Real-world behavior of series-connected electrolytic capacitors

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
Having read hundreds (almost identical) textbook and tutorial texts as well as fora threads on the
subject, it seems that real-world behavior in this situation is not really an issue. Technical info from
the leading manufacturers of e-caps have lately at least adopted the insight that the voltage division
after charging is not determined by the ratio of the capacitances involved (as theory dictates it), but
by their respective leakage resistances. It would be much more interesting (and useful) though, to
establish, which physical entity/entities determine the initial voltage division, and learn how the
charging process progresses from there. References to relevant literature would be welcome -
preferably with as little math as possible: my interest is in the phenomena, not primarily calculation.
If you can come up with a description/explanation yourself - all the better. Please note, I am not
in the habit of burning midnight oil on e-mailing - so don't expect immediate response. Thanks,
Ray
 

crutschow

Joined Mar 14, 2008
28,158
If you apply a step voltage to two electrolytics, their respective voltages immediately after the step will be inversely related to their relative capacitance values.
If you apply a long term DC voltage, then their respective voltages will be determined by their relative leakage resistances.
 

Delta Prime

Joined Nov 15, 2019
1,187
Hello there welcome to AAC!
:) You have to understand how the capacitor is made in the real world. Now! this is only for electrolytic capacitors.
With "No" mathematical equations.
Which is really really hard for me.
Stand back... here I go!
Capacitors stores electrical energy directly, as an ELECTROSTATIC FIELD
created between two metal "plates".
The method of construction used in electrolytic capacitors is they're rolled foil with two very thin layers of paper, one that forms an insulator
separating the rolled pairs of layers and the other, a layer of tissue between the foil plates, soaked in an electrolyte that makes the tissue conductive.
It would seem that the soaked tissue places a short circuit between the plates,but the real dielectric layer is created after construction is complete,
in a process called "Forming". A current is passed through the capacitor,
and the action of the electrolyte causes a very thin layer of aluminium oxide to build up on the positive plate. It is this layer that is used as the insulating dielectric. imagine a "Battery" the positive terminal of the battery is connected to a switch that connects the positive side of the battery to the positive side of the electrolytic capacitor. When the switch is closed completing the circuit, electrons begin to flow from the negative battery terminal, and appear to be flowing around the circuit. Of course they can't because the capacitor has a layer of insulation between its plates, so electrons from the negative battery terminal crowd onto the right hand plate of the capacitor creating an increasingly strong negative charge. The very thin insulating (dielectric) layer between the plates is able to efficiently transfer this negative charge from the electrons, and this charge repels the same number of electrons from the left hand plate of the capacitor. These displaced electrons from the left hand plate are attracted towards the positive terminal of the battery, giving the impression of current flowing around the complete circuit. "But"! A large number of electrons have gathered on the right hand plate of the capacitor, creating a growing negative charge, making it increasingly difficult for electrons flowing from the negative battery terminal to reach the capacitor plate because of the repulsion from the growing number of negative electrons gathered there. Eventually the repulsion from the electrons on the capacitor's right hand plate is approximately equal to the force from the negative battery terminal and current ceases. Once the battery and capacitor voltages are equal we can say that the capacitor has reached its maximum charge. If the battery is now disconnected by opening the switch, the capacitor will remain in a charged state, with a voltage equal to the battery voltage, and provided that no current flows, it should remain charged indefinitely. In practice a very small leakage current will flow across the dielectric, and the capacitor will eventually discharge. This process however can take seconds, hours, days, weeks or months, depending on individual circumstances.Placing capacitors in series effectively increases the thickness of the dielectric, and so reduces the total capacitance. Because the total capacitance is inversely proportional to the distance between the plates voltage across each capacitor will be inversely proportional to the capacitance, with the total voltage being shared out between the capacitors, the smallest capacitance having the largest voltage across it and the largest capacitance having the smallest voltage etc. Capacitance depends on four things!
The area of the plates.
The distance between the plates.
The type of dielectric material.
Temperature.
The capacitance value is varied by either;
Changing the area of the plates.
Changing the thickness of the dielectric.
Capacitance (C) is DIRECTLY PROPORTIONAL TO THE AREA OF THE TWO PLATES that directly overlap, the greater the overlapping area, the greater the capacitance.
Capacitance is INVERSELY PROPORTIONAL TO THE DISTANCE BETWEEN THE PLATES.
if the plates move apart, the capacitance reduces.
The electrons on one plate of the capacitor affect the electrons on the other plate by causing the orbits of the electrons within the dielectric material (the insulating layer between the plates) to distort. The amount of distortion depends on the nature of the dielectric material and this is measured by the permittivity of the material.
Permittivity is quoted for any particular material as RELATIVE PERMITTIVITY, which is a measure of how efficient a dielectric material is. It is a number without units which indicates how much greater the permittivity of the material is than the permittivity of air (or a vacuum), which is given a permittivity of 1 (one). For example, if a dielectric material such as mica has a relative permittivity of 6, this means the capacitor will have a permittivity, and so a capacitance, six times that of one whose dimensions are the same, but whose dielectric is air.
Another important aspect of the dielectric is the DIELECTRIC STRENGTH. this indicates the ability of the dielectric to withstand the voltage placed across it when the capacitor is charged. Ideally the dielectric must by as thin as possible, so giving the maximum capacitance for a given size of component. However, the thinner the dielectric layer, the more easily its insulating properties will break down. The dielectric strength therefore governs the maximum working voltage of a capacitor.
It is very important when using capacitors that the maximum working voltage indicated by the manufacturer is not exceeded. Otherwise there will be a great danger of a sudden insulation breakdown within the capacitor. As it is likely that a maximum voltage existed across the capacitor at this time (hence the breakdown) large currents will flow with a real risk of fire or explosion in some circuits.
