"Owen Duffy" wrote in message
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"John KD5YI" wrote in
:


"Owen Duffy" wrote in message
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"John KD5YI" wrote in
:

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I seem to recall that a small loop has less inductance than a large
loop. I also seem to recall that the transformation between them is
a function of their inductances. So, a small loop would transform
the larger loop's impedance to an even lower impedance. Is this not
so?

You can think of it in terms of inductance, but visit what inductance
really means (ie the relationship between inductance, current and
flux).

I gave you an explanation in terms of area of the two loops. If your
feed loop os 20% of the area of the main loop, it intercepts 20% of
the flux, the voltage induced in the feed loop is 20% (1/5) of the
main loop voltage,


I understand this and I agree. It makes perfect sense.


the impedance is transformed by 1/(1/5^2)=25, so a 2 ohm main
loop is transformed to a 50 ohm load at the feed loop feed point.

That is wrong, isn't it? What was I ever thinking?

The main loop isn't anything like 2 ohms at resonance, it is much much
higher, and the voltage is quite high for a given power.

The reduction in flux cutting the feed loop means that feed loop voltage
is reduced proportionately to area, and therefore the impedance (in a
lossless system) would be decreased by the square of the voltage
reduction


This is the part with which I disagree. Can you show me why it is 25
rather than 1/25?

It is, but it is not 2 ohms that is reduced, it is something much much
higher.

Thanks.

Owen


Thanks for clarifying, Owen.

Also, your reply to Richard Clark was particularly helpful to me as I seem
to learn better by example. I now see what you are doing and how you arrive
at your conclusions. It has all been very enlightening to me.

John