How do I calculate the required inductor current (RMS, saturation) for a class D amplifier?

Assuming the following configuration:

  • Speaker impedance (Z) = 4 ohms
  • Power output (P) = 400W

(calculations are rounded and assume no switching losses at the output)

  • \$V_{pp} = 113V\$
  • \$V_{rms} = 113V \times 0.3535 = 40V\$
  • \$P = \dfrac{V_{rms}^2}{Z} = \dfrac{1600V}{4R} = 400W\$

Is the current calculated as follows?

  • \$I_{pp} = \dfrac{V_{pp}}{Z} = \dfrac{113V}{4R} = 28A\$
  • \$I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{40V}{4R} = 10A\$

Does this mean that I need an inductor that is rated for at least 10A RMS and exceeds 28A saturation current?

The Class-D amplifier I'm using is a "H-bridge" output configuration, meaning it generates the required 40Vpp from a single 20V supply. So I need two inductors per speaker output. Does this mean that my required inductor should be rated for 14A saturation/5A current?

Are my calculations correct? I know the ones I posted are valid for a linear amplifier (Class-AB for example) but I don't know if they are the same for class-D, considering a Class-D amp is always switching max V, and thus, I is always at the max during a short period all the time.

In other posts I've seen mention of "core loss" and "V/us". What does this mean?

(The amplifier in question is a TI TAS5630B. The datasheet doesn't specify a suggested inductor, and the EVM was made with custom ones).


1 Answer 1


You have a couple of false premises in your question.

First of all, the inductor only needs to handle the peak current, not the peak-to-peak current. The peak current is 14A, since the sign doesn't matter.

Second, you only need one inductor per speaker, even in an H-bridge configuration.

Finally, if your power supply to the H-bridge is only 20V, then the peak voltage you'll be able to generate across the load is exactly that — 20V. The peak-to-peak voltage will be 40V and the sinewave RMS voltage wil be 14V. The maximum sinewave power into 4Ω will be 50W. The peak current will be 5A and the RMS current will be 3.54A.

If you really want to develop 40 Vrms across your 4Ω speaker, you'll need a supply of at least 60V to the H-bridge.

  • \$\begingroup\$ Oh, yes! Of course, the voltage across the inductor is never more than Vsupply. I should have realized that. I think I mixed up H-bridge and BTL. Here's an example of the several output configurations this amplifier offers: i.imgur.com/NXm5JaE.png Later I mixed up the figure of 40Vrms with 40Vpp, you're right, I need Vpp/2=113/2=56.2V (ideal, of course). \$\endgroup\$
    – hjf
    Commented Sep 1, 2014 at 15:09
  • \$\begingroup\$ Note that if driving the load from a H-bridge, splitting the inductor into two pieces (one per leg of the load) allows for EMI reduction. \$\endgroup\$ Commented Jan 29, 2015 at 18:35

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