# Wire Gauge Calculator

Use this calculator to work out what AWG gauge wire you need to carry a certain a specified current with a maximum allowable voltage drop across the length of the wire.

Metric prefixes such as u (micro), m (milli), k (kilo), and M (mega) are allowed after the number.

Voltage (DC) | The source voltage provided to the cable. | ||

Voltage Drop | The acceptable voltage drop across the length of the cable, as a percentage. | ||

Cable Length | The length of the cable. | ||

Current | The current you want the wire to take. | ||

Conductor Material | The material of the conductor. | ||

Conductor Resistivity | The resistivity of the conductor. If you want to enter your own value in here, set the "Conductor Material" to custom. | ||

Cross-sectional Area | The required cross-sectional area of the conductor in the cable. | ||

Gauge | The calculated maximum AWG gauge of the cable. The calculated value is rounded down to the nearest integer. |

Given the source voltage and percentage voltage drop we can calculate the absolute voltage drop across the entire wire:

$$ V_{cable} = V_{source} \cdot \frac{V_{cable, perc}}{100} $$

where:

\( V_{cable} \) is the voltage drop across the total length of cable, in \(V\)

\( V_{source} \) is the voltage at the input to the cable, in \(V\)

\( V_{cable, perc} \) is the desired maximum percentage voltage drop across the total length of the cable, as a percentage

From this we can find the total resistance of the cable using Ohm's law:

$$ R_{cable} = \frac{V_{cable}}{I_{cable}} $$

where:

\( R_{cable} \) is the total resistance of the cable, in \(\Omega\)

\(I_{cable}\) is the current through the cable, in \(A\)

We can divide this by the length to get the resistance per meter of cable:

$$ R_{\Omega / m} = \frac{R_{cable}}{L} $$

where:

\( R_{\Omega / m} \) is the resistance per meter of cable, in \( \Omega \cdot m^{-1} \)

\(L\) is the length of the cable, in \(m\)

We can then calculate the cross-sectional area of the cable using the resistance per meter and the resistivity of the conductor (the resistivity is a property of the material the conductor is made from):

$$ A = \frac{\rho}{R_{\Omega / m}}$$

where:

\(A\) is the cross-sectional area, in \(m^2\)

\(\rho\) is the resistivity of the conductor (e.g. copper), in \(\Omega \cdot m\)

We can then find the diameter of the cable from this cross-sectional area:

$$ d = \sqrt{\frac{4A}{\pi}} $$

where:

\(d\) is the diameter of the cable, in \(m\)

Now we just need to convert this cable diameter into an AWG gauge. To convert from an AWG gauge to a cable diameter we can use the equation:

$$ d = 0.127e^{-3} \cdot 92^{\frac{36-n}{39}} $$

where:

\(d\) is the diameter of the cable, in \(m\).

\(n\) is the AWG wire gauge

note: \(0.127mm\) is the diameter of wire gauge #36.

However, we need to solve this for \(n\), so re-arranging the equation gives:

$$ n = 36 - 39\frac{log(\frac{d}{0.127e^{-3}})}{log{92}} $$

All done! Remember that the above equation will likely give you a number with decimal places. AWG gauges are typically supplied in integer jumps, so it's best to round down to the nearest integer (a lower gauge is a larger diameter wire, giving you some safety factor).

These equations do not consider the thermal effects of the power dissipation in the cable due to the cables resistance. They are also only suitable for DC currents, as they do not take into account the skin effects.