What the calculator does
It answers the four sizing questions in order. How many kilowatts-peak of panels produce your daily energy at your site, once real-world losses are taken off? How many panels of the rating you can buy is that, and what will they actually yield? How large a battery bank carries the load for the number of days you choose, at the depth of discharge the chemistry tolerates? And what inverter rating covers your peak load with a starting margin?
It reports the approximate roof area too, taking about 210 W per square metre for current mono-crystalline panels, panels only, without walkways or row spacing.
Formula
Panels = ⌈Array × 1000 ÷ Ppanel⌉ Installed kWp = Panels × Ppanel ÷ 1000
Daily yield (kWh) = Installed kWp × PSH × (1 − losses)
Battery (Wh) = Edaily × 1000 × days ÷ (DoD × ηinverter) Battery (Ah) = Wh ÷ Vbank
Inverter (kW) ≥ Peak load × 1.25
where Edaily is the daily energy in kWh, PSH the peak sun hours (kWh/m²/day on the array plane, numerically the equivalent hours of 1 000 W/m² sun), DoD the usable fraction of the battery and ηinverter the efficiency of converting battery DC to AC. The 1 − losses term is the performance ratio of the system.
Worked example
A house in Lahore uses 300 units a month, so 10 kWh/day. Planning on 5 peak sun hours, 20 % losses, 550 W panels, a 48 V lithium bank for one day's autonomy at 80 % depth of discharge, a 95 % efficient inverter and a 3 kW peak load.
- Array = 10 ÷ (5 × 0.8) = 2.5 kWp.
- Panels = ⌈2 500 ÷ 550⌉ = 5, giving 2.75 kWp installed and 2.75 × 5 × 0.8 = 11.0 kWh on an average day, about 4 015 kWh a year.
- Battery = 10 000 × 1 ÷ (0.8 × 0.95) = 13 158 Wh, which is 274 Ah at 48 V; in practice two 48 V 150 Ah lithium units, or one 300 Ah.
- Inverter ≥ 3 × 1.25 = 3.75 kW, so a 4 kW or 5 kW hybrid inverter. Against that 3.75 kW minimum the array-to-inverter ratio is 2.75 ÷ 3.75 = 0.73; with a 5 kW unit it is 0.55, leaving room to add panels later.
Roof area: 2 750 W ÷ 210 W/m² ≈ 13 m² of panels.
Where the losses go (typical, hot climate)
| Loss | Typical | Why |
|---|---|---|
| Cell temperature | 8–12 % | output falls about 0.35 % per °C above 25 °C; cells run 20–35 °C above ambient |
| Soiling and dust | 2–6 % | higher in dry seasons and near roads; cleaning restores it |
| Inverter conversion | 2–4 % | best at 30–80 % of rating |
| DC and AC wiring | 1–3 % | see the voltage drop calculator |
| Mismatch, shading, ageing | 2–5 % | panels degrade 0.4–0.6 % a year |
Assumptions and limitations
- Average days. Peak sun hours vary by month; a grid-tied system can be sized on the annual average because the grid fills the gaps, an off-grid one must be sized on the worst month or it will fail in winter and the monsoon.
- Orientation and tilt are in your PSH figure. Use a value for your panel plane (roughly latitude tilt facing south); flat or badly oriented arrays receive less.
- Battery sizing is by energy, not by power. Check that the bank's maximum discharge current (its C-rate) can supply the peak load; the battery runtime calculator reports it.
- Inverter margin is a rule of thumb. Motor starting for a 1.5-ton air conditioner or a water pump can need more than 1.25 × running load; check the inverter's surge rating against the actual motors.
- Not a financial model. Net-metering and net-billing rules in Pakistan have been changing; check the current NEPRA regulation and your distribution company's terms before counting on export credit.
Frequently asked questions
How do I find my peak sun hours?
From a solar resource database such as the Global Solar Atlas (World Bank / Solargis) or NASA POWER, which give monthly irradiation for any location. The global horizontal figure is a reasonable start; the value on a tilted plane is a little higher in winter and lower in summer.
Should the array be bigger than the inverter?
Often, yes. Panels rarely produce their rated power because of temperature, so an array 10–30 % larger than the inverter (a DC:AC ratio of 1.1–1.3) uses the inverter better through the day. Do not exceed the inverter's maximum DC input voltage or current.
Do I need a battery at all?
Not for a purely grid-tied system, which exports surplus and imports at night. Batteries are for load-shedding backup or off-grid sites; set autonomy to 0 to size the array alone.
Why divide by inverter efficiency for the battery but not for the array?
The inverter loss on the solar side is already inside the system losses. Energy that goes through the battery is converted to AC by the inverter a second time on the way out, so the bank must hold that little extra.
References
- IEC 61724-1:2021, Photovoltaic system performance — Monitoring — definition of performance ratio and reference yield
- IEC 62548-1:2023, Photovoltaic (PV) arrays — Design requirements — array design, voltage and current limits
- World Bank / Solargis, Global Solar Atlas — irradiation data by location
- NEPRA, Alternative & Renewable Energy Distributed Generation and Net Metering Regulations, 2015 (as amended) — distributed generation and metering rules in Pakistan