RE: MMC resister problem

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Lau, Gary by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Gary.Lau-at-hp-dot-com>

>> Original poster: "Lau, Gary by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<Gary.Lau-at-hp-dot-com>
>> 
>> I fully agree that battery packs can achieve reverse-charges on cells, but
>> this is very different than a string of caps.  A NiCd cell is not a
>> capacitor and is not at all a linear device.  A "bad" cell had radically
>> different V-I properties than the remaining good cells.  Even good cells
>> have different V-I characteristics depending upon their state of charge.
>> But in MMC strings, we don't have a "bad" cap that is responsible for the
>> imbalance.  As far as we understand it, all caps are electrically the same.
>>  The observed results of course suggest otherwise, but I've not yet seen a
>> model suggested that can explain the effect.
>> 
>> While iron-core inductors exhibit hysteresis, I wasn't aware that caps had
>> this property?
>> 
>> Gary Lau
>> MA, USA


>Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<evp-at-pacbell-dot-net>
>	If the capacitances aren't identical, when the series is shorted (or
>discharged through a resistor), the smaller capacitors will have
>reversed polarity if the voltages across each capacitor have been
>equalized by a resistor shunted across each one.  Just ohm's law. 
>Consider a large capacitor and a small capacitor in series with the
>series string charged to some voltage.  If the voltages on each are the
>same (again by means of a resistor divider) then the larger one will
>have more charge than the small one.  On discharge the current will be
>the same for each capacitor, and the smaller capacitor will end up with
>reverse voltage.

I think the assumption that the voltages are being equalized by virtue of
the bleeder resistors is invalid.  A 0.15uF cap with a 10 Meg resistor
across it will have a time constant of 1.5 seconds.  About 180
charge/discharge cycles will have occurred in this timeframe.  For the case
where there are no bleeder resistors, the residual charge problem is most
evident.  When using reasonable-valued bleeder resistors (or none), the cap
voltages will divide based just on the capacitance values.

>	In general, capacitors also have some hysteresis due to a process
>called dielectric absorption.  This manifests itself as a rise in
>terminal voltage after the capacitor is shorted and then open
>circuited.  For good dielectrics (quartz, polystyrene, teflon) the
>effect is very small.  For ordinary oil-filled paper capacitors the
>recovery can be several percent of the original voltage.  It is fairly
>common practice to wrap a shorting wire around the terminals of HV
>filter capacitors, to prevent serious shock.  The effect is often
>modeled by considering the capacitor to be composed of one perfect
>capacitor in parallel with a number of different series RC circuits of
>smaller capacitance and different time constants.  I have measured very
>good capacitors intended for use in precision analog differentiators,
>and even for them current flow (very small at the end) can be measured
>for hours.
>
>Ed

Agreed.  I didn't think the definition of hysteresis included
time-dependant effects, hence my confusion with the term.  Maybe my
definition is too narrow.  I'm still not sure if this dielectric
absorption-effect is pertinent to the issue at hand.

Regards, Gary Lau
MA, USA