Hi DF96, you're right, I did a lot of handwaving and not much math. I wrote "distortion" and "signal-to-noise" when, in fact, I was simply estimating the Ripple-Current gain. I "derived" the transfer function from ripple current input, to loudspeaker current output, by appeal to intuition. I'll derive it again here, this time with less handwaving. A block diagram is attached.

The power supply rectifier(s) charge the power supply filter capacitance Cfilt. Unfortunately, the amplifier builder did not heed Randy Slone's warnings, and so an unwanted series resistance R3 is present. An unwanted noise voltage Vunwanted = (Irect * R3) is present. Notice that if R3 = zero then Vunwanted is also zero. (I've assumed that Z(Cfilt) << Z(R3) for mathematical simplicity)

The transfer function from supply_rail, to amplifier_output, is modeled as a gain block whose voltage gain is "Av". This, in fact, is the amplifier's power supply rejection ratio: Av = PSRR. The loudspeaker connected to the amplifier output converts voltage to current: Iout = Vout / Rload. We now have everything we need to write the expression for the Ripple-Current gain

- Ripple-Current gain = Iout/Irect = PSRR * (R3 / RL)

So if you want your amplifier's Ripple-Current gain to be -90 dB or less, and if your amplifier has got a PSRR of -45 dB like the Blameless amp in Self's book, then you want (R3/RL) to be -45 dB. (R3 < (RL / 178)).

A similar block diagram and a similar analysis applies when the amplifier is operating Class AB or Class B, and the rail currents are time-varying. Now the rail resistor R2 enters the model, and a different noise current is injected at a different place. An expression for the Ripple-Current gain in this second case, is left as an exercise

I observe,

*en passant*, that it may be easier to reduce the magnitude of rectifier current pulses (case 1), than to reduce the magnitude of Class-AB rail current pulses (case 2). If so, the (R2+R3) ripple current gain expression may be the more useful of the two.

. . .