Re: Series connection of Mosfets/IGBTs

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

Original poster: "Robert Jones" <alwynj48-at-earthlink-dot-net> 

Hi,

Comments in the text,
----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Thursday, February 19, 2004 7:05 AM
Subject: RE: Series connection of Mosfets/IGBTs


 > Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com>
 >
 > Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz>
 >
Snip
 > But in a pulsed SSTC, we can take advantage of the transient behaviour to
 > ram large amounts of energy in very quickly. I'm trying to understand the
 > math at the moment (the above was a gross simplification that would bring
 > physicists out in a rash) so I can design for a given worst-case peak
 > transient current in the inverter.
 >

Assuming no break out and assuming its all correctly tuned. I think the
output would consist if the steady state response (SSR) and the transient
response(TR).  The SSR will be a steady sine wave and the TR will be a
modulated sine wave i.e. the classical Tesla coil output. At the moment the
input is applied the two responses at the output are equal and opposite and
cancel to zero. One quarter cycle of the envelope of the TR later the TR
will be zero so the output will be equal to the SSR. One more quarter cycle
later they will be approximately equal and have the same sign hence sum to
approximatly twice the SSR.  This will repeat until the TR decays to zero
leaving just the SSR.  If you assume the Q of the primary and secondary are
equal and its all matched up for max power transfer (same power loss in
primary as in the secondary and the primary turns are selected to get the
correct power input) for the SSR. Then the peak reactive power in the output
is Q time the input power at steady state.  The next question is how does
the input current vary. Again that will consist of the SSR and the TR so
again they alternatively cancel and add. So the peak current will be twice
the SSR current.

Hope you get the jist of my very quick explanation and I hope its correct. I
ignored the harmonics of the input and several other things. I also assumed
max power transfer was when the Qs are equal which may not be correct. I
think the key result is what ever the Qs are so long as they are not too
low(>25) the peak output is approxaimatly twice the SSR and the peak input
current is approximatly twice the SSR input current. Note I think that it
produces a peak of Q/2 or Q times the input voltage across the primary C
!!!! and Yes I know this does not consider spark loading.


Bob