OK, I see I have explained badly. My apologies. I have already built an M2, using a transformer similar to the one you describe. (I like the amp, BTW.) That is not what I am talking about doing here.Analyze the M2 circuit schematic (found in post #1 of this thread) again. You'll discover that "the power demand of the M2 amp" is far greater than 28 watts per channel. Since M2 operates in class A, the power it draws from the power supply is a LOT more than the power it delivers to the loudspeaker load.

You can do a "sanity check" on your calculations by visiting the First Watt website and downloading the owner's manual for the M2. In the specifications table you'll find a line-item that tells you the power an M2 draws from the AC mains. Hint: it's a lot more than 28 watts per channel.

*you calculated 28 watts * 2.5X safety factor = 70 watts for the power transformer. But I suggest that 28 watts is far too low.

If this helps: I used a 400 volt-ampere toroidal transformer to build my (stereo) M2x. That transformer had 2 x 18VAC secondaries at 11 amperes each. I actually planned to use a 300 VA transformer, 2 x 18VAC at 8.3 amperes each . . . . . but those were out of stock on the day I bought. So I paid a little more money and got the 400VA because it was available immediately.

I have two transformers. Each is 160VA, 25V - 0 - 25V center-tapped secondaries. So

*dividing*by the usual 2.5 safety factor, I have something like 60 - 70 watts per channel to play with.

I am considering trying a mini-M2 using these transformers. (With some sort of different front end.) A PSU with such a transformer is going to give +/- 32V or so after rectifying and smoothing. Maybe a little more, but probably not less. This is far more rail voltage than I need.

If I bias at around 800mA then I'll continuously burn 51W on the heatsink. But this should be within the capability of the PSU.

At 800mA bias I have should have around 10W output power in class A. I calculate that as follows. Bias of 800mA means the amp leaves class A when peak current on the output is 1.6A (push-pull, so twice the bias, right?). Into an 8R load that is 8 x (1.6)^2 /2 = 10.24 W. (Dividing by 2 for RMS rather than peak current.)

Does that sound right? (I'm not an engineer, in case it wasn't already obvious.)

My question: If the last calculation is correct then my guess is that I should have more power available in class AB, but how can you calculate/estimate it?

Thanks

Nigel