I'm trying to reduce some switching noise I'm seeing (using LISN & Spectrum Analyzer) on my input power cable. My SMPS operates at 1.2MHz. I have room on the PCB and I'd like to place a SMT common mode choke with a couple of filter capacitors on the input and output of the choke. I was hoping to make a sort of Pi filter.

The device will use 12V from a vehicle cigarette lighter at approximately 75mA. I believe the output resistance of the car battery is relatively low.

I've looked at this Murata Noise Suppression PDF and I see a few examples of possible solutions but no Pi filters.

Any suggestions? Am I going at it all wrong?


1 Answer 1


A "typical" two-stage power supply EMI filter looks like this:


simulate this circuit – Schematic created using CircuitLab

The inductors attenuate common-mode noise (and also offer some differential attentuation via their leakage inductances). C1 and C4 shunt differential-mode noise. If there's an 'earth' point available you will also see two more capacitors between CM1 and C4, connected from each rail to the earthing point. These provide more common-mode attenuation.

Sometimes you will see a single stage, sometimes two. As long as your filter is attenuating the necessary noise components and are OK thermally, I don't see a pi filter being 'wrong' necessarily. (For lower frequency conversion, it's easier to get the attenuation via the inductors - just add more turns / more core etc.)

  • \$\begingroup\$ Thanks. Do you have any reference for calculating inductance for common-mode and capacitors for differential-mode noise? \$\endgroup\$
    – Samee87
    Nov 17, 2015 at 18:10
  • 1
    \$\begingroup\$ EMI filter design is often empirical - you know what your switching frequency is, but other parasitic noise sources are often difficult to quantify until you actually measure the noise with a spectrum analyzer. Once you know what you're dealing with in terms of noise, look at the impedance vs. frequency specifications of the capacitors you intend to use. As for the common-mode inductors, more inductance is generally better but you pay a penalty in terms of copper loss from the winding resistances (i.e. lots of turns). \$\endgroup\$ Nov 24, 2015 at 14:47

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.