"John KD5YI" wrote in
:

....
Owen -

I've had this nagging idea for a couple of years: What if the
(single-turn?) loop passes through a toroid which already has, say, 10
or so turns on it? That should give an impedance step-up of 100 or so.
Would that not be a good way to at least get the matching closer? In
fact, the tuning capacitor could be connected to the high impedance
side to reduce the capacitor's value.

Your comments will be appreciated.

I doubt that the transformer you propose would be near as efficient as the
transformer formed by a smaller feed turn. Though it is tempting to explain
a transformer using turns ratio, the case of the smaller turn is explained
by the ratio of main loop flux intercepted by the feed loop, and that is
determined by area... so deforming the feed loop to change its area will
adjust the transformation.

Owen

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

...
Owen -

I've had this nagging idea for a couple of years: What if the
(single-turn?) loop passes through a toroid which already has, say, 10
or so turns on it? That should give an impedance step-up of 100 or so.
Would that not be a good way to at least get the matching closer? In
fact, the tuning capacitor could be connected to the high impedance
side to reduce the capacitor's value.

Your comments will be appreciated.

I doubt that the transformer you propose would be near as efficient as the
transformer formed by a smaller feed turn. Though it is tempting to
explain
a transformer using turns ratio, the case of the smaller turn is explained
by the ratio of main loop flux intercepted by the feed loop, and that is
determined by area... so deforming the feed loop to change its area will
adjust the transformation.

Owen


Thanks for your reply.

So, as I understand it, there is a large main loop that does the signal
capture. Then there is a smaller loop kind of co-axial with the large loop
and the loops are not electrically connected. Yes? The small loop is a
signal pick-off? Similar to a transformer but with no core?

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?

John

"John KD5YI" wrote in
:

....
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, 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.

Now you can expand that and write it in terms of inductance... but Mike
can measure the loop dimensions with a ruler and calculate area.
Measuring inductance of the components is fraught with problems.

So, if Mike works out what the radiation+loss resistance is of his main
loop, divides it by 50, takes the square root, it gives him the portion
of the main loop area to be occupied by the feed loop. Initially, one
might make the area of the feed loop 20%-50% larger and squash it to fine
tune the transformation ratio.

Owen

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

...
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.


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

(snipped section with which I agree)

Owen

"John KD5YI" wrote in
:


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

...
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

On Sat, 30 May 2009 07:19:26 GMT, Owen Duffy wrote:

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.

Hi Owen,

By your latest comment, it appears you have jumped off the rails
completely. Let's look at the spec:

On Wed, 27 May 2009 13:36:53 -0400, Michael Coslo
wrote:
It's Octagonal, .75 inch copper tubing, around 8 feet in diameter. It's
designed to just reach into the 80 meter voice segment with it's
capacitor setup.
which through any of several methods reveals radiation resistance
equals 0.27 Ohms. The need for large conductors, high voltage
components, and loop (which, for 80M would be half a meter) coupling
to obtain the proper match are all found in their proper
relationships.

Upper end is either 17 or 15 meters.
where the ratio of the physical dimensions to the wavelength (roughly
half wave) renders a much higher value.

Reggie covered all of this years ago. Unfortunately, he buried the
best of his legacy in a drift of bantering making it an extreme chore
to mine Google for the relevant postings.

73's
Richard Clark, KB7QHC

Richard Clark wrote in
:

On Sat, 30 May 2009 07:19:26 GMT, Owen Duffy wrote:

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.

Hi Owen,

By your latest comment, it appears you have jumped off the rails
completely. Let's look at the spec:

On Wed, 27 May 2009 13:36:53 -0400, Michael Coslo
wrote:
It's Octagonal, .75 inch copper tubing, around 8 feet in diameter. It's
designed to just reach into the 80 meter voice segment with it's
capacitor setup.
which through any of several methods reveals radiation resistance
equals 0.27 Ohms. The need for large conductors, high voltage
components, and loop (which, for 80M would be half a meter) coupling
to obtain the proper match are all found in their proper
relationships.

My reckoning:

I ran up a model in NEC of a vertical circular loop of 2.4m diameter,
made of 19mm dia copper conductor, and located with its centre 3m above
average ground.

The feedpoint impedance is 0.13+j168 at 3.5MHz.

If that was resonated with a lossless capacitor, the parallel load
impedance would be about 220k+j0.

To transform that to 50+j0, the impedance ratio is 4400:1, the voltage
ratio is 4400^0.5:1 =66:1, the areas need to be in a ratio of 1:66, the
diameters (for circular feed loop) 1:66^0.5 =1:8.1. The feed loop would
be about 12% of the diameter of the main loop or about 0.3m.

....
Reggie covered all of this years ago. Unfortunately, he buried the
best of his legacy in a drift of bantering making it an extreme chore
to mine Google for the relevant postings.

Reg did leave us a calculator for this problem (MAGLOOP4) but without
(AFAIK) documentation of the method used. MAGLOOP4 comes up with slightly
different figures to my NEC model, slightly higher loss, slightly lower
main loop // Z, slightly larger feed loop.

Owen

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


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

...
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

On Sat, 30 May 2009 21:54:44 GMT, Owen Duffy wrote:

Reg did leave us a calculator for this problem (MAGLOOP4)

Hi Owen,

but without
(AFAIK) documentation of the method used.

Hence my statement of his work being buried in his banter in this
group. Using MAGLOOP as a keyword might help uncover some of his
narrative documentation.

MAGLOOP4 comes up with slightly
different figures to my NEC model, slightly higher loss, slightly lower
main loop // Z, slightly larger feed loop.

Which by several methods hews closely to a common answer (and by far
better accuracy than most guesses).

73's
Richard Clark, KB7QHC