9
\$\begingroup\$

In some pages online, it is mentioned that "whenever possible, the targeted noise frequencies should fall within the bead’s “resistive band,” meaning the range of frequencies in which the resistive impedance dominates the reactive impedance."

Does that mean that if I have noise that is between 50 MHz and 200 MHz, I would need my ferrite bead's X R Z plot to be resistive in this frequency band as shown in the image below, where the ferrite becomes more resistive than inductive for frequencies > 25 MHz?

enter image description here

Moreover, would higher resistance or lower resistance be better at filtering noise? Meaning, if one ferrite is 110 Ω @ 25 MHz while another one is 50 Ω @ 25 MHz, which would be better at suppressing noise?

\$\endgroup\$
1
  • 4
    \$\begingroup\$ Your interest seems to be noise suppression. Often, ferrite is used in combination with filter capacitors to increase attenuation. This gets complicated, because a large-value capacitor may resonate with the ferrite at a lower frequency where inductance dominates. Killing noise at all frequencies can be frustrating, like a game of whack-a-mole. \$\endgroup\$
    – glen_geek
    Commented Aug 31, 2022 at 20:07

5 Answers 5

16
\$\begingroup\$

If you are using the ferrite bead primarily to filter unwanted RF, both inductance and resistance are helpful. Filters can be made either way, with inductors or resistors. Higher impedance provides more filtering. The benefit to using the ferrite in its resistive region is that the RF energy will be dissipated as heat by the ferrite. If you use it in its inductive region, the inductor may become part of a resonant circuit that causes radiation somewhere else.

If you jump off of a tall building, you can come to a stop safely by hitting huge stacks of empty cardboard boxes. Or by landing on a trampoline. The resistor is more like the cardboard boxes. The inductor is more like a trampoline. That may be a poor analogy. But the basic idea is energy dissipation versus energy storage.

\$\endgroup\$
1
  • 4
    \$\begingroup\$ That trampoline analogy is amusing! It really conveys the concept that, once you safely bounce, you still have the problem to safely land somewhere! :-) \$\endgroup\$ Commented Sep 1, 2022 at 11:49
7
\$\begingroup\$

There isn’t a simple answer. Beads not only are sized differently, but are made from a variety of materials that have different characteristics. Some work best for blocking low frequency but don’t perform as well at higher frequencies, while others have wider coverage but trade off current carrying ability.

Generally you will choose a filter that trades small size/cost vs. impedance vs. current handing capability. So you will choose lower impedance for power traces (and thus, a larger bead) and smaller, higher impedance beads for signals.

That said, Murata has a wealth of information on choosing the right filters (including beads) for your application. Try here: https://www.murata.com/en-us/products/emc/emifil

(I don’t work for Murata. Just find their info useful.)

\$\endgroup\$
3
  • \$\begingroup\$ Is the impedance for power low and for signal high for input impedance matching purposes? \$\endgroup\$
    – Shannon
    Commented Aug 31, 2022 at 21:05
  • 3
    \$\begingroup\$ No. Low impedance on power beads is just for reducing IR drop. \$\endgroup\$ Commented Aug 31, 2022 at 21:09
  • 1
    \$\begingroup\$ @Shannon The lower impedance on power beads is more due to core saturation than anything else. The beads suggested for "power lines" only have few windings to allow them to sustain a lot of low frequency current without saturating. Core saturation on a bead turns it essentially into a 0-Ohm-resistor. \$\endgroup\$
    – tobalt
    Commented Sep 1, 2022 at 8:43
4
\$\begingroup\$

Focus less on what you want to block, usually one doesn't know very well and even if you know, blocking that can cause other problems.

Focus instead more on your signal bandwidth that the bead should leave alone. Then select a bead with no attenuation in the signal bandwidth.

As an example, for power distribution and low frequency, the bandwidth is low, so you can use the super-lossy types, which become resistive already at 5-10 MHz. But you also could use any other types that only become resistive e.g. at 30 MHz. So no big mystery here with the selection.

The only exception is high frequency beads for signals up to several 100 MHz. These attenuate in the GHz range but are not very effective at mundane EMI suppression near 100 MHz.

How much impedance?

Again look at your signal trace. If it is a 50 Ohm trace and you put down an impedance between about 20-100 Ohm, you will damp it pretty effectively, so any resonance will have a hard time building sufficient energy. If you put a 1 kOhm impedance instead, it essentially looks like an open circuit for EMI. New modes form at higher frequencies due to the new shorter trace length. The good thing about higher impedances is, that they are of course better at steering the EMI energy to where you want it. E.g. that makes it harder for EMI energy to enter/leave devices and work well in pi filters and such.

\$\endgroup\$
4
\$\begingroup\$

There's a bunch to it, but you do want the bead to be lossy aka resistive in the frequency region of interest.

Lossy means the energy is permanently removed from the circuit (like what a resistor does) rather than only temporarily removed from circulation only to be stored and then released again at a later time (like a capacitor and inductor) because resonance.

And the frequency of the energy when it is released back into the circuit will be the same as when it entered the circuit because capacitors and inductors are linear components. Not what you want if you are trying to suppress the frequency in the first place.

If only that energy were released at a different, much lower frequency like DC. But alas, things don't work that way.

\$\endgroup\$
3
\$\begingroup\$

The specific requirements in your circuit depend on how much noise rejection you need, among other factors. Also, I usually get a pretty good guess from simulations/models but then adjust values empirically.

In general, a ferrite bead is used because it absorbs noise due to its effective resistance at the noise frequency. An inductor can be part of a noise rejection network as well, though it typically has low effective resistance and does not absorb noise, it reflects it back up the chain.

So, both inductance and resistance can be helpful, but when looking at ferrite beads, it’s often the resistive part that is most useful.

You don’t really need high resistance to inductive impedance, you just need good noise rejection. The ~100 ohms resistive and reactive that you have at 25MHz with that ferrite bead is a pretty good start. I would order another bead in addition, with 500 ohms or 1k ohm so if noise was higher than I wanted, I could swap that out to quickly see if it’s a limiting factor.

Also, pay attention to DC rating. You want something rated well above the DC current you need. At least 2x, maybe 5x, as this has a significant impact on effective inductance and resistance as well.

Good luck!

\$\endgroup\$
1
  • \$\begingroup\$ Even 2x is nothing. If I remember correctly when looking at curves, even 10x barely approached the rated specs. We're talking like -30%. \$\endgroup\$
    – DKNguyen
    Commented Sep 1, 2022 at 14:45

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.