Everybody seems to get this wrong. Nelson Pass explains a hypothetical case very clearly, but when it comes to his own amp, he is wrong too (OK, just very slightly wrong, you could call it rounding error compared with the liberties most manufacturers take).
I think what's happening is there are two definitions of Class A. The textbook definition (linear operation...to his credit Nelson Pass uses that one), and the stretched to cutoff definition, and most people use that.
Start with my Aragon 8008BB idling at 160W, with heatsink temperature at 132 degrees F (I'm sure it shouldn't be hotter than that). First of all, since this is a stereo amplifier, that means the idle is 80W per channel.
The RMS output maxes around 45V, which means the rails must be about 67V, about right for a 250W into 8 ohms amplifier (which it is, the 200W spec into 8 ohms is conservative in order to allow the doubling into 4 ohms). Update: the PS schematic says 70 volts, I will correct the following analysis for that.
Assuming that all the quiescent power is dissipated in the output transistors (which would be a limiting case maximizing bias current and therefore Class A operation), this means the current running through from the top rail to the bottom rail for each channel is 80W / 140V or 0.57 amps. If you simply summed all the transistor currents, you'd get 1.14 amps, but I don't think that's a helpful way to think about this, since half are in series with the other half.
In Class A operation one side goes from 0.57 amps to 1.14 amps while the other goes to zero. So the peak output current in class A is 1.14 amps. That makes the Class A peak to peak power 1.14 * 1.14 * 8 or 10.4 watts peak to peak, or 5.2 watts RMS. Very far from the 26wpc Aragon claims.
I suspect this is because cutoff to zero doesn't actually happen on one rail side until the other side is conducting considerably more than 2x the standing bias current. So Aragon/Mondial/Klipsch hooked up scope, and observed the level where cutoff to zero was happening on one side. But this is non-linear operation to cutoff, and wouldn't work without a push-pull feedback amplifier design.
But it gets worse. Not all power is dissipated in the output transistors. The actual bias specified for the amp (in an unconfirmed memo at DIYAudio) is 12 mv across 0.33 ohm resistors on the inner channel, 8mv across 0.33 ohms on the outer. That would mean, for the inner channel, the bias current per transistor is 0.036 amp, or 0.22 for all six on one side. That would mean 0.44 amp peak class A current, or 1.6 peak to peak class A watts, or 0.8 RMS Class A watts. BTW, the power transistor dissipation for that channel would be 30.8 watts, with 20.5 watts for the other channel, for a total consumption (not including other circuitry) of 51 watts. It is hard to imagine the remaining 109 watts dissipated elsewhere, so I'm sure my amplifier must be drawing more bias current than that!
My amp bias seems may be similar to the Palladium II according to the memo, which is specified as 25mV for the inner channel, which would correspond to bias per rail side of .456 amp, and therefore (assuming +/- 70V rails) per-channel dissipation of 64 watts in the power transistors of that channel, and 51 watts in the outer channel having specified bias voltage of 20mV.
Update: it looks like my amp is indeed close to the Palladium specification. In the outer channel I noticed bias voltage around 22 mV after 5-10 minutes of warm up when the total power dissipated was 166 watts. Unfortunately, not thinking them important, I didn't write these numbers down, so this might be a somewhat inaccurate recollection. Later, after playing the amp for a half hour, and 20 minutes of stabilization, the recorded bias voltage was 14mV with power of 132 watts. I didn't measure the 6 hour warm up bias, where the quiescent dissipation reaches 160W, and possibly having the cover off for measurement affected the results, but 20mV +/- 2mV would be consistent with these findings. The palladium specifications would account for 115w of quiescent power, the where does the remaining 45w go? Power supply transformers and filter caps, and two large driver boards could easily account for that. Solid state preamps can use as much as that. Also I may have larger rail voltage than spec, since my AC power is around 123V.
