There was some discussion on this question

which I don't see as being conclusively resolved:

  • "turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics."
  • "No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator."
  • 'Kind of like "you can't make a BJT from two diodes"'
  • "it is a process that a tinkerer cannot do"

So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not? Does it not survive the same voltages? What happens to the reverse-biased cap when a large voltage is placed across the combination? Are there practical limitations other than physical size? Does it matter which polarity is on the outside?

I don't see what the difference is, but a lot of people seem to think there is one.


As posted in one of the comments, there's a sort of electrochemical diode going on:

The film is permeable to free electrons but substantially impermeable to ions, provided the temperature of the cell is not high. When the metal underlying the film is at a negative potential, free electrons are available in this electrode and the current flows through the film of the cell. With the polarity reversed, the electrolyte is subjected to the negative potential, but as there are only ions and no free electrons in the electrolyte the current is blocked. — The Electrolytic Capacitor by Alexander M. Georgiev

Normally a capacitor cannot be reverse-biased for long, or large currents will flow and "destroy the center layer of dielectric material via electrochemical reduction":

An electrolytic can withstand a reverse bias for a short period, but will conduct significant current and not act as a very good capacitor. — Wikipedia: Electrolytic capacitor

However, when you have two back-to-back, the forward-biased capacitor prevents a prolonged DC current from flowing.

Works for tantalums, too:

For circuit positions when reverse voltage excursions are unavoidable, two similar capacitors in series connected “back to back” ... will create a non-polar capacitor function ... This works because almost all the circuit voltage is dropped across the forward biased capacitor, so that the reverse biased device sees only a negligible voltage.

Solid Tantalum Capacitors Frequently Asked Questions (FAQs):

The oxide dielectric construction that is used in tantalum capacitors has a basic rectified property which blocks current flow in one direction and at the same time offers a low resistance path in the opposite direction.

  • \$\begingroup\$ See "Mechanism" addition at end of my answer. \$\endgroup\$ – Russell McMahon Nov 7 '11 at 22:08

11 Answers 11



  • Yes "polarised" aluminum "wet electrolytic" capacitors can legitimately be connected "back-to-back" (ie in series with opposing polarities) to form a non-polar capacitor.

  • C1 + C2 are always equal in capacitance and voltage rating
    Ceffective = = C1/2 = C2/2

  • Veffective = vrating of C1 & C2.

  • See "Mechanism" at end for how this (probably) works.

It is universally assumed that the two capacitors have identical capacitance when this is done.
The resulting capacitor with half the capacitance of each individual capacitor.
eg if two x 10 uF capacitors are placed in series the resulting capacitance will be 5 uF.

I conclude that the resulting capacitor will have the same voltage rating as the individual capacitors. (I may be wrong).

I have seen this method used on many occasions over many years and, more importanttly have seen the method described in application notes from a number of capacitor manufacturers. See at end for one such reference.

Understanding how the individual capacitors become correctly charged requires either faith in the capacitor manufacturers statements ("act as if they had been bypassed by diodes" or additional complexity BUT understanding how the arrangement works once initiated is easier.
Imagine two back-to-back caps with Cl fully charged and Cr fully discharged.
If a current is now passed though the series arrangement such that Cl then discharges to zero charge then the reversed polarity of Cr will cause it to be charged to full voltage. Attempts to apply additional current and to further discharge Cl so it assumes incorrect polarity would lead to Cr being charge above its rated voltage. ie it could be attempted BUT would be outside spec for both devices.

Given the above, the specific questions can be answered:

What are some reasons to connect capacitors in series?

Can create a bipolar cap from 2 x polar caps.
OR can double rated voltage as long as care is taken to balance voltage distribution. Paralleld resistors are sometimes used to help achieve balance.

"turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics."

This can be done with oridinary electrolytics.

"No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator."

Works OK if ratings are not exceeded.

