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I understand that ferrite beads provide high impedance for high frequency signals. However, I am very confused on what that actually means. I've been reading in multiple posts and articles that ferrite beads turns high frequency noise into heat. Is that actually true? I don't understand how high impedance relates to turn noise into heat.
My suspicion is that it's actually false and our interest in using ferrite beads is to increase the impedance in our main path, and so we should provide a lower impedance path to ground for high frequency signals. I mean, in other kind of filters (like bypass capacitors) we don't actually try to vanish noise from our PCB, but lead it into a not important low impedance path.

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Yes, it is true, ferrite beads do convert noise into heat.

However, it is simply because it becomes resistive at the frequency of interest.

So it has resistance. And high resistance means high impedance too. And even high is relative. Ferrite beads are rated to have few tens to few hundred ohms of impedance at 100 MHz, where part of it is resistive and the rest is inductive. It's just higher impedance in series than say a bypass cap can shunt to ground, from milliohms to few ohms.

Also everything else that has resistance turns any current into heat, even wires and PCB traces.

So it is not false, but a ferrite beads also have inductance. And so if you put bypass caps on both sides of a ferrite bead, you have a pi filter construction, which has low impedance to ground from both sides and high impedance between both sides.

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  • \$\begingroup\$ What would happen if we forced the high frequency noise to go through the ferrite bead (by not providing a lower impedance path)? Would it just turn into heat, or would it still exist in the output? \$\endgroup\$ Commented Mar 16 at 1:56
  • \$\begingroup\$ @GabrielCorrales Like I said, the ferrite has resistance and inductance, so any high frequency AC current passing throgh it would see some impedance and would also dissipate power into heat in the resistive part of the impedance. You can simplify a ferrite bead to be an inductor with wiring resistance. \$\endgroup\$
    – Justme
    Commented Mar 16 at 2:07
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Ferrite beads are resistive at high frequencies so they do dissipate power as a result of impeding high frequency noise, just as a series resistor would.

At low frequencies they are inductive, so there is a significant reactance component to the impedance. See the below curve from a datasheet.

enter image description here

The DC resistance of this part is only 120mΩ, but of course that will dissipate heat as well, depending on the DC component of the current passing through it.

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  • \$\begingroup\$ Spehro, These graphs I've seen before, and seem suspicious - could "X" reactance (green) really dive to zero @ 100 MHz, and not become capacitive at higher frequency? That red "Z" curve looks like a resonant peak @ 100 MHz. Do you know why the green curve doesn't continue above 100 MHz? \$\endgroup\$
    – glen_geek
    Commented Mar 16 at 2:48
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    \$\begingroup\$ @glen_geek You are completely right. The reactance will go capacitive above 100MHz, so the reactance will become negative, and thus be outside of this graph. That's why the line doesn't continue. \$\endgroup\$ Commented Mar 16 at 7:35
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Well, yes, but that's not typically an application of a ferrite bead, to provide a heating solution. They dissipate current for high frequency noise. That dissipation does technically generate heat due to characteristic impedances.

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It's both, and it depends on the frequency, the ferrite material, and the geometry of the winding. The choke will usually have an impedance that's both resistive and reactive (normally inductive).

The resistive part represents an opportunity for energy at that frequency to be turned into heat in the choke, while the overall magnitude of the impedance represents the choke's ability to "block", i.e. reflect, that frequency. Energy that's reflected may just end up dissipated somewhere else, or it may take another path, e.g. finding its way to ground through a bypass capacitor. That's a broader design question. Burning noise as heat in a ferrite is generally not a bad thing, anyhow, unless there's enough of it to cause thermal issues.

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