I started a thread about calculating Class A power from Class AB amplifiers at DIYAudio, one of my favorite sites. It was inspired by the fact that I calculate the Class A power from my Aragon 8008 BB as 3.58 watts per channel into 8 ohms (as described in an earlier post).

(Note: the full Class AB power is 250 watts, and about 500 watts into 4 ohms, and even more into a still lower impedance, so the amp has plenty of power available, but the question is specifically how much "Class A" power the amp can produce with all transistors operating in a linear range, therefore having the most quality)

Meanwhile, Mondial and Klipsch (the first and second manufacturers of Aragon amplifiers) claimed it had a maximum of 26 watts Class A power.

Much lively discussion ensued. It turns out my calculation of the Class A power into 8 ohms was correct. Furthermore, if my amp used the lower bias suggested in a memo from technical support, it would have 0.8 watts of Class A power into 8 ohms.

And it turns out that Mondial and Klipsch were correct also, at least for one channel of an Aragon 8008 BB...but misleading. It's quite possible to get 26 watts of Class A power from a factory specified Aragon 8008 BB, but you must use a higher impedance than 8 ohms, probably more like 50 ohms. So this is not really a relevant specification, but rather a way of crafting a technically correct but deliberately misleading specification. Also, given the lower bias, it appears likely their "Class A" power was actually peak power, not average power. So two tricks were used to craft a misleading specification. And possibly two more tricks, read on.

The most reduced formula for Class A power calculation is 2RII where I is the fixed quiescent idle bias current, and R is the load resistance. Since this I could be confused with the current-available-in-class-A, which is twice as much as the idle bias current, I've now decided a better description is 2RIb^2 ("two R Ib squared"). This is an algebraic reduction of the more transparent calculation (based on P=I^2R, Ia=2Ib, and Pa=Pp/2):

2*Ib * 2*Ib * R / 2

This formula yields "average" power (sometimes incorrectly called RMS power, see the link above for some discussion of that), that's the reason for the final 2 divisor (from Pa=Pp/2). Peak power would be 4RIb^2. So obviously increasing R gives higher power and reducing R gives lower power.

But the increasing the R value runs into a limit when the full Class A current available to the load (2I where I is idle current) causes the voltage across the R to reach the maximum voltage available from the amplifier. In the case of my amplifier, that occurs around 65 volts. Given maximum current available in Class A, 0.946 amps, and maximum voltage, 65 volts, it is straightforward to calculate peak then average power:

Pp = 0.946 * 65 = 61.49W

(R = e/i = 65 / 0.946 = 68.71 ohms)

Pa = Pp / 2 = 30.75

OK, so my 8008 BB seems to exceed the factory "spec" of 26Wpc maximum Class A power (into specially chosen resistance) by just a tad. But the amplifier idle bias current recommended by tech support is more than 2 times lower. So an amplifier with that level of bias could not meet factory specification even with any specially chosen resistance. There are two possible interpretations:

1) The factory idle bias current was actually much higher than specified in the memo (shown in earlier post 12mv inner channel and 8mv outer channel), and just short of what my amplifier has.

2) The factory "maximum class A power" specification was actually a peak power specification as well as specially chosen resistance specification.

I think I believe interpretation (2). Assuming that is correct, what would would the bias current need to be?

Pp = 26W, therefore 2Ib = 26 / 65 or Ib = 26/130

Ib = 0.20a

Therefore emitter resistor voltage for 6 transistor pair 8008bb would be

0.33 * 0.2 / 6 = 11mV

Just a tad below the "inner channel" specification of 12mV emitter resistor voltage recommended by Aragon technical support. But it seems the outer channel specification of 8mV would never make it, nor would the same values get there with the 4 transistor pairs in an Aragon 8008 ST:

0.33 * 0.2 / 4 = 16.5mV

So even with two tricks used to get the maximum specification, it only applied to the inner (left) channel on an Aragon 8008 BB, unless the bias was set higher than specified in the memo (which is possible, it's quite possible that tech support recommended an especially conservative bias to prevent subsequent failures for people who already had one problem).

****

My Aragon 8008 BB has 3.58 watts Class A power into 8 ohms, and 1.79 watts into 4 ohms, and therefore meets the "first watt" requirement popularized by Nelson Pass, whereas a factory specified unit would not.

My amplifier is biased more like an Aragon Palladium amplifier (just slightly more, actually). But since those amplifiers are bridged monoblocks, and mine is not, that actually gives me another advantage in Class A power at 8 ohms and less, since each half of a bridged amplifier sees only half of the total load.

