Re: SSTC theory

["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>
 >
 >  >Then required bandwidth resulted too narrow, and so the coupling
 >  >between the coils become too loose. This doesn't look as a realistic
 >  >design...
 >
 > I had real trouble trying to calculate this one too. Doing it as a
 > quasi-disruptive design, using conservation of energy, I ended up with 100uH
 > and 20nF for the primary components, and the primary capacitor would ring to
 > over 44kV.

Another problem with too high Q systems. The voltage over the primary
capacitor may rise excessively.

 > Using the L-match approach for a CW design, starting with a secondary top
 > impedance calculated from (power input/breakout voltage) I couldn't find a
 > solution at all :( If I wanted a reasonable loaded Q for the primary, the
 > inductance tended to infinity :((

A simple L-match for very high resistance changes results in impractical
elements. In that case:
Power input (peak, at the driving frequency): 310*(4/pi)*600=237 kW
Breakout voltage: 300 kV
If all the power goes to the output: Iload (peak)=237/300=790 mA
Rload=300000/0.79=380 kOhms
Input resistance: 310*(4/pi)/600=0.659 Ohms
An L-match network operating at 100 kHz and converting 380 kOhms
to 0.659 Ohms:
Q: 759 !
C: 3180 pF
L: 0.796 mH
This circuit would take forever to reach the steady state.

It's better to make a double L-match, performing two impedance
conversions by the same factor:
Q: 27.5
C2 = 115 pF
L2 = 21.9 mH
(These convert 380 kOhms to 500.4 Ohms)
C1 = 87.6 nF
L1 = 28.9 uH
(These convert 500.5 Ohms to 0.695 Ohms)
Looks much better. Reaches 300 kV in 22 cycles (simulated).
The stored energy in C2 reaches 5.18 J.

 > I haven't figured out how to turn the handle on Antonio's method yet :(((

Me too ;-) If the design above looks reasonable (even with the
quite too large C2), a doubly tuned band-pass design shall be similar:
Rload=380 kOhms, Rin=0.659 Ohms
Voltage gain = sqrt(380000/0.659)=759
Only the bandwidth remains free: Using 5 kHz and the formulas for the
doubly terminated filter:
Ca:  85.5 nF
La:  29.7 uH
Lb:  21.3 mH
kab: 0.035
Cb:  119 pF
Almost the same values, and 5.3 J in Cb.
Note the low coupling.

Antonio Carlos M. de Queiroz