Solar Cable Sizing Calculation: Ampacity and Voltage Drop

Size PV cable with two checks and use the bigger result. Ampacity: at 30°C the cable must carry 1.25 × 1.25 = 1.56 × Isc. After derating it must still carry 1.25 × Isc. Voltage drop: A = 2 × L × I × ρ ÷ ΔV. The rules below come from Sandia’s guide to the 1999 NEC.

Ampacity is how much current a cable can carry without getting too hot. Isc is the short-circuit current printed on the module data sheet. This guide shows both checks step by step, with three worked examples. All example numbers are made up for practice.

Where do these NEC numbers come from?

The code rules and table values here come from Sandia report SAND2001-0674, “Photovoltaic Power Systems and the National Electrical Code: Suggested Practices” (March 2001). It covers the 1999 NEC. In that edition the rules sit in section 690-8 and Tables 310-16 and 310-17. Later editions use new numbering, such as 690.8 and Table 310.16.

Before you build a permit set, read the matching sections in the NEC edition your local authority (AHJ, the permit office) enforces. NFPA offers free read-only access to the NEC after you sign in.

What are the two cable sizing checks?

Run both checks. Then pick the larger cable. A cable can pass one check and fail the other. On a long run, either check can be the one that decides.

Why do you multiply by 1.25 twice?

Sandia gives two reasons.

First 1.25: sunlight above test conditions. A module’s short-circuit current is nominally 10 to 15% higher than its operating current. Irradiance (sunlight power) goes above 1,000 W/m² around noon on most days. Sandia says it can approach 1,200 W/m² for as long as four hours, and that this can raise short-circuit current by 20%. Because it lasts three hours or more, the NEC treats the current as continuous.

Second 1.25: continuous current. The NEC requires continuous current to be multiplied by 125% before you size cables or fuses. This keeps them at no more than 80% of their rating.

Sandia says both factors apply at the same time. The result is 1.25 × 1.25 = 1.5625, which it rounds to 1.56. It adds that there are only two 1.25 factors, so do not apply the first one twice.

What is the step-by-step method?

This is Sandia’s procedure, in plain words.

1. Circuit current. Multiply the module short-circuit current by 1.25. Use this value for all later steps.
2. Fuse or breaker rating. The overcurrent device must be rated at 1.25 × the Step 1 current.
3. Cable at 30°C. The cable’s 30°C ampacity must be at least 1.25 × the Step 1 current, with no other derating.
4. Derated cable. Pick a size from the NEC ampacity tables. Then derate it for temperature and conduit fill. The derated ampacity must be above the Step 1 current.
5. Match the device. The derated ampacity must be equal to or above the device rating from Step 2. If it is not, use a larger cable.
6. Check the terminals. Most devices have terminals rated for 75°C or 60°C. Check that the cable still works with those terminals.

Step 6 matters because ampacity values for 90°C insulation can only be used if the equipment terminals are also rated 90°C. Sandia says to use the 75°C column to pick the size and the 90°C column for derating.

Then run the voltage-drop check from the section below.

What ampacity values does Sandia print?

These are 90°C copper values at a 30°C ambient. Sandia prints them in its worked examples. Read the full tables in your NEC edition.

Sandia picks conduit values from Table 310-16 and free-air values from Table 310-17. So Table 310-16 is the conduit table, not a free-air table. The areas in square millimetres come from the AWG diameter formula, d = 0.127 mm × 92^((36 − n) ÷ 39), where n is the AWG number. They are rounded.

How do temperature and conduit fill reduce ampacity?

Sandia says ampacity tables assume a 30°C ambient. PV cables work in hotter places, so you derate them.

• Temperature factors for 90°C insulation, as printed by Sandia: 0.91 at 40°C, 0.87 at 45°C, and 0.58 at 61 to 70°C.
• Conduit fill: Sandia uses 0.8 for four current-carrying conductors in one conduit.
• Hot places: Modules often run 30 to 40°C above air temperature when no wind cools them. Conductors near the modules may reach 65 to 75°C. Sandia says derating factors of 0.33 to 0.58 apply there, depending on cable type, how it is installed and its insulation rating.
• Insulation limits: Cable with 60°C insulation has no ampacity above 55°C ambient. Cable with 75°C insulation has none above 70°C. So Sandia suggests using only 90°C cable.

Multiply the factors. For 45°C and four conductors in conduit: 0.87 × 0.80 = 0.696. Derated ampacity = base ampacity × 0.696.

How do you check voltage drop?

Voltage drop comes from Ohm’s law. The cable resistance is R = ρ × length ÷ area. A DC circuit has two conductors, so the current travels the length twice.

