Designing an optimized Magnifier

To: tesla-at-pupman-dot-com
Subject: Designing an optimized Magnifier
From: "Antonio Carlos M. de Queiroz" <acmq-at-compuland-dot-com.br> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)
Date: Thu, 29 Jul 1999 08:04:25 -0600
Approved: twftesla-at-uswest-dot-net
Delivered-To: fixup-tesla-at-pupman-dot-com-at-fixme
Hi all:

I post below what I have obtained working a bit the ideas that I found 
in the papers that I mentioned a few days ago. A set of formulas that
allow, at least ideally (losses ignored, lumped model assumed as valid),
the design of a perfect magnifier circuit, are presented.

The magnifier is here modeled by the circuit shown below (ascii art):

       gap
    +--o o--+ k12 +----+---L3---+
 +  |       |     |    |   +    |   +
Vc1 C1      L1    L2   C2 Vc2   C3 Vc3
 -  |       |     |    |   -    |   -
    +-------+     +----+--------+

The capacitor C1 is initially charged to a moderately high voltage,
and when this voltage is high enough it discharges through the primary
of the transformer L1L2, that has a coupling coefficient k12. The
system enters then a complicated oscillatory transient, that aventually
produces a very high voltage at the top terminal of the resonator
coil, or third coil, L3. C3 represents the sum of the self-capacitance
of L3 and the capacitance of its top terminal. C2 represents the
output capacitance of the transformer, added to other distributed
capacitances at that point.

When C2 is negligible, this system behaves exactly as a normal Tesla
coil, with two oscillatory modes, as I discussed in previous posts. 

If C2 is significant, however, a third oscillatory mode appears, and
the transient is more complex. The ideal design would eventually put
all the energy that was stored in C1 in C3 only, with zero voltages
at C1 and C2, and zero currents at L1, L2, and L3 at that moment.

The problem is at first sight very complex, but it has a closed form
solution, that is described (with some errors in the formulas that I
had to figure out) in [1][2].
The idea is that the ideal relation among the three oscillation
frequencies is one that makes them to be ratios of integer numbers
k, l, m, so that l=k+1, k+3, k+5, ..., and m=l+1, l+3, l+5, ...
Ex: k,l,m = 1,2,3; 1,2,5; 2,3,4; 2,3,6; etc.

The design formulas, as functions of these three integers, are:

w*w*L1*C1 = (2*m*m*k*k+(m*m-l*l)*(l*l-k*k))/(2*k*k*l*l*m*m)
w*w*L2*C2 = (l*l)/(k*k*m*m)
w*w*L3*C3 = 1/(l*l) (one/lowercase L squared)
L2/L3     = ((l*l-m*m)*(k*k-l*l))/(2*k*k*m*m)
k12*k12   = ((k*k-l*l)*(l*l-m*m))/(k*k*(l*l+m*m)-l*l*(l*l-m*m))

Where w*k/(2*pi) is the lower oscillation frequency in Hertz.
The complete energy transfer occurs after pi/w seconds, at the
k+1 cycle of Vc3.
As for every (k,l,m) there are 8 unknowns to be computed from 5
relations, 3 values are arbitrary. For example, w, C1 and C3, 
fixing the voltage amplification factor, that is sqrt(C1/C3), 
and the energy transfer time.

When m is much greater than k and l, the circuit behaves as when
C2=0, and the formulas still work correctly, with the "magic" values
of the coupling coefficient obtained from the several combinations
of k and l.

Example (nothing specially optimized chosen):

Let (k,l,m)=2,3,4, and let's fix L1, L3, and w:
w=2*pi*200000 (transfer in 2.5 us)
L1=0.1 mH
L3= 20 mH
Results in:
L2=5.47 mH
C1=896 pF
C2=16.28 pF
C3=3.52 pF
k12=0.463

A simulation of this circuit shows perfect transfer in 2.5 us, and
159.5 kV in C3 for initial 10 kV in C1, corresponding exactly to the
ideal factor of sqrt(896/3.52)=15.95. 

Thanks to Barry (B2) for mentioning [1] and sending me the paper [2],
that contains informations that were missing in [1].

References:

[1] F. M. Bieniosek, Review of Scientific Instruments 61 (6) p. 1717,
June 1990
[2] F. M. Bieniosek, Proc. 6th IEEE Pulsed Power Conference, p. 700,
1987

Antonio Carlos M. de Queiroz