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Background:

I am using a shielded inductor in a switching power supply, and am investigating an EMI issue.

My hypothesis is that since the inductor is saturated, the magnetic field is no longer contained in the ferrite core, and the inductor becomes equivalent to an air core inductor. I know that air core inductors radiate their magnetic field much more than ferrite core inductors.

Question:

Once a ferrite core inductor saturates, does its magnetic field get radiated significantly more?

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    \$\begingroup\$ "Radiated" is a bit strange word for it, but your core will capture fewer and fewer field lines the deeper you put it into saturation. Saturate it fully and your field lines will look like a normal air core. \$\endgroup\$ – winny Sep 20 '17 at 15:41
  • \$\begingroup\$ @winny this is the answer to my question. \$\endgroup\$ – DavidG25 Sep 20 '17 at 15:42
  • \$\begingroup\$ +1 for @winny, but why does it saturate? In a DC-DC you really want to avoid that... \$\endgroup\$ – peufeu Sep 20 '17 at 15:45
  • \$\begingroup\$ Listen to what @peufeu says, don't saturate your inductor. It's highly unlikely it's the source of your EMI problem. \$\endgroup\$ – winny Sep 20 '17 at 16:02
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    \$\begingroup\$ If it's under-sized and does saturate, maybe you get current spiking wildly until the regulator manages to turn off the MOSFET, on every cycle... loss of efficiency and worse di/dt... \$\endgroup\$ – peufeu Sep 20 '17 at 16:18
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Once the core is saturated, it doesn't look like a magnetic core anymore when you try to increate the magnetic field. The inductor becomes essentially air-core for any additional current.

While over-saturating a shielded inductor will cause the magnetic field to leak outside the shield more, this shouldn't cause much radiation by itself. More likely the sudden decrease in inductance causes large current spikes somewhere. These can cause radiation, depending on how they are routed, and therefore how much they act as antennas.

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