So mine is actually better than an Aragon Palladium. It's an Aragon Platinum!

(Note: the full Class AB power is 250 watts, and about 500 watts into 4 ohms, and even more into a still lower impedance, so the amp has plenty of power available, but the question is specifically how much "Class A" power the amp can produce with all transistors operating in a linear range, therefore having the most quality)

Meanwhile, Mondial and Klipsch (the first and second manufacturers of Aragon amplifiers) claimed it had a maximum of 26 watts Class A power.

Much lively discussion ensued. It turns out my calculation of the Class A power into 8 ohms was correct. Furthermore, if my amp used the lower bias suggested in a memo from technical support, it would have 0.8 watts of Class A power into 8 ohms.

And it turns out that Mondial and Klipsch were correct also, at least for one channel of an Aragon 8008 BB...but misleading. It's quite possible to get 26 watts of Class A power from a factory specified Aragon 8008 BB, but you must use a higher impedance than 8 ohms, probably more like 50 ohms. So this is not really a relevant specification, but rather a way of crafting a technically correct but deliberately misleading specification. Also, given the lower bias, it appears likely their "Class A" power was actually peak power, not average power. So two tricks were used to craft a misleading specification. And possibly two more tricks, read on.

The most reduced formula for Class A power calculation is 2RII where I is the fixed quiescent idle bias current, and R is the load resistance. Since this I could be confused with the current-available-in-class-A, which is twice as much as the idle bias current, I've now decided a better description is 2RIb^2 ("two R Ib squared"). This is an algebraic reduction of the more transparent calculation (based on P=I^2R, Ia=2Ib, and Pa=Pp/2):

2*Ib * 2*Ib * R / 2

This formula yields "average" power (sometimes incorrectly called RMS power, see the link above for some discussion of that), that's the reason for the final 2 divisor (from Pa=Pp/2). Peak power would be 4RIb^2. So obviously increasing R gives higher power and reducing R gives lower power.

But the increasing the R value runs into a limit when the full Class A current available to the load (2I where I is idle current) causes the voltage across the R to reach the maximum voltage available from the amplifier. In the case of my amplifier, that occurs around 65 volts. Given maximum current available in Class A, 0.946 amps, and maximum voltage, 65 volts, it is straightforward to calculate peak then average power:

Pp = 0.946 * 65 = 61.49W

(R = e/i = 65 / 0.946 = 68.71 ohms)

Pa = Pp / 2 = 30.75

OK, so my 8008 BB seems to exceed the factory "spec" of 26Wpc maximum Class A power (into specially chosen resistance) by just a tad. But the amplifier idle bias current recommended by tech support is more than 2 times lower. So an amplifier with that level of bias could not meet factory specification even with any specially chosen resistance. There are two possible interpretations:

1) The factory idle bias current was actually much higher than specified in the memo (shown in earlier post 12mv inner channel and 8mv outer channel), and just short of what my amplifier has.

2) The factory "maximum class A power" specification was actually a peak power specification as well as specially chosen resistance specification.

I think I believe interpretation (2). Assuming that is correct, what would would the bias current need to be?

Pp = 26W, therefore 2Ib = 26 / 65 or Ib = 26/130

Ib = 0.20a

Therefore emitter resistor voltage for 6 transistor pair 8008bb would be

0.33 * 0.2 / 6 = 11mV

Just a tad below the "inner channel" specification of 12mV emitter resistor voltage recommended by Aragon technical support. But it seems the outer channel specification of 8mV would never make it, nor would the same values get there with the 4 transistor pairs in an Aragon 8008 ST:

0.33 * 0.2 / 4 = 16.5mV

So even with two tricks used to get the maximum specification, it only applied to the inner (left) channel on an Aragon 8008 BB, unless the bias was set higher than specified in the memo (which is possible, it's quite possible that tech support recommended an especially conservative bias to prevent subsequent failures for people who already had one problem).

****

My Aragon 8008 BB has 3.58 watts Class A power into 8 ohms, and 1.79 watts into 4 ohms, and therefore meets the "first watt" requirement popularized by Nelson Pass, whereas a factory specified unit would not.

My amplifier is biased more like an Aragon Palladium amplifier (just slightly more, actually). But since those amplifiers are bridged monoblocks, and mine is not, that actually gives me another advantage in Class A power at 8 ohms and less, since each half of a bridged amplifier sees only half of the total load.

So mine is actually better than an Aragon Palladium. It's an Aragon Platinum!

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