Re: Marx using inductors (fwd)

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Original poster: Steven Roys <sroys@xxxxxxxxxx>



---------- Forwarded message ----------
Date: Sun, 25 Jun 2006 17:15:32 -0700
From: Jim Lux <jimlux@xxxxxxxxxxxxx>
To: High Voltage list <hvlist@xxxxxxxxxx>
Subject: Re: Marx using inductors (fwd)

At 09:59 AM 6/25/2006, you wrote:
>Original poster: Steven Roys <sroys@xxxxxxxxxx>
>
>
>
>---------- Forwarded message ----------
>Date: Fri, 23 Jun 2006 10:54:27 +0800
>From: Peter Terren <pterren@xxxxxxxxxxxx>
>To: High Voltage list <hvlist@xxxxxxxxxx>
>Subject: Re: Marx using inductors (fwd)
>
>Never made a Marx myself but I would have thought it wise to have the
>driving voltage referenced to ground.  Otherwise any sparks or leakage
>currents from the output to ground will blow diodes in your multiplier. So
>you need to use only half of your NST to keep the ground reference.


You can build a marx that is charged by a bipolar supply symmetric around 
ground (e.g. a doubler driven by a NST)..

  The whole Marx thing is to charge in parallel, discharge in series.  For 
the upper stages, they're floating relative to ground anyway, so it makes 
no difference where the charging voltage comes from.

Conceptually, think about each stage having two caps in series, with the 
usual charging resistor/inductors and a third one, connecting center taps 
to center tap and finally through a resistor to ground. The spark gaps go 
from + end of one stage to - end of next stage, in the usual fashion.  If 
your power supply puts out, say, plus and minus 5kV, the caps will all 
charge to +5 or -5kV, relative to ground, as you'd expect.  Then, fire the 
gaps, and the whole thing erects as usual.  Draw a schematic with the bank 
erected and you'll see what's going on.

Now.. here's the tricky part.  You can do away with the resistors in the 
middle (and hence, the need to have two caps in series), except for the 
very bottom stage.

Jim