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

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