I'm measuring the output with the load attached. The voltage divider has already been accomplished INSIDE the ampl
You are measuring with no load attached and measuring the effect of loading the amplifier with the parallel combination of the resistor and the headphone.
So, I concede you are correct for your measurement scenario. Which is an important consideration in the broader application, and of course the output impedance of the amp has to be considered.
I was looking at the specific case of measuring the output with the load attached. And you're right, if the power specification for the amp is stated at or near the unloaded voltage output limit, then you have to care about the drop across the output impedance of the amp.
BUT, If my initial condition is that I have a bit of headroom over the amp power spec and I can drive x volts into some load, if I change the load and adjust the input signal to the gain stage such that the output voltage is again x volts, then the power to the headphones remains constant.
I was assuming output power was measured terminated into load.
Power specification (I.e. 5 Watts @ 8 Ohms) would be measured with the amp terminated into an 8 Ohm load.
The power spec at full output should be measured and should read(for example):
5 Watts @ 8 Ohms, 1kHz, THD = 0.1 %
Or describe output power with a set of curves.
There are a few reasons why output power from a power amp does not double as the load impedance is halved, the effect of output impedance is one of them.
You are missing my point:
Changing the parallel resistor from 10 to 15 Ohms changed the output (assuming input is fixed) into the headphones only
slightly.
Again,
Read my posts again.
Assuming input is a fixed voltage
10 Ohm parallel resistor. 368 milliWatts
15 Ohm parallel resistor: 380 milliWatts
As I was saying: the difference is trivial. To put it another way, it wouldn't be audible. OTOH, but it is there.
Basically, I was debating Terja's calculations.
He thought that output would be 5 Watts into both an 8 Ohm and a 9.17 Ohm load. His calculations were erroneous.
I don't know what else to say.
You can bet that a 5 Watt tube amp uses an unregulated power supply.
The output tubes have finite output impedance.
The output transformer has finite out impedance.
After working for many years as an Electrical Power Engineer, etc, etc, there is a chance that I may actually know what I am talking about.....