2
\$\begingroup\$

I find myself in the situation of being forced to use filter connectors to solve an EMC issue very late in the project. Normally, this should of course be solved in circuit design, but as projects and schedules go, this is where we are. Now, to simulate the influence of the filter parasitics onto the main circuit, I'm turning to the manufacturer data sheet to get the capacitance and inductance values for the different filter topologies (C, PI):

enter image description here

Source: Glenair EMI/EMP filter connectors (PDF)

To my amazement, Glenair is not listing the inductance values of the PI-filter. Could it be because the inductive element is so small (coil or ferrite bead for each pin), that the filter response is dominated by the capacitive elements anyway? Then why make a PI filter in the first place?

Any insight here is much appreciated.

\$\endgroup\$
2
  • \$\begingroup\$ How will you model the "GROUND"? as zero inductance? is this a plane? or a meter of wire, thus approx. 1microHenry? \$\endgroup\$ Apr 22 '19 at 13:31
  • \$\begingroup\$ Ground, I assume, is the filter connector shell itself, which is connected to equipment chassis ground, so yes, this impedance would be close to 0. \$\endgroup\$ Apr 22 '19 at 16:35
0
\$\begingroup\$

A filter can be made more effective if it has a series element to drop the undesirable frequency energy. For an EMI Pi filter, the inductor is usually a ferrite material. The effectiveness of the ferrite depends on "lossy" characteristics that are not easily modeled. The designer is expected to use the supplied attenuation curves.

\$\endgroup\$
3
  • \$\begingroup\$ Attenuation is not the main issue at hand, but phase margin is. The objective is to avoid instability of the main circuit and clarify how much margin exists. \$\endgroup\$ Apr 22 '19 at 7:45
  • 1
    \$\begingroup\$ The ferrite won't be effective at less than 10s or even 100s of megahertz. It probably can assumed to be negligible at the frequencies in your phase margin analysis. \$\endgroup\$
    – Mattman944
    Apr 22 '19 at 8:16
  • \$\begingroup\$ That's a valid point, I guess. The buck converter switching frequency is less than 1 MHz, so assuming it's a ferrite bead, it would probably be negligible. (But then I would go into splitting hairs mode, and question the manufacturer's designation of the filter. Would it not rather be a distributed capacitance with a HF ferrite bead in between?) \$\endgroup\$ Apr 22 '19 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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