The charge (Q) on a capacitor depends on a combination of the above factors, which can be given together as the Capacitance (C) and the voltage applied (V). For a component of a given capacitance, the relationship between voltage and charge is constant. Increasing the applied voltage results in a proportionally increased
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
If you apply a step voltage to two electrolytics, their respective voltages immediately after the step will be inversely related to their relative capacitance values.
If you apply a long term DC voltage, then their respective voltages will be determined by their relative leakage resistances.
Thank you for your response. I'll answer to all responses together - please refer to my
reply to Delta prime.
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
Hello there welcome to AAC!
:) You have to understand how the capacitor is made in the real world. Now! this is only for electrolytic capacitors.
With "No" mathematical equations.
Which is really really hard for me.
Stand back... here I go!
Capacitors stores electrical energy directly, as an ELECTROSTATIC FIELD
created between two metal "plates".
The method of construction used in electrolytic capacitors is they're rolled foil with two very thin layers of paper, one that forms an insulator
separating the rolled pairs of layers and the other, a layer of tissue between the foil plates, soaked in an electrolyte that makes the tissue conductive.
It would seem that the soaked tissue places a short circuit between the plates,but the real dielectric layer is created after construction is complete,
in a process called "Forming". A current is passed through the capacitor,
and the action of the electrolyte causes a very thin layer of aluminium oxide to build up on the positive plate. It is this layer that is used as the insulating dielectric. imagine a "Battery" the positive terminal of the battery is connected to a switch that connects the positive side of the battery to the positive side of the electrolytic capacitor. When the switch is closed completing the circuit, electrons begin to flow from the negative battery terminal, and appear to be flowing around the circuit. Of course they can't because the capacitor has a layer of insulation between its plates, so electrons from the negative battery terminal crowd onto the right hand plate of the capacitor creating an increasingly strong negative charge. The very thin insulating (dielectric) layer between the plates is able to efficiently transfer this negative charge from the electrons, and this charge repels the same number of electrons from the left hand plate of the capacitor. These displaced electrons from the left hand plate are attracted towards the positive terminal of the battery, giving the impression of current flowing around the complete circuit. "But"! A large number of electrons have gathered on the right hand plate of the capacitor, creating a growing negative charge, making it increasingly difficult for electrons flowing from the negative battery terminal to reach the capacitor plate because of the repulsion from the growing number of negative electrons gathered there. Eventually the repulsion from the electrons on the capacitor's right hand plate is approximately equal to the force from the negative battery terminal and current ceases. Once the battery and capacitor voltages are equal we can say that the capacitor has reached its maximum charge. If the battery is now disconnected by opening the switch, the capacitor will remain in a charged state, with a voltage equal to the battery voltage, and provided that no current flows, it should remain charged indefinitely. In practice a very small leakage current will flow across the dielectric, and the capacitor will eventually discharge. This process however can take seconds, hours, days, weeks or months, depending on individual circumstances.Placing capacitors in series effectively increases the thickness of the dielectric, and so reduces the total capacitance. Because the total capacitance is inversely proportional to the distance between the plates voltage across each capacitor will be inversely proportional to the capacitance, with the total voltage being shared out between the capacitors, the smallest capacitance having the largest voltage across it and the largest capacitance having the smallest voltage etc. Capacitance depends on four things!
The area of the plates.
The distance between the plates.
The type of dielectric material.
Temperature.
The capacitance value is varied by either;
Changing the area of the plates.
Changing the thickness of the dielectric.
Capacitance (C) is DIRECTLY PROPORTIONAL TO THE AREA OF THE TWO PLATES that directly overlap, the greater the overlapping area, the greater the capacitance.
Capacitance is INVERSELY PROPORTIONAL TO THE DISTANCE BETWEEN THE PLATES.
if the plates move apart, the capacitance reduces.
The electrons on one plate of the capacitor affect the electrons on the other plate by causing the orbits of the electrons within the dielectric material (the insulating layer between the plates) to distort. The amount of distortion depends on the nature of the dielectric material and this is measured by the permittivity of the material.
Permittivity is quoted for any particular material as RELATIVE PERMITTIVITY, which is a measure of how efficient a dielectric material is. It is a number without units which indicates how much greater the permittivity of the material is than the permittivity of air (or a vacuum), which is given a permittivity of 1 (one). For example, if a dielectric material such as mica has a relative permittivity of 6, this means the capacitor will have a permittivity, and so a capacitance, six times that of one whose dimensions are the same, but whose dielectric is air.
Another important aspect of the dielectric is the DIELECTRIC STRENGTH. this indicates the ability of the dielectric to withstand the voltage placed across it when the capacitor is charged. Ideally the dielectric must by as thin as possible, so giving the maximum capacitance for a given size of component. However, the thinner the dielectric layer, the more easily its insulating properties will break down. The dielectric strength therefore governs the maximum working voltage of a capacitor.
It is very important when using capacitors that the maximum working voltage indicated by the manufacturer is not exceeded. Otherwise there will be a great danger of a sudden insulation breakdown within the capacitor. As it is likely that a maximum voltage existed across the capacitor at this time (hence the breakdown) large currents will flow with a real risk of fire or explosion in some circuits.
The charge (Q) on a capacitor depends on a combination of the above factors, which can be given together as the Capacitance (C) and the voltage applied (V). For a component of a given capacitance, the relationship between voltage and charge is constant. Increasing the applied voltage results in a proportionally increased
Thanks very much, Delta prime, for your elaborate explanation and friendly welcome. Think I owe all an apology - you foremost, though for providing me
with a crash course in e-caps -:).

