Maximum voltage gain in a Tesla coil

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

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

Hi:

Is Marco Denicolai still in the list? He would find this interesting.

After a discussion with Jim Lux, I wrote an "optimizer" program for
the lumped model of a Tesla coil during the energy transfer transient:

     +--R1--+        +--R2--+
  +  |      |        |      |   +
Vc1 C1     L1 <-k-> L2     C2 Vc2
  -  |      |        |      |   -
     +--<---+        +-->---+
       Il1             Il2

The program calculates exact solutions for the transient that starts
with an initial Vc1, and tries then to optimize the circuit following
several criteria.
One of the possibilities is to generate the maximum possible
voltage gain. Departing from a coil with the usual tuning relation:
T=(L2*C2)/(L1*C1) = 1
that has a voltage gain equal to:
A=sqrt(L2/L1)=sqrt(C1/C2)
the program changes one of the capacitors and k, trying to generate
the largest possible Vc2 after an unespecified time.
It's well known that it's possible to detune the coil, increasing
C1 or decreasing C2, sacrificing complete energy transfer and
efficiency, and obtain a small increase in gain before the
detuning forces a fall. Adjusting k too allows a slightly
larger gain. The objective is to obtain the maximum gain with
the same coils. (The new maximum gain sqrt(C1/C2) is not reached.)
See the paper by Marco in the Review of Scientific Instruments,
73,8, Sept. 2002, and its references, where this possibility is
demonstrated.
The largest voltage gain reported is with T=0.541136 and k=0.545659,
what generates Vc2/Vc1=sqrt(L2/L1)*1.1802.

I was imagining that my program would find this solution. But to my
surprise it found another, better: T=0.479222 and k=0.593349, that
results in Vc2/Vc1=sqrt(L2/L1)*1.19950, at the first negative peak.

This same solution produces a slightly larger positive peak 4
cycles later (*1.19957), and further optimization increases this
peak to *1.2307 with T=0.4603 and k=0.6054.

Similar things happen with smaller ks, but then the gain over the
tuned design is smaller.

Useful? Probably not. But curious.

The program can also optimize the circuit in the presence of losses,
represented by R1 and R2, finding solutions with maximum voltage
gain, maximum efficiency, and complete energy transfer.
It was interesting to see that solutions with complete energy
transfer (at some point all the remaining energy is in C2) are
possible in the lossy case. The differences between the optimized
circuit and the lossless circuit are negligible, however.
Example:
C1=4.5 nF
L1=1 mH
L2=30 mH
C2=15 pF
k=0.18033
The voltage gain is 5.477.
Adding 20 Ohms as R1: A=4.5766.
Optimizing for maximum voltage gain, changing k and L1:
k=0.18018, L1=1.0165 mH. The result is A=4.5813.
The solution with complete energy transfer is inferior:
k=0.1671, L1=0.988 mH, A=4.5367.

Antonio Carlos M. de Queiroz