# Input filter LC for switching power supplies

I have the following LC low-pass filter for a power supply.
I want the filter to cut-off unwanted noise caused by a DC motor.

simulate this circuit – Schematic created using CircuitLab

How can I calculate that the value of L1 is correct and the filter values will not cause any issues to the power supply? For instance, will the filter allow the power supply to respond properly to high step currents like 0.2A step?

A test at the bench shows that it operates properly.

L1 = Unshielded axial wirewound 95mOhm DC resistance 10uH 10% 1.5A max
C1, C2 = Aluminium electrolytic capacitor leaded 22uF 20% 100V DC
U1: CUI P7824-500

The manufacturer suggests an inductor value of 120uH.
Having a 10uH instead, will that cause any harm to the power supply? such as overshoot peaks?

• Everything starts with the signature of the pulsating current absorbed by the dc-dc converter and the ripple specs. See my answer giving guidelines on how to determine the value of the $LC$ filter. Be aware that interaction between the filter and the input impedance of the dc-dc converter needs to be assessed (see my answer) Commented Aug 14, 2023 at 17:08

With these types of filter I find that it's best to assume it will attenuate high-frequency noise based on a 1st order response (as if the inductor were replaced with a resistor) then, you put the inductor in parallel with the resistor to keep the DC current path happy.

This also reduces the Q-factor of the LC circuit and prevents a sudden application of input voltage (a step input) from causing an almost 2:1 increase in the peak voltage at the converter's input. Your voltage converter may not be rated for this.

How can I calculate that the value of L1 is correct

I would regard noise from the input supply as being attenuated as a first order RC. But, you need to understand what that noise is or what it looks like on an oscilloscope. Then I would try to recreate the circuit (and noise impulses for instance) in a simulation tool. They are of course free these days.

If you used a 10 μH and 22 μF, it would have a very large resonance at 10.7 kHz and, it's this resonance that can cause secondary problems, so you need a parallel resistor to the inductor; it rapidly boils down to the main filtering being due to the RC part of the circuit with the inductor providing a low-impedance path for DC power.

I would also suggest, that 10.7 kHz might be too high and, you might wish to use a higher value of inductance. If you used 100 μH and 22 μF, the resonant frequency would be about 3.9 kHz (for instance) and, if you could suffer 3 Ω in series with the inductor, the resonant peak would be well under control.

But, if 3 Ω is too much in series with the power feed then use an equivalent parallel resistor to lower the Q-factor.

Here's what you current design would do if a step input of 50 volts is applied: -

Regarding the spectrum you would see this: -

As you can see, to keep the peak under control, I had to use a 1 Ω resistor parallel with the inductor. Notice how we have a basically a first-order response now with a little peak. But, you need the inductor to give you a low impedance path to DC (if required).

For instance, will the filter allow the power supply to respond properly to high step currents like 0.2A step?

The initial current for the step demand is sourced from the 22 μF but, if sustained you might have problems. This is why I strongly suggest you use a simulator to resolve issues and try out different loading scenarios.

Here's what the response of the CUI specified filter looks like with a source that has 0.5 Ω (a reasonable value) series resistance: -

If the source was 0.1 Ω then it's a bit peakier: -

• Thank you for the answer. The DC/DC power supply is rated at 90V DC. What would the value of the resistor be? Is it okay to put a 47K in parallel with the inductor? Why do I need a resistor in parallel? Commented Aug 14, 2023 at 15:57
• I think you meant that the use of resistor would absorb the energy of the inductor? Commented Aug 14, 2023 at 15:59
• I use LTSpice . Commented Aug 14, 2023 at 16:00
• So, you either need a very low resistance in parallel with the inductor or a TVS right? Commented Aug 14, 2023 at 16:25
• I added the part number of the power supply. Manufacturer recommends an inductor value of 120uH. Will it be okay to use 10uH instead? I have them in stock. Commented Aug 14, 2023 at 16:32

That's probably fine.

