[Prev][Next][Index][Thread]

RE: more rsg/cap queries



>Original Poster: Robin Copini <rcopini-at-merlin-dot-net.au> 
>
> what happens if, while the cap is charging on the down side of the
>incoming AC waveform,(or the upside of the reverse) - and does not
>discharge due to insufficient voltage across the presented gap - it is
>presented with the now 'reversed' polarity of the AC waveform after
>having gone through it's zero point? I mean, all stupid questions
>allowed of course, it is seems to me that for at least a very short time
>the cap and transformer are presenting opposing polarities. Is this
>right? if it is what happens during this time?
>
>Regards
>
> Robin Copini,
> Adelaide - South Australia.

Your question is one that had me wondering and concerned before I fully
(hah!) understood this stuff.  I too feared that if the cap voltage was
less than the gap discharge voltage as it started going toward zero, that
that partial charge on the cap was wasted energy.  Not so!

Think of the analogy of pushing someone on a swing, the amplitude of the
swing representing the voltage on the cap.  Ideally one would be pushing
the swing UPWARD away from it's midpoint (zero volts) and hoping to
discharge it at it's peak amplitude.  But in the event that the "gap
discharge" amplitude is not met, that partial amplitude is not wasted.
Instead, you (or the transformer) would continue to push it, now
DOWNWARD, through zero, and continue pushing, all the time adding more
energy to the swing.

The DOWNWARD push represents the transformer working to reverse the
charge polarity on the cap.  True, it is opposite in polarity to the
open-circuit voltage that the transformer wants to output, but to the
transformer, it's just pushing current - it doesn't matter what it's
terminal voltage is.

It might be helpful to see some of the simulation waveforms I've posted
on my web site: http://people.ne.mediaone-dot-net/lau/tesla/tesla.htm (What
capacitor size is best?)  It shows the cap voltage using a static spark
gap, and there numerous instances where the charging peak doesn't hit the
discharge voltage, so it swings downward all that much harder.

Regards, Gary Lau
Waltham, MA USA