Richard Clark wrote:
On Tue, 28 Mar 2006 12:05:01 -0500, John Popelish
wrote:


But a choke with two wires wound through it is two inductors that also
have mutual inductance between them, and if that doesn't define a
transformer, what does?


Hi John,

Then you think of it as an air core transformer with series driven,
bucking sections. Now, what kind of practical transformer does that
define? 1:1 does not automatically spring to mind unless it is
isolating one circuit from the other. However, it is not the
transformer that does that, it is the choking ferrite and only in the
service of snubbing common mode currents.

With this in mind, do we add a characteristic of loss to the
definition? A lossy air core transformer with series driven, bucking
sections.

Air core? It is a ferrite core transformer with two one turn
windings. One winding is the shield passing through the holes in the
cores and the other winding is the center conductor passing through
the windings. (view with fixed width font, like Courier)

|half of dipole
. |
Center cond.-----MMMM-+

I mainly wanted to know if it would work on 160 & 80m, I dont care
if it acts like a choke/balun only at higher frequencys as long as
as it doesnt interfere with 160 & 80s swr. Normally i just use coax
hooked to each leg cut to frequency, without a balun. Its allways
worked in the past years, i just thought it would be alot easier to use
the ready made balun than try and makeup a weatherproof
hanger

In article . com,
wrote:

I mainly wanted to know if it would work on 160 & 80m, I dont care
if it acts like a choke/balun only at higher frequencys as long as
as it doesnt interfere with 160 & 80s swr.

It should not interfere. If the ferrites don't have enough reactance
to make it act as an effective choke on 160/80, then it'll just "look
like" a foot or two of coax to the signals.

--
Dave Platt AE6EO
Hosting the Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

wrote:
I mainly wanted to know if it would work on 160 & 80m, I dont care
if it acts like a choke/balun only at higher frequencys as long as
as it doesnt interfere with 160 & 80s swr. Normally i just use coax
hooked to each leg cut to frequency, without a balun. Its allways
worked in the past years, i just thought it would be alot easier to use
the ready made balun than try and makeup a weatherproof
hanger.

So what ohm reading do you read between two points on one of the cores?

John I will have to take it down and measure it, im going to do it
anyway because the balun will not load up like it should even with
wire on each leg cut to the proper length. I dosent dip with the dip
meter at the proper freq either, after i take it down ill post the
ohmeter results.

On 28 Mar 2006 11:02:44 -0800, wrote:

I dont care
if it acts like a choke/balun only at higher frequencys as long as
as it doesnt interfere with 160 & 80s swr.

Hi OM,

There is a world of improvement of understanding BalUns slated for
you, but
Its allways worked in the past years
suggests you should skip that line of inquiry and simply go back to
working your scheds. If you encounter problems, then is the time to
ask questions.

73's
Richard Clark, KB7QHC

On Tue, 28 Mar 2006 12:59:52 -0500, John Popelish
wrote:
With this in mind, do we add a characteristic of loss to the
definition? A lossy air core transformer with series driven, bucking
sections.

Air core? It is a ferrite core transformer with two one turn

Hi John,

If there's a transformer in the sense of windings; then it is an air
core, the ferrite is wholly transparent to the transverse currents.
You could remove the ferrite and it wouldn't make a bit of difference
in that sense of transforming.

this current mismatch would cause the transformer to
produce more or less voltage across the windings
In fact, nothing of that sort happens - at least not by your
description. The ferrite is simply bulk resistance inserted into the
common mode path. That is why common mode current is suppressed. The
same thing occurs in the coiled transmission line choke, but the
resistance is replaced by reactance. Again, common mode current is
snubbed by encountering this too.

The transformer property is in the isolation of the balanced circuit
from the unbalanced circuit through this resistive characteristic.

