Do capacitors provide ESD protection even if they are already charged?

I've read about ESD protection with capacitors and it seems that if one knows its limitations it could in some cases be a cheap alternative for or addition to designated ESD protection circuits.

The basic idea is that the charge of the ESD source (i.e. Human Body Model) will be shared across the source's capacitor and the "protective capacitor". The final voltage could then be estimated using simple calculations. ()

But what if the "protective capacitor" is already fully charged when the ESD event occurs? For example, let's say it is a charged input capacitor of a voltage regulator and a ESD event occurs at its input connector. Does it then still provide any protection?

Any clarification would be appreciated.

But what if the "protective capacitor" is already fully charged when the ESD event occurs?

It still provides protection; if more charge is put into the capacitor from an ESD surge then, the capacitor charges up more and the voltage increases. Of course, if the voltage rises above the maximum rating of the capacitor then it will likely damage the capacitor. And, if the voltage rises too high it might become too high for the circuit it is intended to protect.

But, basically, it is a great way of protecting against ESD surges. Not so great with indirect lightning but good for ESD level energies.

• Thanks so much, ok i think possibly got it. For example, if i am using a 100V rated capacitor with 100nF capacitance it can hold 10000n Coulomb charge. So when i use it in a 5V environment it has already 500n Coulomb charge in it. So 9500n Coulomb are "reserved" for an ESD event? Sorry that's crude calculations but is the basic idea right? – CatastrophicFailure Nov 20 '20 at 8:56
• That's the basic idea but it might be better to think of the HBM as a 100 pF charged to (say) 8 kV then if you connected it in parallel with 100 nF, what does the voltage become. So C changes to 100.1 nF and V changes to 7.992 volts. Or if precharged at 5 volts, a decent approximation of the final voltage would be 7.992 volts + 5 volts = 12.992 volts. – Andy aka Nov 20 '20 at 9:24
• Awesome - all ambiguities eliminated! That was very instructive. Thanks again... – CatastrophicFailure Nov 20 '20 at 9:46

If the capacitor is ideal, the equation still holds that change of charge causes change of voltage, no matter what the initial voltage is.

In practice though, ESD pulses are fast, so the ability to absorb an ESD pulse depends on capacitor ESR, ESL, which are basically determined by capacitor type and size.

So a standard electrolytic capacitor would absorb the surge poorly due to high ESR and ESL, compared to small ceramic capacitor, even if some types of ceramic capacitors lose some of their rated capacitance under DC bias.