Low-pass LC filter on both sides of a motor

Question

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:

• Electronicbase.com
• Sedra Smith Micro Electronic Circuits (pages 1077 to 1088)

Answer 1

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.