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It is pretty well established that folded dipoles have much greater acceptable-SWR bandwidth than ordinary dipoles.

Radiation efficiency is simply the quotient of radiated power to the antenna feedpoint input power.

Seeing that folded dipoles are able to achieve the greater bandwidth without introducing a designed-as-lossy element (a technique not unheard of with large-bandwidth antennas), what is the effect on radiation efficiency of using a folded dipole as opposed to a regular dipole? Are there any considerations affecting folded dipoles that would not affect a regular dipole antenna erected in the same physical location which would have a noticable impact on the radiation efficiency?

For the purpose of this question, assume otherwise identical conditions; identical height over ground, identical ground, identical possible parasitic elements, identical feedline, etc. Also, note that I am not asking about the radiation pattern of the folded dipole; I am only concerned with the antenna's radiation efficiency here, unless some other factor has a noticable impact on the radiation efficiency as compared to a regular dipole.

A great answer would look at this from both the perspective of a same-wire-length antenna (meaning approximately double the folded dipole's physical length) as well as a same-physical-space antenna (meaning effectively half the radiator length).

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  • $\begingroup$ The Icom AH-710 has a resistor in it responsible for its absurdly wide bandwidth, so maybe isn't a good example for your question. $\endgroup$ – Phil Frost - W8II Jun 19 '14 at 0:40
  • $\begingroup$ Specifically, it seems to be a "terminated folded dipole" or a "T2FD". $\endgroup$ – Phil Frost - W8II Jun 19 '14 at 0:45
  • $\begingroup$ @PhilFrost Good point, I had missed that. I deleted the example, but I think the question still stands. $\endgroup$ – a CVn Jun 19 '14 at 7:22
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    $\begingroup$ I think "much" greater bandwidth is exaggerating quite a bit. A folded dipole has higher bandwidth, but not a lot. It's about the same increase in bandwidth you'd get by making a dipole from a similarly thicker conductor. $\endgroup$ – Phil Frost - W8II Jun 19 '14 at 11:38
  • $\begingroup$ The link for "designed-as-lossy element" goes to a page that doesn't mention lossy elements. $\endgroup$ – Kevin Reid AG6YO Dec 5 '14 at 4:18
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For all practical purposes, the radiation efficiency of a folded dipole versus an ordinary dipole is the same. Consider, they are essentially the same antenna.

schematic

simulate this circuit – Schematic created using CircuitLab

The only difference is that in the folded dipole, we've replaced the feedpoint with a short. Since the Thévenin equivalent resistance of a voltage source is 0Ω, this doesn't make a lick of difference to the currents in the antenna. The currents are the same, the fields are the same. Everything is the same, except that the voltage source now sees only half the current.

So then, what is there that could affect radiation efficiency? There is nothing. There might be some difference in ohmic losses, depending on if you allow the folded dipole to have twice as much copper or not, but this is a very small contributor to loss.

More significant for terrestrial antennas is ground losses in the Earth, but having established that the fields around a dipole and a folded dipole are the same, how could the losses be different? They aren't. The same reasoning applies to any other kind of loss.

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  • $\begingroup$ To amplify Phil's answer, efficiency is the ratio of the radiation resistance (as distinguished from driving point resistance) to the sum of radiation and loss resistances. Since radiation resistance is a function of antenna aperture (i.e., geometry) and the apertures of the dipole and folded dipole are identical, there is no difference in efficiency. $\endgroup$ – Brian K1LI Nov 1 '18 at 14:25
  • $\begingroup$ @BrianK1LI Agreed, but now someone is going to argue that the radiation resistance of a folded dipole is 4x that of the not folded variety, and then we'll have to argue about the multiple possible definitions of radiation resistance. $\endgroup$ – Phil Frost - W8II Nov 1 '18 at 14:29
  • $\begingroup$ That, Phil, is why I distinguished radiation resistance from driving point (aka, feedpoint) resistance. As you know and have commented before, radiation resistance depends only on the amount of power flowing through a surface around the antenna for a given amount of current, so will be nearly identical for a dipole and a folded dipole. $\endgroup$ – Brian K1LI Nov 1 '18 at 22:52
  • $\begingroup$ Oh I know, I'm just saying, someone is going to disagree :) $\endgroup$ – Phil Frost - W8II Nov 2 '18 at 0:30
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Beg to differ. The effective conductor radius is increased to something like half the wire spacing. That means that there is less loss. An alternative view is that the impedance transformation properties of the antenna imply that less current flows on either conductor than would be the case for an single-wire dipole. The resistance of the wires is therefore of relatively less concern.

Practically speaking, the differences are indeed small, specially at HF, but they become more important at VHF or UHF.

On the receive side, the susceptibility to noise can be different I believe.

