Thanks for the comments. Once again, I scanned a posting too hastily,
and somehow missed the "transmitter" part. My comments were appropriate
for a receiving antenna, not transmitting. So let me try to answer both
Tom's and Mark's postings more appropriately.

None of us has a constant current source with which to drive an antenna,
but generally a source with a fixed amount of power and a finite source
impedance. If we did have a constant current source, then yes, adding
more and more ferrite cores would result in more and more power being
delivered by the source, a larger and larger fraction of which would be
dissipated in the ferrite. A 1 megohm resistor would get a bundle of
power from our source, and a 10 megohm resistor would get 10 times as much.

If we have a tuner, we can adjust our source impedance over some range.
Provided that the feedpoint impedance is within that range with the
ferrites in place, we can deliver all our power to the ferrite-antenna
combination. I believe that the fraction of the power applied to the
antenna which ends up in the ferrites monotonically increases as we add
ferrites (assuming we don't move the previously added ones). If the
ferrites were all at the base, the equivalent load circuit would be just
two impedances in series -- the ferrite impedance and the antenna
feedpoint impedance, and it would behave as Tom said. But putting the
ferrite cores anywhere but the base changes the antenna current
distribution, which has a potentially complex effect on the feedpoint
impedance other than just adding the transformed impedance of the core.
This means that not only does Tom's Z3 increase as we add ferrites, but
Z1 changes also.

Roy Lewallen, W7EL

K7ITM wrote:
On Oct 16, 3:40 pm, Roy Lewallen wrote:
...
Regarding the ferrite absorbing energy from the antenna -- the amount
absorbed will be maximum when the ferrite's impedance is the complex
conjugate of the antenna's. For example, if the vertical is resonant and
grounded with no feed system, you'll get maximum ferrite heating when
the ferrite's impedance is around 36 + j0 ohms. If you add more ferrite,
the amount of power absorbed from a passing wave and delivered to the
ferrite will decrease, approaching zero as the ferrite impedance
increases to a large value.
...

I'm puzzling over this, Roy. It seems like this assumes some source
impedance driving the antenna, but maybe I'm missing something in your
analysis.

My thought-process is to treat the antenna as an impedance Z1, the
ferrite an impedance Z2, and the source an impedance Z3, the three of
them being in series. I suppose thinking of the antenna as a constant
impedance as you change its environment with ferrite might not be
quite right, but to the degree that approximation is correct, then I'd
expect maximum ferrite dissipation (absorption) would occur when its
impedance, Z2 is equal to the complex conjugate of (Z1+Z3). On the
other hand, if I feed the antenna with a constant current source, the
ferrite dissipation increases indefinitely as the resistive component
of its impedance increases.

Am I missing something?

Cheers,
Tom

On Oct 16, 8:19 pm, Roy Lewallen wrote:
Thanks for the comments. Once again, I scanned a posting too hastily,
and somehow missed the "transmitter" part. My comments were appropriate
for a receiving antenna, not transmitting. So let me try to answer both
Tom's and Mark's postings more appropriately.

None of us has a constant current source with which to drive an antenna,
but generally a source with a fixed amount of power and a finite source
impedance. If we did have a constant current source, then yes, adding
more and more ferrite cores would result in more and more power being
delivered by the source, a larger and larger fraction of which would be
dissipated in the ferrite. A 1 megohm resistor would get a bundle of
power from our source, and a 10 megohm resistor would get 10 times as much.

If we have a tuner, we can adjust our source impedance over some range.
Provided that the feedpoint impedance is within that range with the
ferrites in place, we can deliver all our power to the ferrite-antenna
combination. I believe that the fraction of the power applied to the
antenna which ends up in the ferrites monotonically increases as we add
ferrites (assuming we don't move the previously added ones). If the
ferrites were all at the base, the equivalent load circuit would be just
two impedances in series -- the ferrite impedance and the antenna
feedpoint impedance, and it would behave as Tom said. But putting the
ferrite cores anywhere but the base changes the antenna current
distribution, which has a potentially complex effect on the feedpoint
impedance other than just adding the transformed impedance of the core.
This means that not only does Tom's Z3 increase as we add ferrites, but
Z1 changes also.

Roy Lewallen, W7EL



thanks for the replies...
so for talking purposes:

Z1 is the antenna feedpoint Z and we will define "antenna" as the
exposed wire after the end of the ferrite tube.

