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ESL of ceramic capacitor seems too high, at 8.8uH

treez

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Hi,
We wish to calculate the ESL of the following 1nF, 1kV capacitor...
RDER73A102K2M1H03A

The attached shows an impedance vs frequency graph.

We wish to calculate the ESL at 2.8MHz, and with DC Bias of 600V on it. (it is a capacitor connected from drain to ground in a flyback converter.)

The ESR at 2.8MHz is 1.1R
The Capacitance at 600V is 600pF

So using Z = SQRT( A^2 + B^2).
...where A = 1.1R and B = [wL - 1/(w*C) ]

...we get ESL = 8.8uH.

This seems very high for a ceramic capacitor?
 

Attachments

  • RDER73A102K2K1H03B_EC-data.pdf
    48.3 KB · Views: 11

albbg

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First of all, to calcultate the ESL you have not to consider the capacitance under the DC bias, but the nominal value, since the Z graph is referred to that condition.
Then you have to remember that during the calculation we loose the information about the signs since we are applying square and square root operators. In particular we have:

Z^2 = ESR^2 + [wL-1/(wC)]^2

that is [wL-1/(wC)]^2 = Z^2-ESR^2

You did wL-1/(wC)= sqrt(Z^2-ESR^2) thus L =[sqrt(Z^2-ESR^2) +1/(wC)]/w this is not correct

Since we know Z is negative (same sign of 1/(wC)) and ESR is less than Z it must be:

-wL+1/(wC)= sqrt(Z^2-ESR^2)

we can now calculate L as:

L =[-sqrt(Z^2-ESR^2) +1/(wC)]/w

numerically:

L = [-55+57]/(2*pi*2.8MHz)=113 nH
 

treez

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Thanks,
You are correct, i made it out as Z = R + j {wL - [1/(wC)] }

But now we're saying that Z = R + j { [1/(wC)] - wL } ?
 

albbg

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No, Z = R + j {wL - [1/(wC)] } is correct, but when you square you loose the sign informations, because:

{wL - [1/(wC)] }^2 = { [1/(wC)] - wL }^2
 

FvM

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You can't calculate ESL this way. Use either SRF or inductive branch of impedance curve. SRF gives ESL of about 5 nH.
 

Easy peasy

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the length of wires and any loop area in an SRF text fixture will severely affect the measured ESL for ESL < 200nH, this also applies in the layout where the part ESL can be doubled or more by the layout.
 

albbg

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You can't calculate ESL this way. Use either SRF or inductive branch of impedance curve. SRF gives ESL of about 5 nH.
You are right. I didn't realize the frequency was quite low. In any case the mathematical method is correct, the problem in this case is that you can't have |Z| measurement accurate enough. At a frequency of 2.8MHz C=1nF ==> |Z| = 56.8 ohm, while a 5nH inductor have a |Z| = 0.09 ohm so the series will have |Z| = 56.7 ohm
 

mtwieg

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One look at the impedance plot indicates clearly that at 2.8MHz the ESL is negligible. Not sure what else to say.

And ESL will not depend on DC bias.
 

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