Re: Reason why DC mag field disturbs communications on a train

On Sat, 21 Mar 2009 22:43:42 +0000, Guy Macon <http://www.GuyMacon.com/>
wrote:




Jamie wrote:

Guy Macon wrote:

Jamie wrote:

the 100Khz is coming form the 350KW oscillator tube that generates the
AC required in side of a gas pressured vessel where walt Crockcroft
multiplier is employed that can yield up to 5Mvolts.

You have to love those Cockroft-Walton voltage multipliers; no
single component has to withstand the entire output voltage.

Did you know about the trick of using larger capacitors (or same-size
capacitors in parallel) in the lower stages to reduce output ripple?

Yes, and I don't see where that is a trick. I thought that was common?

It probably is, but I have seen physics students build them from
the original 1951 paper by James D. Cockcroft and Ernest T. S.
Walton with equal value capacitors. I probably should look up the
original paper and see if it mentions larger capacitors near the
bottom of the stack.

THIS particular multiplier design has equal potentials across ANY one
segment in the string. ALL the way thru any number of stages.

There are, however, multiplier configurations that push more power, and
there are those that pump more voltage per stage, and there are those
that fire out a center tapped dual voltage output.

Depending on exactly how the stages are configured, voltage at each
component in a stage may or may not grow.

With the C-W configuration, it is the same across all components in all
stages.

That is why it is the most common choice in design. No need to use
different diode and cap elements along a string of multiplication stages.
The amount of energy stored in any one stage should also be equal. The
capacitance should be increased as much as possible, but one should not
store more energy than one plans to use. So the best way to reduce
ripple or PARD here is finding the right frequency of operation, and the
right amount of energy storage to make everybody warm and fuzzy inside.
There is a happy point.

So, you take your diode drops and you subtract that from your
transformer HV side out (secondary) from your step up transformer voltage
to get you per stage multiplication factor to get near what your output
voltage should be.
.