Re: cockcroft-walton question

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

Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>


 >  > About 15 years ago I posed this question to Beau Meskin, Pres. and
Chief EE
 >  > at Plastic Capacitors, Inc., in Chicago, and he told me to be sure to
use
 >  > all equal values in each stage other wise you might form some type of
 >  > capacitive voltage divider and some stages would see over-voltages.

 >  >  > The usual situation is to use equal valued C and PIV, to reduce
ripple,
 >  > you  increase C at the bottom stages of the stack,
 >  >  > so they go, e.g.
 >  >  > N*C, (N-1)*C,...3*C, 2*C, 1*C
 >  >  >

 > In spite of what the Plastic Capacitors engineer said, both experience
 > and simulation say otherwise.  The bigger the capacitors the better, of
 > course.  I can see his point but think he's wrong.
 >

Interesting point about effective dividers, though... Consider a big
transient propagating back into the CW stack. It will divide evenly if all
the caps are the same, but, if you've used the "low ripple" 1,2,3,4 sizes,
then the top stage will take the most voltage (because the C is smallest)...

With modern switch mode drivers and fast recovery HV diodes where you can
raise the pump frequency fairly high, at fairly low cost,  I suspect that
the "equal C" approach is far more popular, as it gives you the lowest
ripple, for a given output power, for a given cost.

The tradeoff analysis might be a bit different in days of yore, where the
relative costs of rectifiers and diodes were different.  Certainly, if you
had to go out and buy (or build) thermionic rectifiers for each stage, and
worry about the filament supplies, etc., minimizing the number of stages,
etc. would be very attractive.