Re: The death of a classic - First look

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

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

Hi Antonio,

On 16 Aug 2002, at 19:25, Tesla list wrote:

> 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

You're right - I overlooked the low duty cycle. 

Regards,
Malcolm