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  • $\begingroup$ What do you mean by "effective" conductor radius? Effective, for what purpose? Do you mean to say that two wires spaced 1 inch apart have the same resistive loss as one wire a half inch thick? $\endgroup$ – Phil Frost - W8II Dec 5 '14 at 3:17
  • $\begingroup$ Great G4ZLZ... I was just about to reply the same. At least when you look at the side (ie. where you actually see the two conductors) it's easy to understand that the effect diameter is the separation. If you look at 90°, it's just the wire thickness. So logic would dictate it's somewhere in between. Now, these changes are very small compared to the wavelength (at least on HF), so do not expect a huge difference. Have a look at simulated data here: w8ji.com/folded_dipole.htm (compare one of the first two, with the last one (single wire) $\endgroup$ – jcoppens Dec 6 '14 at 20:50
  • $\begingroup$ By what mechanism would a folded dipole have increased noise susceptibility? How can any antenna differentiate noise from signal, for that matter? $\endgroup$ – Phil Frost - W8II Dec 12 '14 at 2:12
  • $\begingroup$ On folded dipole increasing noise--- this happens because a folded dipole has a wider bandwidth (between 3:1 SWR thresholds) than a regular dipole. This is especially true on 160 and 80 meters where the resonant part of a dipole is usually sharp and more narrow. The natural notch effect of this on a regular dipole cuts out a lot of the noise but with a folded dipole you are capturing a lot more signal in the bandwidth. $\endgroup$ – K7PEH Dec 13 '14 at 22:20
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    $\begingroup$ @K7PEH Given that the receiver has a filter much, much narrower than the antenna bandwidth, I can't imagine that would make any significant difference, unless the receiver is overloaded or has absolutely terrible filters. Like, really really terrible. $\endgroup$ – Phil Frost - W8II Dec 15 '14 at 13:19
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The folded dipole antenna has lower efficiency than a standard dipole antenna. The efficiency is a factor of both radiation efficiency and impedance matching.

The radiation efficiency of a folded dipole is lower than that of a standard dipole. This is because it is folded; between the two parallel wires of the structure, there is a capacitive effect. This means some part of the electromagnetic energy is stored between these two parallel wires.

The impedance of a folded dipole also differs from a standard dipole, but the impedance can be matched by using a matching network or a transformer.

About the capaticitive energy stored in parallel wire, you can think about a typical explanation of a dipole antenna. If there are two plates parallel, then it acts like a capacitance and if you put them like a dipole antenna, the rods of the antenna gets + and - poles. Then the antenna radiates. That's because the electric field is not stored between two plates, but the field vector is between one pole to another pole. This provides an antenna to radiate. For example, if the angle between two rods are 120 degrees, then it is more like a capacitance then a standart dipole antenna. And a capacitance means the energy is stored in the capacitance. This is a well known principle in small antenna and mobile phone antenna design. In a transmission line, the electric field vectors are between the two plates, and energy is still stored between the plates.

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  • $\begingroup$ Why does a capacitive effect mean lower efficiency? And isn't the parallel conductor more of a transmission line than a capacitor? $\endgroup$ – Phil Frost - W8II Dec 12 '14 at 2:04
  • $\begingroup$ I'm still not sure why the capacitance between the the parallel conductors of the antenna would negatively affect radiation efficiency. Sure, energy can be stored in a capacitance (or an inductance), but that doesn't mean it goes away. Can't a matching network contain a capacitor and yet, to the extent that the capacitor does not introduce additional dielectric or resistive losses, have no impact on radiation efficiency? $\endgroup$ – Phil Frost - W8II Dec 19 '14 at 3:56
  • $\begingroup$ Capacitance in a matching network and in an antenna structure is very different. Capacitance part of the antenna means that it doesn't radiate the energy. The energy is stored so the radiation efficiency decreases. If you have no capacitive part in your antenna and you match it by a SMD capacitance for example, you have no negative effecet on radiation efficiency, and positive effect in matching efficiency. $\endgroup$ – Communicationantennas.com Feb 21 '15 at 21:14
  • $\begingroup$ If energy isn't being radiated because it's being stored (forever, I guess), then what happens when the transmission is stopped? If it's not being stored forever, then for how long is it stored? Where's it go after it's done being stored? $\endgroup$ – Phil Frost - W8II Feb 22 '15 at 4:35
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    $\begingroup$ If the analysis at the head of this discussion is correct (which I believe is the case), and the voltage/current is exactly the same in the fed and un-fed halves of the folded dipole, then the capacity between the lines is irrelevant. You can't store energy unless the voltage is different on the two 'plates' of a capacitor. $\endgroup$ – Geoff Grayer G3NAQ Oct 31 '18 at 21:27

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