Wire Gauge Calculator (AWG Wire Size)

What is a wire gauge calculator?

A wire gauge calculator answers the question an electrician actually asks at the panel: what size conductor do I need for this run? You know the load, you know how far it has to travel, you know the supply voltage, and you know how much voltage you are willing to lose along the way. What you do not know is the gauge.

Most voltage-drop tools work the other way round. They ask you to nominate a wire size and then tell you how many volts it loses, which means finding the right conductor is a guessing game: try 12 AWG, see 6.6% drop, try 10 AWG, still too much, try 8 AWG, finally acceptable. This page inverts the equation. It solves directly for the conductor area your design requires, then walks up the AWG ladder and hands you the smallest standard size that clears your limit — no trial and error.

The result is the cheapest, thinnest, easiest-to-pull conductor that still delivers usable voltage at the far end.

Sizing for drop is not the same as sizing for ampacity

Two completely separate questions decide a conductor’s size, and confusing them is the most common wiring mistake on long runs:

• Ampacity asks whether the wire can carry the current without overheating. That is a safety question, governed by NEC Table 310.16, insulation temperature ratings, ambient temperature, and bundling.
• Voltage drop asks whether enough voltage survives the journey to run the load properly. That is a performance question, and the NEC addresses it in informational notes rather than enforceable rules.

On a short run, ampacity almost always decides the answer — 12 AWG copper is good for 20 A, and 30 feet of it loses a negligible amount. On a long run, voltage drop takes over completely. A 20 A circuit reaching 200 feet out to a detached garage is perfectly legal on 12 AWG from a heating standpoint, yet the load at the end may see only 107 V. This calculator sizes for voltage drop. Always check ampacity separately and use whichever size is larger.

The consequences of ignoring drop are real: induction motors starved of voltage draw more current to make the same torque and cook their windings, incandescent output falls off steeply, LED drivers behave erratically, and resistive heaters lose output with the square of voltage — a 5% voltage loss costs nearly 10% of the heat. The volts you lose are burned as heat inside the wall.

How does the calculator work?

The tool uses the circular-mil form of Ohm’s law and the conductor properties published in NEC Chapter 9, Table 8 — the same basis as the classic voltage-drop approximation, rearranged to solve for area.

Step 1 — the resistivity constant. $K$ is the resistance of a conductor one circular mil in area and one foot long, in ohm-circular mils per foot, at 75 °C:

$$K_{\text{copper}} = 12.9 \qquad K_{\text{aluminum}} = 21.2$$

Aluminum’s constant is about 64% higher, which is exactly why aluminum needs a noticeably fatter conductor for the same job.

Step 2 — the phase multiplier. Current has to make a complete circuit. DC and single-phase AC travel out on one conductor and back on the other, so the one-way length counts twice. In a balanced three-phase system the line currents are 120° apart and partly cancel, so the multiplier drops to $\sqrt{3}$:

$$n_{\text{DC}} = 2 \qquad n_{\text{single}} = 2 \qquad n_{\text{three}} = 1.732$$

Step 3 — the voltage you are allowed to lose. Your percentage limit becomes a number of volts:

$$V_{\text{allowed}} = V_{\text{source}} \times \frac{P_{\text{max}}}{100}$$

The default is 3%, the branch-circuit target in the informational notes to NEC 210.19(A) and 215.2(A) (5% total across feeder plus branch circuit).

Step 4 — solve for the area. This is the inversion. Rearranging the voltage-drop formula for the conductor area gives:

$$CM_{\text{required}} = \frac{n \times K \times I \times L}{V_{\text{allowed}}}$$

where $I$ is the load current in amps and $L$ is the one-way length of the run in feet.

Step 5 — climb the ladder. The calculator scans the standard AWG sizes from smallest to largest and stops at the first one whose tabulated area is at least $CM_{\text{required}}$:

Note that the gauge numbers run backwards: a smaller number means a thicker wire.

Step 6 — report the real numbers. The chosen conductor is almost always bigger than strictly required, so the actual drop comes in under your limit. The calculator converts the area to metric and recomputes the true drop:

$$A_{\text{mm}^2} = CM \times 0.0005067$$ $$V_{\text{drop}} = \frac{n \times K \times I \times L}{CM_{\text{selected}}}$$

Worked examples

Example 1 — a 20 A branch circuit, 100 ft out

A 20 A single-phase load sits 100 ft (30.48 m) from a 120 V panel. Copper conductors, 3% target.

Allowed loss:

$$V_{\text{allowed}} = 120 \times \frac{3}{100} = 3.6$$

Required area:

$$CM_{\text{required}} = \frac{2 \times 12.9 \times 20 \times 100}{3.6} = 14{,}333.33$$

The first tabulated size at or above 14,333.33 circular mils is 8 AWG at 16,510 CM. In metric that is:

$$A_{\text{mm}^2} = 16{,}510 \times 0.0005067 = 8.37$$

And the drop it actually produces:

$$V_{\text{drop}} = \frac{2 \times 12.9 \times 20 \times 100}{16{,}510} = 3.13$$ $$P_{\text{actual}} = \frac{3.13}{120} \times 100 = 2.60$$

So 8 AWG copper, 8.37 mm², losing 3.13 V — 2.60% of the supply, comfortably inside the 3% target. Note that 12 AWG would have been fine for ampacity alone; the distance is what forces two gauge steps up.

