Close-up of dark blue solar PV panels mounted across curved terracotta Mediterranean roof tiles

How to Size an 11kW Off-Grid Solar System in Europe with GSIKE: PV, LiFePO4 Battery & 48V Cable Math

Continuous power 11 kW / 11 kVA · main AC output 11 kW + secondary 5.5 kW · surge 22000 VA · pure sine wave 230 V / 50 Hz · 48 V battery bus (51.2 V LiFePO4) · dual MPPT 2×5500 W = 11 kW · PV 60–500 V DC · solar charge 160 A / AC charge 120 A / discharge 220 A · price EUR 509.99

Short answer: sizing an 11 kW 48 V off-grid system starts on a spreadsheet, not in a catalogue — you convert every appliance to watt-hours per day, then work backwards to the battery, the PV array and the cables. For a European house that uses 10–15 kWh per day, the result is a LiFePO4 bank of roughly 244–732 Ah (depending on how many nights of autonomy you want), a PV array of 4.3–6 kWp split across two 5500 W MPPT trackers, and 48 V copper cables of 25–50 mm² for currents up to 220 A. The GSIKE 11 kW 48 V inverter is built to those exact limits: 11 kW continuous and 22000 VA surge are hard ceilings you cannot push past by adding more panels.

Check the GSIKE 11kW 48V inverter: specs, price and reviews
Neat row of wall-mounted 48V LiFePO4 battery cabinets lined up in a clean European garage
A row of 48 V LiFePO4 cabinets in a garage: overnight, this is where the house draws its power once the sun is down.
Open European distribution board showing thick copper busbars, heavy DC cables and miniature circuit breakers
Thick 48 V copper feeders, busbars and DC breakers inside a distribution board.

First, the formulas and the voltage choice — 12 V, 24 V or 48 V?

The most expensive mistake in an 11 kW installation is sizing parts to a brochure rating instead of to the real load profile of a home, farm or workshop in Europe. Before you calculate anything, you need to know that the GSIKE unit runs on a 48 V bus (physically a 51.2 V LiFePO4 block), not on 12 V or 24 V.

The voltage-selection rule repeats across every solar sizing guide and is worth memorising: 2–2.5 kW suits 12 V, 3.5–4.2 kW suits 24 V, and 6.2–11 kW belongs on 48 V. At 11 kW a 12 V bus would pull around 900 A, which is physically impossible without busbars as thick as water pipes. The 48 V bus drops peak current to about 220–250 A — large, but already manageable with standard battery welding cable and single-battery lugs.

The same five formulas run through every step of this guide. You can drop in your own appliances and re-check the whole system in a spreadsheet.

Daily energy [Wh] = appliance power [W] × hours of use per day [h].

Battery energy [Wh] = voltage [V] × capacity [Ah].

Usable Wh = nominal Wh × depth of discharge (for LiFePO4 we use 0.8).

Nominal capacity [Ah] = daily Wh × autonomy days ÷ DoD ÷ 51.2 V.

Cable voltage drop: ΔU = 2 × one-way length [m] × current [A] × copper resistivity ÷ cross-section [mm²], where copper resistivity ρ = 0.0175 Ω·mm²/m.

PV charge current: I = PV power × efficiency ÷ bus voltage. At 11 kW and 48 V that is about 249 A theoretically, but the inverter caps solar charge at 160 A, so above a certain array size the surplus simply flows straight to midday loads instead of into the battery.

Step 1 — The watt-hour balance for a day: home versus farm

A typical European detached home without whole-house electric heating (it burns gas, wood or runs a heat pump only in season) uses roughly 10–15 kWh per day. A small farm, workshop or outbuilding with pumps, compressors and night lighting already sits at 15–25 kWh per day. The two tables below add up common appliances, and the last row totals everything into watt-hours.

