At 100 watts in a 50-ohm system, the current flowing through an inductor would be about 1.41 A, thus a 26-AWG wire should be plenty since they are chassis-rated for 2A and for just a few windings that isn't a very long wire. This answer indicates that "power-handling requirements will dictate the wire diameter (to handle the current) and the end-to-end spacing (to handle the voltage)" but does not point out how those should be calculated.

I can naively use Ohms Law to calculate current and voltage assuming a 50-ohm system to size the inductor wire appropriately, but if there is more to it then that then I would like to understand why and how to calculate it.


Does the inductor's wire gauge need to be adjusted (for heat or other reasons) based on any of:

  • Frequency?
  • The presence of a ferrite core?
  • Wattage?
  • Inductor reactance?
  • Other considerations?

For our specific application, we are creating a 2m/70cm diplexer for U/V satcom and want to design it to support 100W; the inductors will be part of highpass and low-pass filters in the diplexer. Of course we can follow existing diplexer designs that already do this but we would like to understand the details on the subject.

Thanks and 73!

  • 2
    $\begingroup$ Two quick pointers: Inductors aren't in a 50 ohm system, they're usually paired with something capactive, so current increases by the Q factor, up to 100 x. Your simulation of the diplexer will tell you the current to expect. Then there's the skin depth - in copper, 440 MHz, the current only flows in the outside 3 microns of the wire, and in an inductor probably only in the outside half of that. So your 26 AWG wire will be reduced to about 2% of its DC value, the wire resistance is 50 x DC. At 16 AWG it's just 15x. Put this resistance in your simulation, and see if the wire will overheat. $\endgroup$
    – tomnexus
    Nov 27, 2022 at 6:01

1 Answer 1


First: yes, it's mostly about $I^{2}R$ loss, but if your inductor is part of any kind of resonant circuit, you can't assume that $I^2 = \frac{P}{50}$. The current circulating between reactive components can be much higher than the input/output current, and that still causes real loss.

If you want to model it on a computer, you can use an app like K6STI COIL to take an inductor design (length, diameter, wire gauge, materials) and turn it into a Q-factor, taking into account wire gauge, skin effect, proximity effect, and dielectric loss; then take some kind of circuit simulator (SPICE-type) app to model the overall behavior of the diplexer with those values plugged in, and read the component dissipations off of it. I was going to write a lot more detail here, but I kept having to throw my answer away because the program I wanted to use (SimSmith) was not quite suitable, and I figured you deserved some answer by this point.

The alternative is just to build it and test at a lower power (say, 1-5 watts).

  • If you have a through-line wattmeter, a dummy load, and a radio to use as a signal generator, you can measure the loss of the whole circuit at various frequencies — just take one reading at each frequency with the wattmeter between the radio and the diplexer, and another with the wattmeter between the diplexer and the dummy load, and the ratio between them will give you a dB loss, which you can scale up to any power. A little common sense (and comparison with existing components of known ratings) will give you an idea of whether a given amount of loss runs a risk of generating a problematic amount of heat.

  • If you have a dummy load and an IR camera or non-contact thermometer, you could also investigate thermals that way. The temperature rise (above ambient) of each component should scale about linearly with power. Of course, make sure that the dummy load is big enough — it has to dissipate a lot more power than your diplexer does!

  • If you're concerned about arcing you can use a voltmeter with an RF probe to get the peak voltage between points of interest; the voltages will scale with the square root of power, so you can test at 1W and then multiply by ten.

  • If your inductors are air-core (which I would expect) then you're pretty much set; if they're ferrite-core then bear in mind that saturation and heating can both cause interesting nonlinear effects that won't show up at low power. If these show up, they don't mean that you need fatter wire; they mean that you need a bigger (and/or different composition) core.


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