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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, 2013 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\$ Sep 13, 2015 at 19:33

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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, 2013 at 17:43
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    \$\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, 2013 at 19:11
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    \$\begingroup\$ @MattYoung care to elaborate? seems to be a pretty common practise... \$\endgroup\$
    – akohlsmith
    May 19, 2013 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, 2013 at 3:34
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    \$\begingroup\$ I don't know why people are downvoting this; it's correct. @MattYoung They won't blow up. This is how bipolar capacitors are made. \$\endgroup\$
    – endolith
    May 9, 2020 at 4:04
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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, 2013 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, 2013 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, 2013 at 15:14
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    \$\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\$ Feb 17, 2015 at 4:00
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    \$\begingroup\$ The LM1875 datasheet shows in the split supply schematic a 2.2uF unpolarized/bipolar cap as the input cap. In the suggested board layout thereof they then show two 4.7uF polarized caps with anodes connected together as the implementation of this unpolarized/bipolar cap. So if someone needed another source confirming/suggesting this use... TI/NatSemi has it too. \$\endgroup\$ Sep 13, 2015 at 19:25

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