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Does a balun/choke have to "work harder" in an off-center fed dipole than when the feedpoint is at the center? Or do the standing waves that develop keep the antenna balanced electrically even if the physical feedpoint is driving unequal conductors? And if that's the case, does the situation change if the OFCD antenna is non-resonant (i.e. compared to the same length doublet when fed right in the middle)?

[And also to be clear, I'm not talking about a Windom/G5RV/etc. where a portion of "feedline" is intentionally incorporated into the antenna — assume any necessary balun/choke is incorporated directly at the feedpoint for the purposes of this question.]

We've touched on aspects of this through a few other Q&A's here but I'm not finding anything specific to the "balance" of an off-center feed:

(And for full disclosure, this is admittedly closely related to my How to best ditch the counterpoise of a random wire antenna? question but seems sufficiently standalone to ask separately.)

Basically it's made me realize that my understanding of "balance" and why/when a balun is necessary is still pretty weak — I can imagine how in a simple resonant dipole fed right at the midpoint, a choke might still be have to "fix up" a bit of mismatch between the two halves especially due to real-world environment (nearby conductors, material/construction imperfections, etc.) to fully avoid current on the shield.

But with an off-center feed it seems that there's a great deal of (intentional) "mismatch" between the two sides of the antenna already while the design is still on paper. But yet we talk of the impedance "at the feedpoint", rather than the impedance "into the short side" distinct from "into the long side". Do the currents on the two sides of an off-center feed still naturally balance out despite their totally different lengths — at least as much as they would in any real-world environment? Or does the choke end up having to dissipate more and more current as the feedpoint moves closer to an end?

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  • $\begingroup$ I've retitled this question from its original "Understanding 'balance' in an off-center fed dipole or doublet" — hopefully that helps clarify what I'm asking? $\endgroup$ May 18 at 19:58

4 Answers 4

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The voltage distribution on a dipole comprises that of a standing wave, and is determined by how far the ends of the antenna are away from the feed points, since the voltage at the ends is always fixed at maximum RMS amplitude.

The ends of the antenna set up the conditions which determine the positions of the voltage and current nodes and anti-nodes of the standing wave. For an antenna which is split in the middle and current fed, the same current flows through both antenna elements and the radio receiver / transmitter.

The current flowing through a balun will be that of the standing wave at that point. So at a current node, balun RMS current will be at a maximum and RMS voltage at a minimum.

Ignoring the effect that antenna impedance has on the voltage across the feed points, if the antenna is the right length for the frequency of operation, then from a distance when you look at voltage distribution for both halves at any instant in time, they will line up and appear as one continuous sine wave in space along the antenna elements.

An off center fed dipole is balanced in terms of voltage and current at the feed point. If the ends of the antenna are the right distance away from the feed points, then the ends of the standing wave line up, and at any instant in time the voltage has the same magnitude at each terminal of the feed point.

If either feed point is the wrong distance away from each end, then the standing wave doesn't line up and the voltage won't be the same, and the antenna won't be balanced.

In reality, there is a difference in potential across the feed points which is the voltage which produces the incident current into the antenna at a rate as determined by the real portion of the antenna impedance, according to ohm's law and just like in a resistor.

The figure below shows a snapshot in time of the voltage of the standing wave along the elements of two off center fed dipoles with different lengths.

enter image description here

Each element has its own impedance which adds up to produce the total impedance seen from both elements at the feed points.

So the two elements of an unbalanced antenna each have a different impedance, resulting in a different voltage at the feed points.

In terms of the standing wave on the antenna, both antennas are not balanced as the waveform on each half isn't the same, and the radiation pattern will be produced accordingly.

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  • $\begingroup$ Sorry, gotta confess I'm having trouble understanding this answer. I think you're saying that a resonant dipole is balanced (i.e. will not induce current on the shield) regardless of feedpoint, while a non-resonant dipole is not balanced even if the feedpoint is at the center? $\endgroup$ May 18 at 23:18
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    $\begingroup$ No, resonance has nothing to do with balance, for dipoles split in the middle, the two halves each have their own impedance and standing wave, if the standing waves don't match at the feed point then the transmission line thinks it's not balanced. $\endgroup$
    – Andrew
    May 18 at 23:23
  • $\begingroup$ I posted another question re this ham.stackexchange.com/questions/20927/…, if i'm wrong someone please tell me why so we can all learn something. : ) $\endgroup$
    – Andrew
    May 18 at 23:26
  • $\begingroup$ But, only at a small number of specific frequencies. As soon as you move away from an OCFD's design frequency, the wire isn't the "correct length" anymore, and at the band edges the difference in impedance between the two elements is substantial — unlike a dipole/doublet, which is balanced regardless of frequency (so long as there aren't big metal objects in the near field at least). $\endgroup$ Jun 9 at 21:32
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The point of the balun is not to balance the antenna but to balance the currents in the two conductors of the feedline.

