In many circuits I see the analog supply and the digital supply separated by a ferrite bead.

I am aware that it is done because the devices that are connected to the digital supply create noise and hence the ferrite bead blocks the noise entering into the analog part which is noise sensitive.

How do we calculate the value of the ferrite bead and what are the considerations involved? Instead what if we connect 0hm resistor thereby connecting the analog and digital supply over a single point such that the amount of noise entering the analog side is considerably less?

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    \$\begingroup\$ How do we calculate the value of the ferrite bead Who says it is exactly calculated? To even calculate its value we would need to know the amount of noise (over frequency) at the digital side and how much noise is allowed on the analog side. The impedance (over frequency) of both supplies will be a factor. \$\endgroup\$ Oct 13, 2020 at 13:45
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    \$\begingroup\$ What if we took a more practical approach: if we suppress the noise more than needed, that's a good thing right? So why don't we just put the largest ferrite bead we can fit and can also afford (not too expensive) and hope that that's good enough. My guess is that is what is often done. Then, when a prototype is made, we can try a smaller or larger ferrite bead and see what works or doesn't work. Not everything in product design is defined by formulas! \$\endgroup\$ Oct 13, 2020 at 13:47
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    \$\begingroup\$ Welcome to the voodoo side of electronics design. This type of thing is more trial and error than you might imagine. \$\endgroup\$ Oct 13, 2020 at 14:09
  • \$\begingroup\$ Also realize that for this (choosing a ferrite bead), "trial and error" can be OK. Just assume something will be "good enough", then prototype and test it. If the initial choice was not good enough, there is still the chance to change it as it is just a ferrite around a cable. \$\endgroup\$ Oct 13, 2020 at 14:17
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    \$\begingroup\$ @Bimpelrekkie Instead of writing all those rants in the comment section, why don't you just write it as an answer where it belongs. \$\endgroup\$
    – pipe
    Oct 13, 2020 at 17:05

3 Answers 3


Separating the planes with a ferrite bead can prevent high frequency noise on the digital ground from conducting into the analog ground, but it has a serious drawback. If the analog plane has changing currents (which it probably will because of ADC's or DAC's) then any current traveling back to the digital plane will create common mode noise through the inductor (because it is blocking current and any current through the inductor will create a short term voltage between the two planes).

A better way would be to tie the planes together and not have separated planes (in almost every case this is better but not all) and have a sufficient ground, usually problem current's from digital electronics can be avoided by proper placement of the components, as currents from ground pins return through the plane back to the source (usually a cable on the board or a regulator) so don't place the analog components on this path (or any current path on ground with a switching load). This scheme may not work if there are many cables that interface with the board so keep that in mind (because of ground loops, analog connectors\shields that connect to ground on the PCB and another location could carry currents, so either break the ground loop on one end of the shield OR...)

Isolate between digital and analog ground, if the digital environment is especially noisy and the analog measurements are sensitive then the best way to avoid noise is to isolate, then no current can flow between analog and digital sections.

  1. Make some sort of estimate of how much noise is on your digital side, and at which frequencies it exists.
    a. This can be based on experience with other similar boards.
    b. This can be based on measurements of a prototype.
    c. You may be able to estimate ripple on things like DC-DC converters or other things that switch in a known way.

  2. Estimate how much noise is allowed on the analog side.

  3. Determine the required attenuation based on the estimated noise sources and allowed noise.

  4. Determine how much DC current will be flowing through the bead. The impedance of many beads reduces significantly with DC loading.
    a. Pick a bead that will still have reasonable impedance at your estimated current levels.
    b. The impedance vs frequency and load current is usually found in the datasheets for the parts.
    c. Larger beads are able to handle more load without saturating.

  5. Usually picking the bead that has the highest impedance at the required frequencies and working current will give you the most filtering. This usually means reading some datasheets and looking at curves.

  6. A bead by itself doesn't make that great of a filter. Combine the bead with some high frequency capacitors to ground. Select the capacitors based on your required attenuation.


We don't calculate the value, we buy a value that the vendor offers. Whatever inductance it has is again determined by the supplier.

You can buy beads on their own and pass a wire through one but it's not an exact science at all. The general idea is simply to 'help' decouple one supply from another. Any advantage gained is useful.

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    \$\begingroup\$ I don't understand this answer at all. "- How do you select a resistor? - Oh, just buy the one that the vendor offers.". Doesn't make much sense when the "vendor" offers a large range of values. \$\endgroup\$
    – pipe
    Oct 13, 2020 at 17:10
  • \$\begingroup\$ I'm not going to vote you down, but I've just started a design where I've orderd a lot of different ferrite beads to experiment with to find the best one for minimizing noise. I don't know what frequency of noise I will find, but I'd rather have options to experiment with. \$\endgroup\$ Oct 13, 2020 at 19:24
  • \$\begingroup\$ I'll have to rephrase it to reflect the 'choice by experiment'. Out of interest, how widely varying are the ferrites you're considering ? \$\endgroup\$ Oct 13, 2020 at 21:08

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