Richard Clark wrote:
On Tue, 28 Mar 2006 22:33:18 -0500, John Popelish
wrote:
(snip)
Make that, "impedance (mostly inductive, if the ferrite is well suited
to the frequency)" and I agree.


Hi John,

No, the principle of the impedance is overwhelmingly resistive, not
reactive. One core (#75) that snuggly fits over RG58, at about 10MHz,
exhibits 25 Ohms resistance to common mode current.

Yep. Type 75 operated at 10MHz is very lossy.

My BalUn uses 50
such whose combination presents 1250 Ohms resistance. This value
varies over frequency, but easily finds a 1000:50 ratio over the
entire MF/HF Ham spectrum.

Thank you for some specific facts. I had no idea that you were using
a ferrite whose permeability falls to half the low frequency value
(about 5000 at 200kHz) at only 700kHz and is not recommended for flux
coupling applications higher than about 750kHz. It also has a very
low bulk resistivity of 300 ohm cm. At 3 MHz, its permeability is
only about 150 (and plunging).

Type 43 material has a permeability of about 500 at 3 MHz and falling
much more slowly.

I also have two engineering samples boxes full of various material
types and geometries and have measured them all. Some exhibit
reactance, but are vastly overwhelmed by their intrinsic resistance.

(snip)
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.

Quite the contrary. Inductance, or more properly to the application,
reactance, may come as a consequence of these cores, but that is
really just a minor component to the vastly greater resistance.

If the cores don't have relative hi permeability (causing mutual
inductance in the two conductors passing through) they can't produce
any impedance in series with the common mode current. That is also
transformed over to the two conductors by mutual inductance (as a
loaded secondary reflected to those conductors as a primary).

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

That power, actually current, was forced into a common mode because of
one of several problems:
1. Massive imbalance in the radiator design,
2. Massive catastrophe with the radiator.

If you are destined to "lose a lot of power" into such resistance, you
have far greater problems you are ignoring. However, let's return to
this "loss of a lot of power." You have an antenna with a nominal 50
Ohm load resistance (presumably balanced to the common). You have a
common mode choke with 1000 Ohms resistance in parallel (to the common
of the common mode). Now, why would the power choose the path of 10
to 40 times more resistance to invest its calories in?

I will be generous and pour 1.5KW into a balanced 50 Ohm load to
develop 275V that is also across the 1000 Ohms of the BalUn. 76W into
50 beads. Now this verges on serious power for this load I will
admit, but it also presumes key-down conditions which would only be
found for RTTY or SSTV. By my bench measurements, I figure that the
core's can tolerate as much power as the bulk equivalent carbon
composition resistor - or least this is my rule of thumb. This means
about 1 or 2W per bead to offer enough heat to cause pain, but perhaps
not enough to blister. This is another benchmark from my days in the
Navy.

If the radiator is perfectly balanced, the impedance of the ferrite
(transformed to the primary side) sees half of the voltage that is
between the center conductor and the shield. So if the coax carries
275 volts, half of that or about 1378 volts appears across the bead
bead common mode impedance. So a 1000 ohm common mode choke limits
the common mode current to 138/1000 = 138 mA. And if that impedance
is dominated by resistance, the beads will absorb 138*.138=19 watts.
50 beads will stand that indefinitely.

I don't think the O.P.'s balun has 50 beads, though.

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.


Your count is four terminals (not conductors), there are six. You
need to come to terms with this shortfall in your count as it explains
the utility and design of BalUns.

If the other two are the ends of the loss resistance, they can be
transformed back to the two conductors.
(snip)

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.


This "rate of change" is spurious, call it frequency dependant, but
then frequency dependency is neither here nor there at this moment.
More to the matter trying to turn this BalUn into a power transformer
is doomed in this analysis as it has absolutely no impact to

It may not be the approach you are familiar with, but I think it is
valid. The transformer just has a lot of core loss, if you use a
ferrite optimized for a much lower frequency. And that loss shows up
as if it were a resistor connected across the ends of the two
windings. If a low loss ferrite (at the operating frequency) is used,
then the impedance across the windings is dominated by inductance, as
one normally expects with a transformer. But the analysis handles the
whole range of cases.

On Wed, 29 Mar 2006 10:45:05 -0500, John Popelish
wrote:

It may not be the approach you are familiar with, but I think it is
valid. The transformer just has a lot of core loss, if you use a
ferrite optimized for a much lower frequency. And that loss shows up
as if it were a resistor connected across the ends of the two
windings. If a low loss ferrite (at the operating frequency) is used,
then the impedance across the windings is dominated by inductance, as
one normally expects with a transformer. But the analysis handles the
whole range of cases.

