I don't understand why 14 AWG XHHW-2 wire would work fine when used in ladderline for high-power HF transmission, but the same wire would exhibit higher losses when used in a twisted pair configuration.

Can anyone explain to me why ladderline has lower loss than twisted pair?

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    $\begingroup$ Am I gathering correctly from your previous question that you're referring to a cable type that has XLPE as insulator? $\endgroup$ Jan 19 at 23:29
  • $\begingroup$ @Constantine Yes. From what I understand, XHHW-2 wire has XLPE as insulation. With that said, I read just yesterday that LDPE (low density polyethylene) has extremely low dielectric loss...per a technical note from eXsource, "Additive-free low density polyethylene has a very low dielectric loss, amongst the lowest of any plastics, and virtually at the limits of measurement resolution." So now I'm looking for some 14 AWG LDPE wire and want to try that as twisted pair. $\endgroup$
    – Mark K1LSB
    Jan 20 at 1:08
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    $\begingroup$ Looked for an hour now, hard to find good data that actually goes above 400 Hz. Best I can find is studies on aged material, which go up to 10 MHz and show a behavior where loss grows with frequency much more than linearly (which means that the dielectric loss tangent is not a constant wrt frequency, but grows): see figure 8b of ncbi.nlm.nih.gov/pmc/articles/PMC9182923 and assume you're the "best" curve there (as your cabling is new) $\endgroup$ Jan 20 at 8:42
  • $\begingroup$ @Constantine That's very revealing info, thanks! I was unaware that XLPE dielectric losses increased with age. I'll stay away from that stuff. $\endgroup$
    – Mark K1LSB
    Jan 21 at 13:43
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    $\begingroup$ That was not the point I was making (every long chained polymer will age, and this was really exposed to harsh conditions). The point was that this specific dielectric, aged or fresh, has high attenuation at HF $\endgroup$ Jan 22 at 0:04

2 Answers 2


You should also check out this answer to a related (but not duplicate) question over on electronics SE; in relevant part:

The main limiter of the bandwidth is from the attenuation, which increases as the frequency of the signal increases (mainly due to the skin effect).

Skin effect is a culprit here, though there may be others. Because ladder line has a higher characteristic impedance, there is less current and less opportunity for skin effect losses at a given power input.

With the legs of the transmission line now much closer together, greatly increased capacitive coupling (and dielectric losses as a result) are probably a contributor also, but I'm starting to get out of my depth on that topic and invite others to edit this as appropriate.

The upside to TP is that it's far less sensitive to environmental considerations: you could get away with running TP near metallic objects that would more drastically impact ladder line.

  • $\begingroup$ I'm pretty sure that dielectric losses are minimal at HF. $\endgroup$ Jan 19 at 19:18
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    $\begingroup$ How much dielectric loss you get at which frequencies depends on the material. Pretty sure a power line insulation material was tested up to a couple MHz. Whether it's been done up to a 2m band - uncertain. I bet losses in the dielectric are rather high multiple orders of magnitude above the intended operational frequency range; not like polymers are easy to make low-loss when their chains get long and especially not when you artificially stretch the material, introducing polar regions, when trying to extrude it around a conductor. $\endgroup$ Jan 19 at 23:28
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    $\begingroup$ Dielectric loss is at least partly why coax is worse than ladder line at HF frequencies. But even then, it doesn't matter until you have high SWR. Also, at high frequencies, the diameter of the coax affects propagation mode, so if it's high enough for coax to become wave guide, the dielectric losses are much much much worse for those frequencies. $\endgroup$
    – user10489
    Jan 21 at 15:56

There are two types of losses in the wire (twisted pair or ladder line), and both appear as resistive losses:

  • Radiation, from radiation resistance. This causes energy to escape from the wire.
  • Heat from loss resistance.

Radiation is at least partially canceled by opposing currents in the two wires, but coupling with near by conductors and other imbalances can cause common mode current that does not cancel and radiates instead. This can be mitigated by twisting the pair of wires. Obviously, twisted pair is already twisted. But you can also twist ladder line to reduce these losses and it works just like twisted pair, but typically with a lower and less consistent twist density.

Loss resistance causes the wire to heat up. These are sometimes called $I^2R$ losses, and obviously are proportional to the square of the current. This can be partially mitigated by increasing voltage and reducing current. These losses occur in two places: In the dielectric material between the wires and in the metal in the wires.

Careful selection of the dielectric (to select a material with low ESR) can reduce loss in the insulator. Some plastics have lower ESR than others. The lowest ESR is in materials like glass, ceramic, and air. So ladder line reduces losses by adding gaps ("windows") in the dielectric spacers between the wires. Reducing the thickness of the insulation of the wires (or removing it entirely) can also reduce these losses, with the risk of shorts and dielectric breakdown and arcing at high voltages.

The metal in the wires also causes resistive losses. The resistance of the wire is proportional to $\rho L/A$ where L is the length of the wire and A is the cross sectional area of the wire through which current is flowing and $\rho$ is the conductivity of the metal. Increasing the thickness of the wires can increase the cross sectional area. Using a high conductivity metal also helps. (The top three are silver, copper, and aluminum.)

Unfortunately this is complicated by skin effect. The same opposing fields that reduce radiation losses cause the opposing currents on the wires to attract and common currents to repel, so the currents tend to concentrate on the outside of the wires, and prefer the outside surfaces facing each other. The closer the wires, the higher the concentration on the facing surfaces, and the smaller the cross sectional conduction area. This can be somewhat mitigated by wire shape. For instance, flat wires with the flat sides parallel would have lower resistance, but a small imbalance in the alignment might cause the current to be even more concentrated on the closer side than it would with a round wire, so adjusting shape is of questionable use.

So, in summary, ladder line and twisted pair are really the same thing and have the same losses for the same reasons. But twist density can decrease losses from imbalances and increase usable frequency. Twist density can also increase physical wire length, increasing losses. Increasing distance between wires can decrease losses from skin effect but increase losses from imbalances and might increase needed dielectric material to keep the wires stable. Increased amount of dielectric material or material with high ESR can increase losses.

All of these factors (along with cost) need to be carefully balanced to make an efficient transmission line. The balance will be different for different design frequencies and power levels.

Note also I'm ignoring issues like impedance mismatch, eddy currents, high SWR... which are either caused by issues external to the twisted pair or are easily avoidable in well designed wire.


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