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Suppose I have a capacitor with a certain ESR, capacitance, and voltage. Then I double with width of the barrier between the two plates. The voltage will double, the capacitance will halve. What happens to the ESR?

In other areas, capacitance and ESR are inversely proportional. For example if the plate size is cut in half, the capacitance halves and the ESR doubles. But does ESR transform the same way in this case too?

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2 Answers 2

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No, this would not* change the ESR. The relationship between capacitance and ESR is due to space constraints; you can fit more parallel plates (thus lowering ESR) inside a capacitor when the plates are close together, which also results in higher capacitance and lower voltage rating.

In your example, you're making the capacitor take up twice as much space, so the rule of thumb doesn't apply.


*At least outside of some unusual cases where we're talking about high-frequency ESR in high-κ dielectrics, where dielectric losses dominate and adding more dielectric material would probably change things. I can't say in what direction, though, without more effort than I'd really like to go to on a sunday.

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  • \$\begingroup\$ If only the dielectric is twice as large, the overall capacitor might not have changed much at all. Often the dielectric is the minority of the space in the capacitor. Supposing doubling the voltage in this case makes practically no difference in the package size, will the ESR remain the same? \$\endgroup\$ Commented Jul 7 at 21:07
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    \$\begingroup\$ @JosephSummerhays What I stated still holds. The ESR, particularly at low frequencies, is a property primarily of the metallic parts of the capacitor alone, and what you've described doesn't alter the metallic parts. Now, if you stated you were making the plates thinner to fit more dielectric, that would be another matter, but the implication (as I read it) was that you're just moving the plates further apart. \$\endgroup\$
    – Hearth
    Commented Jul 7 at 21:13
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    \$\begingroup\$ @JosephSummerhays There may be a minuscule adjustment to the ESR from the additional bus length needed to connect the plates to the now-slightly-more-distant leads/termination, but that would be a truly tiny correction. \$\endgroup\$
    – Hearth
    Commented Jul 7 at 21:23
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ESR due to conductors remains constant, or nearly so: note that, in a practical capacitor, the electrodes being farther apart means more distance from the common points to individual electrodes. Probably a small effect (in stacked/rolled types e.g. ceramic caps, the end metallization / schoopage is much thicker so relatively negligible; in electrolytics and lead-in film (usually film-in-oil), the spiral might be slightly longer?), but also depends how it's mounted (e.g. a stacked part is twice as tall, so the electrodes are twice as long).

Dielectric losses are another matter. It depends on the relaxation rate, resistivity and other bulk properties. For relaxation dominant materials (probably, PET is an example), the loss tangent remains constant; C halving means ESR doubles.

For resistivity-dominant materials, the time constant varies with length, so we expect a lower break frequency and thus higher ESR / lower C.

Type 2 ceramics may be an intermediate case, I don't know; another consideration is ferroelectric domain size, which may vary with layer thickness, at least for thin enough layers. The behavior of these may manifest as an equivalent of the above effects.

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