Re: DRSSTC design procedure - draft

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

Original poster: "Antonio Carlos M. de Queiroz" <acmdq-at-uol-dot-com.br> 

Tesla list wrote:
 >
 > Original poster: "Steve Conner" <steve.conner-at-optosci-dot-com>
 >
 >  >I recommend running some simulations to see if I didn't make a
 >  >mistake...
 >
 > Here's my first simulation, it is a transient run using the exact parameters
 > that Antonio posted.

This, I presume:
Ca: 276746.6471913645 pF
La:   33.7819780991 uH
Lb: 224997.2741393207 uH
kab:    0.1152414634
Cb:   41.0000000000 pF
Rb: 638535.0921930474 Ohms
Ra:    1.2732395447 Ohms

This design puts 6 J at Cb at steady state, not 8 as I had said.
 >
 > http://scopeboy-dot-com/tesla/drsstc/antonio_drsstc.gif
 >
 > It seems to work exactly as Antonio said, delivering 23 Joules. And the
 > switching is very good, there are no hard switching events.

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

 > Now I need to investigate its sensitivity to streamer loading, which is a
 > job for another day.
 >
 > It performs very poorly when I run it with the streamer load model I use,
 > which is 25*L pF of cap in series with 103000/L ohms of resistance (L is
 > streamer length in metres) The primary current goes over 2000A. But this is
 > totally different to what Antonio developed it with - 638k in parallel with
 > 10pF, for what (according to the delivered energy trace on my sim) would be
 > a 2 metre streamer. So that would be 5L pF in parallel with 1276/L kOhms.
 >
 > Antonio, where did you get the 638k//10pF figure from? Do you think it is
 > realistic? Do you think _my_ model is realistic? Does anyone else have a
 > view?

The 10 pF was somewhat arbitrary. I should have used more capacitance
for a streamer with 1 meter. The 639 k is the resistance that
if connected to the output reflects 1.27 Ohms at the input, or 600 A
with a square wave with 600 V of peak voltage (or 4/pi*600 V of peak
voltage on the fundamental harmonic). In that design method this is
just at what loading level you choose the input impedance to be
"maximally resistive".
The equivalent series load to Cp=10 pF and Rp=639 k at f=52401 Hz would
be:
Q=2*pi*f*R*C=2.1
Cs=Cp/(1+1/Q^2)=12 pF
Rs=Rp/(Q^2+1)=144 kOhms

If you use Cs=25 pF and Rs=103000 Ohms of load, the equivalent parallel
load
would be:
Q=1/(2*pi*f*Cs*Rs)=1.18
Cp=Cs/(1+1/Q^2)=14.6 pF
Rp=Rs*(1+Q^2)=246 kOhms

With this load, supposing the same 41 pF of load capacitance (decrease
the top load) to keep the same input frequency (52401 Hz), 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.

To obtain long sparks, I imagine that the best, without huge input
currents, would be to store all the energy at the output capacitance
before breakout, as done in a conventional Tesla coil, and turn
off the driver as soon as the input current exceeds a certain limit.
This would fall into my lossless design.
Trying 20 J with Ca=0.28 uF, Lb=225 mH, and Cb=45 pF, I get,
with mode 31:33:35:
Ca= 280.0000000000 nF
La=  36.6939598420 uH
Cb=  45.0000000000 pF
Lb= 225.0000000000 mH
kab=   0.1205497549
Output frequencies:  47072.74,  50109.69,  53146.64 Hz
This results in, with square wave input at 50109.69 Hz:
Maximum VCa (V)=     6276.84957 (   5.51584 J) at 578.69787 us
Maximum ILa (A)=      552.82838 (   5.60719 J) at 573.89739 us
Maximum VCb (V)=   991803.80546 (  22.13268 J) at 164.57646 us
Maximum ILb (A)=      -13.99169 (  22.02384 J) at 498.84988 us
If the terminal breaks down at a good fraction of 1 MV, I
think that this may work. Note that the element values are
almost the same of the first design. Nothing is very critical.

Antonio Carlos M. de Queiroz