Z2 is the Z of the wire passing though the ferrite

Z3 is the source Z which I will stipulate is 50 Ohms

OK as we add ferrite to the antenna, Z1 changes because the antenna is
getting shorter as the ferrite is getting longer. i.e. if there is 7"
of ferrite, then there is only 12" of exposed antenna and it is
elevated over the ground plane so Z1 is going up. In the end case,
when the ferrite is 19" there is no antenna Z1 becomes infinity.

Then looking into the base (thinking as lumped elements), we have Z2 +
Z1. Since Z1 is infinity, the base must look like infinity, but this
does not pass common sense.


In other words, what is the Z looking into a 19" wire that is inside
19" of ferrite. Thinking in lumped element terms, it would be very
high and little power will flow. Thinking in distributed terms there
will be some relatively low Z looking into the base, power will flow
and the ferrite will dissipate heat. The base Z would be related to
some property of the ferrite like the property of free space has a Z
of 377.

What is that propery and what would a typical Z be?

Mark

Mark wrote:
On Oct 16, 8:19 pm, Roy Lewallen wrote:
Thanks for the comments. Once again, I scanned a posting too hastily,
and somehow missed the "transmitter" part. My comments were appropriate
for a receiving antenna, not transmitting. So let me try to answer both
Tom's and Mark's postings more appropriately.

None of us has a constant current source with which to drive an antenna,
but generally a source with a fixed amount of power and a finite source
impedance. If we did have a constant current source, then yes, adding
more and more ferrite cores would result in more and more power being
delivered by the source, a larger and larger fraction of which would be
dissipated in the ferrite. A 1 megohm resistor would get a bundle of
power from our source, and a 10 megohm resistor would get 10 times as much.

If we have a tuner, we can adjust our source impedance over some range.
Provided that the feedpoint impedance is within that range with the
ferrites in place, we can deliver all our power to the ferrite-antenna
combination. I believe that the fraction of the power applied to the
antenna which ends up in the ferrites monotonically increases as we add
ferrites (assuming we don't move the previously added ones). If the
ferrites were all at the base, the equivalent load circuit would be just
two impedances in series -- the ferrite impedance and the antenna
feedpoint impedance, and it would behave as Tom said. But putting the
ferrite cores anywhere but the base changes the antenna current
distribution, which has a potentially complex effect on the feedpoint
impedance other than just adding the transformed impedance of the core.
This means that not only does Tom's Z3 increase as we add ferrites, but
Z1 changes also.

Roy Lewallen, W7EL



thanks for the replies...
so for talking purposes:

Z1 is the antenna feedpoint Z and we will define "antenna" as the
exposed wire after the end of the ferrite tube.

Z2 is the Z of the wire passing though the ferrite

Z3 is the source Z which I will stipulate is 50 Ohms

OK as we add ferrite to the antenna, Z1 changes because the antenna is
getting shorter as the ferrite is getting longer. i.e. if there is 7"
of ferrite, then there is only 12" of exposed antenna and it is
elevated over the ground plane so Z1 is going up. In the end case,
when the ferrite is 19" there is no antenna Z1 becomes infinity.

Then looking into the base (thinking as lumped elements), we have Z2 +
Z1. Since Z1 is infinity, the base must look like infinity, but this
does not pass common sense.


In other words, what is the Z looking into a 19" wire that is inside
19" of ferrite. Thinking in lumped element terms, it would be very
high and little power will flow. Thinking in distributed terms there
will be some relatively low Z looking into the base, power will flow
and the ferrite will dissipate heat. The base Z would be related to
some property of the ferrite like the property of free space has a Z
of 377.

What is that propery and what would a typical Z be?

Mark





Ferrites are notoriously non-linear, and thinking of them in
linear terms is liable to lead to disappointment. Look at
the manufacturer's data before you come to any conclusions regarding
how any of them behave.
73,
Tom Donaly, KA6RUH

Tom Donaly wrote:

Ferrites are notoriously non-linear, and thinking of them in
linear terms is liable to lead to disappointment. Look at
the manufacturer's data before you come to any conclusions regarding
how any of them behave.
73,
Tom Donaly, KA6RUH

Good advice. In antenna applications, we need to strive to keep the flux
density low enough that ferrites behave essentially linearly. If we
don't, harmonic generation will result.

Fortunately for us, the flux density decreases as frequency increases,
all else being equal. Also, many ferrites which are commonly used for
common mode chokes (current baluns), EMI filters, and the like will get
hot enough to explode before the flux density gets anywhere near saturation.

