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 would like to see a numerical example for comparison.
I will try a numerical example with some of my ideas. Initially we
agree:

 > 1: Decide what length of spark you want to produce.

1 meter, or 39".

 > 2: Use Freau's equation (spark length=1.7*sqrt(power)) to calculate the
 > power you will need to do this. Then divide it by two because DRSSTCs are
 > twice as efficient :)

Power = (39/1.7)^2/2 = 263 W

 > 3: From the power, and knowing the breakrate which is usually 100 or 120,
 > calculate the bang energy.

Energy = 263/120 = 2.2 J

 > 4: Now the fun starts. In a DRSSTC the primary and secondary resonators ring
 > up together. When breakout occurs they both empty into the streamer load
 > together. The split of energy between the primary and secondary is
 > approximately (1-k):k. So use the tightest coupling possible, we will assume
 > that you managed to get k=0.33. Knowing this we can size our primary and
 > secondary for energy storage.

Now we diverge. I canīt find a good solution with high k.
Trying a solution with the fastest possible energy transfer (excitation
at the arithmetic mean of the resonances):
I will use a large toroid, with, say, 50 pF of capacitance. I will
assume that the energy in the self-capacitance of the secondary coil is
included in this too, and that it also contributes for the bang energy,
what is probably not true. The output voltage has to reach:
Vmax = sqrt(2.2/(0.5*50e-12) = 297 kV
I will assume that the input voltage is 310 V (220 V rectified), and
that the driver can switch, say, 200 A.
I need a voltage gain of 297000/(310*4/pi) = 752.

My design procedure would be:

1) I choose the mode, that is the ratio of the two resonances and the
driving frequency, based on the k that I want to use.
The best solutions are ratios of successive odd integers, with
sinusoidal (or square wave from a bridge) input.
Mode 31:33:35 results in k=0.12 and energy transfer in 8.25 cycles.
(The formula for k is a bit complicated to list here, but is
implemented in my sstcd program.)

2) I know that the voltage gain is proportional to sqrt(Ca/Cb).
(For mode x:x+2:x+4, x and odd integer, it is:
  Av=sqrt(Ca/Cb)*sqrt(x*(x+4))/2 )
The maximum input current decreases if the inductances increase
(probably a square root relation too, but I didn't verify).
With all the design formulas implemented in my sstcd program (I still
didn't write a page with the formulas...), I just have to adjust
Ca until the voltage gain is high enough and then adjust Lb so
the current is ok.

In the example, I get the final element values:
Ca= 105.0000000000 nF
La=  28.9927583937 uH
Cb=  50.0000000000 pF
Lb=  60.0000000000 mH
kab=   0.1205497549
That results in:
Output frequencies:  86478.14,  92057.37,  97636.61 Hz
Voltage gain= 754.7350528497
Bang energy (square wave input):    2.2185676457 J
Maximum VCa (V)=-3242.78978 (0.55207 J)  at 43.44887 us
Maximum ILa (A)=  196.61827 (0.56041 J)  at 46.05492 us
Maximum VCb (V)=297670.20709 (2.21519 J) at 89.61579 us
Maximum ILb (A)=    8.57035 (2.20353 J)  at 92.31185 us

It is possible to design a system with higher k, that would transfer
energy completely in less cycles, but with lower voltage gain, if
excited at the central frequency, and then excite it at one of the
resonances to obtain an arbitrarily large gain (in the absence of
losses). But there is always a design as the one above that
requires less input current to transfer the same energy in the same
number of cycles (or I think so...).

Antonio Carlos M. de Queiroz