Re: Equivalent lumped inductance and toroidal coils

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

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

Gerry Reynolds wrote:

 > ... integrate the differential current element over the total
 > wiring path to evaluate the H field at a sampling of points
 > within the cross section of that turn.

Ok.

 > The H field at the sample point would be:
 > H(r,phi,z) =
 >   integral [wire beginning to wire ending] {(Ids X R}/4piR^2}

Ok.

 > Once the H field is sufficiently sampled, the total flux
 > threading that turn can be evaluated by integrating B.dA

Ok.

Yes, curl E = -dB/dt at every point across the area enclosed by
the sample turn.  Integrate both sides over the area.  On the
left, area_integral{curl E} becomes line_integral{E} along the turn
(because the overall effect of lots of little rotations across an
area adds to an overall circulation around the perimeter).

 > The total flux thru a turn from all the current elements would
 > then include self and mutual coupling ...

The difficulty comes with the self inductance - some of the sample
points approach the source wire so R tends to zero in the
denominator in the integral for H.   Antonio mentioned a trick
by Maxwell to get around this:

 > Self-inductances can be calculated as the mutual inductance between
 > two toroidal coils placed at a certain distance, that depends on the
 > radius of the wire (Maxwell's gmd method)

The tedious bit is integrating over the area of the sample turn.
Wouldn't it be easier just to integrate around the perimeter
instead?   You can do this by using the vector potential A, for
which curl A = B at every point in space.   Doing that same
trick of integrating a curl field to get the total rotation
around its perimeter, you get

  area_integral( curl A) = line_integral(A) = area_integral(B).

So if you just work out the line integral of A around the sample
turn, you get the same answer as the area integral of B across
the turn.

But that means you have to work out A from the source integral
instead of B.  Do that with

  A(r,phi,z) =
    integral [wire beginning to wire ending] {(Mu * I * ds}/(4*pi*R)}

for every point on the perimeter of the sample turn.

Then you end up with a line integral over a line integral instead
of a surface integral over a line integral.  If you write out this
double integral in full, you get the mutual inductance coefficient as

  M = Mu /(4*pi) *
       line_integral[s along source wire]{
           line_integral[t along sample turn]{
              ds.dt/R(s,t)
           }
       }

This is Neumann's integral which is the starting point for the
closed form solutions and for the brute force computations. (As a
non-mathematician I tend to favour the latter!)  Situations where
the integral is tractable in closed form are hard to find.  The
trick is to find a coordinate system or substitution against which
the integral can be solved.

Of course, in the numerical solutions, you still have the problem
of R=0 when the source and sample elements are the same.  Your only
recourse is to replace this contribution to the integral with the
closed form value for the self inductance of the element, or to
apply Maxwells trick.

(Use of the vector potential here instead of B saves a bit of work.
As a general rule, try to solve problems using potentials instead
of field strengths because it's usually less work to calculate.
And make full use of the various vector identities and integral
laws wherever you can to turn volume integrals into area integrals,
and area integrals into line integrals, etc.)

Bart wrote:

 > As for measurements, I agree that it would be great if Jarad could
 > perform some of these since his coils are already wound. If he
 > decides to do so, I hope he would seek measurement advice prior to
 > taking measurements. That's something I always like to do to ensure
 > everyone is on the same page.

I agree.  Plenty of expertise to draw on here in the list so if
someone keeps the list informed of what they're doing, they're sure
to be alerted pretty quickly if something's gone awry!

It would be nice to see how the velocity factors go (for reasons
discussed in a nearby thread), what the mode spectrum looks like,
what the Q factors and inductances are, and so on.  But I don't see
any application for these coils.

Shaun Epp wrote:
 > In this setup, a toriodal secondary with an air core,  most
 > magnetic flux generated by the primary will be lost...
 > In a toriodal transformer, the core confines and directs the fields,
 > but in a toriodal secondary, there is nothing to make the magnetic
 > field cut across all of the turns of the secondary.

Yes, we'd expect the flux linkage to be poor.  But I think the field
outside the toroidal winding is supposed to be zero everywhere (even
without a magnetic core) due to contributions from different parts of
the winding magically cancelling one other out.  So at low frequencies
(uniform current) the mutual coupling between toroids should I
think be poor.  (That's if memory serves correct.  I don't have any
decent EE textbooks to refer to, just a couple of old undergrad books
and an EE handbook that's full of typos. One of these days I'll have
to buy a copy of Jackson.)

I haven't had time to do any more calcs with that toroidal inductance
this week (too many list posts!) so I'll have another go with it
this weekend.
--
Paul Nicholson
--