ΔV (volts) = 2 × L × I × ρ ÷ A
A (mm²) = 2 × L × I × ρ ÷ ΔV

L = one-way cable length, in metres (use the routed path)
I = circuit current, in amps
ρ = resistivity of copper (this guide uses 0.0175 Ω·mm²/m)
A = cable cross-section, in mm²
ΔV = allowed voltage drop, in volts = system voltage × drop % ÷ 100

The value 0.0175 is an assumption we use for the examples. Use the resistance value that your project or the NEC tables give for your conductor temperature. Sandia’s own example takes resistance from NEC Chapter 9, Table 8, at 75°C. For 400 ft of 8 AWG at 36 A, with 0.778 ohms per 1,000 ft, the drop is 0.778 × 400 ÷ 1000 × 36 = 11.2 V. That is too much for a 12 V battery system. Sandia moved to 2 AWG, which cut the drop to 2.8 V.

Sandia also notes that some well-designed systems use conductors larger than the NEC requires. This cuts steady-state voltage drop and raises system efficiency. If the current-carrying conductors are oversized, the equipment-grounding conductor must be oversized in proportion.

After you find the area in mm², go up to the next standard size. Never round down.

Worked example 1: residential string

Inputs (made up): 20 modules of 330 W in two strings of 10. Module Isc is 9.5 A. String voltage is about 360 V. The one-way run is 15 m, at 30°C, in conduit.

1. Circuit current: 9.5 × 1.25 = 11.88 A (exact 11.875).
2. Ampacity needed at 30°C: 11.88 × 1.25 = 14.84 A.
3. Try 10 AWG, which is 40 A in conduit. Step 3: 40 A is above 14.84 A. Pass. Step 4: there is no derating at 30°C, so 40 A is above 11.88 A. Pass.
4. Fuse or breaker check (Step 5): the device rating is 14.84 A. 40 A is above it. Pass.
5. Voltage drop: 2 × 15 × 9.5 × 0.0175 ÷ 5.26 = 0.95 V, which is 0.95 ÷ 360 = 0.26%.

A 12 AWG cable (3.31 mm²) would drop 1.51 V, or 0.42%. Sandia does not print a conduit ampacity for 12 AWG, so read that value in the NEC table for your edition before you choose it.

Worked example 2: hot rooftop conduit

Same string as example 1, but now the ambient is 45°C and four conductors share the conduit.

1. Combined factor: 0.87 × 0.80 = 0.696.
2. Derated ampacity of 10 AWG: 40 × 0.696 = 27.84 A.
3. Step 3 uses the 30°C value: 40 A is above 14.84 A. Pass. Step 4 uses the derated value: 27.84 A is above 11.88 A. Pass.
4. Fuse or breaker check (Step 5): 27.84 A is also above the 14.84 A device rating. Pass.

Here Sandia’s own example is a good stress test. It has two circuits with Isc = 40 A, four conductors in conduit and 45°C.

• Step 1: 1.25 × 40 = 50 A. Fuse: 1.25 × 50 = 62.5 A, so the next standard size is 70 A.
• 6 AWG (75 A) derates to 75 × 0.8 × 0.87 = 52.2 A. That is above 50 A, but below the 62.5 A fuse rating. It fails Step 5.
• 4 AWG (95 A) derates to 95 × 0.87 × 0.8 = about 66 A. This works with the 70 A fuse.

Worked example 3: long combiner-to-inverter run

Inputs (made up): combined Isc of the strings = 50 A, 800 V system, one-way run 60 m, 40°C, voltage drop target 1.5%, which is 12 V.

Voltage-drop check. A = 2 × 60 × 50 × 0.0175 ÷ 12 = 8.75 mm². The next AWG size up from 8.37 mm² (8 AWG) is 6 AWG at 13.3 mm². Its drop is 105 ÷ 13.3 = 7.89 V, or 0.99%.

Ampacity check. At 30°C you need 1.25 × 1.25 × 50 = 78.13 A. 6 AWG gives 75 A in conduit. That fails. 4 AWG gives 95 A. That passes. Derating at 40°C: 95 × 0.91 = 86.45 A, which is above 1.25 × 50 = 62.5 A. Pass. Fuse or breaker check (Step 5): the device rating is 1.25 × 62.5 = 78.13 A. 86.45 A is above it. Pass.

Result: 4 AWG, 21.15 mm². Its voltage drop is 105 ÷ 21.15 = 4.96 V, or 0.62%. On this run the ampacity check, not voltage drop, set the size. You only see that if you run both checks.

How do voltage and array size affect the cable?

