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I understand that using a ferrite bead attenuates high frequency currents and can help reduce EMI. However, I'm confused about how to select the frequency range that should be attenuated. For example, is this a function of the frequency range of currents that the load IC will draw from the supply? If so, how does one go about determining this frequency band?

Or is the bead more about suppressing currents that are picked up from interference from other devices? Again, if this is the case, how does one determine what frequency band requires attenuation?

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A ferrite bead is a lossy inductor, ie a conductive coil (or straight wire) with ferrite magnetic core material inside/around it.

Like every inductor, its impedance curve has two parts that form an inverted-V shape:

  • Below the self-resonant frequency (SRF) it behaves like an inductor: impedance rises with frequency.
  • Above SRF, parallel parasitic capacitance dominates, and it behaves like a capacitor.

Note that higher inductor value (more coil turns) implies higher interwinding capacitance, thus lower SRF, ie a higher inductor value can be worse at filtering out HF noise.

Like every inductor that has a magnetic core (not air) its properties depend on the core material properties, which also depend on frequency. The typical inductor you'll find in a DC-DC converter has a core optimized for low losses. Ferrite beads are optimized for high losses. The material has high hysteresis. This means it turns high frequency current into heat. This important property allows a ferrite bead to be much better at filtering high frequency noise than an inductor, because it stays lossy even above its SRF. So in case you wonder, this is why inductors and beads are two different components.

Here's an impedance graph for a bead:

enter image description here

The red curve is the imaginary/inductive part of the impedance, and the blue one is the real/resistive part of the impedance. At low frequency it behaves like an inductor, but above a few tens of MHz this one behaves much like a resistor.

I understand that using a ferrite bead attenuates high frequency currents and can help reduce EMI.

What it really does is add series impedance wherever you put it in the circuit. On the example above, on the schematic on the right:

  • At HF, the capacitor provides a low impedance path to GND for current drawn by the load (pictured as a DC current source, but could be a chip)
  • At HF, the bead provides a high impedance path to the power supply

This forms a current divider: load current will be shared between the two paths in inverse proportion of impedance, which means HF current will go through the local decoupling cap, and low frequency/DC current will go through the bead. This avoids injecting noise into the supply. It also keeps the HF current loops local between the chip and decoupling caps, which avoids turning power traces into antennas.

These two components also form a voltage divider. Noise on the power supply is blocked by the high HF impedance of the bead, and shorted by the low impedance of the cap, before it gets to the load. So you can expect HF noise on the supply to be attenuated before it reaches the load.

Note this LC lowpass filter can ring, as shown by the peak on the transfer function graph.

If you think in terms of impedance and current/voltage dividers then it's much simpler: this LC filter is a current divider when you consider the input is load current and the output is noise current injected into the supply, and the same LC filter is a voltage divider when you consider the input comes from power supply noise and the output is at the load.

Thinking in terms of current/voltage divider also avoids the mistake of thinking the ferrite bead does the job alone while forgetting about the cap.

However, I'm confused about how to select the frequency range that should be attenuated. For example, is this a function of the frequency range of currents that the load IC will draw from the supply? If so, how does one go about determining this frequency band?

You can guess by considering digital chips will produce noise at harmonics of the clock frequency. Faster rise times mean more harmonics. Or you can plug a spectrum analyzer and have a look.

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I understand that using a ferrite bead attenuates high frequency currents and can help reduce EMI.

Correct. They act as a high resistance series block at high frequencies usually in the range 30 MHz to 3 GHz. Some will be more targeted at lower frequencies and others will be targeted at higher frequencies. Some will attenuate more; some will attenuate less: -

enter image description here

Their choice depends on your application and what you want them to do. You have to decide why you are contemplating using a ferrite bead. That comes first i.e. what is the problem that makes you think a FB is required? Start with the problem you are trying to solve.

Here's a good article on FBs produced by Analog Devices Inc. Here's its equivalent circuit as shown in that article: -

enter image description here

And here's another section of that article that details the impedance properties of a TDK MPZ1608S101A and 10 nF shunt capacitor. It shows different scenarios of connecting them with a damping resistor should the low frequency resonance be too problematic: -

enter image description here

how does one go about determining this frequency band?

This partially comes from experience and partially comes from EMC lab results that indicate emissions or susceptibility to emissions is too high.

I'm confused about how to select the frequency range that should be attenuated.

This also partially comes from experience and also partially comes from EMC lab results that indicate emissions or susceptibility to emissions is too high.

is the bead more about suppressing currents that are picked up from interference from other devices?

A ferrite bead can be used for suppressing interference that might affect other equipment or for suppressing conducted emissions from other equipment affecting your circuit. An FB can also be used to suppress locally generated EMI affecting other parts of your circuit.

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  • \$\begingroup\$ Another thing to consider is the current rating. Higher currents will cause the ferrite to saturate and fail to work as intended. So, pay attention to the specs. \$\endgroup\$ – Kartman Dec 21 '20 at 10:25
  • \$\begingroup\$ @Kartman that is covered very well in the article I linked into my answer. \$\endgroup\$ – Andy aka Dec 21 '20 at 10:26

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