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Sorry, if this is obvious or stupid, but what is the point of ferrite beads with more than \$Z\approx100\space\Omega\$ of impedance? I mainly see their application as serial filters in either power lines (impedance usually \$Z\approx10\space\Omega\$ or even lower) or signal lines (impedance \$Z\approx50\space\Omega\$).

Therefore, when placing beads with large resistance of several \$100\space\Omega\$ in such situations, won't they essentially work like a rather usual high Q inductor that will nicely resonate with the surrounding capacitances? From what I understand, it is exactly the low parallel resistance that damps resonances...

I am talking about "typical" (to me) beads chips in e.g. 0603 footprints, which have several 100 mA to A rating and give their impedance value at 100 MHz typically, while the impedance could peak at a few 100 MHz.

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    \$\begingroup\$ No, beads are low-Q, intentionally designed to be lossy at high frequencies. Note the frequency at which the nominal impedance is specified. \$\endgroup\$
    – Dave Tweed
    Commented Jul 21, 2021 at 10:33
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    \$\begingroup\$ You are missing @ ?? Mhz, above what frequency that becomes such impedance. For example acquiring analog signals from long wires that may introduce RF noise. \$\endgroup\$
    – Marko Buršič
    Commented Jul 21, 2021 at 10:34
  • \$\begingroup\$ @DaveTweed Does that mean that a 1000R @ 100 MHz bead with e.g 1µH inductance will not resonate with a capacitor in a similar way to a "1µH inductor" ? \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 10:39
  • \$\begingroup\$ @MarkoBuršič Thanks. Added it. Analog inputs typically have a rather large impedance, so is the use high R beads related to this high impedance ? \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 10:40
  • \$\begingroup\$ I didn't say that. But it will be a low-Q resonance, not high-Q. \$\endgroup\$
    – Dave Tweed
    Commented Jul 21, 2021 at 10:50

3 Answers 3

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From what I understand, it is exactly the low parallel resistance that damps resonances...

That inductive reactance at the peak blocking frequency has diminished greatly at the operating frequency (courtesy of Analog Devices): -

enter image description here

Note the red inductive reactance curve above. From about 30 MHz and above the inductive reactance is being replaced by the "loss" of the ferrite material i.e. lossy (resistive) eddy currents in the ferrite structure.

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  • \$\begingroup\$ Thanks, Andy, this is exactly the equivalent circuit which brought me to my question. When \$R_{AC}\$ is high, you have essentially a high Q inductor. When \$R_{AC}\$ is low, you have essentially a nice lossy resistor, which is only shunted at low frequencies by a small L and at high freqencies by a tiny C. Now, consensus seems to be that high R beads are more lossy in all cases?! Does this follow from this equivalent circuit or is it something like experience that teaches you this ? :) \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 11:58
  • \$\begingroup\$ You don't have a hi-Q inductor any more because inductive reactance has hit the ground due to eddy currents and those eddy currents are also quite lossy @tobalt \$\endgroup\$
    – Andy aka
    Commented Jul 21, 2021 at 12:24
  • \$\begingroup\$ I think I see now. When viewing an L(f) curve of inductors, the L is actually constant or even rises up to the SRF. In contrast, L plummets for the bead. So the important point is to check when this happens, e.g. at around 100 MHz in your graphic. This also means, that the equivalent circuit does not predict the behavior in the plot. The \$X_L\$ would rise indefinitely according to the equivalent circuit. \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 12:33
  • \$\begingroup\$ Figure 7 in your link shows what I was getting at. In Method B, they place R in parallel to the bead to damp the resonance, which seems to be equivalent to choosing a 10R bead in the first place. However, HF attentuation is indeed worse with this method. So I think, all the pieces start to align in my head. Thanks again. \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 12:46
  • \$\begingroup\$ @tobalt yes, the circuit in the picture must have frequency dependent values for R and L for it to be regarded as a proper model. If, in your hunting, you end up coming across a model that looks reasonable, please do share. \$\endgroup\$
    – Andy aka
    Commented Jul 21, 2021 at 13:02
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'Typical' ferrite beads tend to be a good reasonable Q inductance up to one MHz or a few MHz, then become increasingly lossy. Their rated resistance is reached well above this, often 100 MHz.

You'll choose a loss resistance depending on what you want to do. You may want to terminate a length of power line so it doesn't resonate (I've been bitten by one of those), or you may want as high an impedance as possible as part of a ladder filter.

However, loss resistance above 1 kΩ or so generally goes with a narrower range of frequencies where that peak resistance is achieved. If you want a flat broadband resistance, then you generally have to stay in the very low 100s of ohms.

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  • \$\begingroup\$ Thank you for the answer. That last paragraph seems to indicate that a bead's characteristic indeed becomes gradually more inductor-like the higher its nominal peak impedance. When an LC resonance frequency then falls slightly out of its high impedance range, (which happens more easily because of the narrow frequency range) one gets ringing. In that case a lower impedance but wider range bead can attenuate better but when the frequency matches, the higher impedance bead attenuates better ? \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 11:29
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    \$\begingroup\$ @tobalt Not the correct interpretation. All beads are inductor-like at low frequencies. All beads get lossy and essentially resistive at high frequencies. If a high impedance is what you're after, then the higher peak impedances are only available over a narrower frequency range. This is nothing to do with an LC resonance. If you are outside the 'good' frequency range, then you get a lower impedance resistive-looking loss. \$\endgroup\$
    – Neil_UK
    Commented Jul 21, 2021 at 11:52
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The purpose of ferrite bead ratings is to know the impedance Z at some reference measurement frequency of for example 100 MHz. You can have ferrites specifically meant for power supply filtering, or meant for data signal filtering. The impedances can easily go from tens of ohms to few thousands of ohms at 100 MHz. So you will need to select a component based on what you are filtering and what you need to pass and what you need to block.

So yes, the impedance will resonate if the surrounding components or parasitics make it possible, but then it also means that you have maybe selected a ferrite bead with too sharp impedance curve for the purpose, and it can be alleviated by proper damping.

You must take into account that there are two kinds of classifications for ferrite beads, they can be classified as inductive or resistive, which kind of says which part, L or R, dominates the impedance Z.

When used in the correct frequency range, ferrite beads have resistance as the dominant part of the impedance, so they are more lossy and thus have less Q than comparable inductors.

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  • \$\begingroup\$ Thanks for the answer. I think, it is a bit beside my question though. In other words, what I am interested in is: "What is a usecase for e.g. a 1000R @ 100 MHz bead, in which a 50 R @ 100 MHz would not be better." \$\endgroup\$
    – tobalt
    Commented Jul 21, 2021 at 11:11
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    \$\begingroup\$ @tobalt OK, that is a simple example with simple answer. If you for example have device with a relatively slow data bus coming out of it, like a serial port, and you need to filter out conducted noise out of the device, a 1000R@100 MHz bead will filter noise out much better than a 50R@100 MHz bead. \$\endgroup\$
    – Justme
    Commented Jul 21, 2021 at 11:20

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