Re: Resonant rise?

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

Original poster: Ed Phillips <evp-at-pacbell-dot-net> 

Tesla list wrote:
 >
 > Original poster: "Lau, Gary" <gary.lau-at-hp-dot-com>
 >
 > Hi Dan:
 >
 > My understanding of the term "resonant rise" is that energy must be
 > continuously applied at a system's resonant frequency, like giving someone
 > on a swing a series of correctly timed small pushes, and this results in a
 > growing amplitude of oscillations, until something breaks down or losses
 > balance the input energy.  While this is what happens in a CW coil, I
 > didn't think that it occurred in a disruptive coil.  There, it really
 > starts with one big primary-push, and that energy is coupled to the
 > secondary, with resonance being a necessary condition but not central to
 > the voltage gain of the system.  Is my understanding or definition flawed?
 >
 > Regards, Gary Lau
 > MA, USA

	Another similar way to look at this is to remember that, in a corretly
tuned coupled circuit, at the peak of the secondary waveform all of the
energy which was originally stored in the primary capacitor is
transferred to the capacitance of the secondary.  This is the source of
the "voltage gain" "sqrt(Cp/Cs)".  If the system is tightly coupled and
this peak occurs after only a few cycles the energy lost in the circuit
resistances doesn't have much effect on the resultant voltage.  Q isn't
really relevent.

Ed

	On the other hand, as Gary points out, if you feed a series resonant
system from a CONTINUOUS AC source (horrible example would be an NST
connected to a "matched" capacitor and without protective gap) the
output voltage will rise to a value of Q times the input voltage. In the
case of a typical NST with a Q of at least ten this would result in an
output voltage of 150 kV for a 15 kV transformer.  Of course, in reality
the insulation would go before this state could arrive...........