I know if you put an electrolytic capacitor backwards they explode and it's quite fun to observe them exploding. I am wondering if you can use two of them (and some diodes) to turn it into a bipolar capacitor so it may be used with AC signals?

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    \$\begingroup\$ Electrolytics capacitors can be used for AC and regularly are! Just that there has to be a net DC bias across them. For situations where you can't guarantee a bias of particular polarity or one that contains the signals wing, there are off-the-shelf non-polar electrolytics you can use. So why do this. \$\endgroup\$ – Kaz May 18 '13 at 16:07
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    \$\begingroup\$ @Kaz: ask TI/NatSemi (see my comment below answer). Probably because you're more likely to already have polarized ones on hand. \$\endgroup\$ – Fizz Sep 13 '15 at 19:33

Yes. You can.

Connect in the following manner


Short both the -ve terminals. The other two +ve terminals can be used as leads of a non-polar capacitor.

I know that if you do like this the capcitance will reduce. I thinks it by 1/2.

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    \$\begingroup\$ It will work until one cap blows up, which won't take long. Do not do this. \$\endgroup\$ – Matt Young May 18 '13 at 17:43
  • \$\begingroup\$ What I have mentioned is same as electronics.stackexchange.com/a/69716/24062 "electrolytic capacitors are connected in series, back-to-back with the positive terminals or the negative terminals connected" \$\endgroup\$ – robomon May 18 '13 at 19:11
  • \$\begingroup\$ @MattYoung care to elaborate? seems to be a pretty common practise... \$\endgroup\$ – akohlsmith May 19 '13 at 2:57
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    \$\begingroup\$ @AndrewKohlsmith Tomorrow night when I am back at my place and have my tools. There's a lot of theory talk going on in this question. I'm going to take some measurements and blow up some caps. \$\endgroup\$ – Matt Young May 19 '13 at 3:34

Using two electrolytic capacitors of identical value back to back is routinely used to get a non-polarized capacitor.

From this document:

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 to either of the original pair. 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.

You do not even need additional diodes. The equivalent model for an electrolytic capacitor is given below. Note the implicit diode in there.

Electrolytic Capacitor equivalent model.

When two such capacitors are connected back to back, the diodes will be in opposition to each other, blocking DC in both polarities.

On a contrary note, see also this answer to another related question on this site, where @Kortuk has a contrary opinion in the comments, against using normal electrolytic capacitors back to back.

While that viewpoint has been put forward in discussion, I have been unable to find any academic study substantiating the concern. That is not to say that such a study or reference does not exist.

I do find several references to applying normal back to back capacitors for non-polarized functioning, and have used this myself for many years in various designs, with never a single failure. The only caveat I adhere to, is to not approach within shouting distance of the rated voltage for each of the capacitors, if using them this way.

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    \$\begingroup\$ You should bias the node in the middle of the caps to some voltage that is guaranteed to be higher or lower than your signal-- and then orient the polarity of the caps accordingly. For example, a pullup resistor to your +V rail works if your signal does not get too close to +V. The reason for this is that two caps are never identical and it is possible that the middle node gets out of whack. This bias removes the issue that @Kortuk raises, and keeps both caps from being reverse biased. That diode in the cap model is not very big and does not come into play for small signals. \$\endgroup\$ – user3624 May 18 '13 at 14:15
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    \$\begingroup\$ Notwithstanding David Kessner's comment, if one cap "deals with" negative voltages and the other "deals with" positive voltages, why does the capacitance become "half the capacitance to either of the original pair"? Either capA or capB is in circuit and surely this means the capacitance is capA or capB and not capA/2? \$\endgroup\$ – Andy aka May 18 '13 at 15:04
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    \$\begingroup\$ @AndyAka My question exactly. Plus I don't see what prevents one capacitor, at any given time, from having a reverse voltage on it. How does the diode help when it's in parallel? DC has to flow since the whole thing charges and discharges. \$\endgroup\$ – Kaz May 18 '13 at 15:14
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    \$\begingroup\$ @Kaz The diode acts as a shunt to reverse polarity signals. On the series computation of capacitance, that's been debated a fair bit on various forums, but oddly enough I don't see any objective and verifiable experimental data one way or another. Perhaps it is time for me to measure the capacitance with a few capacitors. :-) \$\endgroup\$ – Anindo Ghosh May 18 '13 at 15:26
  • \$\begingroup\$ I just hooked up two 470uF electrolytics in anti-series, and my multimeter measured the final capacitance at 244uF. So, it is half. \$\endgroup\$ – Dave Branton Feb 17 '15 at 4:00

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