Class D music amplifier loading?

[Le_Bassiste]

A related thought - if we use a half bridge pwm class D amp to supply a sinewave into a resistive load, what shape will the (smoothed) current draw from the supply rails look like? It won’t be sinusoidal.

if the amplifier can be considered being reasonably lossless (as it is class-d), pin(t) = pout(t) = uout(t) * uout(t)/R. with the DC bus voltage VBUS being constant, ibus(t) = pout(t)/VBUS. thus, for sinusoidal output voltage (and output current), the DC bus current into the amplifier has sin^2 shape.

In reality the impedance of the speaker is frequency dependent, rather than just a plain resistive load, which complicates things.

so what? as long as voltage and current are both sinusoidal, the power ist sin^2 shaped, no matter what impedance (or z=f(f), or cos phi, for that matter) the load has.

Music is not sinusoidal and the impedance of the speaker is frequency dependant.

did i point out that the impedance is frequency dependent? yes
is music non-sinusoidal? yes, in almost all cases.
was that the point in circlotron's question? no. however,
you can generalize what i said above to  p(t) = u(t)*i(t), with i(t) = u(t)/Z,  giving p(t) = u²(t)/Z.
as you see, no matter what impedance you throw at the amplifier's output, the square relationship of output power and output voltage will hold.

[Circlotron]

was that the point in circlotron's question? no. however,
you can generalize what i said above to  p(t) = u(t)*i(t), with i(t) = u(t)/Z,  giving p(t) = u²(t)/Z.
as you see, no matter what impedance you throw at the amplifier's output, the square relationship of output power and output voltage will hold.

After thinking about it a bit further, the power in the load varies according to the square of the load voltage. No surprises there. And seeing the supply rail is a fixed voltage, the current supplied by the supply rail will be directly proportional to the load instantaneous power. So, yes, it seems that the current drawn from the power rails will be sin^2 if the resistive load voltage is a sine. I was getting a bit distracted by the fact that the pwm + filter is similar to a buck regulator so input and output current could differ.

[Circlotron]

That's it.  Nothing is worse, and speakers don't have nearly enough Q at anywhere near low enough of a frequency for regeneration to be a problem.

But seeing speakers are a reactive load, they present a low power factor, if you will. They can suck current on one part of the cycle and return some of it later in the cycle. How low that power factor can get, I'm not sure.

[langwadt]

That's it.  Nothing is worse, and speakers don't have nearly enough Q at anywhere near low enough of a frequency for regeneration to be a problem.

But seeing speakers are a reactive load, they present a low power factor, if you will. They can suck current on one part of the cycle and return some of it later in the cycle. How low that power factor can get, I'm not sure.

but unless there is DC offset or a very asymmetric waveform it all evens out over a few cycles 

[T3sl4co1l]

That's it.  Nothing is worse, and speakers don't have nearly enough Q at anywhere near low enough of a frequency for regeneration to be a problem.

But seeing speakers are a reactive load, they present a low power factor, if you will. They can suck current on one part of the cycle and return some of it later in the cycle. How low that power factor can get, I'm not sure.

That's what I said, the Q is low, like maybe 2 at the most.  Well, piezo tweeters could be more, near or above resonance; but at low frequencies yeh, voice coil resistance is a dominant fraction of the total at any frequency.  In other words:

but unless there is DC offset or a very asymmetric waveform it all evens out over a few cycles

...Like that.

Note that impedance generally goes as sqrt(f) due to skin effect (the steel pole the voice coil moves over, is a diffusion resistance at essentially any frequency of interest), and, the peak at f_s should be small, well matched in a properly designed speaker box.

I have a plot here, that I have no idea what it's from; evidently a fairly large woofer, probably 16 ohms nominal, probably in open air (hence the relatively high peak).  If it's reasonable, then we can interpret it this way...



Note that, even as prominent as the peak looks, it's still only 3x baseline, or about as much Q.

Bizarrely, the seeming asymptote, isn't: notice the plot is semilog, so that's actually logarithmic impedance, not sqrt!

Tim