http://farside.ph.utexas.edu/teaching/em/lectures/node94.html states:

Note that the formula (1100) is only valid for \$l\ll \lambda\$. This suggests that \$R_{\rm rad} \ll R\$ for most Hertzian dipole antennas: i.e., the radiated power is swamped by the ohmic losses. Thus, antennas whose lengths are much less than that of the emitted radiation tend to be extremely inefficient.

Of course, this is a well-known fact. But I don't see the reasoning for that.

Let's plug in some numbers (10 kHz, wavelength 300km; length of dipole \$l=300\mathrm{m}\$ (=100x less than wavelength).

\$ R_{\rm rad} = 789 \left(\frac{l}{\lambda}\right)^2 = 78.9m\Omega . \$

The text above does not clarify what exactly is meant by the swamp of ohmic losses. But on the back of some envelope, let's assume that the wire can't be longer than \$l\$ (by definition) but that also implies that it can't wider (otherwise it would get longer). So as an upper bound we have a metallic cube of length \$l\$. Resistivity of silver is 1e-8:

\$ R = 1\cdot 10^{-8} \frac{l}{l^2} = 33p\Omega . \$

Orders of magnitude smaller than the radiation resistance!

Even if I make the cross section 1000x smaller than the length, the ohmic losses are still just \$33\mu\Omega\$ ... orders of magnitude smaller than the radiation resistance.

I also take the Skin effect into account but it does not change the result significantly:

\$ R_{\rm skin} = \frac{\rho}{2\pi r \delta} = \frac{\rho}{2\pi r \sqrt{\frac{2\rho}{\mu 2\pi f}}} \approx 27\mu\Omega . \$


1 Answer 1

  1. You haven't considered the loss in the matching network - practical inductors have a Q of only a few thousand, these would add significantly to the loss.

  2. You haven't considered the skin effect - at 900 MHz only the outside few microns of the metal are effectively carrying any current.

The most efficient-for-their-size antennas are a bit like two hemispheres fed in the middle, so your idea is as good as it gets.

Edit to add: Since matching networks have limited Q, the unmatched Q of the antenna is of little practical interest.
The main problem with high Q antennas and matching networks is that the available bandwidth is reduced, until they become useless. This is where there has been some interesting research.

Look up the Chu-Harrington Limit - relating the maximum efficiency, bandwidth and dimensions of electrically small antennas. This paper has a graph.
enter image description here
I have in my head a more recent graph with the same limit line, but various newer antennas like the two-hemisphere and others, but I can't find it.

  • \$\begingroup\$ This makes sense but is this really what the text in the link is referring to? This is basic physics do I don't think they'd discuss matching or skin effect. Let's say we don't do matching network, we terminate with the exact impedance (that me magically get). And let's work with 1 Hz (skin depth 5cm but wire length can be 3000km and cross section width maybe 30km) \$\endgroup\$
    – divB
    Commented Aug 21, 2020 at 6:28
  • \$\begingroup\$ @divB try the calculation - efficiency of a dipole (scaled correctly) actually goes down at lower frequency, due to skin depth. \$\endgroup\$
    – tomnexus
    Commented Aug 21, 2020 at 13:30
  • \$\begingroup\$ I wrote that I did the calculation, I don't see this. Skin effect seems to be negligible as compared to the dimensions of the dipole, if I am not mistaken. As I wrote, I would be interested in a universal argument, not just one frequency range. If skin effect is the universal reason why a Hertz dipole is not practical, it would be great to show this (then I would expect there would be a constraint like SkinDepth ~ radius of the conductor) \$\endgroup\$
    – divB
    Commented Aug 24, 2020 at 18:53
  • 1
    \$\begingroup\$ @divB skin effect means that no matter how fat your conductor is, only a shell 634um deep from the surface "counts" for the purpose of figuring resistivity. So long as your wire is more than a couple mm thick, that's going to scale linearly with diameter. But Rrad scales quadratically with length in this regime. \$\endgroup\$
    – hobbs
    Commented Aug 24, 2020 at 19:28
  • 1
    \$\begingroup\$ Skin depth gets thinner with increasing frequency too, but only with sqrt(f) while wavelength gets smaller with f, so you've correctly (re)discovered that higher frequency dipoles suffer less from skin-depth related losses. \$\endgroup\$
    – tomnexus
    Commented Aug 24, 2020 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.