[JohnnyWash1]

I was playing with my models in WinISD and noticed that my cone excursion goes up with increased watts. That makes sense, as more power would mean more excursion. However, the excursion curve didn't change, which I understand as the cone excursion is tied to frequency, and then is amplified by the watts. Given all of this, my incredibly stupid question that I know the answer to but must hear it from somebody else is:


Does an amplifier rated at 650w RMS ALWAYS put out 650w? Or is the wattage controlled directly by the volume knob?

 

[N8DOGG]

The amount of current allowed to flow is based on the size of the signal from the input source. A large signal causes more current to flow and results in more amplification than the smaller signal. The 650W is the maximum that amp is rated to output, so the wattage can vary from 1 watts to 650 watts depending on the signal given.

 

[JohnnyWash1]

So even though it says 650W RMS it is not always putting out 650W?

 

[notnyt]

No. If there is no input signal, it is not amplifying anything, so it is not putting out any power. If there is a small input signal, it will only be a small output signal. The max the amplifier can output as per that rating is 650.

 

[mtg90]

Just think about it, if the amp is on and there is no signal, as in the woofer is not making any noise is the amp using up any power to keep the woofer silent. NO, it must then be leaking those 650 watts.

I would suggest putting a bucket under you amp (or under the subwoofer if using a plate amp) to catch all the leaking amp power.

 

[beastaudio]

reverse the polarity to that bucket and route it to another sub, double the output! I did a paper on going green not too long ago and mentioned this exact approach

 

[JohnnyWash1]

Loving these responses guys, keep them coming!

Like I said, I knew it was a stupid question, but these need to be answered too, sometimes.

 

[A9X-308]

Amplifiers do what their name implies - they amplify a signal, namely the input voltage.


A 650W/8Ω at full power has a voltage at the speaker terminals of 72.11Vrms and would probably have a gain of 60x. So if you put a test tone of 1.2Vrms (72/60) into the input of the amplifier, did not connect a speaker, and measured the voltage at the speaker terminals, you'd see 72Vrms, but almost zero power. The meter will be very high impedance, probably close to 1GΩ.


When you now connect an 8Ω load to the terminals, 650W will be applied to the load. The current in the load will be

I = V/R = 72/8 = 9A.


A solid state amplifier simply supplies the voltage to the output terminals, and the load determines how much current flows*. If the speaker's impedance dropped to 4Ω at some point, it would pull 18A from the amplifier.


Power is voltage x current so 72V x 9A = 648W. (rounded because I'm lazy and used 72V, not 72.11V).

For a 4Ω load 72V x 18A = 1296W.


Now the source, whatever it is has some way of controlling the signal level, in this case the voltage. So if the test tone were reduced to 0.1vrms, the output voltage at the speaker terminals would be 6V (0.1x60), the current in the 8Ω load 0.75A (6/8) and the power supplied to the load 4.5W. So as you can see, the only thing that controls the power into a given load with a given amp is the signal level, and that is determined by the source material and the volume control (or gain) setting.


* All amplifiers have a current limit somewhere, and most do not double their output into half the load as in this example.

 

[JohnnyWash1]

Just what I was looking for! Thank you!