10kW Solar Cable Size: 4mm² or 6mm²? Voltage Drop Guide

01The reference 10kW system this article is built on

Voltage drop percentages are meaningless without the system they were calculated for, so here is the reference array in full. It is a typical 2026 residential-to-light-commercial build: 18 modules of 550 Wp, arranged as two strings of nine into a dual-MPPT inverter.

Module electrical data is representative of the 550 Wp class. Your datasheet controls — particularly I_sc, which sets the design current in section 02, and the V_oc temperature coefficient, which decides whether Configuration B is legal at your site's record low temperature.

02Start with the design current, not the cable

This is the step that most cable sizing articles skip, and it is the step that changes the answer. You do not size a PV cable to I_mp. You size it to a design current that carries a safety factor, because a PV module can deliver more than its STC rating under cloud-edge irradiance and cold-cell conditions.

The NEC factor is the product of two separate 1.25 factors in 690.8(A) and 690.8(B), not a single 1.56 rule — the distinction matters if you ever have to defend the calculation. Note also that bifacial modules require you to use the highest I_sc on the datasheet, which may be the bifacial figure rather than the front-side STC value.

For the voltage drop work in sections 03 to 05 we use I_mp = 13.25 A, because voltage drop is an energy-yield question evaluated at the operating point, not a thermal safety question. The 17.5 A or 21.8 A figure returns in section 06 for the ampacity check. Using the design current for the voltage drop calculation is conservative but slightly overstates the loss.

03The voltage drop formula, and which resistance to put in it

The DC voltage drop across a two-conductor string circuit is:

ΔV = 2 × L × I × R

• L = one-way cable length in metres, measured along the routed path
• I = string current in amperes (I_mp at STC for a yield calculation)
• R = conductor resistance in Ω/m at the operating temperature
• 2 = the return path; both the positive and negative conductor drop voltage

The percentage drop is then ΔV% = ΔV / V_mp(string) × 100.

Two ways to get R, and why they differ by about 18%

Most published calculators use ideal copper resistivity divided by cross-section: R = ρ / A, with ρ₂₀ = 0.017241 Ω·mm²/m. That is physically correct for a solid annealed copper conductor of exactly the nominal area. It is not what you will be supplied.

A certified flexible PV cable has a class 5 stranded conductor, and IEC 60228 permits a maximum DC resistance that is materially higher than ρ/A, because stranding adds length, lay and contact effects. Since you are buying a cable that meets IEC 60228, using the IEC 60228 maximum is both more traceable and more conservative:

Column 2 from EN 50618 parameter tables (five manufacturers agree). Column 3 = 0.017241 / A. Column 5 = column 2 ÷ 1,000 × 1.2751, the temperature correction explained below.

Temperature correction

Copper resistance rises with temperature: R(T) = R₂₀ × [1 + α(T − 20)] with α = 0.00393 /°C. A PV cable on a dark roof in summer runs hot. At a conductor temperature of 90 °C the multiplier is 1 + 0.00393 × 70 = 1.2751, which is the factor used in column 5 above.

90 °C is a deliberate choice, not a rounding convenience. It is the maximum continuous conductor temperature for which EN 50618 PV cable is rated, and a rooftop conduit in a hot climate will approach it. Using 20 °C resistance would understate the drop by roughly 22% at the worst moment of the year — which is exactly the moment the inverter is clipping.

04How much voltage drop is acceptable?

This is one of the most searched questions in the category and the answer is less tidy than people want. IEC 60364-7-712 does not mandate a specific voltage drop percentage. What circulates as "the 3% rule" is a mixture of national practice, IEC 62548 guidance and NEC informational notes:

The binding constraint on your project is the one written into the interconnection agreement or enforced by the authority having jurisdiction. Where nothing is specified, 2% on the DC side is a defensible default and 1% is a reasonable stretch target for high-yield sites.

054mm or 6mm solar cable: results from 20 m to 100 m

Applying ΔV = 2 × L × 13.25 × R₉₀ to Configuration A (V_mp = 373.5 V) gives the grid below. Percentages are of string V_mp.

Every cell is reproducible: ΔV% = 2 × L × 13.25 × R₉₀ / 373.5 × 100, with R₉₀ from Table 3. Example, 4 mm² at 40 m: 2 × 40 × 13.25 × 0.0064903 = 6.880 V; 6.880 / 373.5 = 1.84%.

The practical reading

• Up to about 25 m: 4 mm² keeps you under 1.2%. Going to 6 mm² buys very little and costs copper for no measurable yield.
• 25–45 m: 4 mm² is still inside 2%, but this is the band where 6 mm² starts to be worth pricing. The break-even maths is in section 08.
• 45–70 m: 6 mm² is the right default. 4 mm² is now above the 2% target on yield grounds even where it still passes an ampacity check.
• Beyond 70 m: start with 10 mm², and seriously consider whether the inverter can be moved closer. Cable is usually the more expensive way to fix a layout problem.

