I came across this PCB schematic and I see they use a ferrite bead at the input of the USB 5V power. Full article is here

enter image description here

My questions are:

  1. Why would you use a ferrite like this?

  2. How would I calculate the value for this ferrite bead?

  • 1
    \$\begingroup\$ I have used them to protect against radiated emissions. The ferrite bead can help suppress common mode radiation of the cable. There is often a common-mode choke on the data lines also for the same reason. But this schematic doesn't show any such common-mode choke. \$\endgroup\$
    – user57037
    Aug 22, 2022 at 6:33

2 Answers 2


Most likely for a modest amount of filtering, and to help dampen resonant modes on the cable. Ferrite beads are a convenient source of impedance both inductive and resistive, so can be quite convenient at terminating or dampening resonant modes on transmission lines, cables, etc.

The power lead will be a transmission line with a characteristic impedance on the order of 50-100 ohms, and a resonant frequency inversely proportional to cable length. The maximum cable length then gives the required inductance, and the impedance, the resistance to terminate it against.

Downside: it's not necessarily a great component choice here, because ferrite beads saturate at fairly low currents. That is, an 0805 chip of this value probably loses more than 30% of its impedance (at a given frequency) at 200mA or so. As DC bias rises, impedance (both inductance and resistance) falls. If the only thing powered is the UART, that's undoubtedly fine; if there are loads up to the USB maximum (or beyond, as from chargers, or the various USB-PD standards), it's not as great an idea, and an R || L (that is, a damped inductor) might be better chosen. Or perhaps a much larger ferrite chip (1210 say).

Usually, such lengths (explicit R || L) will not be bothered with, and it's just, good enough. It might not be well behaved at all operating conditions, but it's still better overall than without the bead at all.

DC bias: few manufacturers report this characteristic; Laird however provides this for most of their catalog:
Look for LI0805H121R data and curves, for example. Those characteristics are probably representative of the quoted part, too (there isn't much manufacturer variation here, ferrite is ferrite).

Regarding ESD, this is unlikely, as an extremely high peak current is quickly reached (~10A in under 10ns), under which conditions, the ferrite bead is essentially a short circuit. Even if it stayed linear, the added impedance is a drop in the bucket versus the source impedance (e.g., ~1.5kΩ for HBM). Same goes for EFT. Both of these transients are handled by a combination of good shielding (cable ground must be contiguous with circuit GND plane, or metal enclosure, at high frequency!*) and error detection and re-transmission.

*DC (galvanic) connection can be avoided, but multiple decoupling capacitors must be used, to keep the ESL low. Often you will see even worse than this -- actual isolation impedance (resistance or inductance) applied to the shield ground-return path! This is almost always erroneous. The lesson is, never trust an app note, or example, or etc., further than you already understand it.


How would I calculate the value for this ferrite bead?

You don't need to; you just read the BoM and it tells you what type was used: Item 32 on the BoM is: -

32  100Ω    L3  L0805   1   LCSC    GZ2012D101TF    Sunlord C1015


Just use the data above to look up the data sheet for all the specific characteristics associated with this component such as this: -

enter image description here

Why would you use a ferrite like this?

The most likely reason is ESD suppression but any type of surge suppression is a candidate reason.


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