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GXL Vs. Tefzel amp load capacity

Practical Motorsport Wiring - Club Level

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Trying to find an amperage chart that compares the two. Wanting to switch my wiring from GXL to Tefzel.

There should be no difference for the same wire gauge, however, there may be different specs in terms of what temperature range is allowed, and of course the diameter of the insulation.

Here is a generic automotive wire size chart:



Maxxecu PDM20 wiring chart has this helpful small chart on page 2 that has current rating for the different gauges of wires, with the thin-wall (TXL, R2) and the regular wire (RK, RKUB .. ) this is what i've been using as my guideline.

GXL and TXL have a temp rating of 125C. Tefzel is rated to 150C. Using those values, you can look at a temperature rise chart to figure out the difference in ampacity.

@jannek those ratings in the Maxxecu chart are very high for TXL. Those look like peak values that need to be de-rated to find the actual real world limit.

Some circuits may have a significant voltage drop for longer runs, as well.

An example could be the fuel pump(s).

The spec 55A wire that i'm using has numbers corresponding another document i stumbled upon (https://www.kiwiconnectors.com/data-logging-acquisition-with-racing-cars-bikes/wire-specifications), however as these things do not have any source references mentioned i decided to look at the specs.


well the TE document does not list the ohmic resistance for whatever reasons, but it does list a number of specs the wire complies to. Out of those i picked the ESA spec since its bound to be in more sensible units (metric ) http://escies.org/escc-specs/published/3901012.pdf

This states that the max ohmic resistance for a wire with 19x0.25 strands is 20,9 Ohm/km (TE spec says that the awg18 has 19x0.25wires)

In my setup the longest run which might get high current is to the fuel pump, lets say 5meters. The outputs of Maxxecu have 25A max which is divided to 3 pins, making it 8.3A /pin, voltage loss over the 5 meters for this one would then be 8.3*5*(20.9/1000) = 0.867V, which would be 7.2W.

calculating how much the wire heats up is non trivial so i wont even attempt, but it does not look bad to me.

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