06Ampacity: the check that almost never binds

Voltage drop decides the size on PV DC circuits. Ampacity is the check you still have to do, and it usually passes with room to spare — which is worth stating explicitly, because the opposite is true on almost every other cable type.

Ambient temperature factors relative to 60 °C: 70 °C → 0.91, 80 °C → 0.82, 90 °C → 0.71, 100 °C → 0.58, 110 °C → 0.41. Grouping factors for more than two cables, conduit and burial come from IEC 60364-5-52, not from this table.

Now the actual check. At a 90 °C ambient — hotter than almost anywhere on earth at roof level — the 4 mm² single-cable value becomes 55 × 0.71 = 39 A. Two cables touching on a surface: 44 × 0.71 = 31 A. Both sit comfortably above the 21.8 A NEC design current and the 17.5 A IEC design current. Ampacity is not the constraint here; voltage drop is.

There is one situation where that flips: a long run bundled with many other cables in a tray or conduit in a hot climate. Six or more loaded cables together can push the grouping factor low enough that a 2.5 mm² or even a 4 mm² selection stops being safe. The full derating treatment for bundled and buried runs is in the current carrying capacity guide.

07When 10mm² is right, and when 1500 V is the cheaper answer

Two decisions get conflated in cable sizing: how big the conductor is, and how the array is configured. The second one is usually worth more.

When to step up to 10mm²

• Long runs beyond about 70–80 m where 6 mm² would exceed your voltage drop target.
• Combiner-box output circuits carrying the summed current of several strings — here the current genuinely is higher, and so is the stake.
• High ambient sites with heavy grouping derating, where the ampacity check starts to matter.
• Bifacial modules with a materially higher I_sc than a comparable monofacial module; check the datasheet figure before reusing a previous design.

The 1500 V option

Configuration B in Table 1 puts all 18 modules in one string: V_mp doubles to 747 V while the current stays at 13.25 A. Because voltage drop is a percentage of a now-doubled voltage, every figure in Table 5 halves and you halve the number of cable runs:

• 40 m on 4 mm²: 0.92% instead of 1.84%
• 60 m on 4 mm²: 1.38% instead of 2.76%
• 80 m on 6 mm²: 1.23% instead of 2.45%
• 100 m on 10 mm²: 0.88% instead of 1.76%

The constraint is not the cable — H1Z2Z2-K is rated for DC 1,500 V. The constraint is the inverter's maximum DC input voltage and the string V_oc corrected to your record low temperature. At 18 modules, a V_oc temperature coefficient of −0.25 %/°C and a record low of −10 °C, the correction is 35 K below STC: 891 V × 1.0875 = about 969 V. That is 97% of a 1,000 V inverter's rating, which most designers will not accept; it fits a 1,100 V or 1,500 V inverter comfortably. Colder sites or a worse coefficient push it higher, so run this with your own module data and your own record low before committing to Configuration B.

08What a percentage point of voltage drop is worth over 25 years

Resistive loss as a fraction of delivered power is exactly ΔV/V — the same number as the voltage drop percentage. So a 1.84% drop is a 1.84% energy loss at the operating point, and the conversion to money is direct.

Assumptions, stated so you can change them: annual yield 15,000 kWh for the 9.9 kWp array, electricity at US$0.10/kWh, 25-year horizon, 5% discount rate. The 25-year annuity factor at 5% is 14.094.

Reproduce any cell: percentage points saved × 15,000 kWh × US$0.10 × 14.094. Example, 40 m 4→6 mm²: 0.615 × 15,000 = 92.3 kWh/yr; × 0.10 = US$9.23/yr; × 14.094 = US$130.

We are deliberately not publishing cable prices here. Copper moves monthly, and a static price table in an article is wrong within weeks and misleading from the day it is written. Get the delta from a live quotation and compare it against the NPV figure.

09What this article does not claim

• The reference system is representative, not universal. Substitute your own V_mp, I_mp, I_sc and string voltage; the formula is the deliverable, the table is an illustration.
• The module electrical data is typical of the 550 Wp class. We did not pull it from a specific manufacturer's datasheet, and your datasheet controls.
• The 3% IEC 62548 figure is graded [B] — a single-source reading. The 1–2% DC practice figure is graded [A].
• We publish no cable prices, by design. Compare the quoted delta against the NPV in Table 7.
• This is not a substitute for a stamped design. Where an AHJ, an interconnection agreement or an insurer specifies a limit, that limit wins.