Current limit onset definition?

Atma, I'm even more confused, sorry. What does the 40 amp have to do with?

Atma, did you mean 40? or 60?
I appreciate you have genuine knowledge on these things.

01-07-15: Ptss
Ack, confused, there are no newer DMA180's that I'm aware of. What do you mean?

He means newer models having different model numbers (DMA-200S, DMA-260, DMA-300, DMA-400). Although as I indicated in my previous post the first three of those models are specified at the Spectral site (perhaps incorrectly) as having 2.5 amp fuses for each channel, for 120 volt operation.

Regards,
-- Al

Ralph (Atmasphere), thanks for your comments also. As a point of information, all of the peak current specs Spectral provides at their site for the recent models as well as for the DMA-180 are described as "peak **output** current" [emphasis added]. So I would have to assume that **if** their wording is accurate those specs are based on the short we have referred to being applied to the output of the amplifier, rather than to the output of the power supply. The numbers, btw, are 60 amps in each of those cases, except for the DMA-400 monoblock which is 90 amps.

Giving it the benefit of the doubt (using the power formula), 90 amps squared is 8100 watts into one ohm (into 2 ohms it would be 16,200 watts...). Looking at the specs we see that the 2 ohm power is no-where near that. So the current cannot be anything to do with the output section! It is how much current occurs when the power supply is subjected to a dead short for 10 milliseconds. It is a measure of the power supply capacity and really not anything else.

By the same measure our MA-2 amplifier, which is an OTL, makes about 60 amps.

Hi Ralph,

Yes, but it is also true that 90 amps into 0 ohms is 0 watts (or, more precisely, some relatively small amount of power since no conductor can be exactly 0 ohms, assuming the temperature is not near absolute zero). Isn't it possible **in the case of a solid state amplifier** that the dead short is placed across the output of the amp, and the current through that short which results from application of a 10 ms signal to the input of the amp is measured?

Best regards,
-- Al