The best way to ask my question is to use a thought experiment.

Imagine I have a bucket of ferrite rings. Assume that the permeability
and other material properties of each ring is such that I get about
100 Ohms of resistive loss for one turn through a ring at 144 MHz.

Now assume I have 19" whip on top of a car connected to a 144 MHz
transmitter. If I slip one ring over the whip and let it down to the
bottom feedpoint, it will attenuate the signal by adding 100 Ohms in
series with the feed of the antenna which otherwise would look like
about 30 Ohms. That seems clear and seems to agree with common sense
and observations.


Now if I stack up 20 ferrite rings on the antenna, if we analyze each
ring as a lumped element we would have 2000 Ohms in series and
essentially an open circuit and little current or power will flow.

However this does not pass the common sense test. Common sense says
the ferrite tube will absorb the energy from the antenna and
potentially will get hot and that the feedpoint Z of the antenna will
NOT raise indefinitely with more and more ferrites but rather it seems
it would level out to some value. We need to use distributed rather
than lumped analysis.

Bottom line question... How to determine the feedpoint Z of a wire
that is inside a ferrite tube?


thanks
Mark

Mark wrote:
The best way to ask my question is to use a thought experiment.

Imagine I have a bucket of ferrite rings. Assume that the permeability
and other material properties of each ring is such that I get about
100 Ohms of resistive loss for one turn through a ring at 144 MHz.

Now assume I have 19" whip on top of a car connected to a 144 MHz
transmitter. If I slip one ring over the whip and let it down to the
bottom feedpoint, it will attenuate the signal by adding 100 Ohms in
series with the feed of the antenna which otherwise would look like
about 30 Ohms. That seems clear and seems to agree with common sense
and observations.


Now if I stack up 20 ferrite rings on the antenna, if we analyze each
ring as a lumped element we would have 2000 Ohms in series and
essentially an open circuit and little current or power will flow.

However this does not pass the common sense test. Common sense says
the ferrite tube will absorb the energy from the antenna and
potentially will get hot and that the feedpoint Z of the antenna will
NOT raise indefinitely with more and more ferrites but rather it seems
it would level out to some value. We need to use distributed rather
than lumped analysis.

Yes. The effect of a ferrite ring part way up the antenna won't be the
same as one at the base. A 100 ohm resistor in the center of the
vertical will add 50 ohms at the feedpoint. One 3/4 of the way to the
top will add about 17.5 ohms.

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.

Bottom line question... How to determine the feedpoint Z of a wire
that is inside a ferrite tube?

You could probably do a pretty decent job with an antenna modeling
program, placing a number of lumped loads along the wire.

Roy Lewallen, W7EL

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

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

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

On 16 oct, 23:02, Mark wrote:
The best way to ask my question is to use a thought experiment.

Imagine I have a bucket of ferrite rings. Assume that the permeability
and other material properties of each ring is such that I get about
100 Ohms of resistive loss for one turn through a ring at 144 MHz.

Now assume I have 19" whip on top of a car connected to a 144 MHz
transmitter. If I slip one ring over the whip and let it down to the
bottom feedpoint, it will attenuate the signal by adding 100 Ohms in
series with the feed of the antenna which otherwise would look like
about 30 Ohms. That seems clear and seems to agree with common sense
and observations.

Now if I stack up 20 ferrite rings on the antenna, if we analyze each
ring as a lumped element we would have 2000 Ohms in series and
essentially an open circuit and little current or power will flow.

However this does not pass the common sense test. Common sense says
the ferrite tube will absorb the energy from the antenna and
potentially will get hot and that the feedpoint Z of the antenna will
NOT raise indefinitely with more and more ferrites but rather it seems
it would level out to some value. We need to use distributed rather
than lumped analysis.

Bottom line question... How to determine the feedpoint Z of a wire
that is inside a ferrite tube?

thanks
Mark


Hi Mark,

Good question, but difficult answer. I think the best way is to see
the construction as a transmission line (as you suggested) . If the
ferrite should have no losses, the inductance/m rises significantly.
The capacitance/m rises also (because of the eps.r of the ferrite
material).

So you will end up with a high Z transmission line with significantly
lower propagation speed. A difficulty is: where is the return
conductor? You can assume the return conductor about 0.125 lambda
away when the influence of the ferrite is not that large. Very close
to the feedpoint, the impedance of the ferrite loaded line is strongly
dependent on the distance from the feedpoint.

When the transmission line would be loss free, the impedance at the
feedpoint is strongly dependent on the termination (because all
energy may be reflected back). This is in many (wide band) cases
undesirable. It is the reason that I do not recommend 4C65 material
for current mode baluns at low frequency (the Q of the 4C65 ferrite
material is rather high at, for example, 80m).