Hi John,

Through these last comments, and those that go before, you have
entirely missed the boat of both the dynamics involved (this is not a
magnetic circuit being described, and there are NO magnetic lines
broken in a typical circuit in a balanced configuration), and the
topology. Specifically to this last, you still do not seem to
comprehend that a coax has three conductors and is a six terminal
device.

Roy has also commented to these issues and I think you should review
his correspondence and respond to him.

73's
Richard Clark, KB7QHC

On Wed, 29 Mar 2006 17:01:51 GMT, rocky wrote:

To sort of sum things up, then what should I use for a choke balun on
75 meters, 77, 73 or 43 mix? More the better? I read the W2DU article
and he used 73, but I thought 77 may be better?

Hi OM,

More is better? What do you seek to achieve that is not already
answered by Walt's article? Didn't he specify which type?

#64 material is going to offer about 3 Ohms per bead;
#43 material is going to offer about 8 Ohms per bead;
#73 & 77 material is going to offer about 18 Ohms per bead;
#75 material is going to show offer 28 Ohms per bead.

If you are running power into a matched load, you may want more of the
lower resistance beads. If you are running barefoot, fewer more
resistive beads will work.

Or you could do it the old-fashion way by looping your coax at the
drive point.

73's
Richard Clark, KB7QHC

John Popelish wrote:
. . .
It may not be the approach you are familiar with, but I think it is
valid. The transformer just has a lot of core loss, if you use a
ferrite optimized for a much lower frequency. And that loss shows up as
if it were a resistor connected across the ends of the two windings. If
a low loss ferrite (at the operating frequency) is used, then the
impedance across the windings is dominated by inductance, as one
normally expects with a transformer. But the analysis handles the whole
range of cases.

Broadband transformers, which can operate well over several decades of
frequency, commonly use ferrite cores which are essentially resistive
over most of the operating frequency range. The sign of the impedance is
unimportant to the transformer's operation; all that's necessary is that
its magnitude be adequately high over the operating range (and of course
that the core's permeability be adequately high). The wide band high
impedance requirement is virtually impossible to meet with an inductive
core whose impedance is approximately proportional to frequency, but
easily done with cores whose impedance is essentially resistive.

Roy Lewallen, W7EL

On Wed, 29 Mar 2006 18:50:04 GMT, rocky wrote:

I run up to a KW, dipole and could be using an antenna tuner to stretch
the bandwidth. I was just thinking if it could be made even more
effective by using other bead material. Looks like #75 offers the best
attenuation at 75 M. but if I am getting this right, it will heat up
more under high power and off resonance.

Hi OM,

Think of the beads as series 1W resistors. Ask yourself how much of a
common mode load they will present to the antenna "as a source" (this
means that the dipole will look like a KW source with complex
impedance).

You then have to judge the potential that source will present in
Common Mode to the beads. Elsewhere in this thread I've done a
simplified analysis that suggests 50 #75 beads will get a tad hot
under matched conditions because they are dissipating roughly 1.5W
each. "Hot" is of little consequence until we shatter the ceramic
because of thermal stress (the sintered product contains bulk
irregularities). That stress could be arguably applied through a rain
drop hitting a bead. Then we get into new issues of balance and
symmetry because one part of the bead is cooler than the other. ;-)

If, instead for this matched load, you replace the 50 #75 beads with
80 #73 or #77 beads; then you have roughly the same total resistance,
but the heat dissipation (still 1W suggested load for each) is spread
out more. Again, moving off resonance demands another analysis.

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Wed, 29 Mar 2006 10:45:05 -0500, John Popelish
wrote:


It may not be the approach you are familiar with, but I think it is
valid. The transformer just has a lot of core loss, if you use a
ferrite optimized for a much lower frequency. And that loss shows up
as if it were a resistor connected across the ends of the two
windings. If a low loss ferrite (at the operating frequency) is used,
then the impedance across the windings is dominated by inductance, as
one normally expects with a transformer. But the analysis handles the
whole range of cases.


Hi John,

Through these last comments, and those that go before, you have
entirely missed the boat of both the dynamics involved (this is not a
magnetic circuit being described, and there are NO magnetic lines
broken in a typical circuit in a balanced configuration), and the
topology. Specifically to this last, you still do not seem to
comprehend that a coax has three conductors and is a six terminal
device.

