Re: The death of a classic - First look

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

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

Tesla list wrote:
> 
> Original poster: "Malcolm Watts by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <m.j.watts-at-massey.ac.nz>

>       The *best* way of ensuring that the voltage across each cap is
> equal or nearly so at all times is to closely match each cap in a
> string, something I did right at the outset.

The self-healing feature can result in runaway degeneration of a 
capacitor in a string. Each healing decreases the capacitance, and
so increases the voltage over the capacitor in the following cycles.
The resistors equalize well the DC voltage over the capacitors
during the charging by the power supply, but have little effect
during RF cycles. Imagine this simple case, or two units only:

o-+-L--+-----+
  |    |     |   +
  |    R     C   v1
  o    |     |   -
 gap   +-----+
  o    |     |   +
  |    R    kC   v2
  |    |     |   -
o-+----+-----+ 

The two capacitors are charged to identical voltages v1=v2=v, and then
the gap fires. The resistors can then be ignored. The total capacitor
voltage will oscillate with peak voltage 2v, but the peak voltages
over C and kC will reach 2vk/(k+1) and 2v/(k+1) respectively. If k<1
the changed capacitor will see greater voltage. The resistors
introduce a decay making the oscillations tend to zero,
but the peak values of the oscillating voltages remain close to 
this ratio all the time.

Consider also that the damaged capacitor plates have less metal
left to conduct the current, and so start to heat up.

Antonio Carlos M. de Queiroz