Watts up with that?


I was concerned that my Belles 30 watt Class A amp (SA-30) was not powerful enough for my Montana XP speakers (seven driver 92db at 2 watts due to 4 Ohm). Using the calculation of voltage squared divided by impedance would give you watts, I hooked up my Wavetek digital multimeter across the speaker posts to read AC volts. The meter has a “max” feature so it keeps displaying the highest voltage reading until reset. My speakers have a very flat impedance curve with a low of 3 and a max of 5 Ohms, so I feel pretty safe using the average of 4 Ohms. Upon playing some music at my average listening levels I got a max voltage reading of 2.13 volts. This calculates to just over 1 watt. I then turned up the volume to much louder than I will usually listen and got a max voltage reading of 3.28 volts after a few songs. So with the volume higher than normal, and at the loudest part on the track, I get just under 3 watts being drawn. I still have a lot of watts left! Are my calculations correct? Is this an OK way to measure power? I was thinking I needed a few hundred watts of available power, but it seems I’ve got all I need at just the 60 watts capability (4 Ohm load) of my current amp. Your thoughts please.
koestner

Showing 1 response by zaphodbeeblebrox

Not to change that subject too much, but this sounds very similar to the argument you will get when it comes to those who believe Tube S.E.T. Amps are the best. These low powered Tube S.E.T. Amps only put out 2 to 8 watts but will still player loud and proud with the proper efficient speakers.

Also similar with Pure class A amps, for the most part when people listen to music at a reasonable level they are only pulling a few watts anyway.

It's only after you really crank up the music or play movie soundtrack with lots of full range of sound going on will you start eating up watts. Or if you have some really inefficient speakers that can be hard to drive due to their design.

I could be all wrong about thus, so please feel free to correct me if needed.