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Re: TC Output Voltage
Tesla List wrote:
>
> >From MALCOLM-at-directorate.wnp.ac.nzSun Nov 3 22:48:21 1996
> Date: Mon, 4 Nov 1996 12:50:24 +1200
> From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
> To: tesla-at-pupman-dot-com
> Subject: TC Output Voltage
>
> Hi everyone,
> I think I have figured out a way to measure output
> voltage indirectly (at least using lumped circuit values). Since
> we can measure the primary directly, we can observe the amplitude
> of two successive beat envelopes. The ring-down, ring-back-up of the
> primary will show a drop in amplitude due to losses and if the
> system is prevented from breaking out into sparks, spark losses are
> excluded from these losses. Using the difference in successive peak
> primary amplitudes, we can deduce system Q and compare that with
> theoretical calcs. We can also make a good estimate of the losses
> over one half of a beat hump and thereby quantify how much energy
> went to the secondary first time round, also how much arrived back
> at the primary. These are a couple of random thoughts. I would
> appreciate anyone else who can contribute to this doing so with
> out relegating it to pooh-pooh status. If anyone cares to, the VSWR
> view of things could also be used to calculate Vsec, bearing in mind
> that the helix is topped by a lumped capacitance and that secondary
> current with a top hat is considerably more uniform than that you
> would find in a pure helical resonator.
> A friend of mine who is an RF engineer has examined some
> structures and come to the conclusion that a low Zo is required in
> the secondary. That is, a low L/C ratio in the helix itself. That runs
> completely counter to what we actually do in these machines (Lo
> Zprim, Hi Zsec) with Zsec dragged down by a lumped topload. As a
> result, I have suggested that antenna theory is not really applicable
> to what we are trying to do with our coils. Antennas are of course
> designed to radiate prodigiously whereas we are ideally after none at
> all (v. high radiation resistance).
>
> Any thoughts on the above would be appreciated.
> Malcolm
Malcolm and All,
I spent some more time thinking about this, AND, thanks to Malcolm's
insight above, I think I've developed an approach for not only
estimating Vout, but for estimating power dissipated by the streamers,
and primary:secondary transfer efficiencies!
Assume we have a 2-coil systems where LpriCpri = LsecCsec. We use a
toroid with a ROC that prevents breakout, and can add a small "bump" to
force breakout when we wish. Assume we've placed instruments in the
primary circuit to measure the voltage across the primary capacitor
simultaneously with the current flowing though the primary. Since the
total energy at any instant is Ep = (0.5*Cp^Vcp^2 + 0.5*Lp*Ip^2), the
simultaneous measurements are important - the maximum energy point may
NOT simply correspond to the point of maximum voltage or current!
Now, let's look at the case where we do _not_ quench the gap, but we
prevent breakout from the toroid. Once the gap fires, and all during the
first beat, a portion of the initial energy, E0, is transferred to the
secondary until the primary energy is gone. At this point all the
remaining energy in the system resides in the secondary, as energy E1.
Since we continue to let the gap fire, a portion of E1 now couples back
to the primary circuit until all the energy in the secondary is gone. At
this time the primary circuit will have energy E2.
We can measure E0 and E2. Although we can't directly measure E1, we can
calculate it. Once we know E1, we can then calculate other interesting
things such as the maximum output voltage (no breakout), energy and
power being coupled to the streamers, and various efficiencies related
to streamer energy and power. In order to do these, though, we'll also
need to measure the secondary Q with and without breakout under properly
quenched conditions.
Finding E1:
-----------
Once the gap fires, energy is continually being lost to resistance, gap
losses, and radiation. Because of these losses, a given beat will have
only a fraction (x) of the energy from the prior beat. We will assume
that, while the gap is firing, the fractional loss (per cycle or radian)
stays the same irrespective of the direction of energy flow from
primary-to-secondary or vice-versa. This is consistent with passive
network symmetry, and will hold whether we have linear or exponential
decay.
E0 = 0.5*Cp*Vgap^2 (Measurable)
E1 = x*E0 (Not directly measurable)
E2 = x*E1 = E0*x^2 (Measurable)
Solving for x:
x = Sqrt(E2/E0) = the effective energy transfer ratio per beat
Solving for E1:
E1 = E0*Sqrt(E2/E0) = Sqrt(E0*E2)
This means that if we can measure the initial primary energy (E0), and
the maximum energy returned to the primary at the end of the 2nd beat
(E2), we can calculate the maximum energy (E1) that must have been
transferred to the secondary at the end of the first beat!
Secondary Q's with and without breakout:
----------------------------------------
We also need to capture the waveform of the secondary's output via a
remote pick-up plate, We also need to adjust the gap so that it's
optimally quenched at the end of the first beat (using a synchronous or
vacuum gap). We'll capture the secondary waveforms via a storage scope
for the case where the toroid does not break out, and for the case where
it does. By using cycle counting or back-figuring from the exponential
envelope, we can calculate Qbo and Qnbo (the Q's with, and without,
breakout). Since we are optimally quenching the gap, no energy is lost
back to the primary after the end of the first beat.