Balance for a detached European home (10–15 kWh/day)

Load Power Hours/day Energy/day Notes
LED lighting (12×10 W) 120 W 5 h 600 Wh Evenings
Fridge + freezer (average) 160 W 24 h 3840 Wh Compressor, 3–5× startup
Washing machine + dishwasher 800 W 1 h 800 Wh Water heating element
Pressure pump 750 W 0.4 h 300 Wh Short cycles
Air conditioning / heat pump (season) 900 W 5 h 4500 Wh Mostly daytime
Electronics: router, TV, laptops, chargers 110 W 6 h 660 Wh Needs pure sine wave
Induction hob (lunch + dinner) 2000 W 0.5 h 1000 Wh Peak hour spike
Home total — — 11 700 Wh ≈ 11.7 kWh/day

Balance for a small farm / workshop (15–25 kWh/day)

Load Power Hours/day Energy/day Notes
Workshop + building LED lighting 300 W 6 h 1800 Wh Nights and mornings
Commercial fridge / cold room 400 W 24 h 9600 Wh Continuous run
Deep-well / pressure pump 1500 W 1.5 h 2250 Wh Inductive startup
Compressor / workshop tools 2200 W 1 h 2200 Wh Simultaneous peak
Air conditioning / livestock ventilation 1200 W 6 h 7200 Wh Seasonal
Electronics, control room, battery charger 150 W 8 h 1200 Wh 24/7 monitoring
Farm total — — 24 250 Wh ≈ 24.3 kWh/day

The number that matters is not the daily total but the night without sun. After sunset the PV array vanishes and the battery becomes the only source. For the home above, night load (19:00–07:00) is about 3.5–4.5 kWh; for the farm it is 8–12 kWh. Those are the figures you compare against usable storage in step two.

Remember instantaneous power, not just the daily average. A 2000 W hob, a 2000 W kettle and a 2000 W heater are resistive loads with no soft start, yet they pull their full wattage the whole time they run. Switch all three on while a compressor is running and you approach the 11 kW ceiling in a single hour. That is why off-grid owners spread cookery across the day instead of running two hobs at once.

Step 2 — Sizing the LiFePO4 battery: DoD 80% and autonomy days

The GSIKE inverter pairs with 48 V LiFePO4 blocks (physically 51.2 V), and BMS communication runs over an RS485 cable. Lithium iron phosphate chemistry is designed around a depth of discharge of 80% — that is exactly where it guarantees 6000 cycles at 80% DoD and 25 °C, a design life of around 10 years. It means you never pull more than 80% of a nominal block's energy out in a single cycle.

Two worked examples

Example A — home at 15 kWh/day, two nights of autonomy. Usable energy needed = 15 kWh × 2 days = 30 kWh. Nominal capacity at DoD 0.8 = 30 ÷ 0.8 = 37.5 kWh. At 51.2 V that is 37 500 Wh ÷ 51.2 V = 732 Ah on the 48 V bus.

Example B — minimum variant at 10 kWh/day, one night of autonomy. Usable energy = 10 kWh × 1 = 10 kWh. Nominal = 10 ÷ 0.8 = 12.5 kWh. At 51.2 V = 12 500 ÷ 51.2 = 244 Ah.

Configuration on real GSK blocks

LiFePO4 block Nominal Wh Usable @ DoD 80% Parallel to Cycle life
GSK-51.2V-100Ah 5120 Wh ≈ 4.10 kWh (4096 Wh) Up to 16 6000 @80%
GSK-51.2V-300Ah (314 Ah) 16 076.8 Wh ≈ 12.29 kWh (51.2×300×0.8 = 12 288 Wh) Per BMS design 10-year design life

For example A (732 Ah needed), a practical build is three GSK-51.2V-300Ah blocks in parallel: 3 × 12.29 kWh usable = 36.9 kWh, which covers the planned two nights at 15 kWh/day. The alternative is seven or eight 100 Ah blocks in parallel (each at 4.10 kWh usable), but they take more floor space and need equal-length cables to every block so current shares evenly.