In trying to avoid a long introductory story, I have probably made it too short ...
As I still wouldn't want to bother anyone with my (hi)story, let me try to present
a concise background:
I am not a newbie: in fact produced my first commercial electronic product in
1961 - commissioned; proprietary design. Experienced experimenter since.
I only turn to a forum after I've done my 'desk research' and are left in doubt,
or without an answer at all. Remember: most of my life I had to rely on books.
This publication awoke my interest in series-connected capacitors: "A Myth About Capacitors in Series", by L. Kowalski ('88), found on the Net only a few years ago.
Later A.P. French, in a mathematical approach, arrived at the conclusion, posted
here by Crutschow. The main objection I have is twofold. First, Kowalski declares
the equal charge theorem invalid and concludes that the formula for calculating
total capacitance does not hold as well. So why not investigate other, related
formulae, theorems and the like, but instead go along with the results of a math
approach, not particularly the place to expect critical treatment of such issues ...
Second, (back to reality) the physical entities to determine the current surge after
switch-on (t0) are the respective ESRs of the e-caps. And if the law about the current
in a series circuit being equal all over, holds, it seems odd that not the ESRs, but the
caps values should determine the initial voltage division. I am aware of the replies
and criticism I'm provoking with this suggestion: superficially there seems to be an
inverse proportional relation between the two parameters, but it is easy to find 20%
deviations of that 'rule' in data sheets (mind you, I am not talking tolerances, but specs). Once again: reality, i.e. developments in materials science, but also simple
differences in e.g. size, beats theory !
My own objection against it would be: in a series-connection ESRs do not represent
a voltage divider, as leakage resistances do.
But there is an other problem with 'unruly' ESRs: time constant ! But that is maybe for a next post.
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
Thanks very much, Delta prime, for your elaborate explanation and friendly welcome. Think I owe all an apology - you foremost, though for providing me
with a crash course in e-caps -:).

