ferrite-beadmeasurement
This circuit shows the "current channel" of an energy meter.
IA_IN are the two inputs for a shunt (typically 250 uOhms) while TP1 is the connection to the energy measurement IC, which uses the 3.3 V supply voltage (V3P3) as a reference.
The input uses ferrite beads as a form of filter for the incoming AC signal. I want to know what the effect of saturation will be on the ferrite bead inductors at the input.
The bead is meant to absorb high-frequency noise currents (and dissipate it as heat) to maintain adequate high-frequency separation between shared power rails. There is a common type of ferrite beads that are designed to do exactly that. These beads have high impedance over wide high-frequency bands (low-pass filters).
Typical parameters are
- DC resistance -> should be as low as possible
- rated current -> requirements obviously depend on the load
- impedance characteristics curve -> depends on what kind of noise you primarily want to suppress
Look at the parameters of this ferrite bead for example:
This would actually be a good choice for your design. It has a low DC resistance of 0.8 Ohm (minor losses) and is rated for 200mA. The average microSD shouldn't require more than 100mA (but look at the data sheet to be sure).
This shows impedance VS frequency. High-frequency noise, especially in the range from about 7MHz-3000MHz, will be absorbed.
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).
Best Answer
Two primary things happen when saturated: 1) the effective permeability drops so the relationship between amp x turns and the field is no longer linear 2) you get "field popping" which describes the fact that the core can not longer contain the field and moves outside of the core. You van get very high EMI/RFI in these situation because of the huge disparity in in relative permeability. The field has to rapidly expand (outside of the core) to compensate.
In the case of the ferrite beads, you will lose filtering but in the circuit above only displacement current flows (because of C6 and C12) so only very extreme spikes will cause any saturation.