Cat5e cable has a rated current of 0.577A (wiki page). I am wondering if this is voltage dependent? I am almost certain it is not, but if it is, something might blow up. Say I need to get 2A at 5V across the cable. I boost it to 24V / 0.5A, send it across the wire, and step it down to back to 5V and a bit under 2A. Would that work? If the cable is proper Cat5e could I do this trick with up to 100V as long as I don't exceed 0.577A?

I am worried if the cable heats up at near maximum rated current, for instance, the safe voltage could drop as the insulation softens up or something. I will certainly test this, but I wonder if there is some rule that I don't know about.

  • \$\begingroup\$ Do you also need to get data across the same cable? If not, you can parallel conductors to increase rated current. \$\endgroup\$ May 6 '20 at 13:47

The heating you get in a wire is only dependent on the current you pass through it. The voltage makes no difference.

But you need to check the voltage rating of the cable you are using. For Ethernet cables, it can be very low. A random cable I just looked at is only marked 30V (which was rather surprising since power-over-Ethernet is nominally 48V).

If you are pushing a cable to its limits, then where it is run can also make a difference. A cable enclosed in insulation can get rather hot, while one in free air cools a lot better. Don't exceed the temperature rating of the cable. Again, the random cable I have to hand is only rated for 60°C.

  • \$\begingroup\$ +1 for common sense but your cable rating (30 volts vs 60 volts) are contradictory - do you mean cable 1 = 30 volts and another cable = 60 volts? \$\endgroup\$
    – Andy aka
    May 6 '20 at 8:51
  • \$\begingroup\$ @Andyaka I'm not sure what you mean. The cable is marked 30V. Where does 60V come from? I have just looked at another cable, and it's also marked 30V. \$\endgroup\$
    – Simon B
    May 6 '20 at 12:27
  • \$\begingroup\$ Sorry, looks like I mistook 60 degC for 60 volts d'oh. \$\endgroup\$
    – Andy aka
    May 6 '20 at 12:29

You shouldn't mix up rated and maximum, the are not the same. The wiki page says "Maximum current per conductor". Moreover, I don't believe the wiki covers all cables and also doubt the reliability of that number (1).

In all cases you'd better check the datasheet for the rated and maximum current.
When the rated current would be 0.5A and the rated voltage would be 125V, then, indeed the voltage doesn't matter as long it doesn't exceed this rated voltage.

Stepping up the voltage to reduce the current through the cable may help reduce voltage drop and losses in the cable (but note stepping up and down also gives power losses).

(1) The wiki refers to a AWG table with a (in my eyes) a very conservative current rating. They quite likely used the Maximum amps for power transmission for AWG 24.
Considering 10 mm of a 24 AWG cable carrying 0.5 A, using the resistance per unit length of 84 Ω/km, this piece of 10 mm has a resistance of 0.84 mΩ. Assuming all 8 cables carrying each 0.5 A, gives 8*(1 A)^2*0.84 m&Omega = 1.7 mW per piece of 10 mm CAT5e cable.
I wouldn't worry about it...

  • \$\begingroup\$ Except above a certain voltage, you open up a whole can of worms with the electrical code... \$\endgroup\$ May 6 '20 at 9:20
  • \$\begingroup\$ @Harper-ReinstateMonica Yeah on the second thought 24V is the highest sustained voltage I am comfortable dealing with within my system. I plan to make it with a large safety margin though so I don't have to go fix things when they melt down. That would be even costlier than using a thicker cable. \$\endgroup\$ May 6 '20 at 12:35

Most wiring regulations specify the max temperature for cables.

This depends on the environment where the cable is, such as open air or in a wall, in a conduit, in a conduit that is buried in insulation etc etc

So, check the regulations for your country as getting this wrong can cause a fire.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.