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I'm designing a breakout kit for an open source standalone ECU. The ECU has a 40 pin IDC male socket which has all the pins required to run the engine.

One of the components will be injectors which use around 3 - 4 amps. Each injector actually uses two of the pins to help with the load. The only 40 pin, 1.27mm ribbon cable I can find for a reasonable price is rated at 28awg. 26awg is apparently recommended.

Now since each injector will only fire for a fraction of a second, every second, is there any issue to using 28, or even 30awg wire?

HippoDuck
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  • what does the service manual say? – user3528438 Oct 24 '17 at 22:44
  • There isn't one. It's open source and in early stages. There's only a small community working on it at the moment and we can't agree. – HippoDuck Oct 24 '17 at 22:45
  • then you need to try and see how it works and write the manual, i guess – user3528438 Oct 24 '17 at 22:46
  • I am trying it, it seems to be fine at the moment, but I have no idea what the long term affects will be. – HippoDuck Oct 24 '17 at 22:47
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    You're primarily interested in (a) the heating effect of the wire for safety and longevity and (b) volt drop in the wire. Smaller wires are going to drop more voltage along their length which may or may not effect circuit operation. The heating effect is simply the average current. Simplistically, a wire rated at 2A can carry 4A at 50% duty cycle, 16A at 25% duty cycle...

    Bear in mind the ambient temperature and downrate accordingly if it's hot, uprate if it's cold. Etc. Also, whether the wires are in free air, bundled together, etc. Experiment, and consider actual operating worst cases.

     – Ian Bland Oct 24 '17 at 22:53
  • agreed with Ian. I would add that a thermal camera would be a great tool to check for temperature delta between your cable and the environment. – MAB Oct 24 '17 at 23:03
  • Erratum, I meant 8A at 25% of course :) – Ian Bland Oct 24 '17 at 23:19

3 Answers3

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There are two reasons you want a bigger wire for more current:

  1. Voltage drop due to resistance — i.e. your signal is attenuated too much. You can figure out for yourself whether this is acceptable experimentally, or by calculating or measuring the resistance of your wires and modeling the effect on your circuit.

  2. Overheating (melting insulation and possibly then wire) due to resistive heating. For this, only the average current usually matters. More precisely — you're going to put in a pulse of heat energy exactly as much as the resistive loss you computed for the previous point. Then, there are two questions:

    1. Does this heat immediately raise the temperature enough to melt anything? (Depends on heat capacity of the wire and insulation.)
    2. Will the heat have dissipated sufficiently over the time to the next pulse so that the average temperature doesn't continue to rise until something melts? (Depends on how well the wire is cooled by ambient air etc.)

At the speed of an engine's firing cycle, you almost certainly only have to worry about the average current (heat moves slowly), and the reduced wire gauge will not be a significant issue for heating of the wire. The voltage drop may yet be a problem; you'll have to work that out yourself.

You should also consider including protection devices (such as a resettable fuse) which will open each circuit if the controller (or its drive transistors) fails “stuck on”. This can avoid turning a problem that is temporary or affects one subsystem into a spreading “your wires are melting” problem.

Kevin Reid
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There is absolutely no problem for running 3-4 amps through a 28 AWG ribbon cable conductor with low duty cycle. Certainly if the load is shared between two pins you should not have a problem. Make sure your connector and insulation is rated at the engine temperature plus 30 degrees or so, and if it is, you can easily run 4 amps continuous on two pins.

John Birckhead
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To work out the heating effect in a cable, you need the RMS current.

If you have a flat-topped pulse waveform of 4A, with a duty cycle of 1%, so 10mS every second, then the RMS value of the current is sqrt(duty)*peak = 0.1*4 = 400mA.

The voltage drop may yet bite you with a cable that's too thin.

Neil_UK
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