Re: term understanding: voltage reversal.

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Original poster: "Dr. Resonance" <resonance-at-jvlnet-dot-com> 


Voltage reversal applies to the discharge part of the cycle, ie, the rapid
dV/dt change in the opposite direction which occurs in only 2-5 microSec in
a normal TC circuit.  The charging cycle is tremendously slow as compared to
the discharging cycle in a cap.

Rapid dV/dT creates HF currents produce excessive corona and especially HV
"tracking" across the ends of the dielectric which causes heating effects
and can be fatal for a cap that does not use extended foil technology, ie,
DC filter caps for example.

Dr. Resonance

Resonance Research Corporation
E11870 Shadylane Rd.
Baraboo   WI   53913
----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Wednesday, February 18, 2004 8:06 AM
Subject: term understanding: voltage reversal.


 > Original poster: "Christoph Bohr" <cb-at-luebke-lands.de>
 >
 > Hello All.
 >
 > I came  across something I am no longer sure I understood that right:
 >
 > In pulse cap data sheets there is usually a point called "voltage
reversal".
 > To say it less technical I understand that voltage reversal causes stress
on
 > the cap and is undesireable if you like longelivity.
 > But what exactly is this voltage revesal in a AC, sync gap, TC? is it:
 >
 > 1.: The changing polarity of the carging current, i.e. the fact that I
once
 > carge the cap with the one polarity and during the next half sine wave to
 > the other
 >
 > or
 >
 > 2.: The changing polarity during the HF-"ringing". As the changes happen
 > more often and more rapidly here I feel that this is the main voltage
 > reversal relatet Stress on the cap.
 >
 > I fear this relates to point 1 and 2 as well. I only hoped that if it was
 > point 1, one could reduce the stress on the cap by full-wave rectifying
the
 > xformer output to charge the C1. Would this work? I am not talking about a
 > real DC supply with smoothing and charging inductor, only charging with
 > pulsed dc to avoid the changing in polarity of the charging current.
 >
 > Probably I am missing something but the longer I think about it the more
 > unsure I get...
 >
 > Maybe some of you guys can explain that to me.
 >
 > Best regards
 >
 > Christoph Bohr
 >
 >
 >