Re: Formless secondaries

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

Original poster: "Antonio Carlos M. de Queiroz by way of Terry Fritz <teslalist-at-qwest-dot-net>" <acmq-at-compuland-dot-com.br>

Tesla list wrote:
 >
 > Original poster: "Mark Snoswell by way of Terry Fritz 
<teslalist-at-qwest-dot-net>" <mark-at-cgCharacter-dot-com>

 > > If the energy
 > > transfer occurs in n cycles, there is no significant gain if the coil
 > > has a Q of more
 > > than, say, 10*n.
 >
 > That's interesting -- could you elaborate? ... With typical practical TC
 > values?

Ok. Let's see if the value that I guessed is correct:
Suppose a coil like this:
http://www.coe.ufrj.br/~acmq/tesla/tefp.html

The secondary has L2=28.2 mH, and R2, measured at DC, is 150 Ohms.
The coil resonates at f=290 kHz, and so the ideal secondary Q would be
of 2*pi*f*L2/R2=343. In practice it is significatively smaller, due
to skin losses (small), proximity effect, and form losses. My
measurements agree reasonably with R2=1000 Ohms, and so Q2=51.
The energy transfer occurs in ~4 cycles.
With 5000 V of primary voltage, C1=5.07 nF, and C2=10.55 pF, the
maximum output voltage, without losses, reaches:
V2max = 5000*sqrt(5.07/0.01055)=110 kV
A simulation with the only loss being R2=1000 results in:
V2max = 97.9 kV (10% below the maximum)
For Q2=40, R2 has to be 1285 Ohms. A simulation with this R2 shows:
V2max = 94.6 kV (13% below the maximum)
So, the criterion of Q2 > 10 x number of cycles for energy transfer
seems reasonable.
The number of cycles for complete energy transfer depends on the
coupling coefficient. For energy transfer in n cycles, the
coupling coefficient is k=((2*n)^2-(2*n-1)*2)/((2*n)^2+(2*n-1)^2).
The system can be simulated with the program Teslasim
http://www.coe.ufrj/~acmq/programs

Antonio Carlos M. de Queiroz