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RE: RC filter

Original poster: "Lau, Gary" <gary.lau@xxxxxx>

Hi Sebastiaan:

Sorry for the delay - I found this reply languishing in my Drafts

The easy answer is that the choice of R-C filter caps does not affect
the needed value of the tank cap to achieve pri-sec resonance.

Estimating the power dissipated in the filter R's is more challenging.
There are two components to this (as far as I'm aware).

One component is, by Ohm's law, the current from the NST squared, times
the resistor value.  But the catch here is knowing what the NST current
under actual operating conditions is.  I have a 15/60 NST, no
modifications to the current limiting shunts.  But I've measured in
excess of 150 mA RMS when running.  Such high currents are possible when
Variac-ing up the NST primary voltage and when using high tank cap
values, but is not well understood or predictable.  For my coil, I use
1.7K series resistors, so .15*.15*1700 = 38.2 Watts per resistor.

The second component is determined by realizing that the power in the
bypass caps is dissipated in the resistors each time the main gap fires.
The larger the bypass cap is, and the higher the main gap BPS is, the
more power is dissipated here.  For my coil, I have 450pF caps, and the
bang voltage is about 25kV.  Each of the two bypass caps is charged to
25kV/2 or 12.5kV, 0.5*C*V*V = .0351 joules.  With my 120BPS sync gap,
this will dump 120*.0351= 4.2 Watts into each resistor.  If I were to
use a static gap, this would have a higher BPS, but possibly a lower
charging voltage; not sure if the power would be higher or not.

So I'm guessing my resistors each dissipate about 42.4 Watts.

Regards, Gary Lau

> -----Original Message-----
> From: Tesla list [mailto:tesla@xxxxxxxxxx]
> Sent: Tuesday, March 08, 2005 3:45 PM
> To: tesla@xxxxxxxxxx
> Subject: RC filter
> Original poster: Sebastiaan Draaisma <sebas@xxxxxxxxxxxxxxxx>
> Chiang Mai, den 8 mars 2005
> Hello Gary,
> My name is Sebastiaan and I read your message about rc filters in the
> pupman newsletter that I goth.
> My questions are, how can I calculate the loss of power, coused by the
> filter and do I have to adjust the primary capacity based on this
> Best regards,
> Sebastiaan Draaisma