'Kind of like "you can't make a BJT from two diodes"'

Reason for comparison is noted but is not a valid one. Each half capacitor is still subject to same rules and demands as when standing alone.

"it is a process that a tinkerer cannot do"

Tinkerer can - entirely legitimate.

So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not?

It coild be but the manufacturers usually make a manufacturing change so that there are two Anode foils BUT the result is the same.

Does it not survive the same voltages?

Voltage rating is that of a single cap.

What happens to the reverse-biased cap when a large voltage is placed across the combination?

Under normal operation there is NO reverse biased cap. Each cap handles a full cycle of AC whole effectively seeing half a cycle. See my explanation above.

Are there practical limitations other than physical size?

No obvious limitation that i can think of.

Does it matter which polarity is on the outside?

No. Draw a picture of what each cap sees in isolation without reference to what is "outside it. Now change their order in the circuit. What they see is identical.

I don't see what the difference is, but a lot of people seem to think there is one.

You are correct. Functionally from a "black box" point of view they are the same.


In this document Application Guide, Aluminum Electrolytic Capacitors bY Cornell Dubilier, a competent and respected capacitor manufacturer it says (on age 2.183 & 2.184)

  • If two, same-value, aluminum electrolytic capacitors are connected in series, back-to-back with the positive terminals or the negative terminals connected, the resulting single capacitor is a non-polar capacitor with half the capacitance.

    The two capacitors rectify the applied voltage and act as if they had been bypassed by diodes.

    When voltage is applied, the correct-polarity capacitor gets the full voltage.

    In non-polar aluminum electrolytic capacitors and motor-start aluminum electrolytic capacitors a second anode foil substitutes for the cathode foil to achieve a non-polar capacitor in a single case.

Of relevance to understanding the overall action is this comment from page 2.183.

  • While it may appear that the capacitance is between the two foils, actually the capacitance is between the anode foil and the electrolyte.

    The positive plate is the anode foil;

    the dielectric is the insulating aluminum oxide on the anode foil;

    the true negative plate is the conductive, liquid electrolyte, and the cathode foil merely connects to the electrolyte.

    This construction delivers colossal capacitance because etching the foils can increase surface area more than 100 times and the aluminum-oxide dielectric is less than a micrometer thick. Thus the resulting capacitor has very large plate area and the plates are awfully close together.


I intuitively feel as Olin does that it should be necessary to provide a means of maintaining correct polarity. In practice it seems that the capacitors do a good job of accommodating the startup "boundary condition". Cornell Dubiliers "acts like a diode" needs better understanding.


I think the following describes how the system works.

As I described above, once one capacitor is fully charged at one extreme of the AC waveform and the other fully discharged then the system will operate correctly, with charge being passed into the outside "plate" of one cap, across from inside plate of that cap to the other cap and "out the other end". ie a body of charge transfers to and from between the two capacitors and allows net charge flow to and from through the dual cap. No problem so far.

A correctly biased capacitor has very low leakage.
A reverse biased capacitor has higher leakage and possibly much higher.
At startup one cap is reverse biased on each half cycle and leakage current flows.
The charge flow is such as to drive the capacitors towards the properly balanced condition.
This is the "diode action" referred to - not formal rectification per say but leakage under incorrect operating bias.
After a number of cycles balance will be achieved. The "leakier" the cap is in the reverse direction the quicker balance will be achieved.
Any imperfections or inequalities will be compensated for by this self adjusting mechanism. Very neat.