For example B (244 Ah needed), the emergency minimum is two GSK-51.2V-100Ah blocks in parallel: 2 × 5120 Wh = 10 240 Wh nominal, about 8.2 kWh usable. That is a one-sunny-day option with reserve for one short night; for two cloudy days in a row, add a third block (15 360 Wh nominal, about 12.3 kWh usable).

The bus voltage always stays 48 V — you wire blocks in parallel, never in series. The BMS message travels over RS485, so the inverter knows the state of charge, temperature and current limits of every pack. A licensed electrician should do the physical install: a short on a 48 V bus at 220 A is not dangerous from a voltage point of view, but it is destructive to undersized cable and insulation.

Step 3 — Sizing the PV array: two 5500 W MPPT trackers and the kWp ratio

The inverter has two independent MPPT trackers, each rated to 5500 W, for a maximum of 11 kW of panels. The tracker window is 60–500 V DC. That means you can feed one or two series strings into each tracker, as long as the power on that tracker stays below 5500 W and the string voltage stays inside 60–500 V.

You size the array from the bottom up, starting from daily consumption. Across most of Europe the average effective peak-sun-hours (PSH) sit around 3.0–3.5 hours per day on a year-round mean basis — lower in the far north, higher in the Mediterranean. The ratio is simple: required kWp = daily Wh ÷ peak-sun-hours.

Scenario Daily use Theoretical kWp = use ÷ 3.5 PSH Practical build (with margin)
Minimum home 10 kWh/day 2.9 kWp 3.5–4 kWp, one–two trackers
Average home (example A) 15 kWh/day 4.29 kWp 5–6 kWp, two strings ≤5.5 kW each
Small farm / workshop 24 kWh/day 6.9 kWp 7–8 kWp, two strings near the 5.5 kW limit each
Upper input limit — 2 × 5500 W = 11 kW panels max Never exceed on one tracker

Take a 450 W panel (Vmp ≈ 41.5 V, Voc ≈ 49.5 V, Impp ≈ 10.8 A). On one MPPT tracker you lay two strings of six panels in series and join them in parallel at the input: 12 × 450 W = 5400 W on that tracker, just under the 5500 W limit. Two trackers give 24 panels = 10.8 kWp — close to the upper input edge, but still inside 2 × 5500 W.

What about charge current? At 10.8 kW of panels and 0.95 efficiency, about 217 A would theoretically flow, but the inverter caps solar charge at 160 A. Above roughly 8 kWp the surplus no longer reaches the battery and instead flows directly to midday loads. That is why a home using 11.7 kWh/day does not benefit from buying a full 11 kWp of panels — 5–6 kWp is plenty and avoids wasting half the tracker capacity.

Across a year, 1 kWp in central Europe yields roughly 850–1050 kWh (indicative). Six kWp gives about 5400 kWh a year, averaging 14.8 kWh/day — exactly what our 11.7 kWh home balance needs, plus a small margin to top up the battery.

Step 4 — The cold-weather Voc correction: why winter kills the MPPT window

A panel's open-circuit voltage (Voc) rises as temperature falls. In a European winter at around −15 °C, a string that held 380 V in summer can climb past 500 V and damage the MPPT tracker. That is why you check every series string not at 25 °C but at the lowest roof temperature you expect.

The winter Voc formula is: Voc_winter = Voc_STC × (1 + (T_min − 25) × Voc temperature coefficient). A typical crystalline silicon coefficient is about −0.30 %/°C. At T_min = −15 °C the temperature gap is −40 °C, so the multiplier is 1 + (−0.003) × (−40) = 1 + 0.12 = 1.12.

Take a string of eight 450 W panels with Voc = 49.5 V each. In summer: 8 × 49.5 = 396 V. In winter at −15 °C: 396 × 1.12 = 443.5 V — still below the 500 V limit, so safe. If you laid nine panels in series: 9 × 49.5 × 1.12 = 499 V, almost on the limit; ten panels would already give 554 V and blow past 500 V.

