Equivalent lumped inductance and toroidal coils

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

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

Perhaps we can do some calculations to estimate some of the
properties of a toroidal coil...

A convenient handbook gives the DC inductance of such a
coil to be

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

If the tube radius is small compared to the toroid radius,
this approximates to

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

but let's not use this approximate version.

I use the term Ldc here, rather than just L, to remind us that
that all these inductance formula assume we're working with a
uniform current throughout the coil, which is the case for low
frequency or 'dc' measurements.

But as the operating frequency goes up towards the lowest
self-resonance, the current can no longer be uniform, and so
we cannot expect the voltage induced across the inductance to
continue to be 2 * pi * freq * Ldc * current.

To begin with, we would have to say which current to use: do we
use the current going into one end of the coil, or the current
leaving the other?  Or maybe the current in the middle somewhere.

The answer is we have a free choice: it depends what we want.
In practice we usually pick one of the terminal currents, and
we pick the one nearest to a current anti-node (current max)
of the non-uniform coil current.   For a normal TC this is the
secondary base current.   We can say we 'refer' everything to
the base current.

Now we would like to set up an equation like

   volts_across_coil = 2 * pi * freq * L? * base_current

But what do we use for the mysterious inductance L??.  We know
immediately that Ldc wont work quite right because Ldc was
calculated on the assumption that the current in all the turns
of the coil was equal to the base current.   This becomes less
valid as the frequency is raised towards the lowest resonance,
and as we reach that resonance, we know the current in the coil
will span some number of degrees of a cosinusoidal-like profile.

If we're talking about a base-grounded TC with no extra top load
capacitance, the current profile will span most of 90 degrees, or
pi/2 radians, ie the current node (zero current) is at the top
of the coil.  If some top capacitance is present, the current
profile will span less than 90 degrees of a cosinusoid.

It will probably be clear that we can try an approximation for
L? by modifying Ldc to take account of the average coil current,
This we can do by simply integrating a cosine function over
the estimated number of degrees.

For an unloaded 1/4 wave TC, we then have

  L? ~= Ldc * integral( from 0 to pi/2 of cos(x))/(pi/2)
      = Ldc / (pi/2);
      = Ldc * 0.63662

I usually end up calling L? the equivalent or effective
series inductance, Les.  For a loaded coil, we would use
some angle less than pi/2.

It must be stressed that for a straight solenoid this is quite
a poor approximation because this simple integral takes no
account of the end effects which make the 1/4 wave standing
wave current profile only roughly cosinusoidal.  For typical TC h/d
ratios, the factor is more like 0.85 than 0.65.   For short coils
in which end effects are the most significant, the factor can
exceed unity.

Now, interestingly, for the toroidal coil, we don't have
any end effects other than those we unavoidably introduce
by attaching wires to the coil.  We might therefore expect
that for this case the current distribution along the
standing wave will be closer to sinusoidal than we are
accustomed to seeing, and so the adjustment factor for
the effective inductance might be closer to 2/pi than is
the case for other shapes of solenoids.

The full wave resonance has been mentioned, and a natural
selection for a 'reference' current would be one of the two
current anti-nodes.   Let's leave aside the question of
whether the experimenter will be lucky enough to find
one of these antinodes coinciding with his terminals
tapping into the coil.  The weighting factor for the
effective inductance of this resonant mode now requires
our current averaging integral to go all the way to 2*pi and
the result is zero - a result consistent with the fact that
any resonant mode on the closed loop must have a total
induced voltage around the loop of zero.

We can perhaps try to look at a 1/4 wave quadrant of the
full wave resonance and try to apply our weighting factor
to this piece instead of the whole coil.  We can choose
a quadrant joining a current anti-node to an adjacent current
node and so get a 1/4 wave segment that looks just like the
ealier case.  Then the voltage across the segment is

   2 * pi * Ldc * anti_node_current * 2/pi

which gives the voltage amplitude at the voltage anti-node.

Obviously the Ldc used in this case must now be the Ldc of the
quadrant, not the whole coil of course.  But we must be
a little careful here. Can we safely use Ldc/4 for the quadrant?
The reason for care is that the Ldc of the whole coil is
comprised of the sum of the self-inductance of the four
quadrants, plus the mutual inductance of all 16 pairs of
mutual inductances between the quadrants.  Therefore if we
just cut out a quadrant of the toroidal coil and measure
its inductance, we'll get something less than a quarter that
of the whole coil.   Should we use this lesser (measured) value
for the quadrant's DC inductance, or should we use the whole
coil's Ldc divided by four?

I was about to state that the correct value must be the
total Ldc of the coil divided by four.  The argument is
that when we come to measure the induced voltage across
our 1/4 wave quadrant, the voltage me measure has the
benefit of the 3 mutual inductive couplings to the other
quadrants, each of which is excited by a quarter-wave
cosine of identical average distribution.  Therefore use
of Ldc/4 would take account of this, whereas using just
the self inductance of the isolated quadrant would give
a low prediction for the induced voltage.

But then I got to thinking that although the average
currents in the other 3 quadrants are the same, some
are reversed in polarity and some are reversed in
the spatial distribution.  In the end I couldn't be sure
in my mind how it all added up.

The only sure answer, I suppose, is to code up the familiar
Neumann integral for the whole coil, say on a turn-by-turn
basis, with each turn weighted by the current distribution.
We've done this for all sorts of other coil shapes, but not
I think toroidal coils.

Does Antonio have this case covered?

Perhaps I'll make up the necessary program so that we
can compute these things accurately for toroidal coils.
For anyone interested in doing the same, the relevant
kernel for a turn-by-turn Neumann type summation is
the mutual coupling between two circular filament loops,
with their axes in a common plane but at an angle to each
other.

  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));

where Pn() is the n'th Legendre polynomial.
and the various angles and dimensions are in the image file
  http://www.abelian.demon.co.uk/tmp/pn110704.gif

Then the total DC inductance of the coil is the double
summation over each pair of turns:-

  Ldc = sum{ for i=1 to N; sum{ for j=1 to N; M(i,j)}}

where N is the number of turns and when i = j use the
filament self inductance for M(i,j).

For the effective inductance at high frequency, weight
the sum with the estimated current profile I(x) normalised
to a unity reference current at the anti-nodes,

  Les = sum{ for i=1 to N; sum{ for j=1 to N; M(i,j) * I(j) }}

where Les is the effective series inductance.  For I(j) sinusoidal
and spanning integer multiples of 2*pi, this Les will sum to zero.

For segments of the coil, eg the quadrant discussed earlier,

  Les = sum{ for i=1 to N/4; sum{ for j=1 to N; M(i,j) * I(j) }}

with I(j) in this case being cos( j/(2*pi)).   Using this
quadrant value for Les, the voltage across the quadrant would
be 2 * pi * freq * Les per amp of anti-node current.

I hope I've given a hint here of how the lumped equivalent series
inductance is derived, starting with either a total dc inductance
value, or a primary formula for mutual coupling between elementary
filaments, and then integrating - taking into account the standing
wave current distribution.  The end result is an 'effective'
inductance that correctly relates currents to voltages within
the coil.   A similar process takes place for the capacitance,
giving an effective shunt capacitance Ces. (I don't expect the
corresponding capacitance to be quite so easy to calculate because,
amongst other things, we cannot assume uniform charge around each
filamentary ring.)  The resonant frequency is given in the usual
way by 1/(2 * pi * sqrt(Les * Ces))
--
Paul Nicholson
--