Re: The PING Test

["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>

John Couture wrote:

> What method do you recommend to measure the energy in the
> secondary coil ignoring the streamer loading?

Without hesitation, I would say measure the peak base current,
and take the energy to be 0.5 * Lee * I^2, where Lee is the
equivalent energy storage inductance of the secondary resonator.

Ditto for the primary too, although you can just use the
ordinary DC inductance for that one.
 
> With reference to your equation
>  Efficiency = 100% times(Q_unloaded - Q_loaded)/Q_unloaded
> How would you set up the Second Ping test to find the loaded
> Q factor?  Assume we are looking only for the energy in the
> secondary (useful load) and ignore the streamer loading.

We can get at this directly by actually measuring the peak energy
as described above, then comparing it with the firing energy.

We can't apply a double ping test to this case, because we can't
just turn off the intrinsic losses to get an 'unloaded' condition,
we would just get infinity anyway.  We're dealing here with the
stored energy of the system whose Q we're measuring, rather than
the transfer of energy out of the system, therefore a single Q
measurement will suffice. 

(We're papering over the fact that there are two resonant 'mode'
frequencies in here, each with its own average Q factor.  We can
choose to take (Q1+Q2)/2 or sqrt(Q1*Q2) as the 'overall' Q factor
for our purposes).

Energy transfers from one side of the resonator to the other
every Fr/(2*(F2-F1)) cycles, where F2 and F1 are the upper and
lower mode frequencies determined from the single ping of the
dual resonator.  Now let Q be the overall average Q factor
obtained from the ping.  Then the fraction of stored energy
remaining after each cycle is 1-2*pi/Q. Therefore, during a
one-way exchange of energy, the fraction remaining is

   pow( 1-2*pi/Q, Fr/(2*(F2-F1)))

in other words 1-2*pi/Q raised to the power of Fr/(2*(F2-F1)).
Thus the efficiency John is asking for (what fraction of stored
energy remains after half a beat) is 

   100% * pow( 1-2*pi/Q, Fr/(2*(F2-F1)))

Since F2 = Fr/sqrt(1-k) and F1 = Fr/sqrt(1+k), we can write
the efficiency in terms of Q and k as 

   100% * pow( 1-2*pi/Q, sqrt(1-k^2)/2/(sqrt(1+k)-sqrt(1-k)))

and there is no need for any mysterious unknown efficiency 
coefficients to be factored in.

We've made only one approximation: taking an average Q factor
of the two modes combined.  This is reasonable if the firing
energy is equally divided between the two modes.  This is not
usually the case in practice, and ideally, the dual resonator
should be pinged by actually firing the coil at a level just
below breakout.  Not only does this then take account of normal
gap losses, but the ping can be analysed to extract the relative
proportions of energy in the two modes allowing the two mode Q
factors to be combined with the appropriate weighting.  That would
then take accurate account of the tuning of the particular 
resonator.

So the storage efficiency is easy to get at, either by a 
calibrated measurement of peak Ibase and the firing voltage,
or by firing the coil and measuring the mode Q factors.
 
> I agree some of these questions can only be answered in a general
> way. But it does no harm to make estimates based on the
> information and tests we have available. In this way we can make
> progress and make the necessary corrections as we move along in
> the future.

Absolutely.  I think some of my suggestions, although they may be
theoretically sound, are rather less than practical.  In this
case, it boils down to being able to measure the overall Q of
the dual resonator when it is fired just below breakout.  No
problem if you have a digital scope that can capture the base
current waveform and dump it into a computer file.  A more useful
approach would be to use a simple lamp test to estimate Q, or to
use a cheap analogue scope to measure the relative amplitudes
of consecutive beats, say.

Lamp testing involves coupling a light bulb to the resonator
via an inductive loop, or in series with the secondary base.
Assuming the coil is firing continuously, the lamp registers
the heating effect (ie the RMS value averaged over the thermal
time constant of the bulb) of the current passing through it,
which in turn is proportional to the RMS currents and voltages
(averaged with the same time constant) lurking in the resonator.
Thus the apparent lamp power is proportional to an average of the
energy in the resonator.  The constant of proportionality depends
on the fine details of the lamp coupling and its resistance at
the operating frequency, and is usually unknown.  Therefore
we can only take relative measurents using a lamp test, which
means we need at least two lamp measurements to tell anything
meaningful about the system.

Now if the lamp time constant is much longer than the firing
interval, the lamp power will be proportional to the stored
energy averaged over several shots, and this in turn is
proportional to the overall system Q factor if the lamp is
a low power (torch) bulb weakly coupled.

Suppose we run the system and obtain a lamp power Pu corresponding
to an unknown Q factor Qu. Then we insert a known resistance into
the secondary base and repeat the test to obtain a lamp power Pr
associated with the reduced Q factor Qr.  If Rs is the unknown
total loss resistance of the resonator referred to the secondary
base, then we can say (where ~ means proportional to)

   Pu ~ Qu ~ 1/Rs;
   Pr ~ Qr ~ 1/(Rs + Rb)

where Rb is our (known) extra base resistance.
And then we get

   Pu/Pr = Qu/Qr = (Rs+Rb)/Rs

which leads to 

    Rs = Rb/(Pu/Pr - 1)

>From this, we find the Q factor to be 

    Q = 2*pi*Fr*Lee/Rs = 2*pi*Fr*Lee*(Pu/Pr - 1)/Rb

and the efficiency of energy storage for the duration of a single
pri-sec transfer is

   100% * pow( 1-2*pi/Q, sqrt(1-k^2)/2/(sqrt(1+k)-sqrt(1-k)))

 = 100% * pow( 1 - Rb/(Fr*Lee*(Pu/Pr - 1)),
               sqrt(1-k^2)/2/(sqrt(1+k)-sqrt(1-k)))

This time, k would have to be calculated since we're not measuring
the mode frequencies.   If instead of a lamp test, the rms
secondary base current was measured (where the mean of the root
mean square is taken over several shots), then the term Pu/Pr is
replaced with sqrt(Iu/Ir).

Thus the procedure would be:

 Measure the lamp power ...Pu;
 Add in a measured base resistance ...Rb; 
 Measure the lamp power ...Pr;
 Measure or calculate Fr, k, Lee;
 Compute efficiency with the above formula.

The nice thing about this is that it only requires relative (ie
uncalibrated) lamp powers or base current measurements.

John wrote:
> I agree the estimate is a wild card.

No need now for the wild card, given two lamp tests.
--
Paul Nicholson
--