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Old 28th Mar 2019, 5:55 pm   #1
ms660
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Default Capacitor ESR

I've just come across a formula for converting the leakage resistance of a capacitor to an equivalent series resistance, it's shown below:

Rs = Xc squared/Rsh….where Rs = the equivalent series resistance, Xc = the capacitive reactance and Rsh = the shunt resistance.

Is that accurate or just an approximation?

Lawrence.
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Old 28th Mar 2019, 6:41 pm   #2
G8HQP Dave
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Default Re: Capacitor ESR

It is an approximation, although in some cases it can be quite a good approximation. However, you need to remember that it only applies at one frequency. Both leakage and ESR can be frequency dependent but in different ways as they arise from different causes.

What the formula does is convert (approximately) from the parallel to series form for an impedance. If you had a perfect capacitor in series with a resistor then the formula used in reverse would allow you to calculate an equivalent parallel resistance, but this resistance would tell you nothing at all about the DC leakage (which would be zero for a perfect capacitor).

So use with caution.
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Old 29th Mar 2019, 1:52 pm   #3
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Default Re: Capacitor ESR

Here's where I came across the formula under discussion (page 3):

https://www.americanradiohistory.com...ox-1935-03.pdf

Here can found other editions of the Aerovox Research Worker:

https://www.americanradiohistory.com...Researcher.htm

Lawrence.
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Old 29th Mar 2019, 2:24 pm   #4
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Default Re: Capacitor ESR

The accurate formula is
Rs = Rsh Xc^2 / (Rsh^2 + Xc^2)

If Rsh >> Xc then the second term in the denominator can be ignored so we get
Rs = Rsh Xc^2 / Rsh^2 = Xc^2 / Rsh

They introduced that formula in the article in order to show that capacitor leakage (for a non-faulty component) can be ignored for all except DC purposes. It is not a valid way of calculating ESR from DC leakage.
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Old 29th Mar 2019, 2:26 pm   #5
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Default Re: Capacitor ESR

That's the formula I used when I did a comparison.

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