Could anyone let me know how can I calculate the current rating required for a ferrite bead on speaker path?

The speaker impedance is 32 ohms.

  • 2
    \$\begingroup\$ This question doesn't appear to make sense - what is a "speaker path"? What do ferrite beads have to do with this? \$\endgroup\$
    – Andy aka
    Commented Jul 11, 2014 at 10:34
  • \$\begingroup\$ @Andyaka Speaker path implies the path from Amplifier to Speaker on the board. we have added ferrite beads to suppress radiated emissions. \$\endgroup\$
    – Ash
    Commented Jul 11, 2014 at 15:19
  • \$\begingroup\$ did this work-out OK? Did it suppress emissions sufficiently? \$\endgroup\$
    – Andy aka
    Commented Jul 11, 2014 at 16:01
  • \$\begingroup\$ For the loudspeaker path it's ok. But for the headset we are failing the Radiated Emissions limit for FCC by 2 dbuV/m @ 50-60 MHz. We will increase the ferrite bead impedance at this frequency and test again. \$\endgroup\$
    – Ash
    Commented Jul 11, 2014 at 19:03
  • \$\begingroup\$ How can you possibly have 60MHz going around your audio circuit? \$\endgroup\$ Commented Jul 12, 2014 at 9:16

2 Answers 2


If the speaker is R ohms and is rated for P watts, then the maximum RMS current will be \$I_{RMS(max)}=\sqrt \frac{P}{R}\$

That will suffice for the current rating of the ferrite bead, but if you want to make sure it does not saturate you'll have to use a much higher number for the peak current. PMPO is a ratio of 5~6 higher, so maybe 10:1. There's a bunch of guesswork here because the speaker rating probably takes that into account to some degree (I don't think a typical speaker can take continuous sine-wave inputs at the rated power for very long).

If the amplifier is direct-coupled you can simply look at the power supply rails and divide that by R (and add a bit because the impedance is not constant).


Power is related to current by the equation \$P = I^2 R\$.

Assuming you know how much power is being fed to the speaker, you can solve for the current:

$$I = \sqrt{\frac{P}{R}}$$

This gives the RMS current that any compoenent in the speaker path, including ferrite beads, must handle.


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