RE: DRSSTC design procedure - draft

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

Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com> 

 >Why are your current decreasing so fast? In my simulation it goes to
 >the designed 600 A peak and stays there.

This is meant to be representative of a real DRSSTC. So the system starts up
unloaded, then the streamer load is applied at 200us (around the time when
breakout would occur) and finally the drive is turned off at 300us. It's
probably the missing streamer load that makes the result different. When I
ran the simulation for longer I saw a notch and the primary current
increased again.

 >The equivalent series load to Cp=10 pF and Rp=639 k at f=52401 Hz would
 >be:
 >Cs=12 pF
 >Rs=144 kOhms

I calculated it as Cs=26pF and Rs=117k. One of us is wrong (however it's
probably me)


 >the maximum
 >energy in the load capacitance is just 2.3 Joules to keep the 600 A of
 >input current.
 >This happens because the voltage gain is the square root of the ratio
 >of output and input resistances. There is no way around this.

I'm afraid I don't understand this. I'm a long way from being intuitively
comfortable with the "TC as a filter" design approach. When I need a filter
in one of my circuits at work, I just follow a recipe published by someone
else :P


 >I imagine that the best, without huge input
 >currents, would be to store all the energy at the output capacitance
 >before breakout...
 >Note that the element values are
 >almost the same of the first design. Nothing is very critical.

Well you would certainly think so. But the performance of the new DRSSTCs is
only possible if most of the energy is fed directly from the driver to the
streamer load. For instance on Steve Ward's ISSTC II, he gets something like
40J bang energy using a very thin toroid (I think it's a 4" minor diameter)
The numbers don't add up.


Steve C.