Discussion in 'Antennas, Feedlines, Towers & Rotors' started by N1VAU, Jan 7, 2020.
I heard doublets and ladder line worked 30% better before Al Gore invented the internet.
Maybe you can blame it on the liberals, but not Al Gore specifically. I stopped using OWL about the time that the first switching power supplies and flat-screen TVs started appearing in the early '90s because OWL is inferior to coax for RFI pick-up. Those switching supplies, CFL and LED lighting, solid-state ballasts are mandated by energy-efficiency regulations.
This thread was started by a ham who is complaining about RFI getting into his OWL fed antenna system...
I did this on purpose. With a center fed dipole, 65 feet on each side, I added about 60 feet to one side. Fed with about 80 feet of 300 ohm twin lead, I measured the magnitude of the current coming out of each terminal of my home brew transmatch that, at the time, had a 1:1 bifilar wound to do the unbal to bal conversion. The currents were actually quite equal in magnitude and the phase wasn't that far from 180 degrees apart (measured with an HP 3575A gain/phase meter). This was a long time ago but it convinced me that the balun was in fact doing its job just as K7TRF explained. If the output of the balun is well isolated from the input due to a high common mode impedance, it will work every time Recently with the Nanovna, using the same current transformers as for the earlier test, the currents were within about 0.2 dB and the phase was within about 5 degrees of being out by 180 deg. ( I phased the transformers so I see the difference from 180). The last test above was after I added about the same wire to the other end of the diploe to get close to a 160 meter dipole and the results are about the same on 160 meters as they were on 80 meters.
Now model 2 seperate wires, of unequal lengths, NOT 1 wire with a arbitrary asymmetric connection point.
Post the impedances of those 2 seperate wires, and explain how forcing equal current (or voltage) will allow the back and forth flow of power between the wires.
Can we agree that if a photon enters one wire, a photon has to leave the other?
Well, my sample of one begs to differ.
I run homemade owl, 'bout 4 or 5 inch spacing, to a symetrical doublet.
The owl runs pretty much straight down from the feedpoint.
I detect no difference in my "noise" when I open all the house breakers, this includes a desktop computer 6 feet from the rig.
I find it far, far more effective to eliminate noise at the source,
And far more effictive at eliminating noise by building directional antennas,
Then I do by adding baluns and chokes to my feedline.
After all, there is a antenna attatched to that feedline, and capturing rf signals is what a antenna does!
The dipole (doublet) wires (4 and 5) in post #33 are broken into to two separate wires right where the Ladder Line connects in all three cases I created. The Ladder Line is modeled as two vertical wires (1 & 2) that are parallel, but spaced 4inches apart. There is a 4inch gap between the ends of wires 4 and 5.
Here is a close up view of the wire junctions for the unbalanced case created by purposely unbalancing the doublet:
Wire 1 is connected to wire 5 at the blue dot. The current at the top end of wire 1 (A=0.66A) has to be the same as the current at the left end of wire 5 (Kirchoff). Similarly, wire 2 connects to wire 4 where the current B (=0.44A) s less than current A due to wire 4 being shorter than wire 5. The currents in the two wires at the top of the Ladder Line have a net imbalance of 0.22A, as do the doublet wires.
When you model a 100ft long doublet, a 4inch gap is too small to see unless you zoom way back.
NEC knows nothing about the "impedance" looking into the end of a wire (and neither do you, Rege), but NEC understands Kirchoff, and it accurately computes the current in each wire segment between the green dots in the figures. For example, it understands that the current at the open ends of wire 4 and 5 have to go to zero. It understands that the LL currents into the blue corner dot has to equal the current that flows out of the blue dot on the doublet wire...
The entire goal of putting a CM Choke at the top of the Ladder Line is to force the currents in both wires of the Ladder Line to be equal. It has nothing to do with equalizing "power", which is a bogus idea based on Rege's lack of understanding basic antenna concepts!
The currents in the two Ladder Line wires are equal at the bottom because of the link-coupling. If we force the currents in the two Ladder Line wires to be equal at the top of the ladder line by adding a current transformer (balun), then there is no CM current along the Ladder Line, therefore no radiation from the Ladder Line during transmit, and it will not act as a receiving antenna for RFI, either.
We can agree that if an electron enters one wire, an electron has to leave the other, and that is fundamental in how NEC works! NEC knows nothing about photons.
Absolutely, but that is not always possible.
Then there is the conducted interference. Equipment that generate common mode interference may not be a problem as long as that interference is not radiated, for example because the cables it is attached to is too short to effectively radiate any noticeable amount of energy. Add a feed line and antenna without CM filter to it and it can now both be radiated and picked up by that antenna system.
No, photons do not enter or those wires. Photon is the particle representation of electromagnetic radiation, one half of the wave-particle duality concept. The other half being the planar electromagnetic wave. Both the particle representation and the wave representation is needed to describe the properties of electromagnetic radiation. Inside a conductor, an electromagnetic wave will soon be converted into electric current, ie. moving electrons.
Indeed, and it can work quite well as long as the common mode impedance presented by the CM filter is much higher than the impedance of the path we want the current to take, out through the antenna wires.
Nice illustration by the way
You know exactly what I am trying to say.
The 2 wires of a antenna are transducers, that shuttle power back and forth between them.
For every bit of energy that enters one wire, a equal bit has to leave the other. If the 2 wires do not have equal impedance, you cannot "force" either equal current or voltage into them and expect the transducer to operate correctly.
Kirchhoff's laws say nothing about energy. Kirchhoff's current law says that the sum of all currents entering and leaving a junction must be zero, ie. no electron (or other charged particle) can disappear or pop-up from nowhere.
If you have one long and one short wire (as in the off-centre fed antenna) and push(pull) an equal amount of current into(out of) them over a fixed time, they will simply be charged to different voltage potentials (U=Z×I).