# Field Strength Meter at a dipole

Discussion in 'Antennas, Feedlines, Towers & Rotors' started by N4LQ, Aug 5, 2017.

1. ### AI3VHam MemberQRZ Page

Note that a half wavelength wire has a voltage peak on both ends.

Why doesn't the op measure a voltage peak on both ends of his half wave wire?

How are you feeding the wire?

How long is the wire in feet?

How high above ground is the wire?

Rege

2. ### SM0XHJHam MemberQRZ Page

He wrote in the first post that the wire was end fed. (Since it is not possible to feed an antenna at the end, I assume he means he is feeding the 1/2 WL wire at the end.)

He stated that it was 1/2 WL. That is the important thing here.

Yes, assuming the antenna only consist of that single wire.
In this case I think the antenna is something like this:
https://www.hyendcompany.nl/antenna/monoband/product/detail/10/Hyendfed_40_meter
They are marketed as an end fed 1/2 WL antenna that needs no counterpoise, as it is end fed. (Which of course is wrong, it would violate Kirccoffs's current law to start with.)

If that is the type of antenna being used here, it is perfectly logical that there is no voltage peek at the end of the wire where the feed point is. The end of the wire is simply not the end of the antenna. There must be a considerable amount of current in the feed point for the antenna to work, and then the voltage max/current min will not be there, but somewhere along the feed line or maybe even on multiple places down in the house hold wiring. The best you can hope for is that most of the other side of the antenna consists of the soil , where it at least is converted to heat without causing much harm.

3. ### AI3VHam MemberQRZ Page

That's not how it works.

There will be a voltage maxima at the far end of ANY length of wire.

ANY length. Irregardless of where the antenna is fed.

As you move to the generator, there will be a identical voltage every 1/2 wavelength.

Again, the position of the feedpoint has zero bearing.

If the op has a wire, 1/2 wavelength long, he either reads a voltage peak on each end, or is not describing accurately his setup.

Rege

4. ### SM0XHJHam MemberQRZ Page

That's my point. It will only show a voltage max on the far end of the wire, because the other end of the wire is not the end of the antenna (assuming that the wire is end-fed). The antenna continues down the feed line. The feed point is always in the electrical centre of the antenna. Since the current in the feed point cannot be zero (in which case no energy would enter the antenna), the voltage there (compared to the average voltage potential surrounding the antenna, ie ground), will not be as high as at the opposite side of the antenna, where the wire ends in an open end, and the current is truly zero.
No it hasn't, because the feed point cannot be at the end of an antenna, as no current can flow at the end of an antenna.
(P=U*I, if I=0 which it must be in the end of the antenna then P=0, ie you feed no energy into the antenna. If I != 0, then you don't have a voltage max there either.)

5. ### K9AXNHam MemberQRZ Page

If you apply a stream of 1 volt sine waves to the end of a half wave antenna the first cycle will see 600 ohms and progressively increase to approximately 4700 ohms. If you center feed it, it will also begin at 600 ohms and progressively decrease to 72 ohms. The first cycle in any wire antenna will see 600 ohms before reflections occur.

Maxwell drove two spheres and a wire with the same power and both radiated equally. Electric current doesn't necessarily need wire. Also, if you drive an antenna with 1Kw key down, statistically no electron in the antenna will move more than a fraction of 1/1000 inch from where it started.

If it is fed with a parallel circuit you could contrive that the parallel resonant circuit is part of the antenna but where do you stop. The theoretical end is the last group of atoms that participate in current reversal.

In Steve's case it sounds like he is using a transformer not a tuned circuit. The transformer is part of the antenna so the connection to the wire is downwind of the end thus will measure less than the far end. If he is using a tank to feed it the end of the wire at both ends will read very close to the same.

Regards Jim K9AXN

Last edited: Aug 13, 2017 at 3:48 PM
6. ### KA9JLMHam MemberQRZ Page

You have already made real time measurements. Why don't you tell us ? Theory only works perfect in a vacuum.

RF is magic, FM is that special magic that requires a FM antenna. Digital antennas are even more complex.

Have Fun.

Last edited: Aug 13, 2017 at 4:13 PM
7. ### SM0XHJHam MemberQRZ Page

Well, yes it does. An electric current does need a conductor that it can travel through. (An electric field though, doesn't).
The whole point here is that the end fed antenna is in fact not end fed. The wire(s) that is radiating does not stop at the feed point. Everything connected to that feed point is also part of the antenna. How you match the antenna impedance to the transmitter doesn't take away the fact that equal amount of current must go into the feed line as common mode current, as there is current going into the wire that is referred to as the antenna.

If the feed point impedance is very high then yes, the fed end will have a high AC potential. But never as high as the open end.
In reality such a high impedance is not likely. Objects close to the 1/2 WL wire will in most cases load the wire and make it de-tuned, considerably lowering the feed point impedance. Therefore the voltage at the fed end will also be much lower.

That is very easy, everything that radiates is part of the antenna.

A very easy way of determining if the antenna is what you think it is is to put the antenna up in the air and connect a small battery and remotely operated VNA (like the minivna pro). You cannot hold the VNA in your hand or have it lying on the soil, as that will create enough capacitive loading to affect the results. Take a reading of the impedance. Then connect the screen of the feeder coax to the ground of the VNA and take a new reading. If the feed line is not part of the antenna (ie radiating) the impedance should be exactly the same. If the impedance changed, then the feed line is radiating and is part of the antenna.

Last edited: Aug 13, 2017 at 4:21 PM
8. ### W5DXPHam MemberQRZ Page

At the end of the 1/2WL wire, the current is zero. Therefore all of the energy is in the electric field, i.e. the voltage is maximum. The forward wave on the standing wave antenna encounters an open circuit.

At the feedpoint of the 1/2WL wire, the current is not zero. Therefore all of the energy is not in the electric field, i.e. the voltage is not maximum. The reflected wave from the far end of the standing wave antenna sees an impedance discontinuity, but not an open circuit.

Consider the voltage on a full-wave center-fed Double Zepp. The voltage at the two ends of the antenna is greater than at the feedpoint because there is current at the feedpoint so some of the energy is in the magnetic field, i.e. not in the electric field. The feedpoint impedance is a few thousand ohms, not an open circuit. This happens even when the feedline is not radiating.

9. ### SM0XHJHam MemberQRZ Page

Yes, that is absolutely true. In advertising for end fed antennas the opposite is often used as a argument for these antennas, that since they are 1/2 WL end fed antennas there will be no CM current on the feed line. This is of course utter BS... 1/2 WL of not, if the antenna itself is not electrically symmetric, there will be CM currents on the feed line (or anything connected to the feed point, or even close to it).

10. ### W5DXPHam MemberQRZ Page

Here's what the currents on a 1/4WL of feedline look like when feeding an end-fed 1/2WL wire (a Zepp antenna) assuming a perfect choke at the source. The currents are perfectly balanced out of the choke and perfectly unbalanced at the antenna, i.e. the common-mode current is the same magnitude as the feedpoint current according to EZNEC and Kirchhoff. If one makes the common-mode current equal to zero at the antenna feedpoint, the feedpoint current will also be equal to zero. Note that the common-mode current is a standing wave envelope.