If the currents are equal and opposite, they form magnetic fields that cancel. preventing leakage from the feed line.

If some of the current is flowing the same direction in both wires, then the feed line acts like part of the antenna instead of a transmission line, and it will radiate, and possibly detune the antenna system (unless it happens to also be resonant).

If the antenna is already nonresonant, it can make this issue worse.

In a few cases, the balun (or unun) is positioned on the feed line in an intentional way that makes the feed line part of the antenna.

A parallel feed line actually works the same way coax works... except that the shield in the coax has a chance to couple with the antenna, so you can have a magnetic field and opposing currents between the center conductor and the shield, and another field between the shield and half the antenna.

You can also think of this as (due to skin effect) a current on the inside of the shield coupling with the current on the outside of the center conductor, and another current on the outside of the shield coupling with the antenna. When you sum the currents on the shield, you can have some left over flowing in the same direction as the current on the center conductor, causing the common mode current.

This is a shortened version of the explanation in Sevick's Transmission Line Transformers: Theory and practice

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  • $\begingroup$ "The point of the balun is not to balance the antenna but to balance the currents in the two conductors of the feedline." ? For a series current fed unbalanced dipole the same current flows through both antenna elements and through the radio, and in the transmission line this current is always exactly opposite or 180 deg out of phase regardless of whether or not the system is balanced. $\endgroup$
    – Andrew
    May 17 at 23:09
  • $\begingroup$ "If the currents are equal and opposite, they form magnetic fields that cancel. preventing leakage from the feed line." This only holds true for a balanced transmission line without a shield. For coax leakage isn't determined by whether or not currents are equal and opposite, it is determined by how well the shield works. $\endgroup$
    – Andrew
    May 17 at 23:10
  • $\begingroup$ "If some of the current is flowing the same direction in both wires, then the feed line acts like part of the antenna instead of a transmission line, and it will radiate," Current can't flow in two directions in one piece of wire, it only flows in one direction at any instant in time. For coax, the direction of any current flowing on the outside of the coax has nothing to do with how much radiation there is, there will be radiation for current flowing in both directions. $\endgroup$
    – Andrew
    May 17 at 23:10
  • $\begingroup$ take a look at this, physics.stackexchange.com/questions/286805/… and see if it helps. $\endgroup$
    – Andrew
    May 18 at 1:05
  • $\begingroup$ Andrew: actually, you've got the right reason but your logic of the result is wrong! That's exactly how it works. $\endgroup$
    – user10489
    May 18 at 1:06
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Yes, an off-center fed dipole is unbalanced.

Why? A dipole in free space is balanced. However, put a large conductive object or a long enough wire near that dipole, and it will distort the EM field around that dipole. If the parasitic element isn't centered to the dipole's center, then the EM field distortion will be not be symmetric to the dipole, thus the distorted EM field will unbalance the dipole.

The shield of a coax feed line to an off-center fed dipole usually (always?) can't be centered in the dipole's EM field, thus is an off-center parasitic element.

Hypothetical: It might be possible to bring an off-center fed dipole somewhat back into balance by adding a disconnected parasitic coax on the other side, arranged with a mirrored geometry, maybe terminated with the transmitters output impedance, thus restoring symmetry. Or by feeding the dipole without any feedline (ground isolated miniature transmitter), just a matching lumped load on the other side.

