[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: [TCML] understanding DRSSTC
Em 12/02/2013 07:49, Udo Lenz escreveu:
In my simulations this was close, but not exact. The secondary resonance
is at 283886 Hz and the driving frequeny is 294427 Hz.
The result with excitation at the secondary resonance results in
slightly slower output rise and the full beats in the unloaded case
I wrote a document with some simulations of a DRSSTC, designed with
the method that I have developed:
The central frequency can be made to match the secondary resonant
If you run at this frequency, the operating frequency will be equal to
frequency and this will give you the highest possible power transfer
to the secondary.
The pole frequencies can never coincide with the secondary frequency
due to the
coupling, so that running at their frequencies will require more
to achieve a given power transfer.. I believe, this is what you observed.
But this system has two resonances. Looking at the frequency response of
the system, what I see is:
At the poles the input impedance is never purely resistive, and is
purely reactive in the unloaded case,
But the central frequency has disadvantages: Consider a simple series
If you drive it below the resonant frequency the current phase will
input voltage. A PLL circuit would detect that and increase the
The central frequency has the opposite behaviour. There a drop in the
leads to a current lagging the voltage. The PLL wouldn't lock on to it.
The pole frequencies on the other hand show the "normal" behaviour.
inductive above and capacitive below.
At the central frequency the input impedance is purely resistive, with
any resitive load at the output.
Greater load widens the frequency range where the input impedance is
A PLL controlled by the input current would lock correctly. There is
just a possible problem, because
as in the unloaded case there are full beats of the input current, and
if the driver is not turned off
after the first beat, the PLL will invert the operation of the driver,
causing a great increase in the input
current. In the lightly loaded case this also happens. A PLL would not
lock at the pole frequencies. It
would move the driving frequency to the central frequency automatically.
It's really confuse what to do with a PLL if the system is designed to a
certain maximum number of
cycles per burst and allowed to exceed this number. I am a bit sceptical
about the use of a PLL in these
systems with short bursts. It would be useful only in really long
bursts, because of the number of cycles required to
You could invert the feedback of the PLL, so that it locks onto the
frequency, but I think, this is a fragile mode of operation, since the
phase relation holds only between the poles. A glitch or a ground arc
the PLL off.
lock. And note that the ambient has heavy interferences.
I don't see it disappearing with any load. In the sense that the input
impedance remains always resistive.
What happens with heavy load is that the two resonances become damped
and both move to the
Under heavy arc loading the central frequency will disappear. I believe
this to happen at about Qsec = 1/k In your simulation with k=0.12 that
would be around Qsec = 8.
I've made measurements of arc load at 70kVpeak (at about 200kHz) and
they give a
load resistance of about 100k. With the parameters you used, Qsec
would drop to
about 2 with a 100k load.
central frequency, resulting in just one peak in the voltage gain.
With heavy load, as I said above, the two peaks disappear, and a PLL
would really make the system
operate at a pole frequency, because both pole frequencies are
identical, at the central frequency
My measurement was made under QCW conditions,
so that the arc had time to grow to its final size. With short burst,
I'd expect the arc load
to be smaller. Nevertheless low Qsecs don't seem to be exotic. Under
these conditions you'll
have only one ZCS frequency with a "normal" frequency-phase shift
relation. An inverted
PLL will fail then.
(with different Qs).
If you are using a PLL, compare the frequency where it operates at the
end of a long burst
with arc load with the two pole frequencies of the unloaded system. It's
true that arc load adds
capacitance to the secondary and changes the tuning of the system, so I
would expect the final
frequency to be somewhere below the central frequency. Or you can
operate the system with
small arcs, so the unloaded tuning is preserved.
Antonio Carlos M. de Queiroz
Tesla mailing list