In trying to avoid a long introductory story, I have probably made it too short ...
As I still wouldn't want to bother anyone with my (hi)story, let me try to present
a concise background:
I am not a newbie: in fact produced my first commercial electronic product in
1961 - commissioned; proprietary design. Experienced experimenter since.
I only turn to a forum after I've done my 'desk research' and are left in doubt,
or without an answer at all. Remember: most of my life I had to rely on books.
This publication awoke my interest in series-connected capacitors: "A Myth About Capacitors in Series", by L. Kowalski ('88), found on the Net only a few years ago.
Later A.P. French, in a mathematical approach, arrived at the conclusion, posted
here by Crutschow. The main objection I have is twofold. First, Kowalski declares
the equal charge theorem invalid and concludes that the formula for calculating
total capacitance does not hold as well. So why not investigate other, related
formulae, theorems and the like, but instead go along with the results of a math
approach, not particularly the place to expect critical treatment of such issues ...
Second, (back to reality) the physical entities to determine the current surge after
switch-on (t0) are the respective ESRs of the e-caps. And if the law about the current
in a series circuit being equal all over, holds, it seems odd that not the ESRs, but the
caps values should determine the initial voltage division. I am aware of the replies
and criticism I'm provoking with this suggestion: superficially there seems to be an
inverse proportional relation between the two parameters, but it is easy to find 20%
deviations of that 'rule' in data sheets (mind you, I am not talking tolerances, but specs). Once again: reality, i.e. developments in materials science, but also simple
differences in e.g. size, beats theory !
My own objection against it would be: in a series-connection ESRs do not represent
a voltage divider, as leakage resistances do.
But there is an other problem with 'unruly' ESRs: time constant ! But that is maybe for a next post.
Can I invite members, who are with a manufacturer of electrolytic caps, to either
enter this discussion, or prove wrong my suggestion that the initial voltage division
between electrolytics in series is determined by their respective ESRs, not their
capacitances ? Looking forward to expert opinions, Ray.
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
hi.
Three caps in series.
E
View attachment 243923
Hi Mr. Gibbs, thank you for both LT-SPICE simulations. I am not versed in the art,
but from the graphs you present, what else could one conclude: my assumption was right . (I shall refrain from exposing the consequences here). One remark only: from
my experiments I can confirm, that after switching off, an evening-out current can
cause a negative potential over one of the electrolytes. It is typical for inadequate
modelling (as reported by several experts) that simulation software does not take
into account the risks involved: in this case C3 is likely to get irreversibly damaged,
if not worse ... If you should have any reference from literature, as to the voltage
division following ESR ratio instead of capacitances, I would be obliged.
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
Hi Mr. Gibbs, thank you for both LT-SPICE simulations. I am not versed in the art,
but from the graphs you present, what else could one conclude: my assumption was right . (I shall refrain from exposing the consequences here). One remark only: from
my experiments I can confirm, that after switching off, an evening-out current can
cause a negative potential over one of the electrolytes. It is typical for inadequate
modelling (as reported by several experts) that simulation software does not take
into account the risks involved: in this case C3 is likely to get irreversibly damaged,
if not worse ... If you should have any reference from literature, as to the voltage
division following ESR ratio instead of capacitances, I would be obliged.
P.S.: it would be interesting to check, if the voltage source has been modelled (or
specified by you) in such a way as to be able to function as a path for the charge
redistribution between the electrolytes after switch-off, or that this phenomenon
only takes place internally.
As I said, I am not familiar with sim-programs; I supposed ".tran .2m" has something
to do with the rise time of the test pulse, being 200us. ? Makes no sense, though,
as the pulse width is already shorter. So I'd better stop guessing. All in all the sim
seems to have been set up for a minimum of distracting details. As an audio pro I
know what a square wave looks like after a close encounter with an electrolytic, and
that is not exactly what I see here ... This presentation on the other hand is ever so
effective in showing my point, concerning ESRs !
 

