Re: Measuring self-capacitance directly (Re: flat secondary)

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

Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>

Thanks Antonio, I checked your formula for C0 and it works fine.

For the benefit of anyone trying to follow this thread, I'll just
summarise Antonio's recipe for an equivalent circuit of a
TC secondary:

Take a TC secondary with grounded base and optional topload.

Measure the lowest two resonant frequencies and take simultaneous
measurements of the magnitude of the base current and topvolts
at those frequencies, with the coil driven by some convenient
means from a signal generator.

              1/4 wave             3/4 wave
Measure...
    Frequency:   F1                 F2            (hertz)
 Base current:  Ib1                 Ib2           (amps)
    Top volts:  Vt1                 Vt2           (volts) 

Calculate...
   Transfer Z:  z1 = Vt1/Ib1       z2 = Vt2/Ib2   (ohms)
 Angular freq:  w1 = 2*pi*F1       w2 = 2*pi*F2   (radians/sec)

Feed a signal into the top of the coil and find the frequency in
between F1 and F2 at which the impedance dips to a minimum.

                      1/2 wave
Measure...
   Frequency:             Fp             (hertz)

Calculate...
 Angular freq:        wp = 2*pi*Fp       (radians/sec) 

>From w1,z1,w2,z2,wp, calculate

 a = z1/w1
 b = 1/(z1*w1)
 c = z2/w2
 d = 1/(z2*w2)

And from these, calculate

 C0 = (a*b*c*d*wp^2*(b+d)-(a*b^2+c*d^2))/
      ((a*b*wp^2-1)*(c*d*wp^2-1)*(a*b-c*d)))
 L1 = (a*b-c*d)/(b+d)
 C2 = a*c*(b+d)^2/((a+c)*(a*b-c*d))
 L3 = b*d*(a+c)^2/((a*b-c*d)*(b+d))
 C4 = (a*b-c*d)/(a+c)

Construct the equivalent circuit

 -> Ibase
 o----------+--L1---+---L3---+---o Vtop
            |       |        |
            C0      C2       C4
            |       |        |
            =       =        =

Then this circuit *when unloaded* accurately represents the coil's
three resonant frequencies up to and including the 3/4 wave, and
closely reproduces the relationship between base current and top
voltage over that frequency.  The reproduction is exact at the 1/4
wave and 3/4 wave frequencies, and very accurate at other frequencies
in the range.

Limitations:

This equivalent circuit does not reproduce the input or output
impedance of the coil.  Nor does it correctly represent the 
relationship between the base current (or top volts) and the
stored energy in the inductance (or capacitance).  Also it does
not reproduce the relationship between the input impedance and
Q factor.

A modification to the circuit to approximate the loaded behaviour is 

 -> Ibase
 o----------+--L1---+---L3---+---Ldc-L1-L3---o Vtop
            |       |        |
            C0      C2       C4
            |       |        |
            =       =        =

where Ldc is the DC inductance of the coil.

This LC model is compared with a detailed distributed model of the
secondary, showing the input impedance in 
  http://www.abelian.demon.co.uk/tmp/acmq-zin-3.gif
and the transfer impedance in
  http://www.abelian.demon.co.uk/tmp/acmq-zft-3.gif

[end of summary]

Antonio wrote:
> C0 has no effect in a regular Tesla coil. But would be 
> important in a magnifier.

Absolutely.  That's a point that I've never really considered because
I'm nearly always working with a virtual earth on the coil base.

[zeros of Zft due to mutual M] 
> Ok that they can be at high frequency, beyond the frequency band
> where the model works.

Perhaps.   Requires more thought. I never use the model above about
9/4 waves.

> there is the problem of how are the two inductors coupled to the 
> primary. A 3 coils transformer? Too complicated. The only connection 
> point correctly modeled is the base, and so the model is suitable
> only as model for a third coil of a magnifier. 

Agreed.  I think we'd have to repeat the process by starting with
a transfer impedance relating primary current to secondary top volts,
and then fit in the various measured poles and zeros.

> Can Les be greater than Ldc? 

I think so, net of any gross errors in the math.  Needs one of those
unequivocal experimental confirmations.

> How is the secondary coupled to the primary, if it is not a single
> coil anymore?

A more thorough analysis considers the coupling between the primary 
(approximated by a single lumped inductance) and each individual turn
(or some other convenient current element) of the secondary. The
voltage induced in the primary includes an integral over the current
distribution of the secondary, and the voltage induced across each
secondary turn (or current element) contains a term involving the
lumped Ipri.  The mutual inductance needs to be specified on a turn-
by-turn basis, and the resulting analysis is pretty straightforward. 
You just end up with a big, dense, matrix which is then solved for the
secondary current distribution and the primary current.

There's a rough and unfinished write-up in section 6 of 
 http://www.abelian.demon.co.uk/tssp/pn1401.html

I suspect that we end up with different k factors for the pri-sec
and sec-pri coupling, and that the circuit behaviour will be as if
the overall k was the square root of the product of these two 
separate k factors.  I'm unconvinced that this makes complete sense,
so work is ongoing.

Mods to acmi are in the pipeline to compute inductances weighted by
a given current profile.
--
Paul Nicholson
--