I have a device which has two sections of the PCB completely isolated form each other, but connected by an isolator IC (PN: ????).

I think I need to shield the whole device for reducing EMI because I am having a lot of EMI from the device.

The isolator IC transfers data and clock between the two isolated sections of the board.

One section connects to USB and is powered by the USB host.

The other section is powered by an external power supply and also has connectors for some external input signals that are processed by an FPGA.

What should I do to reduce EMI?

If I use ferrite beads, where should I add the ferrite beads? At the isolators? On pins transferring clock signal, data signal or both? Or on Vcc?

  • \$\begingroup\$ For this level of question, pictures and/or board layouts and schematics would help. Also, list your clock sources, and show your spectrum analysis as well. Also, updating your profile wouldn't hurt. \$\endgroup\$ Nov 7, 2013 at 12:27
  • 1
    \$\begingroup\$ What form does the isolator take i.e. is it magnetic or optical? What frequency is being used. How does the interference cause a problem? \$\endgroup\$
    – Andy aka
    Nov 7, 2013 at 12:28
  • \$\begingroup\$ @RolfOstergaard PCB1 has connectors for input signals which it will process. It is required to be isolated from PC. \$\endgroup\$
    – zud
    Nov 7, 2013 at 14:59
  • \$\begingroup\$ the isolated circuits are on the same board, but totally isolated. Sorry for that. \$\endgroup\$
    – zud
    Nov 13, 2013 at 8:25
  • \$\begingroup\$ I tried to clarify your revised question. \$\endgroup\$ Nov 19, 2013 at 12:25

2 Answers 2


1) Understand what the noise source is.

2) Understand what the antenna is.

3) Understand how the noise gets onto the antenna.

4) Do your fix (which involves reducing the noise source and/or destroying the antenna and/or reducing the coupling to the antenna)

In most cases I have seen, you can make a single board silent enough to pass EMI without having to resort to shielding cans. The trick is a quiet power distribution network (PDN) on the board, which gives you very low impedance over a wide frequency range. Values in the range of 1-100mOhm are common. Use PDNTOOL.COM to check.

Whenever you have multiple boards interconnected, in most cases I have seen you do need a Faraday cage around the whole thing. The trick here is to "short" all cable shields to the Faraday cage right where they exit. In this context, "short" is for all frequencies of interest (where you have problems passing EMI). In your case with two isolated sections, you will have to use AC coupling for everything exiting one section. Make sure that AC short is good at the frequencies you are interested in. Even a small 0603 cap is no good above 1-200MHz.

Filtering of input/outputs for frequencies above the useful frequency range is also required in most cases I have seen. This can usually be achieved using caps and resistors, but depends on your signal types.

As for your idea about using ferrite beads, make sure you understand how they work (impedance versus frequency and tolerances). I have not seen any cases where they were required.

If you have a case, where you can't get the single board silent enough, you can use a shielding can. In this case you could either make the can shield one or both sections of the board. Either way, you would treat the "can area" like a Faraday cage - just as explained above.

Finally, let me stress one thing: understand. If you skip easy on this, make sure you have plenty of time :-/

  • \$\begingroup\$ you are right. I already implemented the slope, it decreased the EMI but still not good enough \$\endgroup\$
    – zud
    Nov 7, 2013 at 17:18
  • \$\begingroup\$ The data are transferred from FPGA to PC via USB cable. The signals from isolator output have high slew rate too. I want to reduce the strength of the EMI on the isolator output side by adding ferrites on the pins. I already added ferrites on the clock and data pins on output side and it looks better. I am not sure if I need to the Vcc and GND too or not. \$\endgroup\$
    – zud
    Nov 8, 2013 at 8:40
  • \$\begingroup\$ Thanks for taking interest in this problem. What I understood is that sharp edges from are the main sources of EMI in digital circuits and they propagate via power plane, ground plane, signal lines and cables. Using ferrites in series with the pins so that ferrites absorb the high frequency energy. e.g., ti.com/sc/docs/apps/msp/intrface/usb/emitest.pdf \$\endgroup\$
    – zud
    Nov 10, 2013 at 21:22
  • \$\begingroup\$ Yes, I think so because there are no other paths for the noise. \$\endgroup\$
    – zud
    Nov 11, 2013 at 8:21
  • \$\begingroup\$ Well that was my question. I am not sure about the way EMI is propagating inside the board. I put ferrites on signals and clock pins and it gave a better result. I wanted to be sure if putting ferrites on power and GND is helpful, because I have to design the board again. \$\endgroup\$
    – zud
    Nov 11, 2013 at 15:46

This issue somewhat depends on what you're doing with the USB ground. See How to connect USB Connector shield? for a fair discussion, and http://forum.allaboutcircuits.com/showthread.php?t=58811 for more.

If the USB shield is not under your control and is tied to ground on board 2, or if you opt do tie it to board 2 on your own, then your choice is already made. You can't tie the shield to board 1.

My choice, in any case, would be to tie the shield to board 2, through a ferrite bead.

  • \$\begingroup\$ could you tell me why is it better to connect the shield to PCB2, not to PCB1? \$\endgroup\$
    – zud
    Nov 7, 2013 at 15:04
  • \$\begingroup\$ just can't think of a reason to float a shield. I'd want it tied to as robust a ground as I could find, so long as it doesn't cause ground loops \$\endgroup\$ Nov 7, 2013 at 15:38
  • \$\begingroup\$ But, I wouldn't necessarily consider clock and data switching noise to be EMI. Those sorts of things should probably be dealt with in terms of board design, rather than hoping a shield will solve your problems \$\endgroup\$ Nov 7, 2013 at 15:40

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.