Hi Almarg,
yes, you are correct - misstatement on my part. Sorry guys!
indeed the linear power supply operates at its designed output voltage (which is over-designed to ensure it will accommodate the max program material voltage) & the extra power dissipated will be (power supply voltage to amp output stage - voltage of program material) * current. And, correct again, the power amp output stage mostly operates at the average voltage of the program material.
yes, you are correct - misstatement on my part. Sorry guys!
indeed the linear power supply operates at its designed output voltage (which is over-designed to ensure it will accommodate the max program material voltage) & the extra power dissipated will be (power supply voltage to amp output stage - voltage of program material) * current. And, correct again, the power amp output stage mostly operates at the average voltage of the program material.