Roy Lewallen, W7EL

Roy Lewallen wrote:
Good advice. In antenna applications, we need to strive to keep the flux
density low enough that ferrites behave essentially linearly. If we
don't, harmonic generation will result.
============================
A question : Is the above the reason why current baluns wound on a
ferrite toroid with ,say, a total of 10 windings , can be best made by
having 5 windings wound in 1 direction and the other 5 in the opposite
direction ?

Frank GM0CSZ / KN6WH

Highland Ham wrote:
Roy Lewallen wrote:
Good advice. In antenna applications, we need to strive to keep the
flux density low enough that ferrites behave essentially linearly. If
we don't, harmonic generation will result.
============================
A question : Is the above the reason why current baluns wound on a
ferrite toroid with ,say, a total of 10 windings , can be best made by
having 5 windings wound in 1 direction and the other 5 in the opposite
direction ?

No.

You probably mean "regressive" winding, where you wind half the turns,
cross the wire to the other side of the core, and wind the remaining
turns in the other direction (but the same sense) around the core. If
you wind half the turns in each sense (half where you pass the wire
downward through the center of the core each turn and half where you
pass it upward), you'll end up with nearly zero impedance and a very
poor balun.

The advantage of the "regressive" winding technique is that it reduces
the end-to-end capacitance of the winding. I've found that with high Q
inductors (but ones operating well below self resonance) it typically
improves the Q by around 10 - 15% or so, which is usually not worth the
trouble. With the sorts of ferrites commonly used for baluns, Q is
typically one or less over the operating frequency range, so
"regressive" winding makes no difference at all. In any case, it makes
no difference in core flux density.

Roy Lewallen, W7EL

Roy Lewallen wrote in
:

....
The advantage of the "regressive" winding technique is that it reduces
the end-to-end capacitance of the winding. I've found that with high Q
inductors (but ones operating well below self resonance) it typically
improves the Q by around 10 - 15% or so, which is usually not worth the
trouble. With the sorts of ferrites commonly used for baluns, Q is
typically one or less over the operating frequency range, so
"regressive" winding makes no difference at all. In any case, it makes
no difference in core flux density.
....

Another advantage is that it may be convenient to have the winding end on
opposite sides of the core... often the case for fitting a balun to a box
with input on one side and output on the other.

Owen

Mark wrote:

thanks for the replies...
so for talking purposes:

Z1 is the antenna feedpoint Z and we will define "antenna" as the
exposed wire after the end of the ferrite tube.

Z2 is the Z of the wire passing though the ferrite

Z3 is the source Z which I will stipulate is 50 Ohms

Ok.

OK as we add ferrite to the antenna, Z1 changes because the antenna is
getting shorter as the ferrite is getting longer. i.e. if there is 7"
of ferrite, then there is only 12" of exposed antenna and it is
elevated over the ground plane so Z1 is going up. In the end case,
when the ferrite is 19" there is no antenna Z1 becomes infinity.

Then looking into the base (thinking as lumped elements), we have Z2 +
Z1. Since Z1 is infinity, the base must look like infinity, but this
does not pass common sense.

Why not? A zero length antenna is an open circuit, which has an infinite
impedance. What would you expect the impedance of a zero length antenna
to be?

In other words, what is the Z looking into a 19" wire that is inside
19" of ferrite. Thinking in lumped element terms, it would be very
high and little power will flow.

That's correct.

Thinking in distributed terms there
will be some relatively low Z looking into the base, power will flow
and the ferrite will dissipate heat.

Can you explain how you reach that conclusion?

The base Z would be related to
some property of the ferrite like the property of free space has a Z
of 377.

The ferrite with wire inside comprises half of a circuit -- I posted
more about this a day or so ago. The only reason current flows into an
open-ended wire like a whip antenna -- that is, the only reason the whip
doesn't have an infinite input impedance -- is that the field created by
the alternating current in the wire couples to some other conductor
which is the other half of the circuit. The field induces a current in
that second conductor which flows into the other terminal. And that
current creates a field which couples into the whip, sustaining current
in it. If you could prevent the field from the wire from coupling to the
ground plane, no further current would flow into the wire and it would
indeed look like an open circuit. The ferrite does essentially just this.

What is that propery and what would a typical Z be?

The ferrite has an intrinsic impedance, as does free space and every
other medium. It's the ratio of E to H fields of a TEM wave in the
material. But what does that have to do with this?

Roy Lewallen, W7EL