Current limit onset definition?

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

No, I don't think so. 90 amps through a 600 watt output section would toast the output devices in a heartbeat. Or less.

The 90 amp figure simply has nothing to do with the output section. It has to do with the power supply.

Ralph,(Atmasphere I think). Why square the 90?
I see nothing about squaring in the power formula.

Why do you say 600 watt output section?

Ptss, the amount of power delivered into a resistive load is:

P = (I squared) x R

where P = power (in watts)
I = current (in amperes)
R = resistance (in ohms)

An alternative formula which produces the same result is:

P = (E squared)/R

where E = voltage (in volts)

Both of those formulas can be mathematically derived from the definition of power (E x I for a resistive load) in combination with Ohm's Law (I = E/R).

Also, the reason Ralph referred to a 600 watt output section is that most of the amplifiers which have been discussed are rated as being able to continuously supply around 600 to 700 watts or so into a 2 ohm load.

Regards,
-- Al