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  • \$\begingroup\$ "Under normal operation there is NO reverse biased cap" It's still reverse biased by one diode drop, no? \$\endgroup\$ – endolith Nov 7 '11 at 17:06
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    \$\begingroup\$ @endolith - No - but no problem with thinking so as I doubt if any of us are 100% sure of what actually happens. Their statement of "like a diode" is more hand waving than good comparison. I suspect that what it means is that a reverse biased electrolytic will "leak" very badly indeed and pass current through to the other correctly bias capacitor and so "pump the system up" until it reaches the correct balanced operating point. Once at a point of balance (as i described in my answer), if the caps are identical, the system will operate each capacitor "perfectly" bias wise. \$\endgroup\$ – Russell McMahon Nov 7 '11 at 21:53
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    \$\begingroup\$ @RussellMcMahon: The essential point is that diodes which are reverse-biased tend to leak by an amount which is non-linearly dependent upon voltage, and it is the reverse current flow rather than the reverse voltage which causes damage. In the absence of forward leakage, the "lifetime" maximum number of electrons that will flow forward through either cap will be limited to the number of electrons that would be required to charge the other cap. If a particular cap happened to leak almost nothing once the reverse voltage got below 250mV, then the cap might sit with a 250mV reverse bias, but... \$\endgroup\$ – supercat Dec 24 '13 at 5:54
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    \$\begingroup\$ ...*it wouldn't really matter*. The only way the voltage can remain is if no current is flowing, and if no current is flowing there's no problem. Until essentially all the reverse leakage charge has flowed that's going to, the leakage would affect the caps' performance characteristics, but in most cases it wouldn't take long to leak 99% of the total charge it's ever going to leak. \$\endgroup\$ – supercat Dec 24 '13 at 5:57

I am aware this has been done successfully for ages, but why it works is worth looking at.

I thought I would set up a quick simulation based on the info given by Russell in his answer. The main point being the "act as if they had been bypassed by diodes" part. It's a very rough approximation but it gives a picture of what might be happening.

Bipolar Schematic

Bipolar Simulation

The I[D1] and I[D2] represent the reverse current through the caps. Initially one of the caps gets a brief surge of reverse current, then it becomes minimal for both. The I[C1] and I[C2] represent the current through the capacitance. This meet expectations of a 0.5uF cap at 100Hz. The capacitive reactance

$$\dfrac{1}{2\pi \cdot 100 \cdot 500e^{-9}} = 3183$$

So peak current will be

$$\frac{10}{3183} = 3.14mA$$

The light blue wave in the third graph is the supply voltage. The dark blue and green waves in the third graph represent the voltage seen across each capacitor (+ terminal with respect to - terminal of each)

As can be seen, both are correctly polarised.

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  • \$\begingroup\$ I used this circuit right now its works but the capacitors heat up, why? i used 1N5804 rather than 1N4148 \$\endgroup\$ – M.A.K Mar 14 '17 at 17:26
  • \$\begingroup\$ @M.A.K Unless I'm mistaken, this circuit would create the equivalent of ripple current as seen at the output of a SMPS. If the caps you used have a high ESR, they'll drop voltage just like a resistor, hence heat. \$\endgroup\$ – GB - AE7OO Oct 20 '19 at 20:37

Yes, it is possible to combine two polarized caps into a effective single non-polarized cap, but with some restrictions. Each individual cap still needs to only see voltages within its specification. The easiest way to do this is to have a supply voltage that is guaranteed to always above or below any voltage applied to either side of the non-polarized cap. The two polarized caps are then connected back to back and a high value resistor connected to the supply:

Note that the total capacitance is the series combination of the two individual capacitors, which is half each if they are equal. In the example above, the total effective capacitance is 235 uF.

The voltage range of each cap must also be carefully considered. The worst case depends on what the external circuit can do. For example, suppose both ends are held at 10V, then the left end suddenly dropped to 0V. The center will be at -5V with 15V accross the right cap immediately after the step. The 1 MΩ impedance on the signal to the supply must also be considered. R1 must be low enough so that leakage thru the caps doesn't add too much voltage, but otherwise as high as possible to not load the signal.

In general, this sort of trick should be considered a last resort. Since bipolar capacitors are usually needed for signals, it can often be arranged to require a lower bipolar capacitance. Multi-layer ceramic caps have advanced significantly in the last decade. If you can make do with 10 uF instead of 100s of uF, a ceramic can can probably do the job.

