Calculation
P = A · G · η
Area, irradiance and efficiency multiply; temperature and inverter losses reduce real output.
P = A·G·η
PV Power Calculator from Area and Efficiency
Area, irradiance and efficiency multiply; temperature and inverter losses reduce real output.
MINTSI
02
Result
Select a target and calculate.
Understand the inputs
- PV peak power P — Electrical DC power at assumed irradiance.
- Module area A — Active module area.
- Irradiance G — Solar irradiance; 1,000 W/m² is STC.
- Module efficiency η — Share of irradiance converted to DC power.
Example
10 m² · 1,000 W/m² · 20% = 2 kW.
Assumptions and limits
Uniform irradiance and cell temperature; shading and system losses are excluded.
Understand PV Power Calculator from Area and Efficiency
This calculator finds the missing quantity among PV peak power P, module area A, irradiance G and efficiency η from the other two. It puts today's solar module efficiency in perspective by comparing it against photosynthesis's much lower natural efficiency.
What does this quantity describe?
The theoretical peak power of a PV area is the product of area, solar irradiance and module efficiency: P = A·G·η. Under Standard Test Conditions, G = 1,000 W/m².
Commercial solar modules today reach efficiencies of about 15-22% in directly converting sunlight to electrical energy. Plants, by contrast, reach only a theoretical maximum efficiency of about 4.6% (C3 plants like wheat and rice) or 6% (C4 plants like corn and sugarcane) even under ideal conditions when converting solar energy into biomass — under real field conditions the actual value usually sits well below that. Modern solar cells are therefore several times more efficient than photosynthesis at raw energy conversion.
Formula and variables
P = A · G · η
P = A · G · ηA = P / (G·η)η = P / (A·G)
| Symbol / input | Meaning |
|---|---|
| PV peak power P | Electrical DC power at assumed irradiance. |
| Module area A | Active module area. |
| Irradiance G | Solar irradiance; 1,000 W/m² is STC. |
| Module efficiency η | Share of irradiance converted to DC power. |
Choose the inputs correctly
Enter two of the three quantities. Peak power requires area, irradiance and efficiency (two of these together with P to find the third). Use G = 1,000 W/m² for STC-based sizing, or actual site irradiance for yield estimates.
How to use the calculator
Select the target quantity and enter the two known values with units. Use the manufacturer's datasheet module efficiency, not a generic rule-of-thumb value, since figures vary considerably by cell technology.
Worked example
Given A = 10 m², G = 1,000 W/m² and η = 0.2 (20%), substitution gives P = 10 · 1,000 · 0.2 = 2,000 W = 2 kW.
Understand the result and units
A peak power of 2 kW is the theoretical output at exactly this irradiance and efficiency. Real systems reach this value only under optimal, perpendicular irradiance and cell temperatures around 25°C; averaged over a year, actual generated power sits well below it.
Area in m², irradiance in W/m², efficiency η dimensionless (0 to 1, or 0-100%), power in watts or kilowatts.
Why are solar cells more efficient than photosynthesis?
Photosynthesis has evolved not primarily for maximum energy yield but to sustain complex biochemical processes, in which only a small share of incident energy actually converts to chemical bond energy. Solar cells, by contrast, exploit the photoelectric effect in specially doped semiconductor materials, purpose-optimised for the highest possible direct conversion of photon energy into electric current. This fundamental difference in conversion mechanism explains why commercial solar modules at 15-22% efficiency clearly exceed photosynthesis's theoretical maximum efficiency of 4.6-6%.
Typical applications
The formula is used for rough area-requirement sizing of PV systems, comparing different module technologies, and roughly estimating peak power for available roof or ground area.
Assumptions, limits and common mistakes
The formula gives theoretical peak power at uniform irradiance and cell temperature. Shading, soiling, temperature dependence of efficiency, and cabling and inverter losses are not included here and further reduce actually usable power.
Common mistake: Do not confuse theoretical peak power with actual annual yield — the latter requires a time-series analysis of irradiance over a full year, not a single irradiance value. Also, do not equate module efficiency with the whole system's efficiency, which includes additional losses from inverters and cabling.
Frequently asked questions
Are solar modules really more efficient than plants at energy conversion?
Measured by raw solar energy conversion, yes: commercial modules reach 15-22% efficiency, while photosynthesis's theoretical maximum efficiency is only 4.6-6%.
What efficiency should I use for an estimate?
Use the module efficiency stated in the datasheet; common crystalline silicon modules currently mostly range between 18% and 22%.
Why isn't this formula enough for an annual yield forecast?
Because it gives only an instantaneous output at a single assumed irradiance. Annual yield needs a time series of irradiance across all weather conditions and seasons.