What's the story with end-fed antenna's?

Discussion in 'Antennas, Feedlines, Towers & Rotors' started by KI7QVR, Feb 27, 2018.

1. K6OFGXML SubscriberQRZ Page

Thank you for the re-post of your article. Those of us that haven't been around through the last millennium appreciate being able to catch up. Nice to be able to absorb a technical subject for us non techies.
Again, thank you.
Steve

2. AF7TSHam MemberQRZ Page

The above is a very misleading statement, implying that if I connect a feedline to a length of wire that physics will somehow force the system to 'grab' an exact match for this length of wire.

Imagine a 100 foot length of coax, with the last 50cm of shield stripped off. Saying 'All antennas are center fed. They have two electrically equal halves.' suggests that the 50 cm of bare center conductor will be automagically matched by 50 cm of shield making a nice 2m antenna.

But this is not what would happen. Instead the entire length of shield would end up carrying RF.

Kirchoff's laws must be understood in the context of small regions compared to the wavelength in question, and an antenna is explicitly large relative to the wavelength.

To invoke Kirchoff, you have to consider a small region right around the feedpoint. And here is it quite clear that the net current into/out of this region must be zero.

Any current flowing into one side of the antenna feedpoint must be matched by current flowing on the other conductors entering the feedpoint region.

If this current flow at the feedpoint is not balanced going in both directions on the antenna, then it _must_ flow as net RF on the feedline. But the antenna need not be symmetric around the feedpoint, and will not automagically make itself macroscopically symmetric around the feedpoint.

To put this another way: current at the feedpoint must be _locally_ balanced, either by equal currents on the antenna elements, or balanced by current going somewhere else. This does not imply _macroscopic_ symmetry of the antenna around the feedpoint.

73
Jon
AF7TS

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I agree 100%.

Circuit theory rules such as KVL and KCL only apply when the components and connections are very small in terms of a wavelength.

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Here's another example:

Imagine a quarter-wave ground-mounted HF vertical. It is mounted over a salt water marsh that extends miles in all directions, with grounding rods and radials in the water so that there is excellent electrical connection. The feedline is coaxial and comes underwater to a point directly under the feedpoint.

Such an antenna will work very well (it needs a bit of matching if fed with 50 ohm coax because the feedpoint impedance is about 35 ohms). How is such an antenna "center fed"?

73 de Jim, N2EY

4. N0TZUPlatinum SubscriberPlatinum SubscriberQRZ Page

Yep, the largest circuit dimension must be much less than a wavelength in order for the circuit theory approximation to be applicable. Otherwise, break out Maxwell's Equations.

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5. W5DXPHam MemberQRZ Page

The wave reflection model does a pretty good job.

6. SM0XHJHam MemberQRZ Page

Yes! That is exactly how an "end fed" works!
The TX chassis and feed line, and anything connected or near them (including your body, which is an essential part for hand-helds) is what is the needed counterpoise. So the end fed antenna is not end fed after all. It is fed between the radiator and the TX chassis/feed line/you/electrical wiring/whatever.

And that current must be exactly 0 if one of the terminals are left open and the transformer is ideal. (In reality it isn't, and capacitive coupling from the secondary to primary side may still make such a construction work, sort of.)

Kirchhoff's current law still applies, because it applies on points only.
Kirchhoff's voltage law cannot be used unless you take into account the R L and C in and between the components, which isn't easy. But in no way can Kirchhoff's laws simply be ignored as soon as you have alternating current. The laws are still applicable, but the components that make up the circuit becomes harder to understand.

I would say it is very much centre fed
It is fed between the vertical radiator and the salt water surface.

I think much of these end fed discussions comes down to what we call an antenna.
In my mind, all parts that are needed to make the antenna radiate as intended is part of the antenna. Take away the salt water and the vertical in the example above doesn't work. Take away the feed line, TX chassis, capacitive coupling to you/the ground, and the end fed doesn't work. All these parts have current flowing through them somehow, and if some of them have high resistance you will have high loss. Therefore it is better to provide a low resistive other half (radials/counterpoise).

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7. G3TXQHam MemberQRZ Page

Don't forget that Kirchoff's current law at the feedpoint can be satisfied with displacement current:

Steve G3TXQ

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8. SM0XHJHam MemberQRZ Page

Displacement current is nothing magical, it is simply the current that seem to pass through a capacitor (through the means of varying electrical fields). A capacitor requires two sides, two plates, to terminals (otherwise no varying electrical field can exist).
If the physical size of one side of the feed point is 0, then the capacitance from that side to the other side is 0 F, and displacement current through that capacitor (and therefor the other side of the feed point to) must be 0.

These type of antennas do work in practice, through common mode currents and capacitive coupling to other objects. There is no doubt they can be quite efficient given the right circumstances, and may very well be the best of all non-optimal antennas we have to choose from in some situations. It is just the name and the description how they work I object to. Providing a low loss counterpoise will always be more efficient than relying on the soil or similar to carry that current.

Last edited: Mar 1, 2018
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9. W5DXPHam MemberQRZ Page

e.g. a 180o inductor has current flowing into the top and into the bottom at the same time.

That looks like an off center fed to me.

10. G3TXQHam MemberQRZ Page

The author of that article claims: "The current flows into a capacity represented by the E-field." In other words there doesn't need to be a physical capacitor.

Steve G3TXQ