Re: Equivalent lumped inductance and toroidal coils

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Paul Nicholson" <paul-at-abelian.demon.co.uk> 

[the R=0 problem]

Gerry wrote:

 > H * 2piR  = I    so H = I/(2piR)

 > Certainly this suggest there is a singularity at R = 0, but
 > this is not the case.

I agree with your calculations, and certainly there is no real
singularity because of course the induced voltage in the actual
wire is finite.

 > But once we enter into the wire the enclosed I gets smaller.

You show that as R tends to zero, the enclosed current tends also
to zero, thus avoiding any singularity.  You are invoking a smooth
integral over a continuous current distribution across the area,
so there is no problem.

The singularity I referred to in the other post is an artifact of
trying to represent a finite current inside an infinitesimal
filament radius.   The current density becomes zero everywhere
except inside the filament, where it is infinite.

 > If you continue to want to represent the wire with a bundle of
 > filaments,  you can start the bundle at the center with a triad.
 > Then no filament will have an R=0.

Avoiding the center of the wire (or any other filament arrangement)
doesn't help because you must now take R to be the radius from
each filament, not from the wire centerline.  So you have an R=0
problem for each filament.

 > If one assumed that J was only a function of (r), would
 > this alter the answer that much?  I suspect not.

Probably not, for TC secondaries and normal primaries where the
wire is relatively thin compared to overall dimensions.  I've tried
in tssp modelling to represent the wire with various arrangements
of filaments around the perimeter of the wire cross section, with
little effect on the inductance.  Now I just have one filament down
the center, which corresponds to what you say in

 > ...stop the H field sampling at the surface of the wire since
 > the current density will drop off once the wire is "penetrated"
 > and not worry about the distribution of current within the wire.
 > For purposes of evaluating the differential current element,
 > assume the current is at the center of the wire...

But if we start to look at inductors formed from thick tubes and
wide sheets, that's another story.   We reach a point where it *does*
make a difference where you choose to place the filament.  So then
you have to switch to multiple parallel filaments for the conductor,
which in turn forces you to a network solution to determine the
current branching, perhaps as you say by

 > ...simulate the multiple current path problem in the past using
 > loop equations and iterating numerically until boundary
 > conditions were met.

 > I'm assuming that the alternate paths you are describing are the
 > multiple current filaments that represent the wire and the varying
 > current density contained within the wire.  Is this complexity
 > really necessary?

There are two levels at which this applies:-
1) When a 'wide' conductor is represented by a bundle of filaments
connected in parallel;
2) When we have more than one distinct conductor in parallel.

We need a 'network' solution to decide the current branching,
between the filaments in (1) and between the conductors in (2).

In practice this R=0 issue for filaments isn't much of a problem.
You just have to take the self inductance of each filament from
a closed form self inductance formula, or perhaps simpler still,
just leave it out!   If you've enough filaments, the inductance
of the complete conductor will be dominated by the mutual terms
between each filament since there are N*N mutual inductance but
only N self inductance terms in the calculations.
--
Paul Nicholson
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