If the ferrite is designed to filter out in the lower region, up to 100 MHz, and the pass-band signal is even way lower, 60Hz, I'm thinking that the ferrite is constantly saturated along with the 60Hz signal. Now let's say 900MHz ISM-band signal is coupling into this, I wonder if the saturated ferrite would actually rectify some of this... what do you guys think?

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    \$\begingroup\$ If it saturates at 60 Hz so its inductance wobbles at 60Hz then it would modulate the RF with 60Hz, not rectify it. Also... ferrites are symmetrical and non polarized, and for rectification you need a polarized/asymmetrical component that behaves differently depending on which way the current flows, like a diode. \$\endgroup\$
    – bobflux
    Commented Feb 22, 2021 at 21:10
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    \$\begingroup\$ @bobflux you can make it asymmetrical with a DC bias current. \$\endgroup\$
    – user16324
    Commented Feb 22, 2021 at 21:24
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    \$\begingroup\$ a tiny ferrite bead 2mm long spans about 1nH and increases that. which is negligible at 60Hz and has no effect. No I don't think ISM signals are getting rectified by it \$\endgroup\$ Commented Feb 22, 2021 at 21:24
  • \$\begingroup\$ @BrianDrummond right, I hadn't thought about that! Interesting. \$\endgroup\$
    – bobflux
    Commented Feb 22, 2021 at 21:32

3 Answers 3


Perhaps not ISM band.
Michael J Rainey made some pretty weird radio circuits, both transmitters and receivers. One of his efforts did use some ferrite toroids as detecting devices, using their non-linear B-H characteristics to convert a 24 KHz radio signal to audio, driving headphones.

A signal generator at 12 kHz provided the bi-polar sine wave current that drives two small ferrite transformers through their non-linear region. Their impedance becomes small on both positive and negative peaks of the 12 kHz "local oscillator" signal. This is called a "Polyakov mixer" for the 24 kHz antenna signal.
Michael Rainey Polakov receiver

  • Ferrite toroids were Fair Rite #5976000801.
  • Local oscillator step-down transformer was 1200:8 ohm (linear) audio transformer.

Those two small ferrite cores saturate at about 0.4 Tesla.
This receiver provided a strong audio signal from nearby U.S. Navy transmitter NAA, a rather high-power source.

Why are two saturating transformers used?
One would not want to listen to the very loud 12 kHz signal from just ONE transformer. With T1, T2 windings oriented properly, 12 kHz from T1 cancels 12 kHz from T2 at the headphones. Yet T1 and T2 saturate at the same part of the 24 kHz signal, directing the RF signal away from headphones. Back-to-back diodes would likely switch more efficiently.

These ferrites are driven hard so that saturation occurs on the peaks of the 12 kHz local oscillator. For a ferrite bead to exhibit non-linear effects, currents (presumably 60 Hz) would have to sweep through a non-linear section of the ferrite B-H curve. A significant DC current might be large-enough to get close to saturation, where B-H is most non-linear.

  • \$\begingroup\$ Thank you all for your thoughts and answers. This one sounds most likely what I was thinking of. Basically the low-frequency signal saturates the core, even if just for a moment but for the high-frequency signal that will look like DC and that the core is saturated for a very long time. For the high-frequency signal, one half of the sine wave will not see the inductance because it is saturated in that direction, while the other side does, at least that's what I was thinking. \$\endgroup\$
    – pgibbons
    Commented Mar 3, 2021 at 20:54
  • \$\begingroup\$ @pgibbons Ceramic magnets are saturated with a "pulse" of current in a magnetizing wire - becoming permanent. The radio example described above swings through the whole +/- BH curve - I wouldn't call it a rectifier. Another similar example is a fluxgate magnetometer. In your example scenario, the AC signal would likely be too small to see much of the BH non-linearity. I can't imagine how Michael managed to wind 100 turns on those very small toroids. \$\endgroup\$
    – glen_geek
    Commented Mar 3, 2021 at 21:59
  • \$\begingroup\$ I'm not sure what was going on in a circuit I was observing, a 1 Watt radio was transmitting and the circuit in question saw a positive voltage (DC) at the same time. The 2 boards share power and some signal lines. It would be no surprise to see interference at such a close distance and strong RF, I was just wondering myself how was it coupling in. I figured if trace picked it up, it would cancel out or be filtered out as AC. I didn't see how it would turn into a positive voltage, thought maybe ferrites were to blame but probably not. There were probably diodes in the circuit somewhere. \$\endgroup\$
    – pgibbons
    Commented Mar 3, 2021 at 22:48

Not all beads behave the same, yet have the Mfg process and equivalent circuit of an inductor.


simulate this circuit – Schematic created using CircuitLab

I haven't seen one yet, but putting an FM signal over a partial saturated DC core with AC ripple might convert a strongly inject line carrier FM into AM. But is that just noise? or signal? If it's signal, you better get the right ferrite.


It is highly unlikely that the ferrite is 'rectifying'. Rectify means something along the lines of 'convert to DC' which a ferrite simply cannot do.

It's possible in the right environment that it could 'pick up' magnetic noise, but remember that its going to attenuate a lot of that. Check the datasheet for the inductor, it should give a graph of attentuation vs frequency (or impedance vs frequency).

I'd say look elsewhere in your circuit first.


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