Re: Equivalent lumped inductance and toroidal coils

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

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

Antonio wrote:
 > I can't access your picture with this address:
 > http://www.abelian.demon.co.uk/tmp/pn110704.gif

No, I took it down because it contained an error.
I've replaced it with another, URL to follow below.

I'm getting sensible looking answers for the toroidal coil DC
inductance through the formula:-

 > Ldc = mu * N^2 * (r1 - sqrt(r1^2 - rt^2))
 >        - mu * r1 * N * (1.2 - ln(pm/rc) + H)
 >
 > where
 >    N = turn count
 >   rc = conductor radius;
 >   pm = mean winding pitch = (inner_pitch + outer_pitch)/2
 >   rt = tube radius of the toroid former
 >   r1 = radius of toroid (measured from axis to centerline
 >        of the tube)
 >    H = a function of the turns count N:-
 >     N,      H
 >     1,      0
 >     2,      0.1137
 >     5,      0.2180
 >     10,     0.2664
 >     50,     0.3182
 >   infinity, 0.3379

but the approximate formula for use when rt << r1 given by the
same handbook,

  Ldc ~= mu * N^2 * rt/(2*r1)

seems to come out with rubbish. Maybe there's a typo error in
the handbook - it wouldn't be the first for their inductance formulas.

Also I tried a turn-by-turn summation of the coupling between circular
filaments, but the kernel function,

  M = pi * mu * Xa * sin(alpha) * sin(beta)
       * sum{ for n=1 to infinity;
              (Xa/Xb)^n / (n*(n+1))
               * Pn( cos( alpha)) * Pn( cos( beta)) * Pn( cos( gamma))}

is misbehaving too.  It converges quickly and reliably to the wrong
answers.  If I've got my sums right, for the toroidal coil the above
formula is implemented as follows:

Referring to
  http://www.abelian.demon.co.uk/tmp/diag1.gif

  theta = longitude around the toroid between the two filaments, 0 - 2*pi;
  gamma = pi - theta = angle between the axes of the two filaments;
  d = tan(theta/2) * r1 = perpendicular distance from a filament center
                          to the meeting point of the two filament axes;
  Xa = Xb = x = sqrt( rt*rt + d*d);
  cos(alpha) = cos(beta) = d/x;
  sin(alpha) = sin(beta) = rt/x;

  M = pi * mu * rt^2/x
         * sum{ for n=1 to infinity;
                Pn( d/x)^2 * Pn( cos( gamma)) / (n*(n+1)) }

Something must be wrong somewhere.  The sum over the Legendre
products is going negative for only a few degrees of theta.  It
comes back positive again as theta increases, and goes up to +1
for theta = pi.

 > I was looking at your formula, and comparing to a formula in the
 > NBS Circular 544, that looks similar

Maybe the circular gives a check on that formula for filaments?

 > A formula for two parallel loops, noncoaxial, is necessary too.

Yes, as you say for diametrically opposite turns.

 > The Circular has one, quite horrible with Gamma functions and
 > hypergeometric series...

We may need it.  But for now I'd be happy to get some realistic
result for just the first quadrant.  We could do with some
inductance measurements of toroidal coils.
--
Paul Nicholson
--