I am building a class D amplifier and need to design passive low-pass filters for H-bridge with cut off frequencies at about 400 Hz. However, I am unsure about few things, this is the simplified equivalent circuit:


simulate this circuit – Schematic created using CircuitLab

Speaker values were measured using an RLC meter set to 100 Hz.

Lets say that the signal is coming into the L1 coil. Should I calculate the cut-off frequency only for L1, C1 pair? If L2 is grounded, do L2 and C2 have a noticeable effect on L1, C1 filter? How can I calculate it then?

I often see a capacitor in parallel with speaker what exactly is its purpose? I am thinking it is to lower reactive power of the speaker, right? To what \$cos\phi\$ should I lower it, and should I use middle range values for voltage and current?

  • \$\begingroup\$ The parallel capacitor does the same as C1 and C2 \$\endgroup\$ – Andy aka May 6 '15 at 20:30
  • \$\begingroup\$ @Andyaka To create a higher order filter? \$\endgroup\$ – Golaž May 6 '15 at 20:31
  • \$\begingroup\$ No it doesn't make a higher order filter - it's still a 2nd order filter. \$\endgroup\$ – Andy aka May 6 '15 at 20:47

Imagine the speaker coil had a centre tap. If you measured the voltage at the centre-tap relative to 0V it would be 0V with a balanced drive. Therefore you can split the problem in half and imagine a grounded speaker with an impedance of 3.3 ohms and self inductance of 3.86mH.

Instead of C1 and C2 where they are, imagine just C1 connected across the half-speaker I have just described. Now proceed to calculate L1.

One word of warning - the 7.72mH self inductance of the speaker seems too high for my liking - I'm thinking that your RLC bridge has told you the effective parallel impedance (or maybe you meant 7.72uH)

  • \$\begingroup\$ Thanks, this clears up a lot! So the purpose of parallel capacitor (C3) is to increase the capacitance of the half filter, to allow for lower inductance values? How can I be sure what RLC meter showed (there is a setting for parallel/series measurement)? Even if the meter showed this as parallel impedence I get 600uH if I calculate it like this: $$Z=\sqrt{ESR^2-\frac{1}{\omega^2 L^2}}=7.72m$$ $$L=\frac{1}{\sqrt{\omega^2(ESR^2-Z^2)}}\approx 600uH$$ \$\endgroup\$ – Golaž May 7 '15 at 13:14

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