Im in a project, where I will have some noisy digital IC's, and a analog sensor circuit that I want to be a little bit noise free -

Im using a buck regulator to take in some power from a Lipo battery.

For when splitting the supply to analog and digital, how /what would i select for connecting the 2 sources to the buck converter? -

Can you use one of those "EMI filter" beads, which have a built in filter, or how would I select an good ferrit bead for this application? Thanks for your time


2 Answers 2


You mean that you want the switching noise on the supply made by the DCDC converter (buck regulator) and the digital circuits not to reach the sensitive analog circuits?

Then I would personally ignore EMI filters!

Well, OK, I would only think about EMI filter after at least implementing the suggestions I give below.


Because these are only reasonably effective at high frequencies, think above 10 MHz or so. Also, the higher frequencies do not tend to travel so much through the wires as through the air!. A wire is about 1 nH (nano Henry) per mm. So a long wire is already an obstacle for any RF signal.

The DCDC converter will switch at a much lower frequency (usually between 50 kHz to 2 MHz) so EMI filters will only suppress part of the noise. For the digital the same is true, it can generate many frequencies also below the range where EMI filters have much effect.

Here are some tips that are more effective:

Choose a ground point and implement a star grounding scheme keeping the digital and analog grounds connected but minimize them influencing each other. If you share the same ground than the return currents cause local, small voltages in the ground. Your sensitive analog IC might not like this. The solution is separate ground wires /connections so that a circuit only "sees" the ground voltages that it has created itself and not the voltages from ground currents from other circuits.

Bypass and decouple your supplies as close as possible to the source of the noise. This keeps the (current) loop for high-frequency signals short and minimizes the chance of noise "escaping".

Use multiple capacitors of different values in parallel for bypass and decoupling. Capacitors are not ideal, to make more ideal ones use multiple, see EEVBlog Dave's video on the subject.

For the analog circuit, consider using a dedicated LDO (non-switching voltage regulator) to make the supply for that circuit only.

What you do is put 0 ohm resistors here and there in the supply lines, then if later it turns out that there is a high-frequency issue you have the chance to add a series resistor or EMI filter (or just an inductor) to solve the issue. Also add (the option) to add a capacitor to ground on both sides of that zero-ohm resistor. No need to add a capacitor just yet but when you do need it, adding one will be easy.

  • 1
    \$\begingroup\$ I especially like your design guidelines. OP, note that it's really common to have a switch-mode power supply generate a supply voltage some 300 to 1200 mV above what your analog device needs, and then use a good LDO (compare datasheets) to regulate the rest. It might also be a good idea to start comparing datasheets and design guidelines for your switch mode SMPS – noise that doesn't exist doesn't have to be filtered out, and at this point, you'd probably want a SMPS that works at a frequency as high as possible (easiest to filter inductively), … \$\endgroup\$ Nov 1, 2017 at 9:10
  • \$\begingroup\$ …, as far as possible away from your analog circuitry (easiest to avoid crosstalk). Oh, and don't give your supply too much of a hard time alone – you're usually not in total control of the noise environment, anyway, so whenever you do analog signals, be sure to consider filtering them intentionally to the bandwidth you're interested in. \$\endgroup\$ Nov 1, 2017 at 9:11
  • \$\begingroup\$ ..and if far away is not possible, at least add a ground wire in between which will act as a shield. \$\endgroup\$ Nov 1, 2017 at 9:18
  • \$\begingroup\$ Thank you for your well written answer! - The part im most worried about is my 2.4GHz transmitter (WIFI), so there I looks like I could use a ferrit - and of course I need a bit of filtering on the output of the switching regulator - Think I will look into an LC filter for that - Thanks once again! \$\endgroup\$
    – M.Egsdal
    Nov 1, 2017 at 9:37
  • \$\begingroup\$ +1 all good advice. However, I would also add it is important to remember that the supply rail is also the current return path for any low signal coming form the digital side to the analog side or vica-versa. If that return path has to go through a regulator and filters to get back to the source supply that can be problematic and actually make the noise situation worse. \$\endgroup\$
    – Trevor_G
    Nov 1, 2017 at 14:34

You can filter the voltage going to the analog IC using an RC (if the current is not too high) or a ferrite and capacitor. Ideally, you would have some idea what you are trying to filter, or what you can allow to pass in order to design a filter. Sometimes a ferrite and capacitor can resonate with fairly high Q. So you have to be careful about that. I've never actually seen this happen, but it seems plausible and I have been told that it can happen.

Another approach that can work sometimes is to power the analog IC with an LDO powered from the buck. This would only work if the analog IC can be operated at a lower voltage. The idea is to let the power supply rejection ratio of the LDO be a filter.

Separating supplies can be good, but I recommend you do not separate the grounds.

It might be helpful if you tell us a bit more about your analog sensor circuit. What is it sensing. How sensitive is it, really?

  • 1
    \$\begingroup\$ Well, one of the more noisy thing I have en a Wifi transmitter (2.4GHz) so I have thought about using something like a murata BLM18HE102SN1 and a ceramic capacitor to remove the high frequency noisy, and mabye also a lower frequency bead in series - \$\endgroup\$
    – M.Egsdal
    Nov 1, 2017 at 8:44
  • \$\begingroup\$ but would EMI filter "beads" also work? - is there a reason why they are labled EMI, that should make it wrong to use for splitting supplies? \$\endgroup\$
    – M.Egsdal
    Nov 1, 2017 at 8:45
  • \$\begingroup\$ Yes. That is what I meant by ferrite. A ferrite bead. RF transmitters can really generate a lot of noise. Make sure you give good thought to the layout and placement and follow all the recommendations of the WiFi vendor. It may be a good idea to add high-frequency capacitors on all analog inputs. This means small caps with low impedance even at 2.4 GHz. \$\endgroup\$
    – user57037
    Nov 1, 2017 at 8:49
  • \$\begingroup\$ An 0402 1k resistor has higher impedance at 100 MHz than most ferrite beads. And higher than all 0402 ferrite beads. So sometimes for micro-current circuits, you can filter with an RC instead of a ferrite bead. \$\endgroup\$
    – user57037
    Nov 1, 2017 at 8:51

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