Solar Wire Size Calculator
Calculate the minimum copper wire cross-sectional area (in circular mils) needed for a solar DC run to keep voltage drop within acceptable limits. Use it when wiring panels, charge controllers, or battery banks.
About this calculator
Undersized wires cause resistive losses that reduce system output and can create fire hazards. The required wire cross-section is calculated from Ohm's law applied to a round-trip DC circuit. The formula is: Wire Area (CM) = √((2 × current × distance × 12.9) / (voltage × voltageDrop)). Here, 12.9 is the resistivity constant for copper in circular-mil·ohms per foot (ρ = 12.9 Ω·CM/ft at 20 °C). The factor of 2 accounts for the full round-trip length of the circuit. Current is in amps, distance is the one-way run in feet, voltage is the system voltage, and voltageDrop is the allowable fractional drop (e.g., 0.03 for 3%). The result in circular mils maps to a standard AWG gauge using a lookup table.
How to use
A 48 V solar system carries 20 A over a 30-foot one-way run, and you want to limit voltage drop to 3% (0.03). Calculate: Area = √((2 × 20 × 30 × 12.9) / (48 × 0.03)) = √(15,480 / 1.44) = √10,750 ≈ 103.7 CM. Looking up the AWG table, 6 AWG copper wire has 26,240 CM — well above the minimum — but the next smaller size, 10 AWG (10,380 CM), falls just short. You would select 8 AWG (16,510 CM) as the smallest standard gauge that exceeds the requirement.
Frequently asked questions
What voltage drop percentage is acceptable for solar panel wiring?
The NEC and most solar installers recommend a maximum of 2–3% voltage drop on any single DC circuit in a solar installation, and no more than 5% total from panels to load. A 3% drop on a 48 V system equals 1.44 V lost to resistance. Exceeding this threshold wastes energy as heat and, over a 25-year system life, can amount to hundreds of kilowatt-hours in lost production. Critical battery-charging circuits are often sized for 1–2% drop to minimize losses.
Why does the solar wire size formula use 12.9 as the copper resistivity constant?
The value 12.9 represents the resistivity of annealed copper in units of circular-mil·ohms per foot at 20 °C (68 °F). This unit system is traditional in North American electrical work, where wire gauges are rated in circular mils (a circular mil being the area of a circle 1 mil = 0.001 inch in diameter). Using this constant keeps the formula self-consistent when current is in amps, distance is in feet, and voltage is in volts. For metric calculations, a different resistivity constant (in Ω·mm²/m) would be used instead.
How do I convert the calculated circular mil area to an AWG wire gauge?
After calculating the minimum circular mil area, you find the smallest AWG gauge whose circular mil rating equals or exceeds your result. Standard gauges run: 14 AWG = 4,107 CM, 12 AWG = 6,530 CM, 10 AWG = 10,380 CM, 8 AWG = 16,510 CM, 6 AWG = 26,240 CM, 4 AWG = 41,740 CM. Always round up to the next larger wire (lower AWG number) — never down. Also verify the wire's ampacity (current-carrying capacity) meets or exceeds your circuit current, especially in conduit or high-temperature environments.