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  • \$\begingroup\$ I'd do it with diodes (to prevent reverse-bias for each capacitor) rather than a bleed resistor, but yeah, you've covered it well. \$\endgroup\$ – Jason S Nov 7 '11 at 16:40
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    \$\begingroup\$ ...but you haven't got the worst-case: suppose both ends are at 0V, and one end suddenly increases to 10V. Until R1 can equalize, this puts the middle node at 5V and reverse-biases one of the capacitors. Which is why I'd advise using diodes. It also keeps the effective capacitance at 470uF. \$\endgroup\$ – Jason S Nov 7 '11 at 16:44
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    \$\begingroup\$ @Jason: Diodes are bad because they make the system non-linear and don't allow for a cap to be discharged easily. \$\endgroup\$ – Olin Lathrop Nov 7 '11 at 17:31
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    \$\begingroup\$ @JasonS: You mean put a diode in parallel with each cap, right? Not a diode to ground. Supposedly, the capacitors in an NP act like diodes already. Why is this? \$\endgroup\$ – endolith Nov 7 '11 at 20:29
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    \$\begingroup\$ @JasonS: In the absence of forward leakage, the total lifetime number of electrons which would ever be pushed backward through either capacitor would be limited to the number required to forward-charge the other. Whether the "diode" mechanism lets those diodes through quickly or slowly doesn't matter; most caps can tolerate a number of reverse-flowing electrons which is greater than the number of forward-flowing electrons that would be necessary to charge them once. \$\endgroup\$ – supercat Dec 24 '13 at 6:06

Please be aware that the Equivalent Series Resistance (ESR) of the paired capacitors in series is doubled. As the case of components veering away from their ideal model (a close to ideal/real world capacitor should have insignificant resistance and impedance), it may have unwanted effects (i.e. smoke-releasing). For example, ICs like LM78xx & LM317 will have poor regulation due to ringing introduced by high ESR filter capacitors

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    \$\begingroup\$ True. Try building the bypass cap for Vref on the ADS7863 ADC using two series-connected 1uF-caps, and you'll be surprised what a beautiful oscillator you've built... (cf. the 470nF cap between pins 11 and 12 in fig. 41 of this data sheet: ti.com/lit/ds/symlink/ads7863.pdf) Not that you want to do this for a production design, but my lab tech did it when he built a test setup, and we were quite amazed when troubleshooting the board. \$\endgroup\$ – zebonaut Nov 8 '11 at 6:55
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    \$\begingroup\$ I learned that fact the hard way (ah, good old days). Electrolytic capacitors are notorious for at least 3 things: they have high ESR, usually the first component to fail (heat/heat cycles, spikes), unforgiving for experimenters (which way is negative again?), and... leaves a nasty mess when they fail. But hey, it's fun seeing that white flak of fireworks from time to time (tip: the larger the capacitance, the more bang). :-) \$\endgroup\$ – shimofuri Nov 8 '11 at 15:21
  • \$\begingroup\$ My favorite one is after I've recapped OLD(I'm talking 30+ years) HV power supplies, I'll take the replaced filter cap(s) outside and hit them with double reversed voltage. Now mind you, these things are HUGE, so... This serves two purposes, first it makes a nice bang(as in the here hold my beer type) and makes sure someone(my son in the past) does not get a hold of it and start playing with it(I caught him doing the same thing(by accident and not safely). \$\endgroup\$ – GB - AE7OO Oct 20 '19 at 20:45
  • \$\begingroup\$ Could someone give me brief explanation about "ringing" in high esr filter capacitor? What actually high esr affects in filter capacitor circuit. \$\endgroup\$ – panoet Aug 10 at 22:40

I built a TTL oscillator to check. The discussion is partly correct.

