I am wondering if I should use a inductor in series of my power supply rail at the input of my PCB to reduce potential noise coming from the cables connected to my PCB like this:


simulate this circuit – Schematic created using CircuitLab

And lets assume the inductor: LPS5030-475MRC 4.7mH, 110mA saturation, 26R internal, 1.1MHz SRF. And I am drawing 30mA which goes on microcontroller so there will be a frequency variation of the voltage drawn.

What I am afraid of, is the possibility for the inductor to oscillate. Each non ideal inductor has a parallel capacitor so it could oscillate. The oscillation would happen to its two terminals, and that is where we measure the oscillation:


simulate this circuit

But in my case, the inductor is not parallel to anything else. If it tries to oscillate, there will be a voltage difference to its two terminals, so wont it short itself and the oscillation would stop before it starts?

I didn't manage to find any sources of such filtering, so I guess its not wise to use the inductor to filter voltage without regulation ( aka without using the inductor on a switching power supply which has feedback), but I want to be sure.


1 Answer 1


You are missing the bigger problem of the inductor resonating with the LDO regulator's input capacitance. I'm talking about the capacitance that is needed at the input of every linear and switching regulator.

That capacitance will be at least 100 nF and completely swamp any parasitic capacitance from the inductor. It can also cause a significant problem when power is applied to the raw input because it will form a resonant low pass filter and create an overshoot voltage at the LDO regulator's input that is up to twice the applied external DC voltage. This does damage regulators and it has happened to me. It will look like this: -

enter image description here

The Vout port connects to the LDO regulator and, the Vin port is the raw input. If I check your values (assuming a 100 nF input capacitance on the LDO regulator) I get 82% overshoot (lowest image in the picture below): -

enter image description here

Calculator image from my basic website. If I increase the capacitance to 1 μF, the overshoot is 54%. A 10 μF produces a 9.5% overshoot. You can test this your self using the calculator or do a simulation.

I reckon, providing you have at least 10 μF you shouldn't have any significant problems. The worries about the parasitic parallel resonance of the inductor are unfounded.


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