I'm trying to spec a power supply for a class D amplifier IC. I have figured out the maximum RMS current draw by taking the RMS power rating of the amp, increasing that to compensate for efficiency losses, then applying: $$P = I^2R$$ Where \$R\$ is the speaker impedance.

However my power supply needs to be spec'd for the peak current, not the RMS current, right? So, assuming the above approach is correct, how do I get the peak current draw of a class D amp?

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    \$\begingroup\$ I'm not sure what best practice is when it comes to speakers, as their impedance is very strongly frequency-dependent; "8-ohm" or whatever the speaker is specified to is an approximation at best, and not a very good one either. \$\endgroup\$ – Hearth May 5 at 22:22

Sadly, there is no such simple relationship. Speakers and the amplifier (and the filters) form a resonant system, and that can introduce much higher voltages into the system than supplied by the power supply.

However, it doesn't really matter:

The peak power consumption isn't something that you need to supply, but a design factor that you can choose. It's up to you, within the limits of not damaging your speakers or the switching transistors, to decide how much peak current your amplifier system needs to be able to draw.

For example, you might decide that you want your power supply to deliver enough power for the max RMS power consumption of the amplifier (that's not a "must"! You can built an amplifier that could draw more power than the supply can offer, and then use it at less-than-maximum-volume). That's easy, you just use \$P=U\cdot I\$ and solve for \$I\$, and get a \$U\$ voltage power supply that can at least supply \$I\$ current continously.

Then you decide that at that (undubitably loud) setting, you still want to be able to, for a short term of \$t=0.5\,\text{s}\$, supply more power, say four times as much. So, that'd be a current factor of four. You decide that you need to supply half of that by having half of the appropriate charge (\$I\cdot t = Q\$) in a capacitor, and overdimensioning your power supply by a factor of 2. (just rough numbers; none of this takes efficiency or the exponential discharge curve of a capacitor into account)

In any case, the speaker impedance doesn't come into play here, anyways: your current goes through the switching transistors of your class-D amplifier, so your amplifier manufacturer is in charge of definining a maximum power and/or current, anyway.

  • \$\begingroup\$ Thanks Marcus. Say I do want my supply to deliver the max rated RMS power. If I simply solve P=UI for the RMS power, will the current figure I get actually be sufficient? Shouldn't I solve P=UI for the peak power, not the RMS power? \$\endgroup\$ – Jeremiah Rose May 6 at 8:46
  • \$\begingroup\$ that's exactly what the second half of my answer is about. \$\endgroup\$ – Marcus Müller May 6 at 11:29

The peak current from RMS is just 1.414 x RMS, but a Class D Amp is really a Buck-Boost SMPS.

Check out and understand all 3 parts to this series. Then try the Analog Power Design tools. https://www.analog.com/en/technical-articles/switch-mode-power-supply-current-sensing-part-2-where-to-place-the-sense-resistor.html

I believe you biggest current spike will be for the Bass speakers. This will be in CCM mode with high duty cycle. But the inductor current must ramp up and down with PWM to achieve this peak current. So at a minimum I suggest 1.5 x RMS and consider the inductor is rated for 2x RMS current without saturating.

If you plan on using a SMPS to power this class D Amp the impedance and ripple at both frequency rates and use some simulation tools like Falstad to model the DCR,L, ESR C and choose SPDT Switch and choose the Ron resistance to test for resonance or export into the frequency domain applet and examine the frequency attenuation of the LPF filter.

But that’s all I’ll say for now.

  • \$\begingroup\$ Thanks Sunnyskyguy for this real world advice. So the actual current draw of the amp could be 1.5-2 times higher than the RMS current? \$\endgroup\$ – Jeremiah Rose May 6 at 8:50
  • \$\begingroup\$ No, it is written about amplifier peak output current, NOT the input one. Amplifier input current flows from power supply to amplifier, amplifier output current flows fom amplifier to speakers. \$\endgroup\$ – Volodymyr Kalinyak May 8 at 6:13
  • \$\begingroup\$ Input current can be smaller if voltage is larger than needed at max current with losses estimated. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 8 at 13:21

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