Dc Wire Size Calculator

Dc Wire Size Calculator — fast, accurate results online. Enter your values and get instant answers.

A
m
V
%

Results

Calculated
Minimum copper cross-section
—
In mm²
Allowed voltage drop
—
In V
Smallest AWG that meets the drop
—
Lower AWG number = thicker wire
Power lost in the cable
—
At the maximum allowed drop, in W

What it is and when to use it

Every wire has resistance, so a current through it causes a voltage drop and wastes power as heat. In low-voltage DC systems such as 12 V vehicles, boats, solar arrays and battery banks, that drop is a large fraction of the supply voltage, so cables must be sized carefully. Too thin, and lights dim, motors run slowly, chargers underperform and cables run warm.

Use this calculator to find the minimum copper cross-section that keeps voltage drop within a percentage you choose, and the smallest AWG size that satisfies it. It covers voltage drop only. Cable heating limits (ampacity), insulation ratings, fuse sizing, temperature derating and local electrical codes must be checked separately, and the larger of the two requirements wins.

The method

The maximum allowed drop is ΔV = V × (percent / 100). Then the minimum cross-sectional area is A = 2 ρ L I / ΔV.

  • ρ: resistivity of copper, taken as 1.72 × 10⁻⁸ Ω·m (near 20 °C).
  • L: one-way cable length in metres. The factor 2 accounts for the current travelling out on the positive conductor and back on the negative one.
  • I: load current in amperes.
  • ΔV: the allowed drop in volts.

The area is converted to a diameter, and AWG size is found from d = 0.127 mm × 92^((36 − n)/39). The calculator rounds the AWG number down to the next whole number, which is the next thicker standard gauge.

Worked example: 20 A load, 5 m away on a 12 V system

Allowed drop at 3%: 12 × 0.03 = 0.36 V.

A = 2 × 1.72×10⁻⁸ × 5 × 20 / 0.36 = 3.44×10⁻⁶ / 0.36 = 9.556×10⁻⁶ m², or 9.56 mm². That corresponds to a diameter of 3.49 mm and an AWG value of 7.43.

Rounding down to a whole gauge gives AWG 7 (about 10.6 mm²). AWG 8 is only about 8.4 mm², which would exceed the 3% target. Power lost in the cable at the maximum drop is 20 A × 0.36 V = 7.2 W.

Common mistakes and how to interpret the result

  • Entering the round-trip length. The calculator already doubles the one-way length, so entering round-trip length oversizes the wire by a factor of two.
  • Sizing only for voltage drop. A short, high-current run may pass the drop test but still overheat, so check the cable's ampacity rating too.
  • Forgetting temperature. Copper resistance rises with temperature, so a cable in a hot engine bay or conduit drops more voltage than the 20 °C figure used here.
  • Confusing AWG direction. A smaller AWG number means a thicker wire, so 4 AWG carries more current than 10 AWG.

Frequently Asked Questions

Why does a 12 V system need thicker wire than a 120 V one?
The same current causes the same absolute voltage drop, but that drop is a much larger percentage of a 12 V supply than of 120 V. Keeping the percentage low at low voltage requires much thicker conductors, or shorter runs.
What voltage drop percentage should I use?
A drop of 3% is a common design target for branch circuits and many DC installations. Sensitive electronics, charging circuits and long solar runs are often designed for 1 to 2%. Follow your local code or equipment guidance.
Does this work for aluminium wire?
No. It assumes copper resistivity. Aluminium has roughly 60% higher resistivity, so it needs about 1.6 times the cross-section for the same drop, and it has different connection requirements.
Why is the current entered instead of power?
Voltage drop depends on current. If you know watts, divide by the supply voltage to get amps: a 240 W load at 12 V draws 20 A. That is why low-voltage loads need such heavy cable.

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