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

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Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

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

> > The model reproduces well the input impedance, and quite well
> > the output impedance too, even without the Ldc-L1-L3 inductance.
> 
> Except for the real part of the input Z, which is zero. Through most
> of the frequency range this is negligible error, but becomes
> significant near f1, f2, and fp.

Certainly, because it is a lossless model.
 
> > As a lossless model, it doesn't model the Q factor, but with
> > a resistor added at the input (series) it would model it too.
> 
> Yes, but with these component values, you can choose a resistor to
> match a given Q, or to match a given Zin, but not both together.

Ok. With one resistor only one maximum or minimum of the input or
transfer impedances would be correctly modelled. The others would
be approximations. With more resistors, the model would match other
points too.
 
> To match both Q and Zin with a single resistor, we would have to use
> different L and C values, derived from the energy storage behaviour of
> the resonator.
> 
> Let me try to explain why, by using a simplified distributed model of
> the base-driven unloaded secondary.
> 
> Consider the coil in n sections, eg
> 
>  Base          V1     V2           Vx              Vn-1     Vn   Top
>  o---------L1--+--L2--+- ... --Lx--+-- ... --- Ln-1--+--Ln--+---o
>                |      |            |                 |      |
>                C1     C2 ...       Cx  ...           Cn-1   Cn
>                |      |            |                 |      |
>               ===    ===          ===               ===    ===
> 
> so that the *peak* voltages across each cap C1..Cn are V1..Vn.
> Treat the coils L1..Ln as lumped self inductances - the mutual
> inductance has no effect on this argument.
> 
> At a point in the RF cycle when there is no current flowing, the total
> stored energy is
>...

I see a problem in this derivation: If the coil follows the lclclc...
model shown, there is no point in the RF cycle where all the energy
is stored in the capacitors. The energy is always distributed among
the capacitances and the inductances. You can only assume that at some
point all the energy is in the capacitance if there is just one section,
falling then into the usual model, with a single L and a single C.
 
> > Fp is the frequency where the input impedance has poles, a maximum,
> > not a minimum.
> 
> Yes, I thought it better to the find the Zin pole by looking for the
> Ztop zero.
> 
> > Measuring from the other side, with the base open, the same poles
> > would appear too,
> 
> Yes, but I meant the base to stay grounded, so that the Zin pole
> becomes a zero at the top.  I just thought it would help to eliminate
> the effect of instrument stray C.  But then you said

Something strange here. Zin is the base impedance with the top end
open, or the z11 open-circuit impedance parameter of the network.
The impedance seen at the top end with the base grounded is the
inverse of the y22 short-circuit admittance parameter of the network.
Some calculation shows that the poles of z11 don't appear anywhere
in y22 (the networks are different, and can't have the same natural 
frequencies *), and that the zeros of z11 are also zeros of y22.
Or simply:
Looking at the top impedance with the base grounded, you see 
poles where zin has zeros, and nothing special where zin has
poles.
See this example, with a simple c-l-c-l-c network (values 1,2,3,4,5 from
left to right):
http://www.coe.ufrj.br/~acmq/tesla/z11xy22a.gif
The plots of z11 and 1/y22 result in:
http://www.coe.ufrj.br/~acmq/tesla/z11xy22b.gif
The same happens with any linear circuit that has shunt impedances
across both ports.
Anyway, to measure at the output, with the base open, is a valid idea, 
since there is more capacitance at the top end (with top load), and so
the measurement is less sensitive to parasitic capacitances. 
 
* This is more complicated to demonstrate than appears to be...

Antonio Carlos M. de Queiroz