In this question Ferrite beads used to filter high frequency noise. Capacitor also can be used to reduced noise and filter out high frequency noise.

Are they have the same effect on filtering high frequency noise or reducing EMI?

When to use ferrite beads and when to use capacitor?


1 Answer 1


Ferrites can be lossy or lossless by design. One can test the bare uncoated bead with ohmmeter to see some DC resistance in range of 50..750 \$\Omega\$.

So capacitors are to lossless ferrites as snubbers to lossy ferrites.

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This losses can be represented as high frequency transformer loaded to low value resistor. The same filter is impossible to mimic with pure capacitor, because instead of absorbing energy, the filter will reflect the high frequency component back.

So the major difference between ferrites and caps is absorb vs reflect.

So when to use ferrite beads ? Practically every filter should be using lossy ferrites and snubbers. The word "filter" is 90% related to use of absorbing filters, as the solutions and environment is mostly digital.

The cases when you can not use ferrites is rest 10%. It is all lossless filters, reflecting the energy back: analog filters in radio is an example. But even in radio, misunderstanding of energy reflection involved in high-Q filters leaves designs incomplete. For example using diplexer with absorber at the output of mixer was not commonly known solution.

  • 1
    \$\begingroup\$ Your DC resistance is high! More than 90 % of the ferrites listed at Digikey have a DC resistance less than 1 ohm, 40 % even less than 100 milliohm. \$\endgroup\$
    – stevenvh
    Jun 13, 2012 at 5:12
  • \$\begingroup\$ It is resistance of ferrite itself, not the resistance of wire put through bead \$\endgroup\$
    – user924
    Jun 13, 2012 at 12:05
  • \$\begingroup\$ @Rocket Surgeon Good explanation! When do we know capacitor is not suitable and we need ferrites to absorb excess energy? \$\endgroup\$
    – pstan
    Jun 13, 2012 at 20:12
  • \$\begingroup\$ @pstan I have added examples to my answer. \$\endgroup\$
    – user924
    Jun 14, 2012 at 1:42

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