Wire Gauge Charts: AWG to mm2 Conversion Charts and Wire Sizes

With various standards around, it is important to know how to convert proper cable sizes from one standard to another.

Also, voltage drop and power losses depend on the cable length, thickness, and current flowing through the cable.

Updated: March 21, 2023.

AWG to mm² Conversion Chart

Some of the most commonly used materials for wires are copper, with a resistivity of pure copper of 16.78 nΩm at 20°C and aluminum, with a resistivity of pure aluminum of 26.5 nΩm at 20°C. Other materials can be used as well, but most of the time, wires are made out of these materials.

Keep in mind that metal resistance increases with temperature, not by much, but if you can't hold your trolling motor cables in your hands due to the heat, decrease the power and get thicker cables as soon as possible.

Chart columns:

- AWG #: American Wire Gauge cable thickness

- Diameter mm: diameter of the cable, given in millimeters. So, if you need, for example, AWG 5 cable, but you are offered cables in millimeters only, you should buy a cable 4.62 mm in diameter (with a 16.77 mm² cross-section area) or next thicker cable.

- Diameter inches: diameter of the cable, given in inches.

- Area mm²: cross-section area of the cable, given in mm².

- Area in²: cross-section area of the cable, given in inch².

- Ampacity of the wires is given in Amps - Ampacity is the current-carrying capacity of the wires and is s usually defined as the maximum current that a wire can carry continuously under the conditions of use without exceeding its temperature rating.

Note: the great thing about Ampacity is that it has the same value, regardless of the voltage applied - generally, it "just" depends on the wire thickness, wire material, and quality of insulation (allowed maximum temperature).

And here is the Ampacity chart of enclosed copper wires:

Personally, wires thinner than AWG20 are too thin to be used in anything except special electric and electronic projects.

Wire Thickness: Circular Mil (kcmil) vs. mm²

A circular mil equals the area of a circle with a diameter of one-thousandth of an inch (one mil) or 0.0254 mm and it has a value of 5.067075×10⁻⁴ mm² (506.7075 μm²).

In Canada and the United States, circular mils (cmil) are used to define wire sizes larger than 0000 AWG (4/0 AWG), starting with 250000 cmil, which is written as 250 kcmil or 250 MCM - 250 thousand circular mil.

Standard wire sizes are from:

• from 250 to 400 in increments of 50 kcmil,
• from 400 to 1000 in increments of 100 kcmil,
• from 1000 to 2000 in increments of 250 kcmil.

The following chart lists solid insulated copper wire Ampacities for wire areas given in kcmil and mm2 and wire diameters given in mm2 and inch2.

Note: stranded copper wires have a diameter 5-7% larger to compensate for gaps between the wire strands and must be checked for each wire.

Ampacities of Wires in Free Air

Wires, cables, and conductors, in general, have higher ampacities when suspended in air due to better cooling. This cooling effect can be even increased when, for example, powering an electric trolling motor and a gentle breeze sweeps across the lake, river, or sea surface.

However, wires suspended freely in the air can also be heated by, for example, the sun. Thus, use these values as such, but when in doubt, always go for thicker wires and keep them protected from the sun and similar heat sources.

12V Wire Gauge Chart

12V voltage is the standard voltage used in cars, trucks, boats, campers, etc., and is equivalent to the nominal voltage of a 6-cell lead-acid battery.

To transfer energy efficiently, wires must be thick enough to keep energy losses to a minimum. Too thick wires can be used, but thick wires are more difficult to work with (even when stranded wires are used), they are heavier and more expensive.

The standard acceptable voltage drop in 12V circuits is 3%, which is 0.36 volts.

Thus, the total resistance of the wire can be calculated as follows:

R (Ω) = U(V) / I(A) = 0.36V / I(A)

where I(A) is the maximum current through the wire.

Note: since cables generally have two wires, if You need a cable that is 10 feet long, calculate the total resistance for 20 feet long wire.

Calculating the Right Wire Gauge

When looking for the right wire thickness, first, we have to define the circuit's maximum current and, using the wire's Ampacity, find the proper wire thickness at the required temperatures.

Note: 140°F (60°C) is already a hot enough temperature to prevent an adult from holding the cable with an unprotected hand for more than a few seconds.

Also, for calculating wire thickness using Ampacity values, the "80% Rule" is also used.

For example, when calculating wire size for 50 Amps circuit, we will go for the wire that features an Ampacity of:

Ampacity = 50 Amps / 0.80 = 62.5 Amps

Since there is no wire with an Ampacity value of 62.5 Amps, we will choose the next best thing:

• T = @60°C/140°F → Ampacity = 70 Amps → AWG 4
• T = @75°C/167°F → Ampacity = 65 Amps → AWG 6
• T = @90°C/194°F → Ampacity = 75 Amps → AWG 6

Note: these are values for "enclosed" wires. Due to additional cooling of air-suspended wires, their Ampacities are higher.