Now, Qnbo tells us how fast energy is being lost due to secondary
resistive losses, radiation, and losses in the RF groundpath, with no
breakout losses. Qbo reflects the combination of the above losses PLUS
energy being lost to the streamers. Among other things, Q is a measure
of the reactive energy stored in a system divided by the energy lost,
and can be calculated on a per-cycle or per-radian basis. Assuming that
we've transferred energy E1 (per-bang) to the secondary, ALL of this
energy will be dissipated or radiated while the secondary rings down -
at a relatively slow rate if we have no breakout, and more rapidly if we
have breakout.
Putting it all together:
-----------------------
Once we determine E1, Qnbo and Qbo, we can now determine the portion of
energy per-bang that ultimately ends up in the streamer versus that
dissipated elsewhere. With no breakout, we lose Ecoil per radian. With
breakout, we lose (Ecoil + Estreamer) per radian. Defining Q's:
Qnbo = Esys/Ecoil (No Streamers, per radian)
Qbo = Esys/(Ecoil + Estreamer) (With Streamers, per radian)
where:
Esys = energy stored reactively per radian
Ecoil = energy lost in coil/groundpath or radiation per radian
Estreamer = energy lost to streamers only per radian
Solving for Estreamer and Ecoil:
Ecoil = (Esys/Qnbo) per radian
Estreamer = (Esys/Qbo) - Ecoil per radian
Estreamer = (Esys/Qbo) - (Esys/Qnbo) per radian
The portion of the total loss attributeable to the streamers is:
Kst = Estreamer/(Estreamer +Ecoil)
Kst = 1 - Qbo/Qnbo (for entire ringdown)
The portion of total loss from sec. resistance, ground, & radiation:
Ksec = Ecoil/(Estreamer +Ecoil)
Ksec = Qnbo/Qbo (for entire ringdown)
Solving for the total energy dissipated during each ringdown:
Per-Bang Energy to Streamer = Kst*E1 = E1*(1-Qbo/Qnbo)
= SQRT(E0*E2)(1-Qbo/Qnbo)
Per-Bang Energy elsewhere = Ksec*E1 = E1*(Qnbo/Qbo)
= SQRT(E0*E2)(Qnbo/Qbo)
An example:
----------
Lets plug in some values for Q that I previously measured on my 10"
coil:
Cp = 0.0205 uF
Lp = 153.7 uH
Cs = 41.5 pF
Lp = 73455 uH
Vgap = 19000 Volts
Qbo = 19 (with breakout)
Qnbo = 206 (with no breakout)
Thus:
E0 = 3.7 Joules (Energy available/bang)
Now, lets assume that we found that the maximum energy transferred
back to the primary at the end of the second beat (E2) is 1.0 Joule.
(This is only an example - I haven't actually measured this...)
Then,
E1 = SQRT(3.7*1.0) = 1.924 Joules/Bang
Vout (Max) = SQRT (2*E1/Cs) = 304 KV
The portions of secondary energy going to streamers and elsewhere:
Kst = 1 - Qbo/Qnbo = (1 - 19/206) = 0.908 (90.8%)
Ksec = 0.092 (9.2%)
The Per-Bang Energy to Streamer = Kst*E1
= 0.908*1.924
= 1.75 Joules/Bang
The Per-Bang Energy elsewhere = Ksec*E1
= 0.092*1.924
= 0.177 Joules/Bang
Percent of Initial energy going to Streamers/Bang = Kst*E1/E0
= 1.75/3.7
= 47.3%
Power to the People:
-------------------
Finally, lets estimate the average power dissipated per second by
multiplying by the break-rate. Assuming 420 bangs/second (PPS), the
output power being dissipated in the streamers can now be determined:
Power into streamers = 420 Bangs/Second * 1.75 Watt-Seconds/Bang
Pst = 735 Watts (average)
Power initially supplied to Primary Tank Cap = 420 * 3.7
Pcap = 1554 Watts (average)
Primary:Secondary transfer efficiency = Pst/Pcap = 47%
Note that the energy transfer efficiency is directly related to the
power transfer efficiency. We could also measure the amount of 60 Hz
power coming into the HV transformers to get another measure of output
sparks versus mains input power. However, since incoming power is
"tainted" by the addition of short-circuit current when the gap fires,
it may not be as useful a measure. The important thing is that ALL the
above variables should be directly measureable or calculable! I think
an extension of this approach should also apply to maggies...
Whew!! That was some _heavy_ stuff... Think I'll go make some ozone now!
As usual, heckling, catcalls, and rotten vegetables are eagerly
welcomed! :?/ :^) :^]
Safe, and more measurable, coilin' to ya!
-- Bert --