I appreciate you taking the time and effort to try to straighten me
out on this, but if there is no magnetic lines broken (whatever that
means) why use a magnetic core? Why wouldn't disks of carbon work
just as well. They are certainly resistive.

John Popelish wrote:
. . .
I appreciate you taking the time and effort to try to straighten me out
on this, but if there is no magnetic lines broken (whatever that means)
why use a magnetic core? Why wouldn't disks of carbon work just as
well. They are certainly resistive.

But resistive impedance isn't the only characteristic of ferrite. Disks
of carbon won't work because they have a relative permeability of one.

Fair-Rite type 73 ferrite has a Q of 1 (R = X) at about 3 MHz, and is
essentially resistive above that; it's a "low frequency" ferrite. But
its relative permeability is 700 at 10 MHz and 70 at 100 MHz. (Note that
the permeability is a complex quantity, and that the permeability is
mostly imaginary, hence the impedance resistive, at the higher
frequencies.) Carbon has a relative permeability of one at all
frequencies. The impedance of a type 73 ferrite core picked at random
from the Fair-Rite data shows an impedance of 23 ohms at 3 MHz, 37 at
10, 51 at 100, and 49 at 200 MHz. Try making a carbon disk that'll give
you that impedance. And if you make your baluns with multiple turns on a
single core as I do, you'll have much poorer coupling between turns on a
carbon core, than on ferrite.

You'll learn a lot by reading the information in the Fair-Rite catalog,
available at their web site. Good, professionally written information on
EMI suppression is useful also. But I don't recommend looking to amateur
publications and web sites for information on ferrites -- too many
people share the same misconceptions you do, and pass them along without
really understanding how transformers work and the properties of
ferrites. Actually, a large number of engineers share those
misconceptions, too. It's often one of the first things I have to
explain to the engineers in my consulting work.

Roy Lewallen, W7EL

On Wed, 29 Mar 2006 15:36:08 -0500, John Popelish
wrote:

I appreciate you taking the time and effort to try to straighten me
out on this, but if there is no magnetic lines broken (whatever that
means) why use a magnetic core? Why wouldn't disks of carbon work
just as well. They are certainly resistive.

Hi John,

Breaking magnetic lines (flux) is a commonplace of fields, motors, and
generators. A single wire that passes through a bead, torus, or core
will build a magnetic field concentrated within that structure when
the circuit is completed outside of it. The flux lines of half the
loop will penetrate the core to reach the other half of the loop. The
core breaks the magnetic line of flux. The dissymmetry of penetration
builds a magnetic field in the core.

However, when the complete current loop is within the same structure,
the flux lines do not fulfill that same function. The flux lines that
do emerge from the tightly bound wires can be said to penetrate the
torus, but here the symmetry creates bucking fields, the net effect is
as though there was no core at all (except to add capacitance).

Both models attempt to stimulate a current within the toroid, the
common mode of the single wire model above is lossy, the differential
mode of the twin line model that followed sees nothing. Superpose
these two for the coaxial solution.

To put this to a test. Load up your rig, through a SWR meter to a
dummy load using two short connection wires (this will undoubtedly
require adapters and such to break out both paths). You should note a
1:1 indication. Place two #75 beads on ONE wire. You should note a
2:1 indication. You have just inserted 40 to 60 Ohms of additional
resistance into the circuit. Now, move the same two beads to
encompass BOTH wires. This should return the SWR meter to a 1:1
indication.

73's
Richard Clark, KB7QHC

rocky wrote:
To sort of sum things up, then what should I use for a choke balun on
75 meters, 77, 73 or 43 mix? More the better? I read the W2DU article
and he used 73, but I thought 77 may be better?

If you are going to have to go out and buy the cores, type 43 (and its
competitors) is way more common and cheap than any of the others. It
is also readily available in long form toroids called shield beads.

For instance, Digikey sells the Steward version of type 43 (called
type 28) beads part number 28B1122-100, with a 13.7mm hole and 28.6mm
long. This part produces about 60 ohms per bead at 4 MHz and costs
$20 for 10 pieces.
http://www.steward.com/

What diameter coax do you want to use a choke BaLun on? Generally the
smaller hole size, the more impedance you get out of a given length core.

Richard Clark wrote:
(snip)
#64 material is going to offer about 3 Ohms per bead;
#43 material is going to offer about 8 Ohms per bead;
#73 & 77 material is going to offer about 18 Ohms per bead;
#75 material is going to show offer 28 Ohms per bead.
(snip)

What dimension cores produce these impedances at 75 meters?