String check (49.5 V Voc panel) Value 500 V limit
Voc of 8× string at 25 °C 8 × 49.5 = 396 V Below 500 V
Voc of 8× string in winter −15 °C (×1.12) 396 × 1.12 = 443.5 V Below 500 V — OK
Vmp of 8× string 8 × 41.5 = 332 V Inside the 60–500 V window
Max panels in series 500 ÷ (49.5 × 1.12) = 9.02 → 9 9 panels = 499 V, on the edge

The rule is simple: size the string length so Voc_winter stays ≤ 500 V, while Vmp on the hottest roof day does not fall below the 60 V lower edge of the MPPT window. In winter, when panels lie under snow, production is low but voltage is at its highest — that design moment is exactly what sets the maximum panels per string. Two parallel strings on one tracker solve the power problem, not the voltage problem; it is the series length you control with the 1.12 multiplier.

Snow-dusted solar PV panels on a pitched roof of a Swedish wooden house during northern European winter
Under snow, output drops but Voc rises — that is the moment that fixes the maximum panels per string.

Step 5 — The 48 V cable cross-section: from 2 m at 1% to 5 m at 3% drop

At 11 kW of discharge the current on the 48 V bus is I = P ÷ (U × efficiency) = 11 000 W ÷ (48 V × 0.92) = 249 A, and the inverter's maximum discharge rating is 220 A. This is a large current, and every square millimetre counts: on a thin cable the voltage collapses, losses turn to heat and insulation can melt. We use copper resistivity ρ = 0.0175 Ω·mm²/m and the formula S = 2 × L_one-way × I × ρ ÷ ΔU, where the factor 2 accounts for current travelling out and back.

Case Working Theoretical S Choose copper
2 m one-way, 1% drop (0.48 V), 249 A (2×2×249×0.0175) ÷ 0.48 36.3 mm² 50 mm²
5 m one-way, 3% drop (1.44 V), 220 A (2×5×220×0.0175) ÷ 1.44 26.7 mm² 35 mm²

In plain terms: with the battery beside the inverter (1–2 m), a 25–50 mm² copper battery cable is right; when the bank sits in a separate room and the run reaches 5 m, you accept a 3% drop and move to 35 mm². Beyond 4–5 m the problem grows — even at a relaxed 2% (0.96 V) the required area does not shrink much, so the better fix is to shorten the route or raise the bus voltage. This matches the published benchmark that a 10 kW 48 V load draws 208 A and a 2 m one-way run at 1% drop calls for 30.8 mm², rounded up to a practical 50 mm².

Place the main DC fuse as close to the battery positive terminal as possible, inside a sealed enclosure. The sizing rule is 1.25 × maximum line current. At discharge up to 220 A that gives 275 A, so a 250–300 A DC fuse is the standard choice. Add a DC isolator so you can safely disconnect the bank for service. Crimp copper ring terminals with a proper tool — no intermediate connectors — because on a 48 V bus at 220 A every loose joint becomes a hot spot.

Step 6 — The AC side, protection and parallelism up to 6 units (66 kW)

The inverter has two AC outputs: a main 11 kW and a secondary 5.5 kW. The connection standard is 230 V / 50 Hz. The main output current at full load is 11 000 ÷ 230 ≈ 47.8 A, so the main feeder needs a type-C overcurrent breaker of 50–63 A plus a 30 mA RCD (type F or A, suited to electronic loads). Protect the secondary 5.5 kW output with a type-C 20–25 A breaker.

Splitting loads makes sense: the heaviest devices (air conditioning, pump, kitchen) land on the main 11 kW circuit, while lighting, sockets and electronics go on the 5.5 kW circuit. That keeps the instantaneous peak easier to control and lets you decide deliberately which circuit must survive as the emergency loop.

The AC input (grid or generator) charges the battery at up to 120 A. During cloudy weeks you top the bank up from the grid or from a generator through the dry-contact. That reserve is exactly what keeps an 11 kW off-grid build alive through long stretches of grey weather.

