Yes, an excellent post by Bombaywalla, and an excellent response by Michael (Swampwalker). Thanks, gentlemen!
I am in full agreement with both posts, aside from what I believe is an inadvertent and minor misstatement in Bombaywalla's post:
Again, though, an excellent and informative post. Thanks!
Best regards,
-- Al
I am in full agreement with both posts, aside from what I believe is an inadvertent and minor misstatement in Bombaywalla's post:
A linear power supply is expensive from a power dissipation perspective - you have to design its max voltage for the max peak voltage of the program material but in normal operation the linear power supply operates mostly at the average voltage of the program material. The difference in the peak & average voltage is dissipated as heat. Of course, you don't know what the max voltage of the program material is so you have to over-design further leading to more heat dissipation.Shouldn't it be the output of the amplifier that has to operate mostly at the average voltage of the program material, not the output of the power supply? With the difference between the average output voltage and the voltage supplied to that stage (which as you indicated has to provide headroom relative to the maximum anticipated output voltage), multiplied by current, corresponding to the heat dissipated in the output stage, not the power supply? Although the heat dissipated in the power supply will also vary with current demand. And although there are a few amplifier designs in which the output voltage of the power supply is actually varied among a number of discrete levels as a function of signal level, some of Bob Carver's older designs being examples.
Again, though, an excellent and informative post. Thanks!
Best regards,
-- Al