At high frequencies sometimes ferrite beads are used instead of inductors as normal/differential mode filters. I draw 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 1MegHz. For. Figure 1 case I can write:

Vout / Vin = Xc / |Xc+Xl|

where Xc is the capacitive reactance and Xl is the inductive reactance. But as you see the denominator is actually the vectorial/phasor sum. It means at particular freq. there will be resonance when Xc=Xl and current will reach maximum ect.

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 Ohm at a particular freq. and by chance the Xc is also 100 Ohm at that freq.; in that case will the denominator be zero or 200 Ohm? Or will it be 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 Dec 6 '18 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 Dec 6 '18 at 20:16
  • \$\begingroup\$ The datasheet should show the real and imaginary parts of the impedance as a function of frequency. \$\endgroup\$ – Evan Dec 6 '18 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 Dec 7 '18 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 Dec 7 '18 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.