How it works
A panel rated 540 W does not deliver 540 W all day. The peak sun hours number is the hours per day of full-strength sunlight that the site receives in total, and the system efficiency covers the losses between the panel and your appliances.
Dividing the daily energy by those two gives the array size in kWp, and dividing by the panel wattage (rounded up to a whole panel) gives the panel count.
A battery should not be emptied completely. The usable fraction (depth of discharge) and the charge-discharge efficiency raise the bank size above the energy you need. The inverter is sized with a 25 % margin over the peak load. Roof shading, tilt, temperature and surge loads are not included.
Formulas used
- Array size (kWp) = daily kWh ÷ (peak sun hours × system efficiency)
- Panels = array size × 1000 ÷ panel watts, rounded up
- Battery energy (kWh) = daily kWh × days ÷ (depth of discharge × battery efficiency)
- Battery Ah = battery kWh × 1000 ÷ bank voltage
- Inverter ≥ 1.25 × peak load
Worked example
10 kWh a day with 5 peak sun hours and 75 % efficiency needs 10 ÷ (5 × 0.75) = 2.67 kWp, which is 5 panels of 540 W (2.7 kWp). One day of backup at 50 % depth and 90 % battery efficiency needs a 22.2 kWh bank, about 463 Ah at 48 V.
The calculator opens with these example values, so you can compare its answer with the text above.
Frequently asked questions
How many solar panels do I need?
Divide your daily kWh by (peak sun hours × system efficiency) to get kWp, then divide by the panel wattage and round up.
What are peak sun hours?
The equivalent number of hours per day at full sunlight strength of 1 kW per square metre. Most of India gets about 4 to 6.
Why is the battery bigger than my daily use?
Because only part of a battery can be used safely (depth of discharge) and some energy is lost in charging and discharging.
Is this enough to order a system?
It is a first estimate. A supplier should confirm the design on site, including shading, roof area, cable sizes and protection.