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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
mechans
3,048 posts
 
My non mathematical answer is that 60 watts is plenty but it kinda depends on the amps output trannies if it is not an OTL. I can tell you that my 80 lb integratedb class A Jadis is also rated at 60 watts and seems relatively impervious to load. If it ios a high quality amp I would suggest you are just fine with that much or even less power.
koestner OP
661 posts
 
Just an FYI, the Belles SA-30 is a solid state amp.
mechans
3,048 posts
 
Yes of course I foorgot as I was writing but your assumption about adequate power is still correct.
almarg's avatar
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almarg
9,670 posts
 
I wouldn't count on that measurement being meaningful without having detailed information on the technical characteristics of the meter, which chances are will not be available. One important parameter being how brief a transient it is capable of capturing. Others would include its frequency bandwidth, and some indication of how it handles non-sinusoidal waveforms (e.g., true rms, or peak converted to sinusoidal rms equivalent, etc.).

That said, the power capability you need will vary dramatically as a function of the dynamic range of the music you are listening to (i.e., the DIFFERENCE in volume between the loudest notes and the softest notes). What I would suggest is that you repeat the measurements with material having the widest dynamic range that you would normally listen to. Many classical symphonic recordings, for example, have vastly wider dynamic range than most rock recordings, and therefore require vastly more power during brief peaks, for the same average volume.

Regards,
-- Al
bombaywalla's avatar
bombaywalla
1,958 posts
 
... hooked up my Wavetek digital multimeter across the speaker posts to read AC volts...
I think that Almarg is on the right track in his 1st paragraph. I was about to write almost the same thing myself.
Pull out the spec sheet for your Wavetek DMM & see what it's bandwidth is.
Most, if not all, DMMs do not have the bandwidth to follow an AC signal & so you are getting a heavily RMS'd reading of a fast moving AC signal. I would, like Almarg, not put much confidence in those readings at all.
Your calc technique is correct tho.
Now, if you can get your hands on an oscilloscope & measure the peak voltages & then use V^2/R, you'll get the true pix....
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