Find the correct AWG wire gauge for current, distance, and voltage drop requirements.
This calculator turns a circuit's current, voltage, one-way conductor distance, and material into a minimum AWG wire size, applying the circular-mil voltage-drop formula and comparing that against standard ampacity minimums for the current entered. It returns whichever wire is thicker of the two requirements, since a conductor sized to carry the current safely can still be too thin to keep voltage drop under a chosen percentage over a long run. Users most often enter the one-way distance to the load without realizing the tool doubles it for the return leg, which is usually what makes the recommended gauge bigger than expected.
A concrete input and expected output from the current implementation.
Input
Current: 20 A, Voltage: 120 V (single-phase), one-way distance: 100 ft, material: copper, max voltage drop: 3%
Expected output
Minimum recommended size: 8 AWG (voltage drop governs). 10 AWG would drop about 4.14% at this load and distance; 8 AWG drops about 2.6%, under the 3% limit. Ampacity alone would only require 12 AWG.
Required circular mils = 2 x 12.9 x 20 x 100 / (0.03 x 120) = 14,333 cmil, which exceeds 10 AWG's 10,380 cmil but not 8 AWG's 16,510 cmil, so 8 AWG is the smallest size meeting the drop target.
For the entered current, one-way distance, and material, it computes required circular mils as 2 x K x I x D / allowed-drop-volts, where K is the conductor's resistivity constant (about 12.9 for copper, roughly 21.2 for aluminum) and allowed-drop-volts is voltage times the chosen drop percentage. It rounds up to the next standard AWG size whose circular-mil area meets that minimum, doubling the one-way distance internally to represent the full circuit path.
Independently, it looks up the smallest standard AWG size whose typical ampacity rating covers the entered current, using commonly published copper or aluminum ampacity figures (for example, 12 AWG copper around 20A). It compares this to the voltage-drop result and reports the larger (thicker) of the two gauges as the final recommendation, since either constraint alone could leave the conductor undersized for the other.
Because voltage drop, not ampacity, is often the binding constraint on long runs. A 20A circuit only needs 12 AWG for ampacity, but at 100 feet a 3% drop limit can require 8 AWG instead, and the tool returns the more conservative of the two results.
Yes, if a material option is offered: aluminum uses a higher resistivity constant, which produces a thicker required gauge than copper for the same current and distance. If there is no material selector, assume the result is copper-only and re-check before running aluminum.
The example and behavioral notes were checked against the browser implementation. Standards and primary references below define the relevant format, formula, or platform behavior.
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