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Capacitor on PCB with trace

I want to know the effect of PCB trace length to a capacitor's frequency response, especially in terms of bypassing and decoupling applications. For instance in the above simplified PCB, we have a capacitor, C1, connected to VDD and GND, near an IC, U1.

I am able to obtain the frequency response of this 0805-sized capacitor (PN CL21C270JBANNNC) from the manufacturer, as below:

Freq response

What is the effect of the PCB traces to the frequency response, particularly to the resonance frequency? Assume, the trace lengths between the capacitor and IC are 10mm, trace widths are 0.25mm, and this is a standard 1oz copper FR-4 board.

My assumption is the manufacturer obtained the frequency response curve with zero or minimal trace lengths. How does the resonant frequency shift because of the traces?

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  • \$\begingroup\$ 10 mm is getting close to 1/2 inch. That is double the maximum length for RF and ultra-fast logic, which need a low ESR capacitor of 10 nF or so within 7.62 mm of the IC. \$\endgroup\$
    – user105652
    Jun 21, 2018 at 1:35
  • \$\begingroup\$ I went into this subject in some detail on the effects of ESL and upt a table of typical ESL values against package sizes. electronics.stackexchange.com/questions/193608/… \$\endgroup\$
    – Peter Smith
    Jun 21, 2018 at 11:44
  • \$\begingroup\$ ESL for tracks is about 5nH per inch on a 0.1mm track; for 0.25mm, that would be about 1nH per inch, which is a lot in a high speed device world (most modern logic falls in this category). \$\endgroup\$
    – Peter Smith
    Jun 21, 2018 at 11:51

2 Answers 2

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My assumption is the manufacturer obtained the frequency response curve with zero or minimal trace lengths. How does the resonant frequency shift because of the traces?

For this particular question, I see that there is not perfect or universal method for the measurement of ESL.ESL numbers are generally obtained by making a measurement of the inductance of a capacitor in a given circuit and comparing that value to the inductance of the same circuit with the capacitor shorted. This method is seriously flawed, since the inductance of a current loop is not the sum of its parts. As a result, ESL numbers published in datasheets are strongly dependent on the method and apparatus used to obtain them. Furthermore, there is virtually no correlation between the ESL numbers published in a datasheet and the connection inductance that determines the high frequency impedance of the capacitors in real applications.

Further info: http://learnemc.com/qotw-180604

The other part of the question is already answered.

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The trace inductance can be computed with free Saturn PCB designer software but basically it around 1nH / mm so you can estimate the drop in SRF with load C by computing the ideal cap only ESL [nH] from the Graph SRF. This is determined by its path L/W aspect ratio, mainly.

SRF series resonant frequency (or self ..).
ESL effective series inductance.
L/W length to width ratio.
nH nanohenry

$$SRF = \dfrac {1}{2\pi \sqrt {LC}} $$.
L=Lc+ 2* Ltrace

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  • \$\begingroup\$ Thanks for the answer. So, to clarify and examplify here: \$\endgroup\$
    – Adam B
    Jun 21, 2018 at 1:00
  • \$\begingroup\$ So what ? You want to try calculate ESL of Cap and traces? \$\endgroup\$
    – Tony Stewart EE75
    Jun 21, 2018 at 1:09
  • \$\begingroup\$ Calculations: I can first compute the cap-only L: L = 1 / ((2*pi*SRF)^2 * C) = 1/ ((2*pi*1.7e9)^2 * 27e-12) = 0.32 nH. New SRF = 1 / (2 * pi * SQRT((0.32e-9 + 10e-9)*(27e-12))) = a SRF much lower than before! So if I was interested in reducing 1.6GHz noise (for GPS applications), then my capacitor is not effectively providing a low impedance path to ground anymore for 1.6Ghz. \$\endgroup\$
    – Adam B
    Jun 21, 2018 at 1:12
  • \$\begingroup\$ Compute the shunt to source impedance ratio for conducted noise reduction as per above answer \$\endgroup\$
    – Tony Stewart EE75
    Jun 21, 2018 at 2:49

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