Ferrite Bead: How much impedance do I need?

Question

I understand that ferrite beads can be used to reduce noise from a power supply input to an IC.

I understand that one selects ferrite beads based on its resistance at the noise frequency one wishes to reduce and I need to select the ferrite based on what frequencies I expect digital switching noise.

My question is: how do I know how large the impedance of the ferrite bead needs to be in the resistive region @ my target freq? This is the part it seems that every guide I can find on the subject excludes.

Does it not really matter precisely as long as the bead has a maximum current rating 50% higher than the expect operating current?

Answer 1

You need to know the behaviour of the noise current: knowing its frequency and its amplitude, you should be able to compute the voltage drop across the ferrite bead at that frequency. Getting to the answer, the impedance should be such that the noise voltage drop on the pin of your IC (at the noise frequency) is much less than the voltage drop at the working frequency (DC, since you're interested in the power supply pin).

Answer 2

Unfortunately, a lot of this selection process is experimental and dependent on your setup. For example, your power supply topology might be really good at picking up 90MHz noise from the local TV station broadcast, etc.

For starters I would select a bead that has enough current rating (as you mention) and has enough impedance at the switching frequency of your power supply (assuming you are using a switching one at some point) to kill enough noise.

As @DavideM mentioned, you should know something about the noise you are trying to block with the bead. You should also know what noise your IC is sensitive to.

If you IC works in radio frequency, then an RF bead with good impedance at 2.8GHz might be useful since that band is pretty noisy.

Keep in mind that you can string beads together for suppression in multiple frequency bands but you have to be careful about their DC resistance.