Re: Equalizing resistors for electrolytic capacitors

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Jan Wagner" <jwagner@xxxxxxxxx>

Hi,

On Sun, 28 Nov 2004, Tesla list wrote:
>    I have a large number of small electrolytic capacitors I am trying to 
> connect in series. Since they have the nominal -10% to +85% value control 
> on them the potential exists for a unequal division of voltages.

If you have many of these electrolytics available, what you can do is 
simply measure their capacitance. For your DC smoothing/supply cap you can 
take those that have about the same capacitance.
Even then, adding a parallel resistor to bleed off the charge from each cap 
would still be a good idea, that way you'd know that in at least X seconds 
the cap bank will have discharged to a safe voltage level.

> I understand than resistors across each capacitor can help solve this 
> problem, but have found no formulas or examples on what values should be 
> used. Is it an arbitrary thing?

This may be wrong, but ;-) : I think the resistance is quite arbitary 
though it has to be "small". Something like
100 * 1/(2*pi*(f_mains/2)*C_singlecap) might be a good place to start, e.g. 
for 50Hz mains and a 10uF cap 64kOhm, metal film type resistor.
There's also the difference in electrolytic cap leakage currents that leads 
to some specific equalizing resistor value that should be used, maybe 
someone knows the correct formula for this...

> Any help you could provide would be greatly appreciated.
>
> So far I have come up with:
>
> I understand the time constant of Tau(R*C) correctly it would be:
>
> 47000 * (330*(10^-6) = 15.51 seconds.
>
> so in about a 78(5 Time constants = 98% charge gone) seconds provided my 
> caps are at full voltage they will only be at 2 volts each with a 47k 
> resistor across each one.

That's if you disconnect the supply.
 V = V0 * exp(-t/RC)
E.g. if t=5*RC, V = V0*exp(-5) = (1-0.9934)*V0

With the voltage supply still on, the voltage accross each cap is about
 V0i = V_supply * Ri/(sum Ri)
so if all Ri={R1,R2,..Rn}=10kOhm then V0 = V_supply * 1/n  is the voltage 
the individual capacitors (n pieces of them in series) see once everything 
has settled. Meaning, if all Ri all the same value, then the voltage 
accross each capacitor is the same == "equalized".

Btw, you should de-rate the cap voltages a bit, if it says 450V working 
voltage then maybe use 380V or less.

cheers,
 - Jan

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