ANTENTOP- 03- 2003, # 004

More info from Cecil, W5DXP on the subject:

"Yuri Blanarovich wrote:

What I was looking for is to see 1. if anyone else MEASURED the current in loading coils, and what results they arrived at (and if we are wrong, then where did we go wrong). 2. If this is right than to have modeling software implement it with least error. I would like to use that for optimizing, say, loaded elements for receiving arrays on low bands, optimizing mobile antennas, loaded multielement beams, etc."

Hi Yuri,
try this out for your argument in the other group. Using EZNEC:

Example 1: 102' CF dipole with loading coils in the center of each arm to cause the antenna to resonate on 3.76 MHz. I get XL=j335 ohms.

Example 2: Replace the above loading coils with series inductive stubs hanging down. Ten foot stubs with six inch spacing between the wires is what I used. What happens to the current across that six inch gap is obvious from the current plot using EZNEC. Hint: There is a step function across that six inch gap just as there will be with a six inch coil.

Then ask: Why doesn't EZNEC treat these two cases the same way?

73, Cecil http://www.qsl.net/w5dxp"

and ...

Yuri Blanarovich wrote:
"There is too much reliance now going on modeling program results, ignoring some realities."

Yuri,
here is a modeling result that you might like. :-) I took a 102' dipole and loaded it in the center of each leg with an inductive stub that made the dipole resonant on 3.76 MHz. I added a one ohm series 'load' to each side of the stub. Drawing one leg of the dipole, it looks like this:

----------R2-+ +-R1----------FP--- ... other half
                 | | 
                 | | inductive
                 | | stub
                 +-+

EZNEC reports 0.85 amps through R1 and 0.57 amps through R2, a difference of 33%. If one could model the inductive loading reactance as an actual physical coil instead of a lumped single point impedance, results would be similar to the above.

Now here is something that might blow some minds. The inductive stub above is ten feet long. That's about 1/8WL on 20m. A 1/8WL shorted stub equals +jZ0. The results of running the above antenna on 20m is that the current through R1 is 185 degrees out of phase with the current through R2. At the time when the current through R2 is flowing toward the end of the antenna, the current through R1 is flowing toward the feedpoint. Wonder what Kirchhoff would say about that. If you replace the stub with a coil of the same reactance, not much changes.

Tell W8JI to stop using lumped circuit analysis when he should be using distributed circuit analysis. :-)

73, Cecil http://www.qsl.net/w5dxp"

W5DXP:
Yuri, my latest posting sheds more light. Apparently, W8JI doesn't realize that there are two superposing currents phasor-adding together to get the net current and the phase distribution between those two current waves are opposite because they are traveling in opposite directions. This is a characteristic of standing-wave antennas.

See what happens when one tries to ignore the component waves?

Because the two currents are traveling in opposite directions, any phase delay through the coil shifts the phase of the two currents IN OPPOSITE DIRECTIONS. Thus the total relative phase shift effect through a 10 degree coil is 20 degrees.

Mark, NM5K wrote:

"Dunno...I finally got up enough courage to wade thru a bunch of that myself. Both had some decent points.. But....Just using my built in "BS" filter only, which  rarely seems to fails me,  and ignoring all other influences, I still have to side with Tom. I still think the current is fairly constant."

W5DXP:

The key to understanding is to realize that the net current is the phasor sum of the forward current and reflected current (on a standing- wave antenna). Assume a 10 degree phase delay through the coil on the frequency of operation. Ifwd-in and Iref-out are on the same side of the coil. Ifwd-out and Iref-out are on the other side of the coil.

Ifwd-in-->    coil    Ifwd-out-->
-----------------------////////////-------------------------
             <--Iref-out            <--Iref-in

Assume that |Ifwd-in| = |Ifwd-out| which satisfies Kirchhoff

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