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Here is what I used for calculating ESR:

enter image description here

Here is a random datasheet which gives the ESR under certain conditions:

enter image description here

Suppose I want to calculate the ESR @ 100 kHz for the 220 µF capacitor:

I apply the following equation:

$$ \mathrm{ ESR = \frac{0.1}{2*PI*100^{3}*220*10^{-6}} } $$

Which is not equal at 0.18 Ohm which is given in the datasheet.

Where is my error?

We actually do not know at which frequency the dissipation factor is given, but I supposed that it is given for 100kHz.

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Suppose I want to calculate the ESR @ 100 kHz for the 220 µF capacitor :

You can't just use the \$\mathrm{tan\delta}\$ value, which is mostly given for 120Hz in the datasheets, for calculating ESR at 100kHz. Because, as it's described in the calculation method, the equivalent series inductance, L, is neglected at frequencies up to 1kHz.

ESR values given in the datasheets are not calculated values. They are measured at the factory/production. And the measurements include the effect of L and r. That's why your calculations and the datasheet values don't match.

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  • \$\begingroup\$ Knowing the ESR and the capacitor at the given frequency, can I find the dissipation factor ? \$\endgroup\$ – Jess Aug 6 at 7:42
  • \$\begingroup\$ @Jess a previous answer I gave you on this subject - it seems I made a decimal place error so I'm going to correct that answer shortly. Bear with me on this. \$\endgroup\$ – Andy aka Aug 6 at 7:57
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    \$\begingroup\$ @Jess for non-polarized ceramic (e.g. X7R, NP0, etc) capacitors, ESR-vs-frequency and Z-vs-frequency graphs are usually given in the datasheets. When it comes to electrolytic capacitors, you may not calculate the ESR but measure it instead. As the frequency increases, the Z of the capacitor increases as the effect of L (a.k.a. ESL) increases. As given in the calculation method, for low frequencies (i.e. <= 1kHz) you can directly use \$\mathrm{tan\delta}\$ and Z. \$\endgroup\$ – Rohat Kılıç Aug 6 at 7:59
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    \$\begingroup\$ No problem Andy Aka ! :) \$\endgroup\$ – Jess Aug 6 at 8:01
  • \$\begingroup\$ Ok I see what you mean RohatKılıç ! Thank you :) \$\endgroup\$ – Jess Aug 6 at 8:02
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We actually do not know at which frequency the dissipation factor is given, but i supposed that it is given for 100kHz.

We do; it's 120 Hz as stated on the front page of the data sheet: -

enter image description here

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  • \$\begingroup\$ Ok ... Well I searched for one half hour into the datasheet and I wasn't able to find it ... \$\endgroup\$ – Jess Aug 6 at 7:37
  • \$\begingroup\$ @jess if you press CTRL-f then enter "Hz" into the box, you can search for hertz. \$\endgroup\$ – Andy aka Aug 6 at 7:42
  • \$\begingroup\$ You re right ! I didn't sleep very well last night ! too much noise in my own neighbourhood :D Knowing the ESR and the capacitor at the given frequency (100 kHz) , can I find the dissipation factor @ 100 kHz ? ie about 25 \$\endgroup\$ – Jess Aug 6 at 7:43
  • \$\begingroup\$ @Jess at 100 kHz D.F. becomes a tad meaningless because the capacitive reactance is really low and may even be cancelled by the parasitic reactance of the capacitor thus, only ESR remains and this will dissipate as per \$I^2R\$ - at low frequencies (where I corrected an error in a previous question), ESR appears to rise but it's just the equivalent series resistance i.e. it combines all dissipation and, at low frequencies this will be dominated by the dielectric losses. Note that in that previous question I recalculated "ESR" to be 1.6 ohms whereas at 100 kHz, the true series..... \$\endgroup\$ – Andy aka Aug 6 at 9:52
  • \$\begingroup\$ .... resistive loss is only 0.16 ohms. In other words an extra 1.5 ohms has "appeared" in series with the true series resistance. That extra 1.5 ohms is representing the dielectric loss but, instead of calculating it as a parallel resistor (which it really is), it gets lumped into ESR where emphasis is placed on the word "effective". I know; it's complex. \$\endgroup\$ – Andy aka Aug 6 at 9:55

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