RE: Capacitor - series?

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Original poster: Terry Fritz <teslalist-at-twfpowerelectronics-dot-com>



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From: Tesla list [mailto:tesla-at-pupman-dot-com]
Sent: Sunday, 8 February 2004 5:18 AM
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Subject: RE: Capacitor - series?

Original poster: "Luke" <Bluu-at-cox-dot-net>

Why would it aply at dc?
Caps don't realy have much current in them with dc.  Except when
originally charging that is.

Luke Galyan
Bluu-at-cox-dot-net


Exactly, and that's the only thing to be considered on DC - when they're
originally charging. When they are charged there's some equilibrium
reached, which can be determined for 'ideal' capacitors. Once charged to
equilibrium there is no further current flow (ideal).

Consider the series of two capacitors, uncharged, connected to a (dc)
battery. There is an initial inrush of current to charge the capacitors.
The same current must flow through both capacitors, and for the same
time in both.

Thus the charge (Q=It) is the same in both capacitors.  If you like,
they both have the same electron imbalance, numerically.

Q also equals C.V ,  so V = Q/C  Therefore the resultant voltage will be
larger on the smaller capacitor & vice-versa.

This is assuming ideal capacitors with perfectly insulating dielectrics
& infinite resistance. Real capacitors fall short of this to some
extent.  Even very high (leakage) resistances in, on or around the
capacitor can alter the results determined from capacitance alone.

For example, placing your multimeter (10M Ohm, say) across one
capacitor, slews theses results dramatically, and will not demonstrate
the theoretical result above ('eventually' your meter will read 0V
across either capacitor.)
Using much larger capacitor values & quick meter readings will tend to
bear it out.

Anyway, for what I assume you want to use it for all this is largely
irrelevant and the ac performance of the combination is what you should
be considering.

(Do this with pen & paper, think of what the electrons are doing,
Remember Q = C.V = I.t, consider the cap resistance infinite, and it may
become clearer.)

Regards,

Phil Chalk.




That's right, but in the example Dan chose that's how it is.

It depends a bit on what the capacitor is for. What you're talking about
is more appropriate at DC, with appropriately sized resistors) but still
not generally considered 'a good idea'.




Can we series caps that are diffrent in working voltage and capitance?
Example if we have a 1 nf 10 kV cap and a 10 nF 1 kV cap? If i series
them il get 11 kV capacitor? What will happen with capitance, will this
work?

Considering this particular example:-

For DC  (e.g. an energy-storage cap)

The series combination would yield a value of about .909nF  On DC, I
seem to recall that theoretically voltage distributes in inverse
proportion to capacitance (Q=CV conserved). In real caps it could be way
different.

That is, on 11kV DC, it would tend toward 10kV across the 1nF capacitor
& 1kV across the 10nF capacitor. To ensure this though, it would be very
wise to put equalizing resistors across the caps, again in inverse
proportion to capacitance - say 10M across the 1nF & 1M across the 10nF.

It serves to jack up the dc voltage rating of your 1nF 10kV cap by 10%
while reducing its capacitance by about 9%.  But it seems like a bit of
a waste of the 10nF 1kV cap.

On ac, 'thinking out loud' :- At 50Hz, the 1nF cap has Xc= 3.18M Ohm
approx. & the 10nF cap Xc=318k Ohm. When passing current at 50Hz (e.g.
as filter cap after a 1/2 wave rectifier) these values begin to swamp
the effects of the equalizing resistors, but the ac voltage still
divides in inverse proportion to the reactances, so the higher voltage
still develops across the smaller capoacitance. So it sounds, on the
surface, O.K. to me.

At 100kHz, (TC tank maybe) 1nF Xc=1.59k Ohm, 10nF Xc=159 Ohm.  Now the
equalizing resistors have virtually no influence, but the voltage will
divide as before.

Sounds OK then to me, but still not usually done. Better I think to use
equal voltage ratings (& ideally 'identical' caps) when seriesing
capacitors.

The particular example you chose with the 10:1 ratios happens to work
out 'OK', other more random combinations would not.  You need to
consider whats going on in each case.

Regards,

Phil Chalk.