I'm thinking that the memo from Tech Support does not necessarily reflect what the factory did. Tech support was possibly giving out more conservative bias setting numbers than required for two reasons: (a) concern that repair shops wouldn't do the proper days long burn-in as done at the factory, and (b) those people who have already had trouble with their amps, which is likely why they would need servicing, should probably use lower bias to increase the safe operating area of the amp. In short, Aragon didn't want to have to see those amplifiers again. On the other hand, they want the amps at the local audio salon to blow away everything else with superior high bias sound quality.
What I'm seeing could also be a result of bias drift or failure of some kind. Drift seems more likely than failure. Looking at the bias circuit in the 8008 schematic, there do not appear to be any electrolytic capacitors which would otherwise be a potential source of partial failure. The bias seems to be mainly controlled by semiconductors, which would either work or be broken and not work at all.
For Nelson's hypothetical amp rail to rail bias of 0.75a which is 100W dissipation across a rail to rail voltage of 130 (+/- 65V), I get the same 18W pp or 9Wrms he does. He gets this exactly correct in the hypothetical example, though he doesn't mention that this hypothetical example must be a mono amplifier, or that his X250 is a stereo amplifier and therefore dissipates only a bit more per channel.
Now lets think about the X250 amplifier which idles at 270W, which means 135 per channel, assuming 65V rails (which is possibly too low). That means 1.04 amp from rail to rail. Peak to peak power is therefore 2.08 * 2.08 * 8 or 35 watts, and RMS power is 17.5 watts. Not far from the 40/20 he claims.
But the rail voltage is probably higher and the rest of the circuitry dissipates enough to make it more inaccurate.
Perhaps he meant the X250.5 amplifier (but forgot to say so). It idles at 350W, which means 175 per channel, which means 1.35 amps from rail to rail, which means 58W peak to peak and 29W RMS, assuming the rails weren't also bumped up a bit from +/-65V. Accounting for higher rails and the rest of the circuitry would bring this right in line with 40/20. But this amplifier wasn't sold until several years after the post I linked above.
One other thing to notice is that because a square law applies, Class A power goes up as the square of the bias current. A 1 amp bias current would produce 4 times as much class A power as a 0.5 amp bias current.
I think what's happening is there are two definitions of Class A. The textbook definition (linear operation...to his credit Nelson Pass uses that one), and the stretched to cutoff definition, and most people use that.
Start with my Aragon 8008BB idling at 160W, with heatsink temperature at 132 degrees F (I'm sure it shouldn't be hotter than that). First of all, since this is a stereo amplifier, that means the idle is 80W per channel.
The RMS output maxes around 45V, which means the rails must be about 67V, about right for a 250W into 8 ohms amplifier (which it is, the 200W spec into 8 ohms is conservative in order to allow the doubling into 4 ohms). Update: the PS schematic says 70 volts, I will correct the following analysis for that.
Assuming that all the quiescent power is dissipated in the output transistors (which would be a limiting case maximizing bias current and therefore Class A operation), this means the current running through from the top rail to the bottom rail for each channel is 80W / 140V or 0.57 amps. If you simply summed all the transistor currents, you'd get 1.14 amps, but I don't think that's a helpful way to think about this, since half are in series with the other half.
In Class A operation one side goes from 0.57 amps to 1.14 amps while the other goes to zero. So the peak output current in class A is 1.14 amps. That makes the Class A peak to peak power 1.14 * 1.14 * 8 or 10.4 watts peak to peak, or 5.2 watts RMS. Very far from the 26wpc Aragon claims.
I suspect this is because cutoff to zero doesn't actually happen on one rail side until the other side is conducting considerably more than 2x the standing bias current. So Aragon/Mondial/Klipsch hooked up scope, and observed the level where cutoff to zero was happening on one side. But this is non-linear operation to cutoff, and wouldn't work without a push-pull feedback amplifier design.
