Hey, sorry about the delay. I'm still interested in this topic. I just had some real life things to deal with.
No worries, it happens to all of us!
I had no idea you could run a chip in mono and get stereo out of them by running two of them in mono. Pretty basic stuff, so I apologize for that.
The channels of an audio chain are fundamentally separate. We just control them in (usually) pairs for convenience.
Class D amplifiers are a bit of an exception, in that many of them take advantage of the fact that they actually have way more power available than they need, letting them actually alternate powering the two channels, reducing the instantaneous current draw. (This is why you must only bridge class D outputs in the ways listed in the datasheet.)
As far as power goes, I'm always talking about continuous power. So if you're running a 24V driver, your max output is going to be 72 watts, right?
Nope. You’re going by the math for DC, but it’s not DC, it’s AC.
Some things to understand:
1. The supply voltage is what limits the maximum power of the amp, as it is the highest voltage that be present on the output. For any given speaker impedance, it is the maximum voltage alone that limits the power. But it’s a sine wave, so the vast majority of the signal is below the peak voltage.
2. Many (most?) class D amps, including the one here, use creative switching to produce both the positive and negative sides of the output waveform from a single-polarity power supply (let’s call this VCC). This means that the output voltage can swing from +VCC to -VCC, meaning a peak-to-peak voltage of VCCx2. So if your VCC is 24V, then your output can swing from +24 to -24 volts, a total peak-to-peak voltage of 48V!
3. In a sine wave, the peak voltage is only reached for a tiny amount of time. Most of the time is less. So we measure using RMS, the value of the AC signal given as the DC voltage with the equivalent heating power. For a pure sine wave, the RMS value is the square root of 2 (about 0.707) times the peak voltage (which in an amp is at most the VCC). So for your 24V VCC, the RMS becomes 24x0.707, or about 17V. And that gives you a maximum average power of 36W into 8 ohms (a current draw of 2.125A). (You can also calculate it as VCC
2/2x(impedance).)
4. Music isn’t a singular, always-full-amplitude sine wave. That’d just be a single loud tone. Nor is music multiple sine waves of different frequencies, but always full amplitude. No, it’s nearly infinitely many simultaneous sine waves of different frequencies
at highly varying amplitudes. Indeed, most of them will be
far below the peak amplitude (since the peak is necessarily the peak
sum of all the frequencies at any instant). The consequence of this is that the average power is far lower than the peak power, easily a factor of half
at full volume.
5. As such, while you will need capacitors to ensure the peak current can be supplied, the power supply only needs to be beefy enough to supply the average current. That’s why you can get away with the small power supplies commonly used in audio gear now.
And that’s all assuming you’re running it at full blast. At typical home, indoor listening levels, you don’t need but a tiny fraction the power. One of the 3116’s at just 7 volts into 8 ohm speakers is enough to be much too loud to have a conversation, and that’s just 3 watts per channel max. Nighttime watching a TV show, for instance, will have the 3116 drawing maybe 30mA at 5 volts, so 150mW total (and at 5V, the 3116’s quiescent current with no signal is about 15mA, so at most 75mW actually going to the speakers!).
A lot of info, but I hope it helps. Of course there are lots of other things that come into play and slightly increase the power needed, but this should help understand why you can’t just apply ohms law to the supply voltage to calculate an amp’s power.