How many watts will this amplifier need to dissipate?

http://datasheets.maximintegrated.com/en/ds/MAX9744.pdf

I'm trying to figure out how much PCB I'd need to dissipate heat from this amplifier when running it at 12V into a 4 ohm speaker.

The datasheet says on page 2:
Continuous Power Dissipation (TA = +70°C)
44-Pin Thin QFN (derate 27mW/°C above +70°C, single-layer board) ...................................................2162mW

So the chip can dissipate 2.1W continuously? But only with thermal vias and a PCB of adequate size, I assume?

Page 5 of this document says a good rule of thumb is 2.37 in^2 per watt of power dissipated for a 40C rise:

But that does me no good as far as calculating PCB size if I don't know how much power the chip is actually dissipating. I also have a PCB size calculator, but it needs JC not JA and this datasheet doesn't seem to supply that, so that's no good to me.

Then there's another snag On page 8 of the amplifier's datasheet there's a graph for 12V into 4ohms, where it shows efficiency (%) & power dissipation (mW) vs output power, which doesn't make any sense. If the amplifier is 85% efficient at 16W output, how is it the power dissipation is only 5mW? If I assume 15% of 16W is wasted, that comes to 2.4W that I'd expect the amplifier has to dissipate. Is the power dissipation axis labeled wrong? Because while 5mW seems absurdly low compared to what I calculate it should be, it wouldn't surprise me if it was double what I calculated.

On a second look, I'm pretty sure they mean Watts, because the graph next to it for an 8 ohm load lists the power dissipation in Watts, and at 10W of output power the dissipation is 1.5W. Which kinda meshes with the dissipation being 3.75W @ 10W output with the 4 ohm load.

But is that power dissipation for both speakers together? Or just one? If it was both together, than my calculation of 16W @ 15% efficiency dissipating 2.4W would make sense, since if you double that it's almost 5W, which is what that first graph says is the power dissipation at 16W into a 4 ohm load with a 12V supply.

But regardless, that brings us back to the first statistic I found in the datasheet:
Continuous Power Dissipation (TA = +70°C)
44-Pin Thin QFN (derate 27mW/°C above +70°C, single-layer board) ...................................................2162mW

If the amplifier can only handle ar ~2W of continuous power dissipation, what of that 5W of power dissipation? Whether that's one or two speakers doesn't really matter, it's well above that 2W limit either way. And a 4 layer PCB only gets you to 3W.

Yet here Adafruit is, selling the same chip on a tiny board, with up to a 12V input and stereo output:

What gives? Will that Adafruit amp melt if run continuously at 12V into a 4 ohm stereo load? I highly doubt it or they'd warn about it. And I've used smaller 10W boards. So something must be up with my math here.

Keep in mind that the heat dissipation is proportional to output power in watts, and you as the user have control of the output power by using a device called a volume control or a fixed voltage divider for the input signal. The gain of the amp will be a fixed value so you have control for device heat dissipation by limiting the amplitude of the input signal.

I'm aware that you can turn down the volume to reduce power consumption and the heat generated, but if they're claiming the amplifier is a 20W per channel amp, then I expect there is a way to get 20W of output power continuously.

Anyway, in looking at that continuous power rating again, I noticed something:
Continuous Power Dissipation (TA = +70°C)
44-Pin Thin QFN (derate 27mW/°C above +70°C, single-layer board) ...................................................2162mW

This bit:
(TA = +70°C)

I think it's a safe assumption that the ambient temperature isn't going to rise to 150F in my intended usage.

Also, there's this:
?JA, Single-Layer Board................................................37°C/W

So, if I assume 5W of power disspation, which I don't actually know is a correct assumption if I am driving two 4 ohm loads, then I can assume a temperature rise of 185C over ambient for which I think 30C (86F) is a reasonable assumption.

So that would give me a total temperature of 215C, which is too high.

Of course this also assumes I am running the amplifier at maximum power with something approaching a square wave at full volume. But it's best to consider the worst case.

If I used a multi-layer board, which I am considering, then the temperature rise would be 27C/W, so at 5W now my total temperature rise above ambient is 135C, ,making the total temperature at 30C ambient... 165C. This is slightly above the allowed junction temperature of 150°C.

