How does distance affect wire gauge?

Voltage drop is proportional to length. Double the run and you double the drop, so a long run needs a thicker conductor — a lower AWG number — to stay within the same percentage. The calculator takes the one-way distance and doubles it for the out-and-back path.

What gauge do I need for a 20 A circuit 150 feet away at 120 V?

Keeping to a 3% drop, the copper conductor works out to roughly 26,500 circular mils, which is between 6 and 8 AWG — you'd run 6 AWG. For ampacity alone, 12 AWG carries 20 A; the extra size is entirely for the distance.

What voltage-drop percentage should I use?

3% is the common target for a branch circuit, 5% for feeder plus branch combined. The NEC states these as recommendations in informational notes, not hard rules, but equipment specs or the local authority may impose stricter limits.

Does aluminum need a bigger size than copper?

Yes. Aluminum's higher resistivity means about 1.6 times the circular-mil area for the same drop — typically two AWG sizes larger. The calculator's material selector applies the right constant.

Does this replace an NEC calculation?

No. It sizes for voltage drop only. The final conductor also has to satisfy ampacity for its insulation rating, temperature and conduit-fill derating, termination temperature limits, and the overcurrent device — whichever requirement is largest wins.

Is voltage drop worse on DC or low-voltage systems?

Much worse at low voltage. A 3 V drop is 2.5% of 120 V but 25% of a 12 V system, which is why solar and RV DC wiring is so heavy for its current.

Wire Gauge Calculator by Distance

Wire Gauge by Distance: Why Run Length Matters

Choosing wire size is not just about matching the conductor to the circuit’s amperage. As the distance from the source increases, resistance increases and the load receives less voltage. Excessive voltage drop can reduce equipment performance, cause motors to overheat, dim lights, and create nuisance shutdowns.

This calculator estimates conductor size using the common voltage-drop relationship for a two-conductor circuit. It accounts for the full circuit length by multiplying the one-way distance by two, then calculates the conductor area required to keep voltage loss below your selected percentage. The result includes required circular mils and an approximate American Wire Gauge (AWG) value.

The default example uses a 120-volt, 15-amp copper circuit running 100 feet one way with a 3% voltage-drop limit. Remember that the approximate AWG result is not a substitute for selecting an actual standard conductor size. Choose the next larger conductor, which means the next lower AWG number, that satisfies the calculation. Confirm the final selection against NEC ampacity tables and local requirements. NEC Article 210.19(A) includes a 3% branch-circuit voltage-drop recommendation in its informational notes, but voltage drop is only one part of safe wire sizing.

How to Calculate Wire Gauge by Distance (step by step)

Step One: Find the Maximum Allowable Voltage Drop

Convert the selected voltage-drop percentage into volts by multiplying system voltage by the percentage and dividing by 100.

maximum voltage drop = system voltage × maximum drop percentage ÷ 100

120 V × 3% ÷ 100 = 3.6 V

Step Two: Convert One-Way Distance to Total Circuit Length

A typical two-conductor circuit carries current out to the load and back to the source, so multiply the one-way distance by two.

total circuit length = one-way distance × 2

100 ft × 2 = 200 ft

Step Three: Calculate the Required Conductor Area

Use the resistivity constant for the conductor material, the load current, the one-way distance, and the allowable voltage drop. The calculator applies the distance multiplier directly in this formula.

required circular mils = 2 × conductor constant × current × one-way distance ÷ maximum voltage drop

2 × 12 × 15 A × 100 ft ÷ 3.6 V = 10,000 circular mils

Step Four: Convert Circular Mils to Approximate AWG

The AWG scale is logarithmic — each three gauge numbers roughly doubles the area — so converting a circular-mil area to a gauge number takes a logarithm:

With circular mils:

36 − 19.5 × log10(10,000 ÷ 25) ÷ log10(92) = 10.2 AWG

A lower AWG number is a thicker wire. Round toward the thicker conductor: 10.2 becomes 10 AWG copper for this example. Then check that 10 AWG also satisfies ampacity for the circuit’s 15 A — it does — and all the installation conditions below.

Wire Gauge Calculator by Distance worked example diagram — A visual read of the worked example above — the same numbers the calculator returns for its default inputs.

Voltage drop by gauge, at a glance

Copper, 120 V, 15 A load, 3% (3.6 V) limit — the longest one-way run each gauge supports:

Gauge	Circular mils	Max one-way run
14 AWG	4,110	~41 ft
12 AWG	6,530	~65 ft
10 AWG	10,380	~104 ft
8 AWG	16,510	~165 ft
6 AWG	26,240	~262 ft

Notice 14 AWG is rated to carry 15 A all day, but past about 40 feet the voltage drop, not the ampacity, is what forces a size increase.

Why voltage drop matters

Motors run hotter and produce less torque on low voltage, and can stall or trip on starting.
Heating elements put out power proportional to voltage squared, so a 5% drop is nearly a 10% loss of heat.
LED drivers and electronics mostly compensate, but at the cost of drawing more current, which widens the drop further.
Long low-voltage DC runs (solar, RV, marine) are the worst case — a few volts lost is a large percentage of 12 or 24 V.

Common mistakes

Using one-way length in a drop formula that expects total length. Current flows out and back; the calculator doubles it for you, but a hand calculation must too.
Sizing for ampacity only and ignoring distance on a long run.
Forgetting aluminum’s penalty — roughly two gauges larger than copper for the same drop.
Treating the AWG output as final. It’s the voltage-drop minimum; ampacity, derating, and terminations can all demand more.

For the plain Ohm’s-law drop across a known resistance, see the amps to volts calculator.