When do you need to parallel up to six inverters? When one unit is simply too small: a whole farm with several machines running at once, a heated livestock building, a small processing workshop, or a home where the instantaneous peak regularly exceeds 11 kW. Six units in parallel give 6 × 11 kW = 66 kW. For a normal detached house, a single 11 kW unit with 22000 VA surge headroom is more than enough — you do not need to build a parallel cabinet on day one.

Complete builds — home and small farm, in numbers

Detached home variant (≈11.7 kWh/day): one GSIKE 11 kW 48 V inverter; storage of two to three GSK-51.2V-100Ah blocks in parallel (8.2–12.3 kWh usable, i.e. one to two nights of autonomy) or one 300 Ah block for a comfortable single night; 12–14 × 450 W panels = 5.4–6.3 kWp on two MPPT trackers; 25–35 mm² battery cable with a 250–300 A DC fuse; circuits split across the 11 kW main and 5.5 kW secondary; a 6–8 kVA generator for the cloudiest weeks. The inverter itself costs EUR 509.99 from an EU warehouse, 2–7 working days to most European countries, free shipping, 30-day returns.

Small farm / workshop variant (≈24.3 kWh/day, paralleled): two 11 kW inverters in parallel (22 kW continuous); three GSK-51.2V-300Ah blocks (≈36.9 kWh usable for two nights); 18–20 × 450 W panels = 8–9 kWp on two trackers; separate circuits for motors and lighting; a 30 mA RCD on each AC leg. Here the deciding figure is the simultaneous peak, not the daily total — which is why you scale the number of units rather than overloading one cabinet with panels.

The Bottom Line — design for the peak hour and two cloudy nights

One closing thought: on an 11 kW 48 V system it is never the daily average that wins — it is the instantaneous peak in the busiest hour and the reserve for two or three cloudy nights. A 15 kWh/day home needs a 732 Ah LiFePO4 bank (three 300 Ah blocks); the minimum 10 kWh/day variant needs 244 Ah (two to three 100 Ah blocks). The PV array is 4.3–6.9 kWp depending on use, split across two 5500 W MPPT trackers, with the mandatory ×1.12 cold-Voc correction keeping winter string voltage below 500 V. The 48 V feeder for 220–249 A is 50 mm² at 2 m / 1% drop or 35 mm² at 5 m / 3% drop, plus a 250–300 A DC fuse near the battery. Stay inside the hard limits: 11 kW continuous and 22000 VA surge serve a whole home and a small farm — they are not rated for full electric space heating of a building.

Frequently asked questions

How many PV panels do I need for an 11 kW 48 V inverter?

The inverter accepts up to 11 kW of panels (two 5500 W trackers), but in practice you rarely need that much. For a 15 kWh/day home, 5–6 kWp is enough: at 3.5 peak-sun-hours that is 4.29 kWp theoretical, built with margin to 5–6 kWp. With 450 W panels that means 12–14 panels split across two strings. Above roughly 8 kWp you waste the surplus, because solar charge current is capped at 160 A.

What LiFePO4 battery do I need for an 11 kW inverter — how many Ah?

You calculate from the daily use and autonomy nights: Ah = daily Wh × autonomy days ÷ DoD 0.8 ÷ 51.2 V. At 15 kWh and 2 nights that is 732 Ah (for example three GSK-51.2V-300Ah blocks). At 10 kWh and 1 night that is 244 Ah (two to three 100 Ah blocks). Size LiFePO4 to 80% DoD, because only then it holds 6000 cycles and a 10-year life.

How do I work out daily watt-hours for a European home or farm?

Give each appliance a wattage and its hours of use per day, then multiply: Wh = W × h. The home example totalled 11 700 Wh (≈11.7 kWh): fridge-freezer 3840 Wh, air conditioning 4500 Wh, induction hob 1000 Wh, the rest being lighting, washing, pump and electronics. The small-farm total reaches 24 250 Wh (≈24.3 kWh) once a 24-hour cold room and a deep-well pump are added.

What 48 V DC cable cross-section do I use at 11 kW?

