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How thick copper cable should I use for 16A current? How thick it should be if I use aluminum wire? Is there any generic equation which I can use?

I am using direct current with 12V but as far as I know it doesn't matter.

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    \$\begingroup\$ re: "direct current with 12V" it matters with AC, due to skin effect. Before we can answer, though, what are you trying to do? A different sort of answer comes from "Just enough capacity to be safe over 5 feet" or "Minimize losses to heat over two inches" or "I need to carry a lot of juice 600 feet away" \$\endgroup\$ – SingleNegationElimination Mar 29 '12 at 14:26
  • \$\begingroup\$ @TokenMacGuy: Primary I want to be safe over one or two feet. Secondary I want to understand what it is about. \$\endgroup\$ – Jakub Šturc Mar 29 '12 at 14:50
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There are some topics you should pay attention concerning the cable thickness selection: the voltage dropout, cable heating, electro-magnetic interference, impedance matching.

The maximum allowed voltage dropout is a good parameter to calculate the cable thickness. In most situations this is the limitation factor and may define alone (without the need to calculate the other factors) the cable thickness.

\$ V_{drop} = R_{cable} \times I \$

\$ R_{cable} = \dfrac {p \times length}{CrossSectionArea}\$

Where p is the material electrical resistivity

For a round cable:

\$ R_{cable} = \dfrac {p \times length}{\pi \times radius²}\$

Joining the equations we have that the thickness should be at least:

\$ radius >= \sqrt{ \dfrac {p \times length \times I}{V_{drop}\times \pi}}\$

Assuming that a voltage drop of 0.5(v) is allowed and you'll be using a 2 meters copper cable (p=1.68×10−8) for a maximum 16(A) current:

\$ radius >= \sqrt{ \dfrac {1.68×10^{−8}\times 2 \times 16}{0.5 \times \pi}}\$

\$ radius = 0.59 mm \$

\$ diameter = 2 \times radius = 1.18 (minimum) \$

From this AWG table we have that a cable of AWG-thickness of 11 has a diameter of 2.3mm, that is enough for your application.

The table also states that this AWG 11 can carry over 12(A) for what they call "Maximum amps for power transmission". This is a lower number than the one we calculated, but we should pay attention that this is a general advice for any cable length, not for a specific 2 meters cable (which I supposed in this calculation).

The cable heating is a little harder to calculate. You'll need information on heat dissipation characteristics of the environment and the cable: temperature, air flow, sunlight incidence, cable thinness, cable material, insulator layer and even the cable color (black cables heat up faster but also cool faster).

In general the cable insulation has a good thermal protection (around 80 Celsius degrees) which is enough for most cases. You should pay attention only if your cable will go through some heat source. In this case a thicker cable with a better insulator can stand up better.

Unfortunately there is differences among AC and DC current for cable selection. If you're going to transport AC current, a thinner cable may limit the maximum frequency you can carry, above which the signal may start degrading due Skin Effect as @TokenMacGuy noted . Lucky this is not the case with DC.

The last topic I said that could interfere with the cable thickness selection is the impedance matching. For more precise circuits, the resistance from the cable can make a huge impact in the power transmission. This problem arises mainly when there is a huge current, more than 100 (A), or the load resistance is really low, as speakers with 8 or 4 ohms. In this case a thicker cable with a lower resistance is desirable.

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  • \$\begingroup\$ Thank you. Especially for pointing out that I should base calculation on voltage dropout. \$\endgroup\$ – Jakub Šturc Mar 29 '12 at 15:27
  • \$\begingroup\$ The general rating for the ampacity of the wire (the reason the table gives 12 A ampacity for AWG 11 wire) is related to self heating, probably something like keeping the temperature rise less than 40 degrees C in a confined environment. The voltage drop requirement is something specific to a particular application, and not something a reference table can tell you about. \$\endgroup\$ – The Photon Mar 29 '12 at 15:28
  • \$\begingroup\$ Alos, if two requirements give different limits, like you need bigger than AWG 11 cable to achieve your voltage drop-out, and you need bigger than AWG 10 to limit heating and avoid starting a fire, you should use the bigger wire. \$\endgroup\$ – The Photon Mar 29 '12 at 15:30
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The minimum size of cable required to carry a given current is generally set by two factors:

  1. How much voltage loss is tolerable?
  2. How much above ambient are you willing to have the cable get?

To compute the voltage loss in a cable, start by computing the resistance per meter; cable manufacturers may specify this directly, or you can compute it based upon the material and cross-sectional area. Then multiply the resistance per meter by the number of meters and the amount of current to compute total voltage loss. This will tell you whether a particular cable will meet the first requirement. Note that increasing the length of the cable by a factor of four will require doing likewise with the cross-sectional area (i.e. doubling the diameter) to keep the same total loss.

To compute the temperature rise above ambient, multiply multiply the resistance per meter by the square of the current to get the amount of power per meter that will need to be dissipated. Then determine the amount of heat dissipated per meter per degree C (cable manufacturers may specify this figure for still air, though ventilation or lack thereof will affect the actual value substantially). Divide the total power by the dissipation per degree C to compute the temperature rise that would result with a particular cable. Note that the required wire size to keep temperature rise within acceptable limits will generally be independent of cable length, but doubling the current will require doubling the diameter (four-fold increase in cross-sectional).

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