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Re: [TowerTalk] Ladderline - what are the facts??/

To: towertalk@contesting.com
Subject: Re: [TowerTalk] Ladderline - what are the facts??/
From: Dan Zimmerman N3OX <n3umh@n3umh.net>
Date: Wed, 6 May 2009 12:35:08 -0400
List-post: <towertalk@contesting.com">mailto:towertalk@contesting.com>
>
> Yes, but consider this.. A long bar of iron.  One winding on one end, the
> other winding on the other.
>

>
> There's no net flux,


First of all, there is net flux due to leakage.  In your case, the flux is
highly nonuniform and so the bar midplane is the *only place* the fluxes are
equal and opposite.


> but the flux has to travel through the core to actually be cancelled
> (whether you conceptualize it as cancelling at the mid point, or sort of
> uniformly along the entire distance, or just as the superposition of the two
> fluxes).


Your example is intuitive, but it only goes to show that there's a lot of
leakage in a long bar with widely seperated windings.  In a toroid with
close windings, the fluxes from differential mode currents are very close to
equal and opposite.

Incomplete cancellation and leakage flux go hand in hand.  There is no core
for which ZERO net flux is set up by real windings, but if there were, there
would really be zero net flux, and the less that excess net flux, the less
the heating from core losses.

I've got a book on transformers here that puts it well when talking about
magnetic circuits: If you ask how good air is as an "insulator" for magnetic
flux as it covers a "wire" of magnetic material, and you draw an analogy to
electrical wire, it's kind of like using salt water as wire insulation.

Yes, most of the current will flow on the wire over short distances, but
there will always be measurable leakage current into the saltwater anyway.
And over long distances, lots of current will take the large-scale saltwater
path rather than the wire path.

The same goes for windings on an iron bar at, say, 60Hz.  With a single
winding on one end, the flux in the bar gets much weaker at one end.  So if
you try to build a common mode choke this way, it's going to suck, because
the differential mode fluxes aren't going to cancel.

You're right that you have to worry about flux leakage issues but if you
control them you get very low core loss.

But we don't have to hash this out theoretically.  All we have to do is
build one of K9YC's recommended chokes with hefty teflon twisted pair
instead of coax and measure the loss.  The form of the common mode conductor
won't be so different.  If anything, common mode turn to turn capacitance
should be less.  So choking impedance should be similar.  The only
difference is that the differential mode vs. common mode seperation isn't
enforced by skin effect of a shield.

Then we just run a few kW through it and see if the core heats up, and we
measure how much power comes out the other end.

If you do this with a long ferrite bar with widely seperated windings it's
going to suck as a common mode choke with little differential effect, but a
teflon pair on a toroid will NOT be a problem.

Let's just go try it.  See if you can get an amount of core heating due
purely to differential mode losses consistent with 1dB loss in a lossy core
material. If you do it at 1.5kW, it should be quite dramatic.  You dissipate
a few hundred watts in even a big toroid in steady state, it's going to be
too hot to touch!


73
Dan
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