Class D

Yes, please read about EPDR, then you can tell George he does not understand it just like every other person who understands amplifiers has told him.

Equivalent Peak Dissipation Resistance - i.e. EQUIVALENT resistance to equal peak thermal dissipation. Not peak current supply. Not minimum impedance. The EQUIVALENT (not actual) resistance that would equal the peak THERMAL dissipation.

This only has meaning for traditional amplifiers that operate in the linear region, i.e. Class A, AB. It is where the (current output * (rail voltage - output voltage)) is at a maximum.  With a linear amp, let's say the current output is 10A, and the voltage difference = 40V at some point in time. That would be 400 watts dissipated across the output devices. That is the thermal dissipation in a linear amplifier.


In a Class-D amplifier, the average (as they are switching) voltage in the output devices, while they are conducting, will be a small fraction of a volt. Let's say it is 0.1V at 10A.  The dissipation in the output devices is now 1W.  Notice that the rail voltage does not enter the equation?  Now practically there is a small contribution from the rail voltage, but there is not a direct correlation like in a Class A/ AB amplifier because obviously they don't work the same way.

Now one thing a Class-D amplifer may due is hard limit current which they will "measure" or at least check for a peak on every single switching cycle.  This provides short circuit protection and protection that the output inductors do not enter saturation. As that current will be fixed, this is why in a Class-D amplifier the output power probably does not keep doubling with halving of the output impedance. The power will probably increase a lot between 8 and 4 ohms, because in this case the major limitation in output power is due to the power supply rail voltage. It will likely not double into 2 ohms because you will run into the current limit before the voltage limit.

What does this mean?  As long as the amp does not have stability issues at 2ohms, which most newer ones will not, and you don't drive the amplifier into clipping, then the class-D amp will be just fine into low impedance loads no matter what some people who make claims, but don't understand the underlying technology make.

+1 on @twoleftears suggestion.

As @audio2design points out, EPDR does not apply to class D amplifiers. However, the efficiency at which a class D amplifier operates is reduced as load impedance goes down since the output transistors will be conducting more current during their switching and on-phases for the same power into the load. 

Class D amps are theoretically 100% efficient if the output transistors had zero on-resistance, infinite off resistance, and switched infinitely fast. But, of course, this isn't the case. Typical class D amps are 90% or more efficient into 8 ohms (at max output power), but efficiency drops by approximately 40% into 4 ohms, and 40% again at 2 ohms.

Under normal circumstances, this isn't a huge problem because an audio amplifier in a normal home music environment is typically operating at a small fraction of it's peak power. But under heavy demand (or test conditions), the lower operating efficiency into lower impedance loads will eventually cause the amplifier to get too hot and shut down. The good news is that modern class D amps all have circuitry to protect the amplifier under these conditions. 

As @audio2design also points out, class D amps will also incorporate maximum current limiters which will also limit maximum power into low impedance loads. Since this circuitry operates almost instantaneously (checking current on every switching cycle), this is more likely to limit maximum power into low impedance loads since music peaks can often be many times higher than average power requirements. But these current limits are generally quite high. For example, the Purifi module, which is rated at just over 200w into 8ohms, has a 25A current limiter, but this is what ultimately limits the peak power into loads below about 2 ohms.

But, of course, this isn’t the case. Typical class D amps are 90% or more efficient into 8 ohms (at max output power), but efficiency drops by approximately 40% into 4 ohms, and 40% again at 2 ohms.

I am not sure you wrote this correctly. I think you mean losses go up 40%, or stated as efficiency, the reduction in efficiency goes up by 40%, i.e. 10% reduction becomes 14%?


Conduction losses are I^2 * R. At the same wattage, half the resistance, current goes up sqrt(2) = 41.4%, so losses must go up 100% due to conduction losses, but realistically total losses are going to be a quiescent component (in this case about 11 watts), plus a linear component and a squared component to come up with a very good model.


For reference as example, the NCORE 500 OEM module is about 95.2% efficient at 400W/8 ohms, about 92.6% efficiency at 400W/4 ohms, but drops down to about 87.5% at 400W/2 ohms. Losses are 20, 32, and 67 watts at 8/4/2 ohms, 400W output.


For the NC500, the current limit is 26A. That puts maximum possible RMS power at 2ohms = (26/sqrt(2))^2 * 2 = (26*26/2)*2 = 676W, not too far from the rated power of 550W as a ratio, and even closer if you add losses above to the 550W.

@audio2design Yes, that is what I meant. Sorry if my language wasn't clear.