Sometimes we can be happy that ferrite has (some) losses. In that case
the forward and reflected wave both suffer losses and the input
impedance becomes less sensitive to load variations at the other side
of the one turn ferrite inductor. So practically spoken, yes, when
you add infinite ferrites, the impedance will reach a finite value.

In a real world, the situation is more complicated, because when using
High permeability cores, you also have high permittivity. The
propagation speed in the ferrite material itself can be that low that
the flux density distribution is no longer uniform. This causes the
effective permeability to change faster then expected based on the
datasheet of the ferrite material. Above a certain size, adding
thicker ferrites (in outer diameter) does not increase the inductance/
loss.

When you would take a long thick ferrite tube, the field will even not
come out of the ferrite because of the attenuation in combination with
the low EM wave propagation speed inside the ferrite. Ferrite
absorbing tiles also make use of both the epsilon.r and permeability.

So finally I did not answer your question about how to determine the
feedpoint impedance. Probably you will need full 3D EM software that
can handle 3D structures of different mu and eps. As far as I know,
all momentum based EM software cannot handle this.

Hope this will help a bit.

Wim
PA3DJS
www.tetech.nl

Wimpie wrote:

So you will end up with a high Z transmission line with significantly
lower propagation speed. A difficulty is: where is the return
conductor? You can assume the return conductor about 0.125 lambda
away when the influence of the ferrite is not that large. Very close
to the feedpoint, the impedance of the ferrite loaded line is strongly
dependent on the distance from the feedpoint.
. . .

The second conductor of the transmission line is the ground plane.

When you would take a long thick ferrite tube, the field will even not
come out of the ferrite because of the attenuation in combination with
the low EM wave propagation speed inside the ferrite. Ferrite
absorbing tiles also make use of both the epsilon.r and permeability.

So finally I did not answer your question about how to determine the
feedpoint impedance. Probably you will need full 3D EM software that
can handle 3D structures of different mu and eps. As far as I know,
all momentum based EM software cannot handle this.

I don't believe that's necessary. As you say, the field doesn't escape
the ferrite. This means that the E field between transmission line
conductors (the whip and ground plane) is zero. Therefore, the impedance
of the line (E/H) is infinite. The loss of the ferrite is adequate to
suppress any reflections from the end of the line. A reflectionless,
infinite impedance line will have an infinite input impedance. This is,
as it should be, the same result I got from a somewhat different
perspective.

Roy Lewallen, W7EL

On Oct 19, 4:12 am, Roy Lewallen wrote:
Wimpie wrote:

So you will end up with a high Z transmission line with significantly
lower propagation speed. A difficulty is: where is the return
conductor? You can assume the return conductor about 0.125 lambda
away when the influence of the ferrite is not that large. Very close
to the feedpoint, the impedance of the ferrite loaded line is strongly
dependent on the distance from the feedpoint.
. . .

The second conductor of the transmission line is the ground plane.

When you would take a long thick ferrite tube, the field will even not
come out of the ferrite because of the attenuation in combination with
the low EM wave propagation speed inside the ferrite. Ferrite
absorbing tiles also make use of both the epsilon.r and permeability.

So finally I did not answer your question about how to determine the
feedpoint impedance. Probably you will need full 3D EM software that
can handle 3D structures of different mu and eps. As far as I know,
all momentum based EM software cannot handle this.

I don't believe that's necessary. As you say, the field doesn't escape
the ferrite. This means that the E field between transmission line
conductors (the whip and ground plane) is zero. Therefore, the impedance
of the line (E/H) is infinite. The loss of the ferrite is adequate to
suppress any reflections from the end of the line. A reflectionless,
infinite impedance line will have an infinite input impedance. This is,
as it should be, the same result I got from a somewhat different
perspective.

Roy Lewallen, W7EL

Thanks for the replies..

so I think you are saying that the 19" wire inside a 19" tube of
ferrite will present a Hi Z a the base and will therefore consume
little power i.e. it looks like an open circuit...

Does that seem resonable to you? It's the the same result as if the
ferrite were copper. i.e a cavity resonantor. I know the answer is
correct for copper but copper is not lossy.

From the other perspective, the ferrite is lossy so it will be
absorbing energy and dissipating heat and therefore the base must look
like SOME impedance that has an R comonent.

So I think the more fundamenal nature of my questoin is about how a
wire radiates in concert with the return current to the ground plane.
What if the wire is surrounded by a lossy medium such that the field
is greatly attenuated before it can get to the ground. How does the
return current flow and how does energy get absorbed by the lossy
medium?

Mark