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  • $\begingroup$ Isn't coax changing the balance — I presume somewhat subtly — just a special case of "environmental effect" though? What I'm trying to understand is the feedpoint itself. Say I have a magic infinitesimally small radio that can also measure any "unbalance" of its output. If I attach it to a dipole in free space at the 50%/50% point I'd expect to see 0% imbalance. But what if I connect it at 33%/66% point? Or even further to 25%/75% or all the way to 1%/99% — does the "unbalance" grow and grow, or just the impedance? $\endgroup$ May 18 at 19:50
  • $\begingroup$ Between the two feed points - unbalanced = impedance and therefore voltage not the same for each side of the antenna, the same current flows through antenna, transmission line and radio ? $\endgroup$
    – Andrew
    May 18 at 23:32
  • $\begingroup$ And yes if the coax doesn't have a balun then it will mess up the balance. $\endgroup$
    – Andrew
    May 18 at 23:35
  • $\begingroup$ We need the help of a ham expert to point us in the right direction ! $\endgroup$
    – Andrew
    May 18 at 23:36
  • $\begingroup$ Even if a coax off to one side does have a balun, if the coax is as long or longer than around 1/10th wavelength, it will mess up the balance. Unless you put another coax on the other side. The half-wave antenna itself is balanced around its center, nothing to do with the location of the feed point. $\endgroup$
    – hotpaw2
    Jun 2 at 23:42
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I'm still looking for more explanation and more understanding but I wondered:

do the standing waves that develop keep the antenna balanced electrically even if the physical feedpoint is driving unequal conductors?

W8JI's Common Mode Currents article says outright:

Non-symmetrical antennas, such as off center fed antennas or end fed antennas, create severe common mode issues. The further the feeder is offset from center, the worse the common mode issue. This means center fed doublets are least problematic, off-center-fed antennas in between, and end fed antennas are the worst possible common mode problems. As a matter of fact, end fed antennas are 100% common mode at the feedpoint!

This is somewhat surprising to me — I mean it does make sense when looked at in the obvious way (the antenna being split unevenly) of course, but not in the case of a resonant antenna (e.g. why would the electrons "want" to go outside and along the coax when the antenna itself is a perfect fit…). But that's probably yet another separate question!

That said, the W8JI description is corroborated by this W5ALT Antenna Balance article:

[…] it is not neccesary to feed an antenna in the center, nor is it neccesary to make it symmetrical. In either of those cases, the antenna currents most likely will not be equal and opposite. When we connect our transmission line to the antenna, the currents in the transmission line will most likely not be balanced, so feedline radiation will occur. In that case we say that the antenna is "unbalanced".


As far as the resonant vs. non-resonant part of my question, there's perhaps some clue in the ARRL Antenna Book, in a "Feeder Unbalance" section on page 26-5 in my 1994 17th edition:

With end feed, the currents in the two line wires do not balance exactly and there is therefore some radiation from the line. […] In addition to this unavoidable line radiation a further unbalance will occur if the antenna is not exactly resonant at the operating frequency. […]

Unfortunately this discussion is in the context of the most extreme (end-fed) case and its unclear how it might change as a function of only partly off-center feeding.

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  • $\begingroup$ [For anyone else following along at home, there's some related reading in the intro to w8ji.com/balun_single_core_41_analysis.htm as well.] $\endgroup$ May 18 at 20:28
  • $\begingroup$ Just because some ham radio expert publishes something on a website doesn't mean what they say is fact. I've looked at that site and in my opinion most of the stuff on there is highly dubious and and in some cases plain wrong. $\endgroup$
    – Andrew
    May 18 at 21:50
  • $\begingroup$ Technically speaking, common mode is even the wrong term for current flowing on the outside of coax. Common mode voltage or current is that which is common to two wires. For coax, if you add together vectorily the current flowing on the inside of the braid and the outside of the center conductor at any one point along the line, the result is the CM current, and in a perfect line current on each conductor is equal and opposite and so the adding together gives zero. $\endgroup$
    – Andrew
    May 18 at 21:52
  • $\begingroup$ Current can't flow in two directions at the same time in the same peice of wire, so even saying that CM current is that which flows in the same direction in a transmission line is wrong. What happens on the outside of the braid of coax has nothing to do with CM. The amplitude of current which flow on the outside of coax is determined by the ratio of antenna element impedance and the impedance of the outside of the coax WRT earth, and antenna balance has nothing to do with this current. All those people who refer to current on the outside of coax as CM current are wrong. $\endgroup$
    – Andrew
    May 18 at 22:02
  • $\begingroup$ Sorry but this comment "Non-symmetrical antennas, such as off center fed antennas or end fed antennas, create severe common mode issues. " is complete rubbish. $\endgroup$
    – Andrew
    May 19 at 0:05

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