This is a follow up to another recent question of mine: Does a balun need to be made with coax? In that question, we establish that a choke balun doesn't actually work by protecting the "third conductor" on the outside of the coax shield, so there's no need for the coax itself to be part of the balun.

The design in question was the sort pictured below, where a pair of regular wires are wrapped around a toroid core. I've seen this kind of design in many places, including in other balun questions on this site, so the idea must be sound.

Given that there is no sort of precise spacing between the two wires, there must not be any sort of precise characteristic impedance, either. The question is, why doesn't this cause problems? Or does it, and you're expected to be using an antenna tuner to compensate?

a howebrew balun
(From http://www.m0pzt.com/baluns/)

  • 1
    $\begingroup$ Spacing of wires in a balun can and does matter. Balun designers usually go to great lengths to control the impedance of the transmission line formed. Raising it above 100 ohms using thin-walled silicone tubing, or lowering it by using flat conductors taped together. In this simple choke balun, however, the impedance is just wrong, it'll transform the 50 ohms into something else. Perhaps they're using it to drive an antenna that needs a tuner anyway. Otherwise, they'll find it affects the impedance a bit, especially at 20 or 30 MHz. $\endgroup$
    – tomnexus
    Commented Jan 16, 2017 at 20:27

2 Answers 2


It more or less doesn't matter because it's so small, relative to wavelength. As such, a lumped element model is valid.

You can make a conjecture to that effect by looking inside your antenna tuner (or really, a lot of HF equipment). Unless it's a fancy kind, there will be wires running whatever way between the components, with no attention paid to the characteristic impedance of those connections.

Think about it this way: imagine a change in voltage propagating down the line. When it encounters the balun, it will encounter some mismatched impedance, and some amount of the energy is reflected back. A very short time later (due to the very small size, relative to the balun) the same change, but reversed, which will send another reflection except opposite in phase back down the line.

If these two reflected waves occurred at exactly the same time, they would entirely cancel. In practice they aren't exactly at the same time, but since the change in phase between the two is so small (remember, tiny relative to wavelength), they almost completely cancel, and the effect is negligible.

The spacing of the wires does matter in a different way: there's some capacitance between them. We can incorporate that into the lumped model like this:


simulate this circuit – Schematic created using CircuitLab

The inductor and the capacitor each have some impedance, with the inductor's impedance increasing with frequency, and the capacitor's inductance decreasing with frequency. The two form a parallel LC circuit, and at the frequency where the impedances are equal, the impedance of the balun as a whole is at a maximum.

It's easy to add more turns to a balun, so achieving a sufficiently high inductance isn't too hard. But each turn also increases the area where the wires are close together, introducing additional capacitance. At some point there's so much capacitance that the associated impedance is so low it effectively makes the balun a short, making it ineffective as a balun.


The wire spacing can matter in a common-mode choke balun.

A bifilar choke wound around a toroid forms a short transmission line whose impedance may be different from that of the line into which it is inserted. Thus, a bifilar choke can introduces some mismatch. (A tightly spaced enameled #14 bifilar winding is very close to 50 ohms. Tightly spaced #14 THHN windings are close to 100 ohms.)

The paragraph above is an excerpt from K9YC's excellent technical publication on this page. See p. 35 of this PDF linked to from there.

  • $\begingroup$ @Phil Frost- W8II Thank you for mentioning wavelength, but that was not specified. On some bands and some spacing --and in the context of the OP-- it very well could make a difference. :-) $\endgroup$ Commented Jan 9, 2017 at 18:52

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