I don't quite understand the concept of the DC current Rating of the ferrite beads.

Let's say we have a ferrite bead with 120Ohm @100MHz and with 3A rated current (and a DC resistance of 30mOhm).

What will happen if the `DC current through the bead increases above 3A? Can I see it as a fuse, that will "break" above the rated current, meaning that after some point no current will pass through it? Or is it totally different?

I have read somewhere on the net that the bead will saturate if the current exceeds the rated value. Does that mean that it will not function normally? Something like not being able to supress the ripple? That its resistance at 100MHz will be lower than 120Ohm?

  • \$\begingroup\$ Saturation will cause it to not function properly and let the HF signals thru. \$\endgroup\$
    – Andy aka
    Sep 29, 2015 at 9:16

1 Answer 1


Two phenomena will affect things here.

Firstly, magnetic effects. An increase in current will cause an increased flux density inside the ferrite core. At a high enough current - likely around the nominal 3A rating, this increase in flux density will approach a point where the material's relative permeability will start to decrease - saturation, which you mentioned in your question. This will result in a lower inductance, and hence a lower AC impedance. There is also the phenomenon of "field popping" described in this SE answer - but it is not directly relevant to your question.

Secondly, thermal effects. It is unlikely that the ferrite will fuse at 3A - although it will at some current higher than this. Before that happens, however, the ferrite will heat up, causing an increase in DC resistance. This will also affect the permeability of the ferrite core, and hence the AC impedance of the part. This effect isn't trivial to predict, but it is actually likely to increase the AC impedance (over a range from room temperature up to high tens of degrees C). Take a look at this Google image search for some typical permeability vs temperature charts for ferrite materials.

So we have three different effects - a decrease in AC impedance, and increase in DC resistance and an increase in AC impedance. So what dominates? This will depend on the exact makeup of the inductor, but in general you will see an overall decrease in AC impedance. This is a bad thing (in general).

  • \$\begingroup\$ Thanks for your detailed and helpful answer! It helped me understand a lot about what is happening. Actually the background behind the question is whether we can use ferrite beads as cheap fuses. As far as I understand this may work but we cannot be sure at what current level the ferrite will fuse, right? I actually did an experiment with a ferrite with 1A rated DC current. I sent through 3.5A and still didn't see any difference; the current through the load resistance didn't Change. I suppose I should go higher in order to see something? \$\endgroup\$
    – nickagian
    Sep 29, 2015 at 13:44
  • \$\begingroup\$ Honestly, there is no way I would attempt to replace a fuse with a small ferrite. The break current would not be a controlled parameter by the manufacturer, and could change without warning. There's no guarantee as to the maximum current the fuse would be able to interrupt, or any guarantee that extremely high pulsed currently wouldn't cause the component to fail destructively. You could end up with blast and charred remains all over your board. Added to which there is no way that using something that is not a fuse as a fuse would result in a standards-compliant product. Just use a fuse. \$\endgroup\$
    – stefandz
    Sep 29, 2015 at 14:27
  • \$\begingroup\$ It was actually just an idea. More or less I agreed with you before asking. But I wanted to see if there was any chance. Anyway I didn't know exactly how the beads are working and didn't understand exactly what this DC current specification meant for circuit design. I mean in a standard application of the bead. \$\endgroup\$
    – nickagian
    Sep 30, 2015 at 6:18

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