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Yagi Voodoo

Watergate

Well-Known Member
Apr 11, 2011
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I have been studying for my ticket, and came across a fact about yagi antennas that flat out baffles me.

I get the theory behind the design, the parasitic elements radiating the magnetic field of the driven element. I will accept on faith that you can somehow attach the driven element to the boom and not have it be a dead short. I still can't fathom that with a loop or a yagi, but all of my electrical experience is in automotive, and therefore DC.

Here is what I don't understand. I have read that if you use a gamma match, you don't need to isolate the driven element from the boom. How then is that still the driven element? Wouldn't the whole array receive signal from the feed line? And was what I read correct? Can the driven element be attached straight to the boom? And if by some magical AC property that is above my head you can do that, would I be able to grill a burger on that antenna while transmitting or does it not radiate heat by being some sort of short circuit?
 

AC is definitely different then DC. The point you are referring to is an RF neutral point and does not radiate. As that point does not radiate anything connected there is not seen by the antenna. The principle is the same for the parasitic elements (directors and reflector).

Correct me if I am wrong, but I believe impedance (effectively AC resistance) at that point approaches infinity.


The DB
 
i still don't quite understand it myself , but i could tune to make the meter happy and the antennas worked . kinda like women ..... no idea why they do what they do sometimes ............. . but thank casper that antennas are much much more predictable and consistent in behavior :D
 
You figure out women, and you'll have statues and monuments around the world.

I look at the world through the eyes of a mechanic. If I see something, I always try to figure out how it works. This is baffling to me. It seems like

1 = center conductor, or positive for a lack of a better term

2 = shield, or negative

1 + 2 should equal 3, short circuit.

But here 1 + 2 = 97 for some reason.
 
You're confusing RF resonace with DC circuit theory. What is a direct DC short does not necessarily constitute a short at RF resonance. Depending on how you feed a dipole antenna you can see an open or short with an ohm meter.

Connecting a dipole directly with coax and you will see a DC short from one element to the center conductor, or the shield. There will be an open circuit shown between the two antenna elements.

Feeding the dipole with a 1:1 current (Guanella) balun and you will see the same thing. But using a 1:1 voltage (Ruthroff) balun and you will see a DC short across any port, however, the antenna will show a low SWR (provided the antenna is resonant) indicating no short for an RF signal. It all depends on the matching method used.

The Gamma match will show an open circuit with an ohm meter since the center conductor is isolated from the shield of the matching system. However, all the antenna elements will show a DC short through the boom.

The reason the boom does not radiate RF is: it is not resonant and thus looks like an open circuit to the RF signal.

You will not be able to grill a burger from your HF antenna. To "cook" food (or flesh) you need a frequency that causes molecules to vibrate and cause thermal heating. Your typical home microwave operates around 2.4 GHz at 500 to 1,000 Watts. Microwave frequencies have a small enough wavelength to cause thermal heating of tissue. The HF Ham Bands would require many MegaWatts to begin to accomplish the same task. (That is why there is an RF safety calculation required by the FCC).

That's not to say you can't get a good RF burn from a 5 Watt signal if you place your finger at a high voltage point of an antenna. It can be done with a small loop antenna using a variable capacitor for tuning. Just don't put your finger on the cap when power is applied to test the theory! RF burns are nastly and can take a long time to heal.

73,
Mike
 
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