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RotoVerter Reactive Power Recovery

Started by contempoarts, February 04, 2013, 07:06:34 AM

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WhatIsIt

Yes Magluvin,

At least somebody understood. You got it right.
This circuit contains more than we can see at first look.
It is much more complex than it looks like.

Cap will resonate with both windings, but that is not true if you look deeper.
If the frequency is 5khz, C = 1uF and low inductance second primary P' is 1mH,
resonance will occur between C and low inductance P'.

But resonance will not occur between C (1uF) and high inductance first secondary S,
because S has more turns (inductance) and won't resonate with C at that frequency!

Reactive power will rather stay in LC parallel tank P' C, because P' provides low resistance path,
while high inductance S is high resistance path.
So, high inductance S will actually block most of reactive power, forcing it to stay in P' C tank.

Because it is one wire output, it won't kill resonance in P' C tank.

What voltage will circulate in P' C tank?

Very high, provided from high inductance S winding.
Parallel tank will produce high current when resonate (parallel LC resonant tank does that),
at high voltage also, from S winding.

So you have high voltage, high current LC parallel resonant tank,
where resonance occur only in P' C,
with ability of S to block most of reactive power, which otherwise goes back to source,
forcing the source to consume more.

High voltage, high current means high power in output winding S'.
What will happen if we use 50Hz, C 100uF, P' 100mH?
How much power will circulate through P' C parallel resonant tank, if the voltage of S is 1000V ???
LTSpice says 48kW.

Original circuit has spark gap, but I removed it and made a joke,
which everyone immediately understood wrong.
Sorry for that.

Now the words of Tesla on that type of circuit maybe has more sense:

But if, on the contrary, the rate of discharging is greater than that of charging,
then a succession of rushes of current is obtained.
It is evident that, if the rate at which the energy is dissipated by the discharge
is very much greater than the rate of supply to the condensers.