DC Wire Size Calculator

Pick a copper wire size for a DC run from current, one-way length, and a voltage-drop budget — checked against NEC ampacity, not drop alone.

DC amps on this cable. From DC-side watts: A = W ÷ V. From inverter AC output watts: A = W ÷ (battery V × 0.90) so inverter loss is included — or paste the DC current from Inverter Size Calculator

The DC bus this cable sits on. Higher voltage means less current and a thinner wire for the same watts.

Battery to load, one way. The formula doubles this for the return conductor.

3% is the usual design target. Sensitive electronics sometimes use 2%; a short, cheap run can live with 5%.

Check only if this current lasts three hours without a break. That raises the ampacity demand by 125%; voltage drop still uses the amps you typed.

Reset

Results

Recommended copper wire: 6 AWG

Voltage drop is what sizes this cable.

Circular mils the drop formula asks for: 21,500

Circular mils of the recommended gauge: 26,240

Voltage drop on the recommended gauge: 0.29 V (2.46%)

Voltage at the load: 11.71 V

Heat lost in the cable: 5.89 W

Ampacity the cable must cover: 20 A

Effective ampacity of the recommended gauge: 65 A

How it works

A DC cable has to clear two independent checks: voltage drop on the round-trip run, and the current the copper (and its breaker) can carry. The circular-mil formula finds the drop-limited size; the pick is the smallest AWG that also passes the NEC ampacity table.

required_cm = 2 × K × I × L / VD VD = system_voltage × (voltage_drop_pct / 100) actual_drop_v = 2 × L × (R / 1000) × I ampacity_need = I, or I × 1.25 if continuous recommended = smallest AWG with cmil ≥ required_cm AND actual_drop_v ≤ VD AND effective_ampacity ≥ ampacity_need

K
12.9 — IAEI copper k-factor at 75 °C, derived from NEC Chapter 9 Table 8 (ohm/kFT × circular mils).
I
DC current on the run, in amperes. DC-side watts divide by voltage; inverter AC watts divide by battery voltage × 0.90 (same as the inverter-size tool). Voltage drop uses this number even when the load is continuous.
L
One-way length in feet. Both formulas multiply by 2 for the return conductor.
VD
Allowed drop in volts: system voltage × (budget percent ÷ 100). At 12 V and 3% that is 0.36 V.
required_cm
Minimum circular mils from the k-factor formula. A gauge still has to beat this number.
actual_drop_v
Round-trip drop using that gauge's Table 8 stranded resistance at 75 °C, not the rounded k-factor.
ampacity_need
Current the conductor must be rated for. Continuous loads (3+ hours at a stretch) are multiplied by 1.25. 14/12/10 AWG then take the 240.4(D) cap of 15/20/30 A.

Assumptions

FAQ

What size wire for 30 amps at 12 volts?

It depends on length. At the default 15 ft one-way and a 3% budget, 30 A on 12 V needs 4 AWG copper (6 AWG drops 3.7%). A 5 ft run of the same current is mostly an ampacity problem: 10 AWG would nearly hold 3%, but NEC 240.4(D) caps it at 30 A, so a continuous 30 A load still steps up.

Is 3% voltage drop a code requirement?

No. NEC 210.19's 3% / 5% figures sit in an informational note — they describe reasonable efficiency, they do not require a size. A 12 V inverter input is unforgiving of sag, so 3% is still the sensible default. Tighten it to 2% for electronics; loosen it only when you accept the extra heat.

Is wire length one-way or round trip?

One way — battery to the load. Both the circular-mil formula and the Table 8 resistance check multiply by 2 for the return conductor. If you type a round-trip tape measure, the drop is counted twice and the wire comes out a size (or two) heavy.

Sources

Related calculators

Some product links on this site are affiliate links; we may earn a commission at no extra cost to you. Read the Affiliate Disclosure.