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Assume a capacitor's voltage limit be v, and the target voltage limit be V.

What is the way of connecting the capacitors to reach the V?

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Connecting two identical capacitors in series, each with voltage threshold v and capacitance c, will result into a combined capacitance of 1/2 c and voltage threshold of 2 v.

However, it is far better to get a single capacitor that meets the higher voltage threshold on its own as combining capacitors in series will also lead to a higher Effective Series Resistance (ESR). In the scenario above, you will double the ESR. High ESR can cause unwanted or catastrophic effects on circuits not designed to handle it.

See more here: What are some reasons to connect capacitors in series? and here Can you make a non-polar electrolytic capacitor out of two regular electrolytic capacitors?

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    \$\begingroup\$ For real life capacitors (not identical ones) You will never get a "voltage threshold of 2" for the reasons outlined in my answer. Despite the OP accepting your answer there is a risk someone may assume it applies to real components so I'm making this comment. In that part of your answer where you have attempted to answer the op, the lack of reality when using real caps means I have to downvote, sorry. \$\endgroup\$ – Andy aka Aug 29 '13 at 8:10
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ON DC

If you have two 100V rated capacitors in series, you cannot assume that the combination of the two will be 200V. Slightly different leakage currents will mean one cap has more voltage across it than the other.

This can be mitigated by adding resistors across each capacitor that have values lower than the effective leakage resistance of the capacitors. If the cap has a leakage R of (say) 10M\$\Omega\$, go for 1M\$\Omega\$ resistors. Any variation will be reduced by about 10:1 BUT, I'd suggest you wouldn't get better than 1.9 x original voltage rating.

ON AC

This is determined by the capacitor values so if one capacitor is smaller than the other, it will receive more AC volts across it.

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The hypothetical case of ideal capacitors which are perfectly identical in leakage current and voltage ratings, is well described in existing answers. In practice, things get a bit more complicated.

If the load in the application has a high duty cycle, active balancing of capacitors connected in series is required, to protect the capacitors from an early demise. This can be done with discrete components, but there are also purpose-built, highly integrated capacitor balancing ICs available.

While general-purpose electrolytic capacitors are sensitive to voltages over rating, non-electrolytic capacitors such as mica or ceramic ones are less so. Most sensitive are electric double layer capacitors (supercapacitors), which are also typically rated for fairly low voltages, making the problem more relevant in moderate voltage circuit designs.

Thus capacitor active balancing ICs are most often used for supercapacitor banks, such as for car audio buffer capacitors. An example is the Texas Instruments BQ33100 Super Capacitor Manager, that can manage from 2 to 5 or even 9 supercapacitors or conventional capacitors in series.

Of course, for conventional electrolytic capacitors, it is simply more cost effective to use a capacitor with a higher voltage rating, or a bunch of high voltage lower value capacitors in parallel.


At a simpler level, for low duty cycle / low load applications, a passive balancing approach can be adopted. This Maxwell Technologies appnote describes the approach, consisting of a resistor in parallel with each capacitor, in a ladder arrangement. The resistors are sized to "dominate the total cell leakage current" by typically 10 times the maximum leakage current of the target capacitors.

From an answer to another question, here is such a passive balancing arrangement:

Schematic

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Capacitors connected in series add their voltage tolerances. (This is true if their capacitance values are identical.)

Note that the equivalent capacitance value of capacitors in series is smaller than any individual value according to the formula:

$$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \dotsm$$

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