At high frequencies sometimes ferrite beads are used instead of inductors as normal/differential-mode filters. I drew Figure 1 and Figure 2 below showing LC and ferrite bead C differential filters:

enter image description here

Let’s assume we want to quantify the voltage attenuation at 1 MHz. For. Figure 1 case I can write:


where \$\small X_C\$ is the capacitive reactance and \$\small X_L\$ is the inductive reactance. But as you see the denominator is actually the vectorial/phasor sum. It means at a particular frequency there will be resonance when \$\small X_C=X_L\$ and current will reach maximum etc.

What if we use a ferrite bead instead of an inductor as in Figure 2; will the denominator be vectorial/phasor sum or scalar sum? Imagine the ferrite bead impedance is given as 100 Ω at a particular frequency and by chance the Xc is also 100 Ω at that frequency; in that case will the denominator be zero or 200 Ω? Or will it be \$\small\sqrt{100^2+100^2}\$?

  • \$\begingroup\$ You still do the phasor sum in the denominator, but the ferrite impedance has a large real component, so it won't cancel completely with the imaginary impedance of the capacitor to create a strong resonance. \$\endgroup\$
    – Evan
    Commented Dec 6, 2018 at 20:12
  • \$\begingroup\$ Will it be real so should it be treated as a resistor? Does that mean the denominator will be sqrt(100^2+100^2) ? \$\endgroup\$
    – cm64
    Commented Dec 6, 2018 at 20:16
  • \$\begingroup\$ The datasheet should show the real and imaginary parts of the impedance as a function of frequency. \$\endgroup\$
    – Evan
    Commented Dec 6, 2018 at 23:13
  • \$\begingroup\$ Are you sure? Where sare uch imaginary parts here?: docs-emea.rs-online.com/webdocs/12a6/0900766b812a6654.pdf I only see one value for one frequency. \$\endgroup\$
    – cm64
    Commented Dec 7, 2018 at 6:26
  • 1
    \$\begingroup\$ I would say that the ferrite component you linked does not give enough information in its data sheet to calculate resistive and inductive values. Most ferrite beads I've seen do provide graphical information about inductive and resistive values versus frequency so, maybe you need to choose something else or speak to Wurth. \$\endgroup\$
    – Andy aka
    Commented Dec 7, 2018 at 8:47

1 Answer 1


Unlike an inductor, a ferrite bead does not have a single, well-defined characteristic that can be used in an equation. Instead, it can be modeled as a network of 4 fundamental components: 2 resistors, a capacitor, and an inductor. More easily, you can consult its spec sheet for its impedance at the frequency or frequency range of interest.

See this good paper on the subject.

Ferrite bead model and characteristics Source: Analog

  • 2
    \$\begingroup\$ The 4-element model captures the basics, but a more accurate (more elements) model may be required for better accuracy, depending on how "lumpy" an actual part is; and many FBs have a "diffusion" characteristic i.e. best expressed with a Warburg element. Note also, this is true at zero DC bias; ferrite beads generally saturate (impedance drops with increasing current) in the 50-200mA range, with lower impedance types doing better (e.g. some ~10R parts saturate at 2A+). So they're good for signals and individual PS loads, but not overall power filtering. \$\endgroup\$ Commented Jan 26, 2023 at 22:48

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