ericgibbs

Joined Jan 29, 2010
14,130
hi.
The .tran .2m , is the run time of the simulation, 0.2milliSec

The 50Vdc Pulse was applied at 10microsec after time=0
The rise and fall time of the pulse was 1nanoSec
The period the 50Vdc was applied is 0.1mSec ie: 100microSec.

The only Cap parameter that was edited was the ESR for the 3 capacitors.

E
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
hi.
The .tran .2m , is the run time of the simulation, 0.2milliSec

The 50Vdc Pulse was applied at 10microsec after time=0
The rise and fall time of the pulse was 1nanoSec
The period the 50Vdc was applied is 0.1mSec ie: 100microSec.

The only Cap parameter that was edited was the ESR for the 3 capacitors.

E
Thank you, Mr. Gibbs
for specifying conditions for the simulations.
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
Thank you, Mr. Gibbs
for specifying conditions for the simulations.
Is there anybody who can shed some light upon the 'mechanism' of redistribution
of charge after switch-off in electrolytic caps ? In the old days of picture tube TVs,
it was referred to as 'polarity reversal'. More to the point:
- Is this an internal affair only, or can a source (in the above simulation examples),
and an eventual load also be a path for the redistribution current ?
If I remember well, there was only 1 cap involved in the TV circuit, so the external
path seems the only logical option there. In case of series-connected electrolytics,
I am not so sure.
- Is the phenomenon lossless (apart from the ESR-losses), or do we see energy
'vanishing', as in the classic two capacitor problem ?
Thanks beforehand.
 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
hi S.
Identical Caps except for ESR.
E
View attachment 243922
Hallo Mr. Gibbs, following up on (unequal) ESRs: the respective time constants of
the caps will differ in the same ratio. One would expect a sort of current-pinching
effect as a result - I mean: temporary, while each cap is in a different point on its
(dis)charge curve. I would appreciate your opinion on this assumption.
 

ericgibbs

Joined Jan 29, 2010
14,130
hi S,
As I am sure you can appreciate, the capacitor models are somewhat limited in their parameters.

This video is related to the way LTSpice models capacitors, you may find it useful as background information.

E

 

Thread Starter

SYNFONIQUE

Joined Jun 5, 2021
35
hi S,
As I am sure you can appreciate, the capacitor models are somewhat limited in their parameters.

This video is related to the way LTSpice models capacitors, you may find it useful as background information.

E

Thank you very much, Mr. Gibbs, for a really enlightening tutorial about simulation
vs. reality ! Some would argue now, finding themselves caught between a rock and
a hard place - theory being the former. My criticism (even doubts) notwithstanding,
I believe, in view of the rapid developments in capacitor technology, designers will
increasingly have to turn to and learn to trust (updated) simulations rather than
(conventional) theory. Btw, as I'm only working on linear PSUs for high-grade audio,
my 'problems' are relatively simple - revolving around the 100Hz mark ...
 
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