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I vaguely remember that 100μF represent the capacity of a capacitor, but I also kinda remember that it has to do with the voltage. So my question is, do 10V 100μF and 16V 100μF capacitors have the same capacity and output (meaning does the 16V 100μF capacitor is the same as 10V 100μF capacitor only capable of working in a higher voltage system)?

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    \$\begingroup\$ Just as a word of warning, capacitors are quirky. In a perfect world, you would only look at capacitance, or maybe capacitance and series resistance. But in the real world, some types of capacitors have other non-linear behavior, such as ceramic capacitors where capacitance depends on voltage, and series resistance depends on frequency. So, it is good to read the fine print if you are relying on these things. But conceptually, you are on the right track as far as ideal capacitors goes. \$\endgroup\$
    – mkeith
    Aug 17 '20 at 5:58
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Consider a glass of water: -

enter image description here

I vaguely remember that 100μF represent the capacity of a capacitor

No, capacity is how much something can be filled without spilling. Capacitance is the term measured in farads and is equivalent to the area of the glass above.

The height in which water is filled is equivalent to the voltage and therefore capacity is capacitance multiplied by voltage. This equals charge i.e. Q = CV.

So my question is, do 10V 100μF and 16V 100μF capacitors have the same capacity

No they have the same capacitance. Capacity is how full you can safely fill the capacitor and that related to charge (Q = CV).

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  • \$\begingroup\$ Thank you! When the capacitors are set to release energy, would they release the same amount of energy (meaning the same voltage and the same amperage)? \$\endgroup\$
    – Aero Wang
    Aug 16 '20 at 10:22
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    \$\begingroup\$ I never was aware of this difference. Is this some official definition, that the capacity is given by capacitance and voltage rating? I can completly follow the logic, but I never heard somebody use the term capacity like that before (in regard to a capacitor). \$\endgroup\$
    – jusaca
    Aug 16 '20 at 12:14
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    \$\begingroup\$ @jusaca it hardly ever gets called capacity (maximum charge that can be held aka C.V) but, the OP called capacitance "capacity" and I used the water glass to point out the difference between the terms. \$\endgroup\$
    – Andy aka
    Aug 16 '20 at 12:22
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    \$\begingroup\$ I see. In german both terms translate to the same word, so I probably never made a difference. \$\endgroup\$
    – jusaca
    Aug 16 '20 at 12:38
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    \$\begingroup\$ I'm probably using the term capacity more like how we use it for a battery. You could argue that \$I\cdot dt = C\cdot dv\$, hence "ampere seconds" is like battery capacity and it equates to capacitance x voltage change. \$\endgroup\$
    – Andy aka
    Aug 16 '20 at 12:46
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Up to 10 V, both capacitors behave as a 100μF capacitor.

Between 10 V and 16 V, the 10 V capacitor may behave as a capacitor for a while, but it may increase its leakage current, it may go off with a bang, we just don't know. The 16 V capacitor will continue to behave like a capacitor.

Above 16 V, neither capacitor can be trusted.

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Assuming (from the high capacitance) you are talking about polarized caps, then yes, both will behave pretty much the same below 10V.

That you remember a correlation with the cap's voltage might be due to the fact that ceramic caps DO change behaviour with voltage. A ceramic cap loses capacitance with increasing voltage. Depending on the dielectric, you might have only something like 20% of the rated capacity at full voltage (at least for X5R, others can be better). So a higher voltage rating will give you more capacitance at your working voltage.

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  • \$\begingroup\$ I see. So do 10V 100μF and 16V 100μF tantalum capacitors behave the same under 10V? The only difference is that I am assuming if the voltage is 12V, the former would break and the latter would withstand, right? \$\endgroup\$
    – Aero Wang
    Aug 16 '20 at 10:00
  • \$\begingroup\$ Correct, up to 10V they will behave pretty much the same. The higher voltage one might have a larger package and therefore a slightly increased parasitic inductance, but in typical applications for these high capacitance values this will probably be of no concern. Be aware that tantalum caps are very sensitive to voltages above their ratings and can literally explode with only half a volt or so above. \$\endgroup\$
    – jusaca
    Aug 16 '20 at 10:06
  • \$\begingroup\$ Thank you! When the capacitors are set to release energy, would they release the same amount of energy (meaning the same voltage and the same amperage)? \$\endgroup\$
    – Aero Wang
    Aug 16 '20 at 10:24
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In most circuits they will behave the same, but as the voltage rating goes up, generally the ESR goes down and the maximum allowed ripple current goes down

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They have the same capacity for voltages below 10V.For voltages above 10V the 10V 100uF capacitor no longer acts like a capacitor more like a resistor. For voltages above 16V the 16V 100uF capacitor no longer acts like a capacitor more like a resistor.

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    \$\begingroup\$ What makes you say that a capacitor will behave like a resistor above it's rating? Depending on the type of cap it will more likely either fail short or open! \$\endgroup\$
    – jusaca
    Aug 16 '20 at 12:16
  • \$\begingroup\$ This doesn't seem right to me. A capacitor doesn't automatically fail once it reaches its rated voltage. Your 10V capacitor may continue to function just fine as a capacitor, at 11V or 12V. The rated voltage is a bit like the best before date on a litre of milk. Don't trust it past this voltage, but don't expect immediate failure. \$\endgroup\$ Aug 16 '20 at 20:01
  • \$\begingroup\$ But that is not guaranteed, you can not rely on that at all. Sure, a lot of caps won't blow up with a few percent of overvoltage, but espacially tantalum caps (which the OT seems to refer to) are very sensitive in this regard and are known to blow up pretty quickly. \$\endgroup\$
    – jusaca
    Aug 17 '20 at 5:45
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    \$\begingroup\$ The last thing we need is someone seeing this and drops in a capacity in place of a resistor and amps up the voltage. \$\endgroup\$
    – Nelson
    Aug 17 '20 at 8:50

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