Re: Magnetizing current in SSTCs, my previous posting

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

Original poster: "jimmy hynes by way of Terry Fritz <teslalist-at-qwest-dot-net>" <chunkyboy86-at-yahoo-dot-com>

I put the pspice file on my site 
(<http://www.hot-streamer-dot-com/chunkyboy86>www.hot-streamer-dot-com/chunkyboy86).

The rise time is limited by coupling, so by lowering the coupling to 0.01, 
I force the rise time to be slow. It also gives more energy transfer 
because it allows more cycles to transfer energy. By raising the impedance 
untill I get to the same energy transfer, I get a bigger M.
  (same energy/(1/2*I^2)*2t).

I think "M" should be redefined as energy/(rms*time), because bigger 
systems using more IGBTs would have lower resistances, and should get 
credit for it. The voltage drop across the IGBTs would be the same in a big 
and small system, otherwise you're using too many(or too few) IGBTs on one 
of them. Using rms current would be more fair to big coils with many IGBTs.

The fact that the loss is proportional to current squared is taken into 
acount by using rms instead of mean current. The only thing that really 
affects that is the rf envelope, because we all have sinusoidal current 
waveforms. Basically, it gives you a better M if you spend the whole burst 
at your peak current with a rms/mean ratio of 1, instead of a spikey 
envelope with a terrible rms/mean ratio.

Using your definition for M, I get M=1.3 -at- k=0.28 and M=.56 -at- k=0.1 (same 
primary impedence). For k=0.28 the numbers are 3.7 joules ~210 amps rms and 
62 useconds. For k=0.1 the numbers are 30 joules 550 amps rms and 168 useconds.

Tesla list <tesla-at-pupman-dot-com> wrote:
Original poster: "Stephen Conner by way of Terry Fritz "


 > I have a matlab model, and a pspice model if you want to play with it.

Yes please (pspice that is because I don't have matlab)



 > M is a better way to look at it, but as you said, the sparks might like
 > quicker rise times alot more. I could get a great M by lowering the
 > coupling to .01 and using high impedance primary system, but the rise
 > time would be horrible.

It doesn't work like that. The bigger M is, the faster rise time you should
get for a given inverter circuit pushed to its transient thermal limits. M
is like "pulse performance per unit silicon". In the above example, using a
high impedance primary will reduce I^2 but it will also reduce the energy
transferred because the load on the inverter is! less. so M will probably
stay the same if not get smaller (worse)

M is actually in ohms and 1-(Rds(on)/M)= efficiency of inverter circuit

 >You could also get higher values for m by using mini coils (1/2 energy/1/2
 >current^2). Although M is a good starting point, there could be some
 >improvements.

If you can think of any (especially if you think my theory is wrong) please
go ahead

 >How did you come up with 23? I get 0.2/(380^2*25e-6)=0.055

That's what I get for dozing through "Using Your Calculator 101"

Steve C.





Jimmy


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