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I try to calculate the ESR of a capacitor.

I checked this datasheet:

http://www.farnell.com/datasheets/2245214.pdf?_ga=2.167304311.817060034.1534770853-676261198.1531404202

I can see from the datasheet for the Aluminum Polymer Capacitors that they specify the max impedance ( Z ) and the tan( δ ) factors

I would like to calculate the ESR for some specific capacitors ( for example 220uF)

Can I just multiply these values in order to get the ESRmax value at 100kHz = tan(δ)* Z ?

Is that correct?

Thanks a lot for your help.

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  • \$\begingroup\$ Some useful information for Aluminum polymer: kemet.com/Aluminum-Polymer \$\endgroup\$
    – Peter Smith
    Commented Aug 20, 2018 at 15:37
  • \$\begingroup\$ In many places you want the ESR of a capacitor, it is best to also check the maker's spec on maximum ripple current. This spec is usually based on the heating effect of the I^2R. If in looking at the two you see something that appears not to compute one from the other always that the worst. \$\endgroup\$
    – Ken Smith
    Commented Apr 26, 2021 at 18:09
  • \$\begingroup\$ If you have access to an impedance analyzer, measuring is the best method. I have found the published ESR curves to be way off for ceramic capacitors. \$\endgroup\$
    – qrk
    Commented Apr 26, 2021 at 19:42

3 Answers 3

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In theory, you can find the ESR at a certain frequency by multiplying the dissipation factor at that frequency, also knows as tan(δ), by the capacitor's reactance at that frequency.

In practice, capacitors aren't completely linear. You can make a decent guess at the ESR, which lets you engineer for the range the ESR might be, but you won't know the exact value. See this graph from Murata for a visual example:

https://www.murata.com/en-us/products/emiconfun/capacitor/2013/02/14/en-20130214-p1

To estimate the ESR for your capacitors, I would look up the tan(δ) value, finding 0.12 at 100 Hz, and I would calculate the reactance at 100 Hz, ignoring parasitic inductance and ignoring the fact that it's nonlinear.

Sticking that into Wolfram Alpha gives me an ESR of about 0.87 Ω:

enter image description here

As an aside, in case you're confused as to why datasheets list dissipation factor under "tan(δ)", the graph below, which I found here, illustrates why the dissipation factor is equal to tan(δ):
impedance graph

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    \$\begingroup\$ Thanks for your answer. Why do I have to calculate the reactance at 100kHz? I thought that the manufacturer provides this max reactance in order to find the max ESR. And in any case, the result of my multiplication with yours should be close I think but the difference is very big. I am still not sure what is the correct thing to do. \$\endgroup\$
    – George A
    Commented Aug 20, 2018 at 16:13
  • \$\begingroup\$ tan(δ) is itself a function of frequency, so it will be different at 100 kHz than at 100 Hz. Since the datasheet only gives us a value at 100 Hz (not kHz), we need to use the reactance at 100 Hz for it to be useful. The impedance (Z) at 100 kHz isn't as relevant. You could guess the ESR using the impedance at 100 kHz as well by using a different formula, but since you don't know the parasitic inductance and since electrolytics are so nonlinear, you will end up with a pretty inaccurate answer. @GeorgeAvgenakis \$\endgroup\$
    – Darius
    Commented Aug 20, 2018 at 16:21
  • \$\begingroup\$ thanks a lot. And after this calculation is there a way to translate the calculated ESR at 100Hz to ESR at 100KHz? (I would like an aluminum Polymer as the datasheet and not an electrolytic) \$\endgroup\$
    – George A
    Commented Aug 20, 2018 at 16:26
  • \$\begingroup\$ The only way to know for sure would be to measure the part physically. You can use the same value at 100 Hz and 100 kHz for your estimate - since it's not electrolytic the ESR will be fairly, although not completely, flat. You just have to keep in mind that it's not an exact value. \$\endgroup\$
    – Darius
    Commented Aug 20, 2018 at 16:50
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ESR of a capacitor occurs when the impedance of its capacitance and inductance are equal and cancel out leaving only resistance. This occurs at SRF (self resonant frequency). The data sheet you provided only gives impedance at 100 KHz which is unlikely to be the SRF but is as close as you are going to get without measuring it. 100 kHz is a common frequency used for specifying the impedance of electrolytic capacitors as they are commonly used for low frequency filtering.

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    \$\begingroup\$ This is incorrect. A capacitor always has an equivalent series resistance. \$\endgroup\$
    – Darius
    Commented Aug 20, 2018 at 15:36
  • \$\begingroup\$ Yes they do, what I meant was the SRF is where the impedance = ESR. \$\endgroup\$
    – EE_socal
    Commented Aug 20, 2018 at 15:56
  • \$\begingroup\$ Thanks for your answer. In my case how do you think that I have to calculate the ESR for the capacitor of 220uF? \$\endgroup\$
    – George A
    Commented Aug 20, 2018 at 16:15
  • \$\begingroup\$ ESR is not something you calculate it is given by the manufacturer. I would use the impedance at 100 KHz on the datasheet. it would help to know how you are using this capacitor and why you want to know the ESR. \$\endgroup\$
    – EE_socal
    Commented Aug 20, 2018 at 16:42
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I computed Xc(f) after your component specs.

\$Zc(f)= R(f) - jXc(f) ~~~~,~~Xc(f)= 1/\omega C\$

I use R(f) since ESR is frequency sensitive and is mainly a percentage of dielectric impedance well below SRF and called "dielectric loss" then above SRF, it is the loss of the inductive electrodes with skin effects.

The lowest impedance is at the series resonant frequency and the shape of the "notch" like shape is due to the Q factor which is somewhat inverse to tan δ.

The important thing to learn is that DF and tan(δ) does not predict the ESR performance at 100kHz.

So if you see caps with only DF or tan(δ) then you cannot guess the ESR at high frequency and most likely, not a low ESR type.**

Volt   uF   tan(δ)  ESR (100kHz)  Xc(100Hz) Xc(100kHz)  Zc
 16V 220μF   0.12  0.022         723       0.723      0.723
 16V 270μF   0.15  0.012         589       0.589      0.590

If you compare aspect ratios, you will find e-caps have lower ESR with taller cylindrical shapes and higher voltage ratings when volume and C(uF) in the same type.

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  • \$\begingroup\$ For the computation of R(f) you used |Z|=sqrt( R^2+Xc^2) ? And then the result is R(f)=ESR(f) ? Thanks a lot \$\endgroup\$
    – George A
    Commented Aug 21, 2018 at 9:56

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