Re: Series connection of Mosfets/IGBTs

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

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

Hi,

I think I may have been heasty. see aditional comments below


 > 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.
 >
 >
I think there may be something wrong with he above because there does not
seem to be sufficient time and current input to get to 2 times the SSR
output voltage.
The problem may be in the phase of the input current SSR and TR or they have
different  frequencies. I would expect the comments about the output would
be about right unless the TR is wrong.

Bob