0
\$\begingroup\$

I have completed a schematic for a PCB and selected a series of decoupling capacitors for usage with a variety Xbee format radios on one board (i.e. one at a time, but changeable). I have included in decoupling capacitors of various sizes as specified by data sheets. As I begin the the physical layout my question is: does the physical order of these capacitors have any importance? (i.e. 8.2pF closest, then 1uF, then 10uF, then 100uF furthest from the power pin OR 100uF closest and 8.2pf Furthest)

\$\endgroup\$

1 Answer 1

1
\$\begingroup\$

Usually you want the one with the lowest ESR to be closest to the thing thats sensitive to noise. This also tends to be the one that is smallest, assuming that the selection was done properly. In this case, the 8.2pF should be closest to the device thats drawing power.

\$\endgroup\$
3
  • \$\begingroup\$ What effect does the ESR have on the pin? I am thinking the ESR is ESR no matter where it is placed... \$\endgroup\$ Commented Jun 10, 2016 at 15:03
  • \$\begingroup\$ It is the closeness to the pin that counts. The small valued capacitors likely have the best high frequency performance. Trace length has impedance including its distributed inductance. The shorter the RF path to the capacitor the less impedance in the traces to components that can best handle that frequency filtering. \$\endgroup\$ Commented Jun 10, 2016 at 15:57
  • \$\begingroup\$ You can think of it like a low pass filter. It's best if you can have a really big ideal low pass filter right next to the pin, with a high capacitance and low ESR. Unfortunately it's difficult if not impossible to have low ESR and high capacity, at least in a reasonable size. So you compromise, and basically have several low pass filters in series. the ones far away are not ideal, but have a lot of capacitance to deal with any big slow power droops. The ones close by cant deal with big droops, but can deal with very small fast ones. \$\endgroup\$
    – BeB00
    Commented Jun 10, 2016 at 16:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.