Cable voltage rating matters too. Module voltage rises as modules get colder. Sandia’s example is a system with a rated open-circuit voltage of 595 V at 25°C. A typical crystalline module has a voltage coefficient of −0.38% per °C. At −30°C ambient, Sandia works out 595 × (1 + (25 + 30) × 0.0038) = 719 V. That is well above what common 600 V cable and switchgear can handle. Sandia notes that UL 1703 listed modules for a maximum system voltage of 600 V in its day. Check the voltage limit in your own code edition and on the module data sheet.

Sandia also says the NEC does not allow paralleling conductors for added ampacity, except for 1/0 AWG and larger under certain conditions. It suggests splitting large arrays into subarrays with a short-circuit current under 64 A. Then each source circuit can use 100 A rated equipment, because 1.25 × 1.25 × 64 = 100 A.

For systems under 50 V, Sandia says the NEC calls for 12 AWG or larger conductors under Article 720. It adds that Article 690 math may give a smaller size, but some inspectors still apply the Article 720 rule to DC circuits.

How do you size AC cable after the inverter?

Sandia says the ampacity of conductors to and from an inverter must be at least 125% of the inverter’s rated current. Use the rated continuous current, not Isc.

The voltage-drop formula changes for three-phase circuits, because three conductors share the current:

ΔV (line-to-line) = √3 × L × I × ρ ÷ A

Example (made up): a 10 kW three-phase inverter runs at 400 V. Its rated current is 10,000 ÷ (√3 × 400) = 14.43 A at unity power factor. The cable must carry at least 1.25 × 14.43 = 18.04 A. Take a 30 m run in 2.5 mm² copper. The drop is √3 × 30 × 14.43 × 0.0175 ÷ 2.5 = 5.25 V. That is 5.25 ÷ 400 = 1.31%. Check the ampacity of the 2.5 mm² cable on its own data sheet.

For single-phase AC, use the same formula as DC, with the factor of 2.

Which cable types does Sandia describe?

Sandia also says copper is recommended for almost all PV wiring because it has a lower voltage drop and resists corrosion. It does not recommend aluminium or copper-clad aluminium, especially in dwellings.

Sandia’s report is from 2001 and does not mention newer cable types such as PV wire. For those, read the maker’s data sheet and the NEC.

What mistakes cause the most failures?

Where does SurgePV fit?

SurgePV supports electrical workflow, bill-of-materials output and proposals. Its outputs support design and documentation. They do not replace approval by the responsible engineer or the authority. Book a Free Demo to see how the electrical workflow and the solar design software fit your projects. For quick checks, you can also use the voltage drop calculator.

For related topics, read solar wire sizing under NEC 690.8, temperature derating of conductors, the DC cable length calculator guide and AC disconnect sizing. The glossary explains short-circuit current, ampacity, wire gauge and conduit fill.

Frequently Asked Questions

What is the correct cable size for a solar panel string?

It depends on the current, the run length and the site conditions. In this guide’s example (Isc 9.5 A, 15 m one way, 30°C, conduit), 10 AWG passes the ampacity check at 40 A against 14.84 A needed, and its voltage drop is 0.95 V, or 0.26% of 360 V. Always run both checks for your own project.

Why do I have to multiply by 125% twice for solar cable sizing?

Sandia explains that the first 1.25 covers sunlight above 1,000 W/m² and other module effects. The second 1.25 treats the current as continuous, so fuses and cables run at no more than 80% of their rating. Together they give 1.25 × 1.25 = 1.56 times the module short-circuit current.

What voltage drop limit should I use for solar DC cables?

Use the limit in your project specification or the equipment instructions. A lower limit means a bigger cable and less lost energy. For 50 A over 60 m at 800 V, 1% (8 V) needs about 13.1 mm² and 3% (24 V) needs about 4.4 mm².

Is a 6 mm² cable enough for a solar string?

Check both ampacity and voltage drop. For voltage drop, 9.5 A over 30 m one way on a 360 V string gives 2 × 30 × 9.5 × 0.0175 ÷ 6 = 1.66 V, or 0.46%. For ampacity, read the rating on the cable data sheet and apply the 1.56 factor and your derating factors.

Can I use the same cable sizing formula for IEC and NEC systems?

The voltage-drop formula is the same, because it comes from Ohm’s law. The current multipliers and the ampacity tables come from the code you work under. The multipliers in this guide are from the NEC as described in Sandia’s report.

What cable type should I use for exposed PV conductors?

Sandia suggests USE-2 for module interconnects. It has a 90°C rating and is sunlight resistant. Sandia says 60°C insulation cannot be used near modules, because conductors there can run hotter than 60°C.