# Low-pass LC filter on both sides of a motor

I'm building an H-bridge to control a load. The design is done, and all that's left is to filter out the switching frequency.

The switching frequency is at 20 kHz, and since the current flowing through the load is relatively high (9.5 A) I'll need to use an LC circuit to low-pass filter the noise. The rail voltage is 24 V.

The motor only needs to operate at about 2 Hz, maximum is 10 Hz - so I've got plenty of room to place my cut off frequency. I plan to place the fc about 1 kHz (does this sound good enough?)

Would it make sense to place two filters, one of each side of the load? It was briefly suggested to me, since it might lower the size of the filter components (I'll need a relatively large inductor, as I understand it.) How does this work? How can I calculate the component sizes for my filter if I do it the way I was suggested? Is it a good suggestion?

Edit: Added figure for clarity. This is the filter I was suggested to create:

Also, the motor in question is a load with an inductance of 6.2 mH and a resistance of 2.35 Ohm.

Resources I've used to read up on:

• How do you mean two filters at each side of the motor, if the motor has only two wires (supposing brush DC)? Commented Aug 24, 2021 at 12:28
• Also consider, if the motor only needs to run at 2-10Hz, you could probably use a much smaller motor and 100:1 or 1000:1 gear-reduction. The motor then runs at 200-1kHz or 2kHz-10kHz for the same shaft speed, but with 100x or 1000x the torque (ignoring losses.) Commented Aug 24, 2021 at 12:38
• Hi! I've clearified what I mean with a drawing. I can't use another motor, it has to be this specific one, and it have to operate at that low frequency. Commented Aug 24, 2021 at 12:57
• Link a datasheet of the motor, as well the other calculations you have found. Also voltage of the H-bridge is missing. Commented Aug 24, 2021 at 13:04
• @MarkoBuršič I've added the rail voltage to my question. The only datasheet of the motor is a PDF document, which I can't add here. Commented Aug 24, 2021 at 13:09

You do calculate the capacitor to have a maximum 1% reactive current of the max. fundamental frequency.

$$I_c=V_c \cdot\omega\cdot C$$ $$C \leq \dfrac{I_c}{V_c \cdot\omega}= \dfrac{0.095A}{V_c \cdot 62.8}$$

I guess this condition would apply for PMSM, IM, BLDC motors but in your case would give a very big capacitor.

The LC resonance frequency to be choosen is: $$10\cdot f_{fundamental} \geq f_{res} \geq f_{PWM}/10$$

Of course you choose the highest possible, so that LC components are as small as possible. In my opinion I would take a risk and choose somewhere 4-8 kHz, else a 2kHz is safe.

EDIT:

$$\tau=L/R=2.6 ms$$

If you don't want to alter the response time of the motor, then choose an inductance of 1/10-th of the motor. Let say 600uH.

simulate this circuit – Schematic created using CircuitLab

$$f_{res}=\dfrac{1}{2\pi\sqrt{LC}}$$

$$C=\dfrac{1}{4\pi^2L f_{res}^2}$$

$$\C\approx 1.5 \mu F \$$ @ fres=5kHz

$$\C\approx 9.3 \mu F \$$ @ fres=2kHz

simulate this circuit

Inductor

Sweep response:

Transient response f=20kHz DT=50%, motor voltage, motor current. Note: No BEMF is simulated (only 6.2mH & 2.3ohm), aka rotor locked.

• How did you get omega = 62.8? I get: omega = 2 * Pi * f = 2 * Pi * 20*10^3 = 125660 Commented Aug 24, 2021 at 12:55
• @HelenaFærch $2\pi\cdot 10Hz$ Commented Aug 24, 2021 at 12:57
• Ah, now your answer makes more sense! But your calculations on the capacitor size is very different from others I have found on the internet. Commented Aug 24, 2021 at 12:59
• @HelenaFærch One thing you should know is that there's no such thing as a "one rule fits all" for sizing up elements. In this case, the value of the capacitor can depend on either the frequency of the lowest harmonic to be attenuated, or the maximum current sustainable by the capacitor, or the maxium allowable voltage ripple, or the step resonse time, etc. In short: just because you've seen one value, somewhere, it doesn't mean that that value is universal. Commented Aug 24, 2021 at 15:18