If the duty cycle is close to 50% then the capacitors do act as if they have built in diodes and limit the negative (wrong direction) voltage excursion. If the duty cycle is not 50% (in my case about 30%) then the negative voltage excursion was about 0 V on one capacitor and 1.1 V on the other.

I would include protection diodes on all but low power symmetrical signal applications. For power applications Schottky diodes might be a worthwhile investment.

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    \$\begingroup\$ During the high and low periods, what were the voltages at the ends of the string and at the middle? \$\endgroup\$ – supercat Mar 29 '17 at 23:04
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    \$\begingroup\$ Finally an answer based on first hand measurement. More believable than a SPICE run with added diodes. \$\endgroup\$ – crokusek Oct 28 '18 at 2:27
  • \$\begingroup\$ Would you please include a diagram of how you would use protection diodes? \$\endgroup\$ – MicroservicesOnDDD Aug 29 at 2:03

The company I worked for built thousands of instruments over the years that used back-to-back electrolytics. There was never a problem.

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    \$\begingroup\$ I used back-to-back electrolytics right now its works but the capacitors heat up, why? i used that with revers diode for each capacitor \$\endgroup\$ – M.A.K Mar 14 '17 at 17:27

Yes, it can be done, and safely, but I'm afraid there are various issues if you simply follow the advice in some of the answers.

For example, if you resistively bias the midpoint to the supply potential the capacitors could be exposed to 1.5 x the supply potential even if the capacitors match. Any mismatch would increase the possible maximum, and depending on specification the mismatch could be substantial: even +/-20% represents a worst-case ratio of 1.5:1).

Back-to-back connection relying on the ionic diodes avoids the above problem, but introduces a different one - at least in theory. It is possible for capacitors to have low-level leakage that is not directly associated with the intended operation; this could cause problems to develop over time if one capacitor leaks and the other does not. I don't have any knowledge of this happening, but applying cheap small-signal diodes in parallel should be more than adequate to suppress the effect, as aluminium electrolytics don't conduct significantly below about 1.5-Volt (though personally I prefer to maintain at most 1-Volt long-term). (As a side-note: other than dodgy connectors, the most common cause of equpment failures that I have seen is due to electrolytic leakage rebiasing circuits away from their intended bias conditions - so, back-to-back or not, don't use electrolytics in places where their maximum leakage will upset your circuit's bias).

A final note on safety: the requirement to use back-to-back bias suggests that there is significant AC signal across the pair. This implies ripple current; be sure not to exceed the ripple-current rating, and also be aware that (depending on type and on design) the ripple-current-rating of electrolytic capacitors is frequency-dependent

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This may seem exceedingly basic observational analysis, but looking at a sine wave as it crosses zero we have two halves, e.g. a 110V AC wave is 220V from (+) to (-) peaks. That means that C1 & C2 are alternately forward and reversed biased to their electrolyte. The forward bias voltage is 110V across the forward biased cap C1, and then across C2 respectively during each of their positive half- cycle(s). Looking at quarter cycles, the caps charge positively over their respective positive first quarter cycle and discharge over the second quarter cycle. 110V charges and discharges one capacitor, and then the other, alternately.

But assuming 110V were being dropped across both capacitors, one forward and the other reverse-biased, then the drop across any one cap would actually only be 55V. Maybe it's not smart, or recommended to reverse bias an electrolytic cap but in the case described the amount of reverse bias is only half, or actually one quarter of the actual applied (220) voltage. Following best practices, using caps rated at least twice the applied voltage and never exceeding half that rating, (1/4 being dropped across each cap) apparently doesn't reach the point of destruction.

