Testing the 20 meter inductive loop antenna

Interesting. What is it meant for?

Regards

Klaus

On a sunny day (Fri, 22 Aug 2014 09:05:33 -0700 (PDT)) it happened Klaus Kragelund wrote in :

Ham radio on 14 MHz.

On a sunny day (Fri, 22 Aug 2014 16:18:12 GMT) it happened Jan Panteltje wrote in :

PS it has been done for longer wavelength too:

Interesting. How does that feed work? Is it inductive coupling to the big loop? (Your feed looks a little wimpy compared to the web link :^) Is the area of the feed important?

George H.

On a sunny day (Fri, 22 Aug 2014 09:37:18 -0700 (PDT)) it happened George Herold wrote in :

Look at the thing as a coil, it is just 2 taps on the L, so basically a transformer matching the 50 Ohm cable to the impedance of the loop. Anyways it should be... to get low SWR. Then the loop is tuned to the right frequency with C.

inner ----- _______( | C _______( === screen ( | coax -----

I get very inconsistent results, just tried a lot of things, including a separate coupling loop I think the iron in the ground and the big shorting alu table frame right next to it has an effect. It it ever stops raining I will take it outside, solder it together, and repeat. There are some other antenna types I need to test too (outside), but this is essentially supposed to go in the attic. Anyways I managed to get it tuned to 14 MHz an hour or so ago.

You can also try an isolated smaller loop inside the larger loop to match 50 ohms. 1/5 the size seems to come to mind. OH, just reread what you wrote, you tried that. I don't know why you have inconsistent results. I will say, the loop needs to be peaked, then set your match for 50 ohms. You may need iteration between adjusting for resonance and impedance matching, to get it right. Mikek

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If it's not soldered together then buidling shack might appear in your coil.

What kind of Q can you get? George H.

On a sunny day (Fri, 22 Aug 2014 16:24:44 -0700 (PDT)) it happened George Herold wrote in :

Well some 100 kHz bandwidth at 14 MHz, B = f0 / Q, so Q = 14.10^6 / 10^5 = 140

To find the tap point. transformer ratio, maybe if the coil and rest of circuit is perfect (it is not), and my idea is right (but is just a wild guess). given infinite Q it is just radiating into free space, and sees 367 Ohms (u0 and all that stuff) in parallel.

So to match to 50 Ohm, the impedance ratio must be 367 / 50 = about 7.3. That means a turn ratio of sqrt(7.3) = 2.7. If the whole coil length is 4 meters, the tap should be at 4 / 2.7 = 1.4 meters, past a whole 1 meter side. mmm So there is more tests to do, seems things are different. The other thing I noticed is that the analyzer shows there is resonance that looks dipole like at 18 MHz, nice smooth curve, very wide. This cannot be tuned (moved) by varying C it seems. So that seems length related and behaves just like a folded dipole? I have some extra pieces tubing and connection pieces to try increase length so that point moves to 14 MHz where I want to be. No idea if it should be there or just should not be there. There is a magnetic loop windows calculation program link on that site i referred to: I tried it and this size should work. But I do not have the source neither the calculation used. So purely experimental, always learning, very curiuos too.

Any ideas that work are most welcome.

On a sunny day (Fri, 22 Aug 2014 15:33:05 -0500) it happened amdx wrote in :

Yes, now it is very sensitive and already changes if I am near the thing. They recommend hanging it from some strings away from any metal. See also my reply to George Harold. It just started raining again, no way am I going out there with laptop an analyzer now. So now I will play a bit with my high end fed design, that also needs to be tested,

while trying to settle the magnetic loop issue down in my mind (so form some understanding). Just learning.

You can buy the high end fed for > 100 euro

but this is more fun. Just got the cores in, we will see.

All this reminds me of the sixties, but then I was driving with big tubes..

1 kV anode voltage. :-)