But it gets worse. Not all power is dissipated in the output transistors. The actual bias specified for the amp (in an unconfirmed memo at DIYAudio) is 12 mv across 0.33 ohm resistors on the inner channel, 8mv across 0.33 ohms on the outer. That would mean, for the inner channel, the bias current per transistor is 0.036 amp, or 0.22 for all six on one side. That would mean 0.44 amp peak class A current, or 1.6 peak to peak class A watts, or 0.8 RMS Class A watts. BTW, the power transistor dissipation for that channel would be 30.8 watts, with 20.5 watts for the other channel, for a total consumption (not including other circuitry) of 51 watts. It is hard to imagine the remaining 109 watts dissipated elsewhere, so I'm sure my amplifier must be drawing more bias current than that!
My amp bias seems may be similar to the Palladium II according to the memo, which is specified as 25mV for the inner channel, which would correspond to bias per rail side of .456 amp, and therefore (assuming +/- 70V rails) per-channel dissipation of 64 watts in the power transistors of that channel, and 51 watts in the outer channel having specified bias voltage of 20mV.
Update: it looks like my amp is indeed close to the Palladium specification. In the outer channel I noticed bias voltage around 22 mV after 5-10 minutes of warm up when the total power dissipated was 166 watts. Unfortunately, not thinking them important, I didn't write these numbers down, so this might be a somewhat inaccurate recollection. Later, after playing the amp for a half hour, and 20 minutes of stabilization, the recorded bias voltage was 14mV with power of 132 watts. I didn't measure the 6 hour warm up bias, where the quiescent dissipation reaches 160W, and possibly having the cover off for measurement affected the results, but 20mV +/- 2mV would be consistent with these findings. The palladium specifications would account for 115w of quiescent power, the where does the remaining 45w go? Power supply transformers and filter caps, and two large driver boards could easily account for that. Solid state preamps can use as much as that. Also I may have larger rail voltage than spec, since my AC power is around 123V.
I'm thinking that the memo from Tech Support does not necessarily reflect what the factory did. Tech support was possibly giving out more conservative bias setting numbers than required for two reasons: (a) concern that repair shops wouldn't do the proper days long burn-in as done at the factory, and (b) those people who have already had trouble with their amps, which is likely why they would need servicing, should probably use lower bias to increase the safe operating area of the amp. In short, Aragon didn't want to have to see those amplifiers again. On the other hand, they want the amps at the local audio salon to blow away everything else with superior high bias sound quality.
What I'm seeing could also be a result of bias drift or failure of some kind. Drift seems more likely than failure. Looking at the bias circuit in the 8008 schematic, there do not appear to be any electrolytic capacitors which would otherwise be a potential source of partial failure. The bias seems to be mainly controlled by semiconductors, which would either work or be broken and not work at all.
For Nelson's hypothetical amp rail to rail bias of 0.75a which is 100W dissipation across a rail to rail voltage of 130 (+/- 65V), I get the same 18W pp or 9Wrms he does. He gets this exactly correct in the hypothetical example, though he doesn't mention that this hypothetical example must be a mono amplifier, or that his X250 is a stereo amplifier and therefore dissipates only a bit more per channel.
Now lets think about the X250 amplifier which idles at 270W, which means 135 per channel, assuming 65V rails (which is possibly too low). That means 1.04 amp from rail to rail. Peak to peak power is therefore 2.08 * 2.08 * 8 or 35 watts, and RMS power is 17.5 watts. Not far from the 40/20 he claims.
But the rail voltage is probably higher and the rest of the circuitry dissipates enough to make it more inaccurate.
Perhaps he meant the X250.5 amplifier (but forgot to say so). It idles at 350W, which means 175 per channel, which means 1.35 amps from rail to rail, which means 58W peak to peak and 29W RMS, assuming the rails weren't also bumped up a bit from +/-65V. Accounting for higher rails and the rest of the circuitry would bring this right in line with 40/20. But this amplifier wasn't sold until several years after the post I linked above.
One other thing to notice is that because a square law applies, Class A power goes up as the square of the bias current. A 1 amp bias current would produce 4 times as much class A power as a 0.5 amp bias current.
woy
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