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  • \$\begingroup\$ When the series cap combination is sitting with zero volts across it, both caps will have an equal voltage on them which will be roughly half the maximum voltage the assembly has encountered. Raising the voltage to that maximum will cause one of the cap's voltage to drop to zero while the other goes up to that maximum voltage. If the voltage were raised above the earlier maximum, one of the caps would go negative, but any leakage that resulted from that would "permanently" increase the voltage on the other cap. \$\endgroup\$ – supercat Dec 24 '13 at 6:01
  • \$\begingroup\$ Supercat: I think someone should determine this by actual measurement. If the V across any one C is V applied and across C Reverse-biased is zero during any half cycle then we have no worries. \$\endgroup\$ – guestimate Dec 25 '13 at 3:42

i did it with four 30uf caps, all negs connected. Take two pos for one side, two for the other. All four caps were 30uf at 150V. Do the math however you want, but the speakers worked well for 33 years until I had to change the components out simply because it was time to re-cap the crossovers and re-foam the woofers.

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Remember, Q=CV. If you have two capacitors in series, and integrate the current through the capacitors, the Q will alternate from one capacitor to the other... However, it is possible to have zero volts across the series combination of two capacitors, but still have a very significant, but equal voltage, on each capacitor. e.g., +10v + -10v = 0v.

But wait, there is still another consideration. Remember the Q=CV you learned in your first electronics class? Well, since the tolerance of electrolytics can very easily be 20 percent, then the same Q on capacitors that vary in C by 20%, or more if the tolerances are in the opposite direction, can cause a very large difference in the voltage peak that each capacitor will see.

The idea of placing a diode across each capacitor to eliminate the possibility of an appreciable reverse current is a very good solution. The voltages on the capacitors will equalize, mostly, very quickly under AC conditions. By doing this, and making sure that you are conservative in selecting the max voltage for the capacitors, you will have a viable solution.

The only other problem you will have is that electrolytics do not like to be fully charged and discharged repetitively. The are much better for controlling ripple than passing AC - the capacitor manufacturers state this in their application notes.

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    \$\begingroup\$ Passing AC does not charge and discharge the capacitor repetitively. \$\endgroup\$ – endolith Aug 24 '14 at 0:17
  • \$\begingroup\$ How could it possibly not. The voltage changes from positive, to zero, to negative, etc., etc., and since V=Q/C, Q must change in amount and polarity, thus charging and discharging repetitively... Simple AC Electronics. \$\endgroup\$ – William Aug 28 '14 at 2:34
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    \$\begingroup\$ Yes, the voltage changes from positive to negative, on both sides of the cap. Capacitors resist a change in voltage across their terminals; the higher the frequency, the more they behave like a short circuit. In an AC-coupling application, the voltage across the cap is constant (DC bias or 0 V); both sides move up and down with the signal together. \$\endgroup\$ – endolith Aug 28 '14 at 13:54

I saw the roll your own bipolar cap made from two back to back electros of the same value with the reverse diodes across them.This was back in 1988 ,when I was fresh out of Uni.I did not like them at the time and vowed not to do this on anything new .The broadcast audio product that used them was manufactured in hundreds and did not fail so my boss did not get me to redesign it .So the reliability was fine in this low ripple current application .What the tech and I did was test the aformentioned caps on the AUDIO PRECISION test set which was at the time the bees knees .None of us could find any distortion coming from the caps.This was not what I expected or what others that watched the tests expected.On the other hand speaker crossover networks often use bipolar eltec caps that have been known to fail and take out the tweeter.On the rare occasions that I replace someones tweeter I discard the bipolar electro and replace with metal film .

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    \$\begingroup\$ When used in the passband, caps have minimal voltage across them and therefore minimal distortion. Both sides go up and down together. If you put low enough frequencies through it, you would see distortion. \$\endgroup\$ – endolith Sep 20 '15 at 18:29
  • \$\begingroup\$ We were using the electro as a coupling cap because we wanted the test to represent the broadcast line amp that was being manufactured.I have and will never use an electro in an audio filter.I saw someone use one in a telephone hybrid once .It wasnt good 15microfarad ,borderline return loss .If someone uses electros for subsonic cutoff they could expect distortion around the cut frequency. \$\endgroup\$ – Autistic Sep 21 '15 at 11:15

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