At 11 kW the current is about 249 A (11 000 ÷ (48 × 0.92)). With a 1% drop (0.48 V) and copper ρ = 0.0175, a 2 m one-way run calls for 36.3 mm², so you choose 50 mm². With a relaxed 3% drop (1.44 V) at 220 A over 5 m, the figure is 26.7 mm², so you choose 35 mm². The formula is S = 2 × L × I × 0.0175 ÷ ΔU. Beyond 4–5 m, shorten the route rather than keep shrinking the cable.

What does DoD 80% mean and how does it change the battery size?

DoD (depth of discharge) is how much of a pack you draw out each cycle. LiFePO4 is engineered to use at most 80% of its nominal energy per cycle — that is where it lasts 6000 cycles. That is why you divide the energy you need by 0.8: instead of the 30 kWh you actually need, you must install 37.5 kWh nominal. Pulling 100% every cycle cuts pack life several-fold.

How do I apply the Voc temperature coefficient in winter?

Open-circuit voltage rises in the cold. With a −0.30%/°C coefficient and a −15 °C minimum roof temperature, the multiplier is 1.12. A string of eight 49.5 V panels = 396 V in summer but 443.5 V in winter. Size the series string so Voc_winter = Voc_STC × 1.12 stays below the inverter's 500 V limit, otherwise the first frost can over-voltage the MPPT tracker.

How many autonomy days should I plan for a European winter?

For one sunny day, one night of autonomy (storage covering overnight load) is enough. For the European reality of weeks of cloud and snow, plan 2–3 nights of reserve. At 15 kWh/day that is 2 nights = 30 kWh usable, i.e. 37.5 kWh nominal and 732 Ah. A 6–8 kVA generator closes the gaps during longer grey spells.

Are two 5500 W MPPT trackers enough for an 11 kW system?

Yes, that is exactly how it works: the inverter has two independent MPPT trackers, each up to 5500 W, for 11 kW of maximum panels. Feed each tracker series strings totalling at most 5500 W with Voc_winter below 500 V. A 5–6 kWp home uses only half the tracker capacity, so the headroom is large.

How do I size the AC breakers and cable on the 11 kW output?

The main 11 kW output carries about 47.8 A, so use a type-C 50–63 A breaker and a 30 mA RCD (type F/A). Protect the secondary 5.5 kW output with a type-C 20–25 A breaker. On the grid/generator input, protect the 120 A AC charging leg with a separate isolator. Bond the whole system to earth per EN standards and connect neutral, phase and protective earth.

When do I have to parallel up to six inverters (66 kW)?

When one 11 kW unit is too small: a farm with several machines running at once, a heated livestock building, a small processing workshop, or a home where the instantaneous peak regularly passes 11 kW. Six units in parallel give 6 × 11 kW = 66 kW. For an ordinary house, one inverter with 22000 VA surge is plenty and needs no parallel cabinet.

Sources and figures used in the math

The figures below are the parameter and data sources behind the calculations. Consumption and climate values are indicative for a European-wide mean.

• GSIKE 11 kW 48 V product page (11 kW / 11 kVA, 22000 VA surge, dual 5500 W MPPT, PV 60–500 V, charge 160/120 A, discharge 220 A, price EUR 509.99): gesike.com.

• GSK-51.2V-100Ah battery datasheet (5120 Wh, DoD 80%, 6000 cycles, RS485, parallel up to 16) and GSK-51.2V-300Ah (16 076.8 Wh) — product attachment.

• European single-phase grid standard 230 V / 50 Hz (EN 50160) and indicative PV yield of 850–1050 kWh per kWp per year across Europe.

• Crystalline silicon Voc temperature coefficient around −0.30 %/°C and copper resistivity 0.0175 Ω·mm²/m — typical catalogue values for installation components; cross-checked against published 48 V cable-sizing guidance (10 kW @ 48 V ≈ 208 A, 2 m at 1% → 30.8 mm², practical 50 mm²).

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