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Re: What happens in series caps? (MMC)

Scott! and Mike. N (I will also post this to the pupman list)

Thanks for a good explanation, I will try to be more presice about, what
I don`t understand about the middle plates:

Let`s say that we have got 2 identical stacks of capacitors. The one of
them is simply capacitors in series, the other has shunt resistors
added.

In the array with the resistors, these resistors will divide the voltage
so that the intermediate plates are lifted to a potential, proportional
to their relative position in the stack. I assume that the plates are
supplied with electrones flowing in the resistors to do this.

In the unshunted stack of capacitors, I would assume that the
intermediate plates would also be on similar increasing potentials, from
zero at the bottom to max. at the top, but how? Where do the electrones
come from, to build the necessary charge. Are they pulled off from the
dielectric?

Since an unshunted capacitor stack is known to be able to leave charge
in the middle capacitors, it is hard to understand, that this charge is
able to _get_ there when the caps gets charged, but unable to get away
again when the cap is discharged.

Cheers, Finn Hammer

Scott Hayward wrote:
> 
> Dear List,
> I'll give this my best shot, if I'm wrong please correct me.  A capacitor IS
> two plates seperated by a dielectric.  These two plates have area, and
> therefore electrons can build up on one plate, and be removed from the other
> plate.  This leaves one plate with a positive charge, because it has less
> electrons than the other plate, which has a build up of negative charges on
> it (electrons).  The limiting factor is the area, because when electrons
> move onto a metal plate they will eventually start repelling each other, and
> at that point no more electrons can "fit" on to the plate.  The more area
> there is, the more charges that will "fit" before the charges start
> repelling each other.  If we have two plates separated by a distance (d)
> with a voltage (v) across them, then we have an electric field.  This
> electric field = v/d =f/q where f=force in newtons and q= charge in
> coulombs.  If there is a charged particle on one plate (eg: electrons on the
> negative plate), then the electric field will exert a force on the electron
> dragging it to the positive plate, just as the earths gravitational field
> will drag an apple from one height to the lowest height possible.  So a
> strong electric field can actually drag electrons from one plate to another,
> building up charge and hence storing energy.  I think in series caps, the
> middle caps are at 0V potential with respect to the charged plates and so
> that creates the electric field.  This then drags electrons off the plate at
> 0V changing the voltage so that the next plate along can have elctrons
> dragged off it.  On this part I am very sketchy.  When you have series caps
> you are effectively increasing the amount of dielectric, and that increases
> the puncture voltage.
> 
> Hope that helps a bit
> 
> Scott Hayward   smash-at-eisa-dot-net.au
> 
> ".......Blowing fuses since 1982"
> 
> >
> > > Untill this afternoon, my understanding of the function of a capacitor
> > > was like this:
> > >
> > > 2 plates, separated from each other with a dielectric, which stops
> > > electrons from passing from one plate to the other. This way, electrons
> > > are blocked from passing _trough_ the capacitor, but they can travel
> > > from one plate to the other trough the external circuit, either as a DC
> > > pulse, or as an AC oscillation.
> > > However, the potential difference between the plates is built by adding
> > > electrons to one plate and/or depleting electrons from the other plate.
> > >
> > > This level of understanding has worked for me up to this afternoon, when
> > > I was picking my nose at a red traffic light :-), and it struck me, what
> > > when there are several capacitors in series? How do the electrons gain
> > > access to all the middle plates? there is dielectric inbetween!, and no
> > > external connection.
> > >
> > > I feel I must have missed out on something very basic, here.
> > >
> > And why would it double the puncture voltage?
> >