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