A complete analysis requires much more information (voltage range, amount of ripple and noise, surge conditions if applicable, environmental, etc.), but these data likely don't have too much consequence for the immediate question.

For purposes of supply stability, we want the input impedance to remain low. At DC, the inductor looks like a short (well, its DCR rather), and at high frequency the capacitor looks like a short (well, its ESR). It's the critical transition range inbetween where impedance can rise higher (LC resonance). To minimize this, we set $$\\textrm{ESR} > \sqrt{\frac{L}{C}}\$$.

Most specifically, for a switching regulator, we need the input impedance, at frequencies within the control bandwidth (so, typically under 100kHz), strictly less than the worst-case negative input resistance. That is, for a buck converter of say 24V 0.3A output, at its minimum regulating input (say 27V), a change in 1V at the input causes an inverse change in current draw (at 100% efficiency: 27V 267mA, 28V 257mA, etc.), so the incremental resistance is (28V - 27V) / (257mA - 267mA) = -105Ω. The resistance gets lower (read: more negative) at higher voltages, so this is the critical case. Evidently, the ballpark ~1Ω filter in your case should be adequate. (Mind: do perform this calculation for the worst-case current the converter can draw. If it's only 0.3A nominal, but you're using a 3A regulator, it can still draw those three amperes.)

Furthermore, we generally want the step response of the filter to have little ripple, so that the regulator isn't working hard to correct that ripple (may have performance consequences, particularly near Vin(min)). Zin being much less than -Rin, say 1/5th or less, provides this condition. Again, the ~1Ω filter seems adequate.

Anyway, electrolytics generally have convenient ESR values (22µFs might be in the 0.5-1Ω range, etc.), which makes them effective this purpose. Occasionally, L can be chosen with DCR large enough to the same end, or resistance placed in parallel with it (which provides damping, at expense of HF performance).

If there are other low-ESR capacitors in parallel with the electrolytic, the electrolytic must dominate over their total value by a modest factor. Say there's 2µF of ceramic at the converter's input; 6.8µF or higher should be chosen. (Again, 22µF is safe.)

This also solves an issue related to hot-plugging or inrush transients, where the LC resonance is similarly excited by the input voltage step. This condition highlights an additional drawback: for commonly used ceramic capacitors, their capacitance drops as voltage rises, greatly accelerating the overshoot as it goes -- potentially the peak voltage can be triple the input step, or more! (Whereas the peak, for a lossless linear capacitor and inductor, is strictly double the step.) Overshoot can also be solved by clamping with a TVS diode, which may additionally solve other surge conditions.

The choice of TVS or electrolytic (or both) varies between applications; again, without those conditions specified here, it probably doesn't matter to the question per se.

Both methods of course (clamping with TVS, snubbing with bulk capacitance) incur higher overall inrush currents (or charge), which may be problematic for switching circuits upstream of the converter; this is a separate issue, and may require a hot-swap or load-switch controller, or current limiter, or fuse, etc.

• The manufacturer of the power supply recommends an inductor of 120uH. Doesn't that mean that they have already counted for overshoot peaks? Commented Aug 14, 2023 at 16:37
• Not just 120uH, but a 680uF electrolytic as well. They also show a MOV, in an abundance of caution (unlikely to ever be relevant, but I suppose there will be the inevitable customer that hooks it up to a long run and needs to endure full induced lightning transients or whatever). It's not clear what type they intended for the smaller capacitors, but I would recommend electrolytic for the above reason. Also, you might not need any inductor at all; note that the question presupposes a need, and you haven't proven that need through specification or measurement of the source. Commented Aug 14, 2023 at 16:40
• Why do I not need an inductor at all? I want to cut-off EMI noise Commented Aug 14, 2023 at 16:41
• No, I didn't say that. "Why might I not need an inductor" depends on what noise level you need to filter (in either direction!). Commented Aug 14, 2023 at 16:44
• Correct. It's unlikely to cause problems. Commented Aug 14, 2023 at 17:16