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Re: How (does) voltage rise(s?).

Original poster: Chip Atkinson <chip@xxxxxxxxxx>

Dang.  I thought I had it figured out. I felt so confident.:-} I'll have
to digest this some more.  Thanks for the response though.  I appreciate


On Mon, 17 Jan 2005, Tesla list wrote:

> Original poster: "Antonio Carlos M. de Queiroz" <acmdq@xxxxxxxxxx>
> Tesla list wrote:
> >
> > Original poster: Chip Atkinson <chip@xxxxxxxxxx>
> > The first question was why one wants to loosely couple the primary and
> > secondary. Why not just have a solenoidal primary enclosing the
> > entire secondary rather than a pancake or conical primary?
> The first reason is insulation. Another reason is that there is no
> practical solution to the problem of complete energy transfer from
> the primary capacitance to the secondary capacitance with coupling
> coefficient higher than 0.6.
> > With the uniform solenoid field the secondary is just subject to turns
> > ratio and the voltage doesn't rise any more than that. There is no
> > "whipping action" where the voltage is more or less free to rise as there
> > is when only the base is driven as by a flat coil.
> There is no "whipping action" in a capacitor-discharge Tesla coil.
> It is just a system of two coupled resonators where the energy that
> is initially in a low-impedance resonator (high currents and relatively
> low voltages) is transferred to a high-impedance resonator, where it
> appears as high voltages and small currents.
> The ideal voltage gain in a system tuned so L1C1=L2C2 is
> sqrt(L2/L1), that would be the turns ratio if the coils had identical
> geometries.
> > The next question in my mind is *why* does the voltage rise? How can I
> > think of it?
> > Here's what I came up with -- the electrons are like a gas. If you have a
> > column of air, say, and drive it at the base you can actually get louder
> > (greater voltage) output if you drive it at the resonant frequency. For
> > example, a didgeridoo.
> This is closer to the case of a SSTC, where a continuous low-level input
> is amplified by resonance, and in this case there is really a whipping
> action. But the drive at the base is just a convenience. Could be
> anywhere.
> > The thing though was that doesn't the oscillating
> > medium have to have momentum to resonate? That's where I figured the
> > inductor of the secondary came into play. An inductor conceptually gives
> > electrons momentum -- they keep flowing in their current direction and
> > resist changing directions.
> >
> > The purpose of the capacitor at the top is mainly to store up this
> > pressure wave of electrons until the pressure (voltage) at the top reaches
> > the breakout voltage. Then once it's broken out, the electrons are
> > suddenly released and produce a more energetic streamer or spark. The
> > reason for not just having a bare wire at the top is that the point effect
> > of the bare wire allows the breakout voltage to be too low. The reason
> > that we don't put a toroid the size of a box car on a 3" coil is that a
> > toroid that big would "soak up" all the surplus electrons and spread them
> > over such a large area that the voltage would actually drop from what it
> > was right where the wire attached to the toroid.
> Ok, but the top capacitance is essential part of the secondary
> resonator.
> Without correct tuning of both sides the energy in the primary system is
> not transferred completely to the secondary system. The energy transfer
> does not occur in a single pulse, but in several full oscillations.
> The faster case is with k=0.6, where energy transfer occurs in a
> single full cycle.
> Antonio Carlos M. de Queiroz