So, assuming that 5W power dissipation is for two channels at once, and again, I am unsure if it is, then I think for the typical case of playing music in stereo, the power dissipation will probably be more like 2.5W, and my temperature rise will be 68C, making my maximum temperature 98C. And 2.5W of power dissipation would require around 5 in^2 of four layer PCB. Which seems, roughly, to be the the size of the Adafruit PCB. Though it appears to be only two layers.

Perhaps you could gain some ideas from this adafruit device based on the same chip. They have some download files including board layout, etc.

Also they mention a built in protection mode used in that chip: "Short-Circuit and Thermal-Overload Protection" So running too hot should not damage the device, just cause it to shutdown?

I linked to that same board in my posts above, wondering if it was okay for their board to be so small when my calculations indicated it would get pretty hot when running on a 12v supply. Their description on that page you linked though says "it's completely cool running", which seems to contradict the power calculations I did, if one is truly running it at close to 20W per channel peak.

scswift:
Of course this also assumes I am running the amplifier at maximum power with something approaching a square wave at full volume. But it's best to consider the worst case.

If it's a class D amplifier then the waveform and volume level won't make an awful lot of difference.

Class D uses PWM to generate the output waveform. MOSFETs only heat up when they're in transition from LOW<->HIGH. The number of transitions per second doesn't vary with the signal being amplified.

So...you have to assume full power dissipation at all times.

That's... counter-intuitive. At first I thought you must be wrong, because the duty cycle has to vary to vary the power into the speaker, but it doesn't really matter what the duty cycle is, does it? As long as it's greater than 0, if the PWM freqeuncy is 300KHz, which it is for this chip, then the mosfet will transition from high to low 300K time a second, regardless of whether the duty cycle is 5% or 95%. All that varies is how long in that period it is in the on or off state.

Interesting.

That raises more questions than it answers though, in regards to heat dissipation. That would mean that at 12V into a 4 ohm load, regardless of how loud the music is, I'm going to dissipate 5W. But that seems to be more than...

Wait...
?JA, Single-Layer Board................................................37°C/W

That's JA, not JC. That's the temperature rise if the chip is only dissipating power through the air surrounding it.

It must be able to dissipate much more heat directly through the heat sink into the PCB. But where's the JC rating? And why does that rating specify that it is for a single layer board, if it's the rating for dissipation to ambient air? And why is there another rating for JA for a multi layer board? None of this makes any sense.

A theoretically ideal Class D amplifier is 100% efficient and doesn't dissipate any heat. Page 15 of the datasheet shows that it's more than 80% efficient across most of the output range, so at 20W output, you'd be dissipating about 1W.

Normal music & voice has a peak-to-average power ratio of around 10, so with 20W (undistorted) peak output, the average output would be about 2W. (But of course, good audio amplifiers are designed to run at full power continuously.

So...you have to assume full power dissipation at all times.

The datasheet does NOT show that. The efficiency is 80% or more above about 1W. Below 1W, the efficiency falls-off.

DVDdoug:
A theoretically ideal Class D amplifier is 100% efficient and doesn't dissipate any heat. Page 15 of the datasheet shows that it's more than 80% efficient across most of the output range, so at 20W output, you'd be dissipating about 1W.

20% of 20W is 4W. Where are you getting 1W of power dissipation?

Also on page 8 of the datsheet, the second chart at the top shows that at 12V, with a 4 ohm load, the PD is 5W. I'm still unsure if that is for both channels at once or for each individual one.

Normal music & voice has a peak-to-average power ratio of around 10, so with 20W (undistorted) peak output, the average output would be about 2W.

Well, that's good to know, if true. That implies that at 16W output for the 12V into 4 ohm case, at 85% efficiency, you've got 1.6W of average power * 0.15 percent losses = 0.24W of power dissipation.

But what the other fellow said about the power losses being tied to the switching of the mosfet and nothing else seems to make sense, and even if the average power output were only 2W, the dissipated power would remain the same because all you're varying is the duty cycle to get that lower power output while the PWM frequency, and thus the number of times per second the mosfets switch and pass through that high resistance state, would remain the same.

