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 2 responses by kijanki

Speakers are only 5% efficient, so that means the majority of the impedance is imaginary in nature and does not do work.

Everybody knows that speakers play on one wall better than on another. It is because 90 degree rotation eliminates imaginary (parasitic) impedance.
And no, you can't remove imaginary components of an impedance curve by physicaly moving your physicaly.

I'm sure I can. For instance when I rotate my incandescent bulb 90 degree counter-clockwise they become dim but when I turn them 90 degree clockwise they're lit again. I'm sure it is this imaginary impedance taking over.