Re: Best secondary size

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

Original poster: Rob Maas <robm-at-nikhef.nl> 

At 07:23 AM 9/15/2003 -0600, you wrote:
 >Original poster: "Malcolm Watts" <m.j.watts-at-massey.ac.nz>
 >
 >Hi All,
 >         I'd have to take issue with this formula:
 >
 >On 13 Sep 2003, at 14:05, Tesla list wrote:
 >
 > > Original poster: "Dr. Resonance" <resonance-at-jvlnet-dot-com>
 > >
 > >
 > > The 240 mm (9.5 inch) dia. sec would offer more performance as output
 > > potential is:
 > >
 > > Vout = Vin x k x sqr (Ls/Lp)
 >
 >In the lossless case, k doesn't matter and may be any value. With a
 >normal lossy situation, the degree of loss must be balanced against
 >the degree of k to get the correct answer. Since the loss in a
 >normally operating sparking coil is a parametric value and changes by
 >the nS, I'd consider it impossible to quantify. Hence the normal
 >practice of quoting the lossless ideal Vo = Vi x SQRT(Ls/Lp).
 >
 >Malcolm
 ><snip>
 >
Hi,

You are right. I get the feeling sometimes coupling and efficiency are being
confused in these discussions.

The coupling between prim. and sec. coil, k,  is defined by

            k = M/SQRT(Lp*Ls)                    (1)

M being the mutual inductance of prim. and sec. coil.

The efficiency, eps, of the system expresses how much of the primary
energy (e.g. in C1) will *eventually* end up in C2:

            Es = eps*Ep                          (2)

There is not really a direct link between k and eps, but there *is* a
link between how many oscillations it takes to transfer the total
primary energy, Ep, to C2 and k. The smaller k, the longer it takes
for Ep to end up at C2.
In [1] the following approximate expression is given for this transfer
time tau

            tau = (1/(2*f_res))*SQRT(1-k^2)/k    (3)

so if k = 0.15 -> tau = (1/(2*f_res))*6.6
    if k = 0.3  -> tau = (1/(2*f_res))*3.2

in case k = 0.15 it takes 6.6 half oscillations for Ep to arrive at C2;
in the ideal case then the spark gap will quench, so the energy
stays in the secundary circuit and will exponentially decay.

Since Ep = 0.5*Cp*Vp^2, and the same for Es, one gets from (2)

            Vs = Vp*SQRT(eps*Cp/Cs)              (4)

or, expressed in inductances,

            Vs = Vp*SQRT(eps*Ls/Lp)              (5)

k does not occur in (4) and (5)

------------------
[1] - Kenneth D. Skeldon, Alastair I. Grant and Slan A. Scott, "A high
potential
       Tesla coil impulse generator for lecture demonstrations and science
exhibitions",
       Am. J. Phys. 65(8), 744-754 (1997)



hope this is helpful,    Rob Maas