Re: rectifier stack experts?

["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 Craig,
          Further to the question I just posted, I performed the 
following tests using a dual bench supply with isolated outputs, 
detailed below:

On 24 Oct 2002, at 9:51, Tesla list wrote:

> Original poster: "Craig Fuller by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <cbfull-at-hotmail-dot-com>
> 
> 
> 
> 
> >Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-qwest-dot-net>" 
> ><m.j.watts-at-massey.ac.nz>
> >
> >On 22 Oct 2002, at 18:44, Tesla list wrote:
> >
> > > Original poster: "Craig Fuller by way of Terry Fritz
> ><twftesla-at-qwest-dot-net>" <cbfull-at-hotmail-dot-com>
> 
> >What do you mean by "potentially"? Consider one half cycle and
> >replace each conducting diode by a piece of wire in your diagram.
> >What you actually get is two reverse-biased diodes in parallel across
> >the transformer/capacitor which are also essentially in parallel. In
> >a f.w. bridge configuration, Vtransformer and Vcapacitor never appear
> >added in series across anything. Very different from centre-tapped
> >f.w. or voltage-doubling half wave configurations I might add.
> >
> >Regards,
> >Malcolm
> 
> I like your simplification of replacing forward biased diodes with wires.  
> It helps to see what's happening more clearly.  I probably could have found 
> a better word to use than "potentially".  The engineering book I was reading 
> did not go into much detail as to why the bridge must be rated Vpeak * 2.  I 
> assumed it was because the diodes and capacitor tend to act together in a 
> fashion similar to how a cascade multiplier works to increase voltage.  
> Another reason I didn't question it was that I once rectified the output of 
> a 15kV NST with a full bridge rated for 25kV, 1A continuous, which worked 
> fine until I added a capacitor.  After that, it lasted about two minutes.  
> Maybe I will do some more research to see if I can find something that goes 
> into more detail.  Maybe you could too and post your findings.  I think it 
> would help everyone of us tinkerers to know that we are needlessly 
> destroying our diodes when it could be avoided.


 0 to +40V
    o-------------------+------------+
                        |            |
                        v           ---
  SUPPLY 1             ---           ^
                        |  A      B  |
                        +--o +  - o--+    SUPPLY 2 set to 40V,
                        |            |    applied with polarity
                       ---           v    shown across A and B
                        ^           ---   (represents a charged
   0V                   |            |     capacitor)
    o-------------------+------------+

SUPPLY 1 was adjusted from zero to 40V to represent the real supply 
waveform across a transformer (although it never hits the capacitor 
voltage of 40V + 2 junction drops so that the bridge fully conducts 
in one direction, the wire-replacement analysis covers that case).

     To really put the heat on, I placed voltmeters with different 
current draws across diodes in opposite legs of the bridge. This 
situation represents one in which a diode is leakier than the others.
In no case at any setting of SUPPLY 1 from 0 to 40V did more than the 
capacitor voltage (40V) appear across any one diode. With the unequal 
"leakage" currents, the burden was unevenly shared but still the sum 
across the two did not exceed the capacitor 40V. I also metered the 
other two diodes with a third voltmeter and the first two still in 
place with the same result.

I admit that analysing this circuit is a bit of a mind-bender. 
Perhaps fast risetimes, parasitic circuit Ls and Cs and/or slow or 
unequal diode recovery times changes this but it is not easy to see 
how. Connected ringing circuits on the other hand are a whole new 
ball game.

Regards,
Malcolm