Re: Transformerless Tesla coil

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

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <twftesla-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
> 
> Original poster: "Jolyon Vater Cox by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <jolyon-at-vatercox.freeserve.co.uk>
> 
> Antonio,
> thanks for putting me straight on the series resonant network connected
> across L1.
> Yes, I can see it now- L1 is connected across two branches of the bridge;
> the branches are two series-resonant tuned circuits are connected in
> parallel and together they will have half the impedance of a single branch.
> 
> Since they are series-resonant each branch will have low impedance and
> becauseb there are two of them in parallel, the total impedance will be
> lower still.
> 
> The only part of the circuit which is parallel resonant is between a and b
> (neglecting C3 and C3', the components C2-L2 -L2' -C2' form a continuous
> path)
> so the only high-impedance points in the circuit are between a and b.
> 
> Am I correct?

Between a and b is where the highest voltage appears. Yes, it can be
said that the impedance there is the highest in the circuit. But note
that the circuit operates in transient mode and not at sinusoidal
steady state. The primary circuit alone resonates at the same
frequency of the resonators: L1*C1=L2*(C2+C3).
The voltage gain is something close to 2*sqrt(L2/L1).

    o------+--C1--+------+----------+----+
           o      |      C2         L2'  C3'
   PSU    gap     L1     +----+a    +----+b
           o      |      L2   C3         C2'
    o------+------+------+----+----------+-o Ground

Antonio Carlos M. de Queiroz