You are missing one path. The two from the source in the form of the
inner shield of the coax, and the center conductor, and the one from
the load in the form of the outer shield of the coax (same shield, but
isolated circuits). Further, there is no flux linkage of the two
conductors coming from the source. Their magnetic lines never break
the cores, whereas the common mode current does break the core which
thus inserts the resistance of the ferrite.

73's
Richard Clark, KB7QHC

On Tue, 28 Mar 2006 17:39:52 -0800, Richard Clark
wrote:

Further, there is no flux linkage of the two
conductors coming from the source.

That was not correctly expressed, the flux between the two are tightly
bound and:
Their magnetic lines never break
the cores, whereas the common mode current does break the core which
thus inserts the resistance of the ferrite.

Richard Clark wrote:
On Tue, 28 Mar 2006 12:59:52 -0500, John Popelish
wrote:

Air core? It is a ferrite core transformer with two one turn

Hi John,

If there's a transformer in the sense of windings; then it is an air
core, the ferrite is wholly transparent to the transverse currents.

I said that. It is a common mode transformer.

You could remove the ferrite and it wouldn't make a bit of difference
in that sense of transforming.

The short length of the two conductors passing through the ferrite is
certainly a poorer transformer (less mutual inductance between them)
if you remove the cores.

this current mismatch would cause the transformer to
produce more or less voltage across the windings

In fact, nothing of that sort happens - at least not by your
description. The ferrite is simply bulk resistance inserted into the
common mode path.

Make that, "impedance (mostly inductive, if the ferrite is well suited
to the frequency)" and I agree.

That is why common mode current is suppressed. The
same thing occurs in the coiled transmission line choke, but the
resistance is replaced by reactance. Again, common mode current is
snubbed by encountering this too.

I agree with this, except that the purpose of the ferrite is to
increase the common mode inductance of the section of coax passing
through it, not add resistance. Some resistance is inevitable,
because no ferrite is lossless, but the intention is for inductance.

The transformer property is in the isolation of the balanced circuit
from the unbalanced circuit through this resistive characteristic.

Try transmitting through such a resistance and you are going to lose a
lot of your power.

You are missing one path. The two from the source in the form of the
inner shield of the coax, and the center conductor, and the one from
the load in the form of the outer shield of the coax (same shield, but
isolated circuits).

I can't parse this. There are two metal conductors entering the
choke, and two exiting it. All currents pass through those 4 conductors.

Further, there is no flux linkage of the two
conductors coming from the source. Their magnetic lines never break
the cores,

I think you mean by this that a normal unbalanced signal in a coax has
no magnetic field external to the shield. It is all between the
center conductor and the shield. And I agree that this is what you
are trying to accomplish by adding this two conductor choke between
the coax and the balanced antenna. Without it, there would be some
magnetic field from a net (uncanceled) current and voltage on the
outside of the shield that would cause the coax to radiate. And the
voltages and currents fed to the balanced antenna would not be equal
and opposite (balanced) but somewhat unbalanced. There would also be
non equal currents in the center conductor and shield. I think we
agree on all that, but have a different picture of how a choke balun
corrects these problems.

whereas the common mode current does break the core which
thus inserts the resistance of the ferrite.

The common mode current causes flux in the core, and the conductors
passing through that flux produce a voltage proportional to the rate
of change of that flux, just as the conductor passing through any
inductor would. The transformer aspect is that since both conductors
pass through the exact same rate of change of flux, there is the same
voltage produced at the ends sticking out of the core, and this
voltage gets algebraically added to what is already there. If the
inductance of each winding is high enough (5 to 10 times the coax
impedance) a very small common mode current is enough to produce a
large enough voltage across the ends to the two conductors to correct
most of the unbalanced to balanced coupling.

Admittedly, there is no need to get this inductance (including mutual
inductance) with the aid of ferrite around the coax. You could just
wind the coax into an air core transformer. But it would be
considerable larger than one made with a high permeability core,
though, probably lower loss.

John Popelish wrote:
Richard Clark wrote:
. . .