Example 2 — a three-phase aluminum feeder

A 40 A three-phase load runs 250 ft from a 240 V source on aluminum, 3% target.

$$V_{\text{allowed}} = 240 \times \frac{3}{100} = 7.2$$ $$CM_{\text{required}} = \frac{1.732 \times 21.2 \times 40 \times 250}{7.2} = 50{,}997.78$$

The first size that clears 50,997.78 CM is 3 AWG at 52,620 CM — 26.66 mm². The real drop:

$$V_{\text{drop}} = \frac{1.732 \times 21.2 \times 40 \times 250}{52{,}620} = 6.98$$ $$P_{\text{actual}} = \frac{6.98}{240} \times 100 = 2.91$$

3 AWG aluminum, 2.91% drop. Two things carried the day here: aluminum’s higher resistivity pushed the size up, while the three-phase multiplier of 1.732 instead of 2 pulled it back down.

Example 3 — landing exactly on a tabulated size

Occasionally the requirement lands precisely on a table entry. Take 6.53 A over 100 ft of copper at 120 V with a 2.15% limit:

$$V_{\text{allowed}} = 120 \times \frac{2.15}{100} = 2.58$$ $$CM_{\text{required}} = \frac{2 \times 12.9 \times 6.53 \times 100}{2.58} = 6{,}530.00$$

That is exactly the area of 12 AWG. Because the comparison is “at least as large as”, the calculator selects 12 AWG rather than stepping up to 10 AWG — 3.31 mm², and the actual drop equals the allowance exactly:

$$V_{\text{drop}} = \frac{2 \times 12.9 \times 6.53 \times 100}{6{,}530} = 2.58$$

At 2.15% of 120 V, the design sits exactly on its limit. Push the current to 6.6 A and the requirement becomes 6,600 CM, which 12 AWG can no longer satisfy — the answer jumps to 10 AWG.

Example 4 — past the end of the table

A 200 A load 500 ft away on 120 V with a 3% target:

$$CM_{\text{required}} = \frac{2 \times 12.9 \times 200 \times 500}{3.6} = 716{,}666.67$$

That is more than three times the area of 4/0, the largest AWG size. Rather than emit a wrong gauge, the calculator tells you the run is past the table: you need kcmil conductors, parallel runs, or — far more sensibly — a higher supply voltage.

Practical notes

• Enter the one-way distance. The multiplier already doubles it for DC and single-phase circuits. Measure the wire as it actually routes — up walls, along joists, around obstacles — not the straight line between panel and load.
• Raising the voltage beats upsizing the wire. Power is voltage times current, so a 240 V circuit delivering the same watts draws half the current and has twice the volts to spend. That is a factor-of-four improvement in percentage drop for free. Example 4 is a 3/0 problem at 240 V and an easy one at 480 V.
• Aluminum costs you about two gauge sizes. With $K = 21.2$ against copper’s 12.9, aluminum needs roughly 64% more area for the same drop. It is cheaper per foot and lighter to pull, so the trade is often still worth it on feeders — just budget for the larger conduit.
• A tighter percentage target is expensive. Required area scales inversely with the allowed drop: halving your limit from 3% to 1.5% doubles the required circular mils, typically three gauge steps. Reserve tight limits for sensitive equipment.
• This is the DC-resistance approximation. It ignores conductor reactance, temperature rise above 75 °C, and power factor — accurate enough for the overwhelming majority of branch circuits and short feeders. For long runs, very large conductors, or a poor power factor, use the AC impedance figures in NEC Chapter 9, Table 9. Treat this as a solid design estimate, not an engineered study.
• Check the result both ways. Once you have a gauge, confirm it with a voltage drop calculator, size for ampacity using an Ohm’s law calculator, and convert your load from watts with a watts to amps calculator.

Frequently asked questions

Is the 3% limit a legal requirement? No. In the NEC it appears as an informational note — guidance rather than an enforceable rule. Some local amendments, utility rules, or equipment specifications do make it binding, and a few impose stricter limits. Designing to 3% on the branch circuit and 5% overall satisfies nearly everyone.

Why does the gauge number get smaller as the wire gets bigger? American Wire Gauge counts the number of drawing dies a rod passes through to reach its final diameter. More passes means a thinner wire and a higher number. Once the scale runs out at 1 AWG, it continues as 1/0, 2/0, 3/0, 4/0 (spoken “one-aught” through “four-aught”), each thicker than the last.

What is a circular mil? It is the area of a circle one mil — one thousandth of an inch — in diameter. Wire tables use it because the area in circular mils is simply the diameter in mils squared, with no $\pi$ involved, which makes comparing conductors trivial.

Can I enter metric? Yes. Switch the length field to meters or kilometres and the current field to milliamps or kiloamps; the calculator converts internally to the feet and amps the NEC constant expects. The cross-sectional area result also switches between mm², cm², and in².

The answer is bigger than my breaker needs — is that wrong? No, that is the point. Ampacity and voltage drop are independent constraints, and on a long run the drop constraint wins. A 20 A breaker protects 12 AWG just fine, but at 100 ft you still want 8 AWG so the load actually receives usable voltage. Always install the larger of the two answers.