I'm using this device VSC8514 and this is its checklist. I just want to know what spec of ferrite bead I should use for this device's power rails.

They have neither mentioned this in the checklist nor in the datasheet.

I went through the Eval board schematic over here. They have used an inductor on the Eval board, but in the checklist they mention that a ferrite bead should be used over a series inductor. Not sure which one to follow.


1 Answer 1


Same as for any ferrite bead application. Relevant reading:
USB 2.0 VBUS Filter
Ferrite beads for high currents
Switching PSU and residual noise on the rails

Current consumption varies from 10 to 605 mA depending on rail and activity (datasheet p.70).

Typical ferrite beads saturate in the 20-100 mA range, so would be contraindicated for some of these. Note a bead's current rating is meaningless: it is ampacity only, a thermal figure, not related to electrical characteristics (or anything beyond DCR at least).

How to find saturation? Look for the characteristic sheet. Some manufacturers don't provide them at all; others tuck them away in proprietary databases. Yes, it's a pain in the ass.

Note that type 2 ceramics are the same way (for the same general reason, actually). The voltage rating is irrelevant to electrical characteristics; often a part is significantly saturated (-30%, say) at a tiny fraction of its rated voltage.

For this reason, using inductors is often preferable, and mandatory at high currents, and for low cutoff frequencies (read: requires large inductances). The value of a ferrite bead is its lossiness, saving on a component; this must be substituted with an R+C or R||L network in the filter circuit. PDN (power distribution network) analysis is probably a bit out of scope here in this question, but are the keywords to look at for further reading. Typically a lossy "bulk" capacitor provides the damping in such a case.

They also show a ferrite bead in Fig. 4-2 (checklist p.5). As this is shown shorted out by a capacitor, I suspect they intend these as placement options, with neither, or just one, used in the end.

  • \$\begingroup\$ Thank you for your answer. Would this bead help: mouser.in/ProductDetail/Murata-Electronics/… \$\endgroup\$
    – user220456
    Feb 21, 2023 at 6:39
  • \$\begingroup\$ Bead saturation is a bit different from inductor saturation because the materials are rather different. Beads are made with highly hysteretic (lossy) material. These naturally have a wide switching field distribution, all the way to rather high currents, so a small fraction remains inductive. Their B(H) curve is an S rather than a Z. They don't saturate as abruptly as inductors and also in contrast to inductors they have mediocre conductivity at HF (eddy losses) \$\endgroup\$
    – tobalt
    Feb 21, 2023 at 7:49
  • \$\begingroup\$ @Newbie what characteristics does it have? \$\endgroup\$ Feb 21, 2023 at 7:51
  • \$\begingroup\$ @tobalt That's not really true -- they don't exhibit hysteresis any more than other soft ferrites, and indeed the material is sometimes specified (e.g. from Fair-Rite, Laird, etc.). The losses are small-signal in nature, effectively resistance of the core. There are no 'Z' curves among ferroics that I'm aware of (but this isn't much to go on and I would be interested to see more detail regarding this). Saturation sharpness is down to geometry, and indeed the innermost part say of a thick hollow cylinder shape saturates first, but this isn't the geometry of a multilayer monolithic type. \$\endgroup\$ Feb 21, 2023 at 7:55
  • \$\begingroup\$ Surely the ferrite cores have also magnetic losses, why else magnetic material at all? One could just wrap wire with a badish metal for loss. For inductors, imagine the Z with its roof and bottom flipped "outwards" ;). This is the ideal (and challenging) case for a lossless magnetic. The "magnetic viscosity" is different for beads. At low F, they have no hysteresis, like inductors. But as F rises, their hysteresis opens up. Saturation is of course also a material property, it can require many dozen Tesla for hardmagnetic materials. but you might be right about the geometry as reason in beads. \$\endgroup\$
    – tobalt
    Feb 21, 2023 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.