When designing an output filter for a simple class-D amplifier I am unsure about sizing the components. Given:

  • cut off frequency of approximately 50kHz
  • the inductor DC resistance can be disregarded for this question
  • the load is 8Ω, for this discussion purely resistive
  • this architecture:


simulate this circuit – Schematic created using CircuitLab

The center frequency for the LC filter is: $$f_c = \dfrac{1}{2\pi\sqrt{LC}}$$

Possible solutions include (but not limited to):

\begin{array}{|c||c|}\hline L & C \\ \hline 220\text{μH} & 47\text{nF}\\ 22\text{μH} & 470\text{nF}\\ 2.2\text{μH} & 4.7\text{μF}\\ 220\text{nH} & 47\text{μF}\\ \hline \end{array}

What are the proper design criteria for choosing a larger or smaller inductor / smaller or larger capacitor?

  • \$\begingroup\$ what is your PWM frequency? \$\endgroup\$
    – Andy aka
    Feb 1, 2014 at 11:25
  • \$\begingroup\$ @Andyaka I don't want to discuss the choice of PWM-frequency vs. cutoff frequency, that design decision is probably worth a question in its own right. The PWM frequency is much higher than the 50kHz. \$\endgroup\$
    – jippie
    Feb 1, 2014 at 15:50
  • \$\begingroup\$ Ah if only it were as simple as that - at where the filter cut-off may be pitched you might want to examine what the speaker impedance looks like (especially if switching is sub 200kHz) - this might cause another resonance peak that is closer to your PWM frequency. At well-above audio frequencies I can envisage the speaker acting just like an inductor - that's my reason for asking dude because you can't rely on the 8 ohms for damping anymore. Also, what power output are you considering or are you OK working out whether inductor saturation will be a problem? \$\endgroup\$
    – Andy aka
    Feb 1, 2014 at 17:15

1 Answer 1


You must consider the load, which the loudspeaker creates for your LC filter. It changes the response of the filter. Below are results of the simulation performed with LTspice for different values of L and C (keeping almost constant the LC product):

Results of simulation for different L and C values (with constant LC product)

As you can see, from the tested values 330nF, 33µH gives the best results (I have added this set of values basing on results of simulation).

\begin{array}{|c|c|c|} L & C & \text{color} \\ \hline 220\text{μH} & 47\text{nF} & \text{violet}\\ 33\text{μH} & 330\text{nF} & \text{turqoise}\\ 22\text{μH} & 470\text{nF} & \text{red}\\ 2.2\text{μH} & 4.7\text{μF} & \text{blue}\\ 220\text{nH} & 47\text{μF} & \text{green}\\ \end{array}

From the point of view of the loudspeaker (R1) we have a parallel RLC circuit. Its Q factor is defined by the formula: $$Q = R_1\sqrt{\dfrac{C_1}{L_1}}$$

For C1=330nF, L1=33µH the Q factor is equal to 0.8

The optimal solution will be the one, for each Q is equal to 1.

  • \$\begingroup\$ Very helpful answer, basically the load dampens the resonance from the LC combination and that needs to be aligned. \$\endgroup\$
    – jippie
    Feb 1, 2014 at 10:09
  • \$\begingroup\$ Yes, exactly. If one wants to provide any theoretical background, it should be necessary to analyze the Q factor of the RLC circuit created by the L1, C1 and R1 (speaker). For the highest value of L1, the circuit behaves almost as a low-pass RL filter consisting of L1 and R1 (the influence of C1 is visible only above 300kHz). \$\endgroup\$
    – wzab
    Feb 1, 2014 at 10:18
  • 1
    \$\begingroup\$ If you might drive different speakers, remember they aren't all 8 ohm loads and most hi-fi speakers use L-C filters as crossovers. So it may be worth re-run the simulation with a variety of different speaker models (4 ohm, 8 ohm, and a typical crossover network) before making a final decision. \$\endgroup\$
    – user16324
    Feb 1, 2014 at 11:23

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