In fact, nothing of that sort happens - at least not by your
description. The ferrite is simply bulk resistance inserted into the
common mode path.

Make that, "impedance (mostly inductive, if the ferrite is well suited
to the frequency)" and I agree.

"Low frequency" ferrites are well suited to applications as HF chokes
and broadband transformers. What you don't want to do is use them for a
high Q inductor in a filter, tuned circuit, or similar application.

That is why common mode current is suppressed. The
same thing occurs in the coiled transmission line choke, but the
resistance is replaced by reactance. Again, common mode current is
snubbed by encountering this too.

I agree with this, except that the purpose of the ferrite is to increase
the common mode inductance of the section of coax passing through it,
not add resistance. Some resistance is inevitable, because no ferrite
is lossless, but the intention is for inductance.

Baluns work fine with a resistive impedance, with the exception of
applications involving large power. In fact, resistive impedance is
desirable because the impedance changes little with frequency, and is
relatively free of resonance effects. (More below.)

The transformer property is in the isolation of the balanced circuit
from the unbalanced circuit through this resistive characteristic.

Try transmitting through such a resistance and you are going to lose a
lot of your power.

You're not "transmitting through" a balun's impedance. Only the common
mode current effectively flows through it, and the power dissipated by
the balun is Icm^2 * R, where Icm is the common mode current and R is
the resistive part of the balun's common mode impedance. If R is small,
dissipation is low. But if R is large, that makes Icm small, so
dissipation is also low. It's really an impedance matching problem when
the balun is resistive -- a very low or very high balun R results in low
dissipation. Dissipation is maximum at some intermediate value of R and
decreases on each side.

A typical balun made with "low frequency" ferrite (e.g. Fair-Rite 70
series) and operating at HF or above (and therefore primarily resistive)
having a common mode impedance of 500 ohms or greater generally won't
dissipate any significant fraction of the transmitted power. However, if
you're running high power, even a fraction of a dB dissipated in the
balun will cause it to overheat. Consequently, people running high power
often resort to type 43 ferrite (a Fair-Rite designation; or its
equivalent from other manufacturers), which is less resistive than lower
frequency ferrites. In extreme cases, high frequency (60 series) ferrite
is necessary. The problem is that it's increasingly difficult to get
adequate impedance with the higher frequency ferrites. Type 43 is often
a good compromise, and it's widely available in many core sizes.

. . .
I think you mean by this that a normal unbalanced signal in a coax has
no magnetic field external to the shield. It is all between the center
conductor and the shield. And I agree that this is what you are trying
to accomplish by adding this two conductor choke between the coax and
the balanced antenna. Without it, there would be some magnetic field
from a net (uncanceled) current and voltage on the outside of the shield
that would cause the coax to radiate. And the voltages and currents
fed to the balanced antenna would not be equal and opposite (balanced)
but somewhat unbalanced. There would also be non equal currents in the
center conductor and shield. I think we agree on all that, but have a
different picture of how a choke balun corrects these problems.
. . .

Common and differential mode currents are physically separated in a coax
cable, and so are the fields from the two components, providing that the
shield is at least several skin depths thick. The differential mode
current and its fields are entirely inside the coax, decaying rapidly as
you go outward from the inner boundary of the shield. By the time you
reach the outer boundary of the shield the fields from the differential
current is negligibly small. So any core you put over the coax doesn't
see or interact with the common mode current or its fields at all, and
you can completely ignore it when analyzing balun action. Similarly, you
can ignore the core when analyzing the differential mode properties of
the system. The common mode current resides in a thin layer on the
outside of the shield, it and its fields never reaching the inside. The
balun provides an impedance to this current just as it would to any
current on the outside of a conductor.

When bifilar wound, the fields from the differential mode current are
primarily between the turns, although some relatively small amount
extends beyond to interact with the core. The net result is nearly the same.

. . .

Roy Lewallen, W7EL