DVDdoug:

So...you have to assume full power dissipation at all times.

The datasheet does NOT show that. The efficiency is 80% or more above about 1W. Below 1W, the efficiency falls-off.

But... the output power depends on the supply voltage, which is fixed.

Once you decide on your supply voltage you should be able to calculate the heat that will be generated by the chip. That heat won't vary with the volume/type of the music because it comes from the PWM switching frequency, not the duty cycle.

You mean the power dissipation, not the output power, don't you?

scswift:
You mean the power dissipation, not the output power, don't you?

Power dissipated by the amplifier chip, yes.

That's what we're trying to calculate.

scswift:
http://datasheets.maximintegrated.com/en/ds/MAX9744.pdf

Then there's another snag On page 8 of the amplifier's datasheet there's a graph for 12V into 4ohms, where it shows efficiency (%) & power dissipation (mW) vs output power, which doesn't make any sense. If the amplifier is 85% efficient at 16W output, how is it the power dissipation is only 5mW?

Two things - there's a misprint in the datasheet (a common occurence!) and it should be
W, not mW for the dissipation. Secondly its a stereo amplifier so you have twice the
dissipation, so 90% efficiency at 20W means 4W dissipated.

Also remember most audio signals are nowhere near maximum amplitude most of
the time, its only the percussive peaks that get up there, so in practice the dissipation
will be a lot less than the maximum.

MarkT:
Also remember most audio signals are nowhere near maximum amplitude most of
the time, its only the percussive peaks that get up there, so in practice the dissipation
will be a lot less than the maximum.

Not in a class D amplifier.

fungus:

MarkT:
Also remember most audio signals are nowhere near maximum amplitude most of
the time, its only the percussive peaks that get up there, so in practice the dissipation
will be a lot less than the maximum.

Not in a class D amplifier.

While I'm inclined to agree with you because what you said earlier seems to make sense, if that's true, then how do you explain the efficiency vs output power graph in the top center of page 8 in the datasheet:

That graph is for a 12V input into a 4 ohm load. The line at the bottom shows the power dissipated vs the output power. The power dissipated clearly increases in a linear manner with the output power, which varies with an increasing duty cycle.

It's not even a little change either, it's a big change in power dissipation. And that directly contradicts your suggestion that the dissipation remains constant.

scswift:
While I'm inclined to agree with you because what you said earlier seems to make sense, if that's true, then how do you explain the efficiency vs output power graph in the top center of page 8 in the datasheet:
http://datasheets.maximintegrated.com/en/ds/MAX9744.pdf

Good question.

I'm guessing it's something to do with where it says "...with an inductor in series..." at the top of the page.

Here class D losses are I^2.R in the MOSFETs, and I^2 is proportional to power. A
good class D amp with discrete MOSFETs will be ~ 90% efficiency at full power.
The dissipation is likely shared between I^2.R, switching and inductor losses.

No single chip calss D amp can have low resistance MOSFETs integrated as these are
vertical current flow devices with the substrate being one terminal.

MarkT:
Here class D losses are I^2.R in the MOSFETs

What do you mean by I^2.R? I^2 is obviously current squared, and R is resistance, but what's with the dot?

fungus:
I'm guessing it's something to do with where it says "...with an inductor in series..." at the top of the page.

I dunno, that sounds to me like they were just simulating a speaker with a resistor and inductor.

I found this document on Class D amplifier design:

On page 6, on the right side, it shows the equations for calculating power losses in a class D amp.

It looks like in addition to the switching losses, (which I think are constant, but I can't say for sure because I don't know what some of those variables represent) there are also the conduction losses that MarkT mentioned, which are given by:
(Rds(on) / Rload) * Pout

And, there are gate drive losses for the mosfets, which appear to also be constant so can just be considered switching losses.

Anyway, given that chart in the datasheet indicates a large drop in dissipated power as the output power is reduced, I think it's safe to say that the conduction losses are the biggest contributor here to the dissipated power, and as those do vary with the volume, a class D amp will in fact draw less current and dissipate less heat when you turn the volume down or the music is quiet.