels of solar irradiation depending on the region. Bavaria and Baden-Württemberg lead with 950–1,050 peak sun hours. Saxony and Thuringia reach 900–950 h, while Schleswig-Holstein and Hamburg trail with 800–870 h. The European Commission provides location-accurate figures via the PVGIS tool – free of charge and available down to postcode level.

Performance ratio in practice: The single most important factor for a PV system's efficiency is the inverter. Modern devices operate at 97–98% efficiency, older ones at only 90–93%. On top of that come temperature losses: silicon solar cells lose output at high temperatures (about –0.4%/°C above 25°C). A system on a dark flat roof without rear ventilation only reaches 75–77% PR in summer, while a well-ventilated pitched-roof system reaches 82–85%.

Economic viability: The payback period of a PV system is around 8–12 years as of 2025. With a feed-in tariff of about 8 cents/kWh and a self-consumption value of 30–35 cents/kWh, battery storage pays off to raise the self-consumption share to 60–80%. For an accurate calculation, maintenance costs (0.5–1% of the investment per year) and module degradation (0.3–0.5%/year) should also be factored in.

Peak sun hours by region and orientation

Typical peak sun hours (h/year)

Bavaria / Baden-Württemberg: 950–1,050 h
Germany's best locations; south-facing orientation, optimal roof pitch
NRW / Hesse: 880–950 h
Average values for central Germany; frequent cloud cover from low-pressure systems
Northern Germany: 800–880 h
Coastal influence and more cloud cover; PV still worthwhile thanks to diffuse light
South-facing: 100%
Optimal yield at 30–35° pitch; reference value for all comparisons
East/West: 75–85%
Loss compared to south-facing; but better self-consumption coverage throughout the day
Flat roof (0°): 85–90%
Suboptimal pitch, but elevated mounting at 15–20° is possible

Calculation examples

10 kWp · 950 h · 0.80 PR – reference value for southern Germany

10 kWp · 950 h · 0.80 PR – reference value for southern Germany
ItemAmount
System capacity10.00 kWp
× Peak sun hours950 h/year
× Performance ratio0.80 (80%)
= Annual yield7,600 kWh

6 kWp · 880 h · 0.78 PR – northern Germany

6 kWp · 880 h · 0.78 PR – northern Germany
ItemAmount
System capacity6.00 kWp
× Peak sun hours880 h/year
× Performance ratio0.78 (78%)
= Annual yield4,118 kWh

Frequently asked questions about the solar panel calculator

Everything about yield, peak sun hours and performance ratio

The formula is: yield (kWh/year) = system capacity (kWp) × peak sun hours (h/year) × performance ratio. A 10 kWp system in southern Germany with 950 peak sun hours and a performance ratio of 0.80 produces 7,600 kWh/year. Peak sun hours vary by location: northern Germany 800–900 h, southern Germany 950–1,050 h.

The performance ratio (PR) is the ratio of the actual energy yield to the theoretically possible yield. It accounts for losses from the inverter (3–5%), temperature (3–8%), shading, cable losses and soiling. Modern systems achieve PR values of 75–85%. A value of 80% is realistic for well-planned systems without significant shading.

Peak sun hours (also called specific yield) indicate how many hours per year a 1 kWp system would need to run at rated power to produce the same yield. In Germany, values range from 800 to 1,100 hours. A good reference is the PVGIS database from the European Commission for location-accurate figures. For most planning purposes, 950 h is a good average value.

In Germany, 1 kWp of solar produces between 700 and 1,000 kWh per year depending on location and performance ratio. Typical average: 950 h × 0.80 PR = 760 kWh/kWp. A 10 kWp system therefore delivers about 7,600 kWh/year – enough for an average four-person household.

An average four-person household consumes about 4,000–5,000 kWh/year. With a 6–8 kWp system and battery storage, a self-consumption rate of 60–80% is achievable. As a rule of thumb: 1 kWp per person. With a heat pump or electric vehicle, you should plan for 10–15 kWp, as consumption increases considerably.

Sources & calculation basis

Our calculations are based on the following official sources (as of: July 2026):

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