# Why is this LC filter where it is, and what function does it perform?

My customer has a motor controller that apparently only works with an LC filter in front of it. This configuration trips out on undervoltage:

simulate this circuit – Schematic created using CircuitLab

This works fine:

simulate this circuit

(I'm leaving out details like a line impedance and precharge, but that's the general idea. I can add them back if anyone thinks they're relevant.)

I've never seen a motor controller require an LC filter like that. It is of interest to me that this filter has a pole at 49 Hz, and this equipment is designed to operate at both 50 and 60 Hz. Tests are being run at 60 Hz. If I've computed the frequency response right, that LC should have a gain magnitude of roughly 2x at 60 Hz.

Is this thing really just a tuned voltage doubler? What other function could this possibly perform?

• I work with 3 phase rectifiers, and usually, an LC filter is on the output to improve ripple. I don't known why the motor controller need it... – Martin Petrei Nov 12 '15 at 17:54
• Come to think of it, I suppose I've seen that too... I just hadn't put it together in this context. I suppose my question should become, "How does one compute the ripple?" – Stephen Collings Nov 12 '15 at 18:00
• When you say "trips out" what trips? (incorrect guess elided) – user_1818839 Nov 12 '15 at 19:08
• @BrianDrummond The controller trips out on undervoltage. It throws a fault and stops running. – Stephen Collings Nov 12 '15 at 19:10

Without the inductor the ripple is greater and the under voltage trip is likely due to the bottom of the ripple tripping some comparator in the controller and it decides to shut down. The inductor will reduce the tops of the ripple AND significantly increase the bottom of the ripple hence the instantaneous voltage seen by the controller is never as low as the case without the inductor.

Speculation based on experience! You could probably simulate it easily.

The reason to create an L-C Filter with regards to an inverter is for power quality reasons on the AC side.

A 6 pulse rectifier, in theory, will draw current harmonics: fund, 5,7, 11,13 .... This is however dependent on continuous DC current. Depending on the output power draw and the size of the DClink capacitor, the operating power point that continuous DC current occurs (and thus harmonic content of the AC current) may be too high with regards to requirements.

This inductor lowers the power at which continuous DC current occurs

• Your statements are true, but they don't explain why the motor controller faults out on undervoltage, I don't think. – Stephen Collings Nov 12 '15 at 18:02
• True and that is partly due to the lack of additional information. One difference between C and LC is the 6th harmonic current in the output ASSUMING the capacitor is large enough:. With a C filter there is no 6th harmonic current in a resistive load, the L forces a V6/6wL component. – JonRB Nov 12 '15 at 18:07
• Equally "undervoltage" at what level is the undervoltage set at. remember the trough of a 400V 3ph system is 489V (no additional drops). Add some diode drop and cable drop... if you are close to that because the DCcap is too small for hte power being drawn then yes you will get undervoltage. the L will help smooth it – JonRB Nov 12 '15 at 18:13

I think I get it now. The gain of the LC shouldn't be computed at 60 Hz. The signal the LC is seeing is the rectified three-phase, which I shall henceforth call the humpty-hump wave.

This signal has a fundamental of 360 Hz, with some energy content at the harmonics of 360 Hz, since it's not a perfect sine wave. At 360 Hz, with the components we're using, the gain of the LC filter is $\frac{1}{1-\omega^2LC} = \frac{1}{-52.7}=-.019$. Negative gain means the output is 180 degrees out of phase with the input, but at this juncture we don't care about that. At higher frequencies the gain is even lower, so less than 1.9% of the total energy content gets through the filter. Instead of having a peak-to-peak ripple $1-sin120^o = 13.4 \%$ of our peak voltage, we get $13.4\%x1.9\%=.25\%$.

So the DC content of the wave (.955 Vpk per this answer) is unaffected, remaining 324VDC. The AC content was 45.5V peak-to-peak, and is now .86V peak-to-peak. So our troughs, instead of being ~300 volts, are now ~323 volts.

Adding a resistive load reduces the AC gain further, and reduces the phase angle, but the effects are relatively small at the loads of interest. I've run simulations that match up with the math quite well.

• first of all, i each phase, you only have a half-wave rectifier. so you'll get your bumpy DC with a 180 Hz ripple, not 360 Hz. and the ripple will be bigger than the 6-phase case that you are showing. second, the LC filter simply tries to be a better ripple filter than the C-input filter, which really just holds the voltage sorta constant and discharges at a rate proportional to the current draw. for the C-input filter, if the current draw was zero, there would be essentially no ripple and your voltage would be held at the maximum value. – robert bristow-johnson Nov 12 '15 at 21:43

The principle for using the LC is the same as the one when it is used at the output of the switching regulators or buck regulators.

You have noisy current riding on the output voltage that has a frequency. This is filtered using the inductor. The capacitor filters out voltage ripples with the capacitor acting as a small battery that kicks in during ripples in voltage or DC rails.

There are many hidden parameters like dcr,frequency ...etc of the inductor and the ESR of the capacitor that basically governs the selection of these specific component.