As one can see, we got two different wire thicknesses for a 50 Amps wire because three maximum temperatures were used - some electricians may consider AWG 4 wire an overkill for a 50 Amps current, especially if the wires are not extra long, but better safe than sorry. So:

50 Amps wire size → AWG 4 wire

When dimensioning circuit breakers, one must use the ones recommended by the appliances that are going to be powered by the electric circuit.

However, cables can be slightly over-dimensioned, although this may lead to higher wire costs. Never, but really never use wires that are thinner than required, in this case, that would be AWG 8.

When calculating wire thickness, one also has to consider the length of the wires by increasing the required Ampacity of the wire by 10% for every 50 feet of the wire.

For more details about calculating wire thicknesses, feel free to check our articles:

• 20 Amp Wire Size
• 30 Amp Wire Size
• 40 Amp Wire Size
• 50 Amp Wire Size
• 60 Amp Wire Size
• 100 Amp Wire Size

Note: if You are unsure, contact a local certified electrician of a company for more information, including local laws and regulations.

How To Calculate Wire Resistance

Wire resistance is very simple to calculate using wire thickness, length, and material.

For example, we want to calculate the electric resistance of a two-wire, 6-gauge, 10 feet long, cable.

R (Ω) = 20 feet * 0.3951 mΩ/feet = 7.902 mΩ

Note: two-wire 10 feet long cable has 20 feet of wire, hence, we multiply with 20 feet and not with 10 feet.

So, a two-wire, the 6-gauge cable features electric resistance of 7.902 mΩ, or 7.902 * 10^-3 Ω, or 0.007902 Ω.

Another example:

For example, if we have 6 gauge wire which is 3 meters long, its resistance is:

R = (1.296 mΩ/m) * 3m = 3.888 mΩ

which is a very simple thing to do.

But, if we want to calculate the wire resistance of the same wire using copper specific electrical resistance of 16.78 nΩm, we get:

R = (16.78 nΩm * 3m) / 13.3018 mm² = 3.78445 mΩ

So, we get two different values - the question is, why?

Wire resistance values in the chart are given for real-life copper wire, which always has some impurities, while copper specific electrical resistance of 16.78 nΩm is given for pure copper.

So, for real-life wires, always use values from the chart, or use real-life copper relative resistance of ~17.24 nΩm (instead of 16.78 nΩm), or simply go for a somewhat thicker wire.

How To Calculate Wire Energy Losses

Energy losses should be kept to a minimum for many reasons, including lower electric bills and decreased danger of wire overheating or even wire fire.

To calculate energy losses and voltage drops, we use the following formulas:

U(V) = I(A) * R(Ω)

P(W) = U(V) * I(A) = I(A)² * R(Ω) = U(V)² / R(Ω)

E(J) = P(W) * t(s)

For example, we want to calculate energy losses and voltage drop for an electric trolling motor with the following features:

• maximum current draw: 50 Amps
• voltage: 12V
• cable: 5ft 6 gauge cable or 5ft 8 gauge cable (10ft total wire length, wires are free in the air)

If we check the charts, we can calculate the voltage drop in the cable:

• 8 gauge wire: 0.6282 mΩ/ft → U(V) = 314.1 mV
• 6 gauge wire: 0.3951 mΩ/ft → U(V) = 197.55 mV

In order to keep energy losses to 3% or lower, the maximum voltage drop should be:

U_max.drop(V) = 12V * 0.03 = 0.36V = 360 mV

Thus, both 8 gauge and 6 gauge wires satisfy this requirement and as such can be used for powering an electric trolling motor in this example.

Also, power losses are:

• 8 gauge wire: P_loss-8-gauge(W) = 0.3141V * 50A = 15.7W
• 6 gauge wire: P_loss-6-gauge(W) = 0.19755V * 50A = 9.8775W
• total power: P_tot(W) = 12V * 50A = 600W

As one can see, the power loss difference between 8 gauge and 6 gauge wire in this example is "just" ~6 Watts, which is "only" 1% of total battery power.

Note: 6 gauge wire features a diameter of 4.1154mm, and 8 gauge wire features a diameter of 3.2636mm - nice braided 6 gauge wire is just as easily worked with as nice braided 8 gauge wire and can withstand much more. Fortunately, electric trolling motors are rarely pushed at 100% throttle, generally keeping their wires relatively cold. And if You push your trolling motor at 100% throttle almost all the time, your electric trolling motor is seriously underpowered.

However, theoretical Ampacities for solid copper 8 gauge and 6 gauge wires is 60 Amps and 80 Amps respectively, which after applying 80% Rule drops down to 48 Amps and 72 Amps. Thus, for 10 feet long wire (5 feet cable) that must support real 50 Amps, one should go for 6 gauge wire, not 8 gauge wire.