I can't understand how to properly calculate AWG of a cable working in DC at high current: I found several tables for AC, but I don't know if/how they can be applied to DC.

My system is rated 70V/100A; I think I should use at least 16mm2, but how to be sure?

Example tables:

http://www.carnevale.bz/wp-content/uploads/2012/cataloghi/Cavi_Speciali/BERICACAVI/BERICACAVI2011_CaviSpeciali.pdf

http://www.bericacavi.com/site/index.php?option=com_content&view=article&id=59

I read about "ampacity", but it too is applicable to AC and I can't understand if it is also useful for DC.

I also can't understand on which length are awg/ampacity tables calculated: shouldn't AWG also depend on line length?!?

On my specific system I have cables long at most 1 meter, and I want to be sure they don't get hot, as they are enclosed in plastic with no air cooling and they must tolerate safely 100A continuously for several minutes.

  • The only thing different (AFAIK) between AC and DC is the skin depth and for the size wires you are talking about it will make no difference. (use the AC values.. the other thing to do is calculate the resistance.) – George Herold Feb 29 '16 at 20:08
  • For long cables, you may have to consider the voltage drop due to the cable resistance, and may have to use a larger cable than the ampacity tables suggest for your current to have an acceptable voltage drop along the cable. – Peter Bennett Feb 29 '16 at 20:24
  • Ampacity has to do with fire safety. Making sure the wires do not get too hot and start a fire. If you read closely you will see that there are assumptions and fine print, depending on the insulation temperature rating and whether the wire is run alone or with other wires in conduit, etc. So, generally speaking, ampacity does apply to you. As others have noted, check overall resistance as well (unless the wires are very short) to make sure you can live with the voltage drop in the wire. Remember that both the positive and negative wire have voltage drop, so you need an x2 factor somewhere. – mkeith Mar 1 '16 at 4:40
  • As said, my cables are at most 1 meter long. – jumpjack Mar 1 '16 at 18:44
up vote 3 down vote accepted

An amp of DC current through a cable will generate the same# heating as an amp of AC current through a cable.

(# Actually slightly less. The resistance of a cable at DC is less than at AC. This is because there is no AC skin effect so the resistance is a few percent less.)

If you select a cable for DC current, using AC cable sizing guides / cable ampacities, your DC cable size will be conservative with respect to heating.

Note that various de-rating factors apply based on the installation environment of the cable. Proximity to other cables (which produce more heat), installation in a enclosed air space (which reduces air circulation i.e. cooling air), etc.

In Australia, the standard AS/NZS 3008 "Selection of Cables" covers all these de-rating factors in great detail.


Voltage drop is calculated using Ohm's Law and the DC conductor resistance.

The DC cable resistance is published in all respectable cable catalogues. This includes AC cable catalogues - the DC conductor resistance is still published.

The resistance is usually given at 20 degrees C, adjust it to your maximum operating temperature using the temperature coefficient of resistance.

If your cable catalogue does not mention a maximum operating temperature, it is approximately 75C for PVC insulation, 90C for XLPE, and 110C for special high-temperature cable i.e. 110C-rated EPR cables.


Finally, for high voltage DC cabling there might be special considerations - normal 0.6/1kV AC power cable might not be suitable for 600V DC. I seem to recall that DC puts a different kind of stress on insulation than AC does. In your particular case, you are only dealing with 70 VDC so I wouldn't worry about it.

  • there's a AC/AC typo in first line of your answer... – jumpjack Mar 7 '16 at 14:50
  • @jumpjack thanks, fixed. – Li-aung Yip Mar 7 '16 at 15:14

The two things to consider with sizing cable are:
1) Self heating due to resistance. Voltage drop. Since you know what your max amperage is, size the cables for that. Calculate the resistance from the cable length, you can use cross sectional area of the wire to estimate the resistance. Or find a table that has done this for you. In the table they have already 'derated' the cables for self heating and other factors. Once you have the resistance you can use P=I^2*R to find the self heating.

2) Voltage drop. If your application requires a certain voltage, then the voltage drop across the cable can become a problem (not in your case since the cables are so short)

You may have to take into account the inductance of the cables if your application requires it. (which is probably negligible in your case because the wires are short).

  • I already knew all this thing. I need clarifications about differences between DC and AC when sizing cables. – jumpjack Mar 1 '16 at 18:46
  • What I'm telling you is to ignore the charts and run your own calculations, with the power equations. Then ask yourself if it matters. – laptop2d Mar 1 '16 at 18:50
  • Basically, there are two issues. Fire, and voltage drop. The ampacity table gives you the minimum size to make sure you don't have to worry about fire (or overheating the insulation). For lower voltage applications such as yours (70V), you may need to worry more about voltage drop in the cable. So start with the AWG from the ampacity table. Then calculate the voltage drop for your AWG and wire length at 100 Amps and see if it is acceptable. Remember that the positive and negative cables both have the same current and the same voltage drop. I can't make it any simpler than this. – mkeith Mar 1 '16 at 23:25
  • I highlight the topics in my question: I found several tables for AC, but I don't know if/how they can be applied to DC - I read about "ampacity", but it too is applicable to AC and I can't understand if it is also useful for DC. - I also can't understand on which length are awg/ampacity tables calculated: shouldn't AWG also depend on line length?!? – jumpjack Mar 3 '16 at 11:14
  • @laptop2d To do my own calculation I'd need to know thermal resistance of copper-insulator and insulator-air and proper formulas, but after months of useless searches I gave up and switched to ready-made tables... with no luck. How to determine which temperature my cables will reach appears an impossible task up to now. :-( – jumpjack Mar 3 '16 at 11:18

Although I didn't find an official table for sizing DC cables, I found a precise explanation about the answer from @Li-aung Yip, "If you select a cable for DC current, using AC cable sizing guides / cable ampacities, your DC cable size will be conservative with respect to heating.":

http://www.mondini.com/system/files/documenti/Manuale%20Tecnico%202012.pdf

This document explains how to calculate current in DC, AC/mono and AC/3 systems:

  • DC: I = P/V
  • AC/1: I = P/(V*cosF)
  • AC/3: I = P/(V*1.73*cosF)

This means that, for same voltage, DC current is greater than AC/1 and AC/3 current; hence using wire sizing tables for AC current is safe also for DC systems, because cable heating is directly proportional to current carried by the cable, and the target is to prevent cable from overheating.

For my specific system, rated 70V/100A and enclosed into an electric scooter, I assume as applicable the table at p.22, column "posa interrata in tubo" (cables enclosed in pipe), sub-column "3 cavi unipolari" (three single-wire cables), as probably the ground thermal resistance is equal or greater than the thermal resistance of plastic case of the scooter. For 100A I get a 25 mm2 section.

"AWG 3" is 26.7 mm2, so the final answer to my question is:

Cable section required for a 70V/100A DC system is: AWG3 / 25mm2

But on an electric scooter there are actually both DC and AC: DC goes from battery to controller, but controller creates an AC current which drives the motor; so, for same voltage, I think that cable size from controller to motor (three cables) can be slightly less than battery-controller cable size. Unfortunately I don't currently know the frequency of the motor current and the anount of CosF.


How do I know which current is in use on my system?

My scooter is rated 5000 W, but I also "manually" calculated the current needed to maintain a 90 km/h speed on flat road; I assumed 0.8m2 frontal area and 0.8 Cd for the scooter+driver system. This leads to 6000W needed (link). Scooter uses a 60V LiFePO4 battery, with actual operating voltage between 56 and 66V; 6000W/66V gives 90A, rounded to 100A.

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