I was checking a schematic for work, and I saw this component that I don't recognize, marked 600R/2A.

Since they are marked L and are used in combination with capacitors, I think they are inductors meant to serve as low pass filters in combination with the capacitors.

Why do they have such an odd symbol and why are they marked the way they are?

I assume the "2A" part to be the maximum current, and 600R to be some sort of resistance value. But what is the inductance here?

I read online that it is a type of ferrite inductor/filter.

I need to copy parts of this circuit into another design but I would rather understand what it does first.

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  • \$\begingroup\$ If it's a work/copying exercise, then dig up the BoM for the circuit/PCB and find the component's data sheet. \$\endgroup\$
    – Andy aka
    Jan 1, 2021 at 15:32
  • \$\begingroup\$ @Andyaka I need to copy part of this circuit to use in another design. But sadly there is no BOM available. All I have is the kicad files. When I open the pcb layout I see that it has a 0805 footprint. \$\endgroup\$
    – Emre Mutlu
    Jan 1, 2021 at 15:34
  • \$\begingroup\$ @Andyaka I read that it propabbly is a ferrite bead. But I read online that ferrite beads have a higher parasitic resistance hence a higher power loss. So how logical is it to use one in a 5V power line? Isnt it more desirable to have a low DC resistance? \$\endgroup\$
    – Emre Mutlu
    Jan 1, 2021 at 15:40
  • \$\begingroup\$ How can you tell it's a good circuit? Maybe you have a PCB and can photograph the part? If you don't have a BoM and you don't have a PCB, then I'd be worried about blindly copying something. Maybe you have access to the original designer. If none of the above, I'd be concerned. That isn't the way to design electronics. \$\endgroup\$
    – Andy aka
    Jan 1, 2021 at 15:41
  • \$\begingroup\$ @Andyaka I think I will replace the ferrite beads in his design with regular smd Inductors, I am thinking 220nH. I just dont see why he used ferrite beads. \$\endgroup\$
    – Emre Mutlu
    Jan 1, 2021 at 21:05

1 Answer 1


Most likely it's a ferrite bead. 600R / 2A indicates that its impedance is 600 Ohms at 100MHz, and its current rating is 2 Amps.

  • \$\begingroup\$ But it would work the same way as an inductor in this case right? So as a low pass filter? \$\endgroup\$
    – Emre Mutlu
    Jan 1, 2021 at 15:32
  • \$\begingroup\$ Ferrite beads are typically pure inductive up to about 1 MHz, then get increasingly lossy, being mostly lossy by 10 MHz. The loss rises to a peak at the rated 600 ohms at some frequency generally well above 10 MHz. You can infer the inductance from the slope of the impedance versus frequency graph in the first MHz that the manufacturer supplies. However, this value is never specified in the headline parameters. \$\endgroup\$
    – Neil_UK
    Jan 1, 2021 at 17:11
  • \$\begingroup\$ @Neil_UK, the spec resistance value isn't always the peak value. For example, Murata just specifies the value at 100 MHz, even if the peak is at a much higher frequency. \$\endgroup\$
    – The Photon
    Jan 1, 2021 at 17:54
  • \$\begingroup\$ @Emre MUTLU, re: "works like an inductor?" - kindof. However, in addition to being resistive at the frequency for which they are designed, they are also made from a magnetic material optimized for small size, and the side effect is that inductance falls very rapidly with instantaneous current, making it difficult to apply in an LC or LR situation. But you can still do it with some care, to filter slightly lower frequencies (1Mhz region) \$\endgroup\$
    – Pete W
    Jan 1, 2021 at 19:19

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