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.

• This question doesn't appear to make sense - what is a "speaker path"? What do ferrite beads have to do with this? – Andy aka Jul 11 '14 at 10:34
• @Andyaka Speaker path implies the path from Amplifier to Speaker on the board. we have added ferrite beads to suppress radiated emissions. – Ash Jul 11 '14 at 15:19
• did this work-out OK? Did it suppress emissions sufficiently? – Andy aka Jul 11 '14 at 16:01
• 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. – Ash Jul 11 '14 at 19:03
• How can you possibly have 60MHz going around your audio circuit? – Vladimir Cravero Jul 12 '14 at 9:16

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.