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
Thanks Al, you're really helping a lot. So is everyone else too. It's probably what I expected, I'm fine listening to jazz and vocals at normal to slightly louder volumes, but when I want to play heavy classical pieces to stir me up, I should be aware that this amp does not have large reserves. I still think the oscilloscope is cool and I may get one to play with.
Your speaker isn't a resistor. The REAL part of the complex impedance value is what is doing "work (making music). So be very careful to use the impedance as a resistive load...it is far from that. Speakers are only 5% efficient, so that means the majority of the impedance is imaginary in nature and does not do work.

A reasonable SPL is near 85 dB with 1 watt at 1 meter with a 1 KHz tone. Seems good to me. But, as you increase volume or decrease the frequency, power requirements go up dramatically. To not clip peaks on music (test tones are not dynamic) you aften time need 10 times the average power.

It is this dynamic power requirement that demands attention. When music moves from 1 watt to two watts average, for instance, you need an amp ten time bigger than the last one! A rule of thumb is every 3dB average SPL increase needs twice the power as the previous level. Most music will NEVER see a 30 dB dynamic range for this very reason. No amp can manage it. With digital you could do it, but should you?

If you listen to "normal" SPL around 83 dB and 93 dB peaks (where I listen on most music, and with typical 10dB dynamic range) with 92 dB SPL rated speakers it looks like you should have decent headroom with 30 watt continuous amps as they usually provide more than the instantaneously.

Rower, thanks for your comment, but I disagree with some of your statements:
Speakers are only 5% efficient, so that means the majority of the impedance is imaginary in nature and does not do work.
Much of the inefficiency reflects real (resistive) impedance, that consumes power but converts most of it into heat, rather than sound.
When music moves from 1 watt to two watts average, for instance, you need an amp ten time bigger than the last one! A rule of thumb is every 3dB average SPL increase needs twice the power as the previous level.
This statement is self-contradictory. An increase from 1 watt to 2 watts IS a 3db increase (as is an increase from 10 watts to 20 watts), and requires twice as much amplifier power (as the second sentence indicates), not an amp that is ten times bigger.
Most music will NEVER see a 30 dB dynamic range for this very reason. No amp can manage it.
I could show you waveform diagrams on my computer of the Sheffield Lab recording of Prokofiev's "Romeo and Juliet," which clearly depict a difference in volume between the loudest notes and the softest notes of approximately 55 db. That corresponds to a power ratio of 316,000 times. At my listening position, the softest notes are around 50 db, and the loudest are about 105 db. My 65W amp and 98 db speakers have no problem at all dealing with that. MANY other symphonic recordings in my collection EASILY exceed 30 db of dynamic range.

The 10 db typical dynamic range you refer to is probably typical of (or even greater than) the dynamic range of the majority of rock recordings that are released these days, but does not apply to a lot of other kinds of material.

Regards,
-- Al
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.
The dynamic range reference is from the AVERAGE SPL, not the minimum. So if your avevrage SPL is 85dB, your peak dynamic range will be 115dB, not a value most systems can manage. Music has to be recorded to be at a comfortable volume WITH the expected dynamic range. If you keep turning down the volume to increase the dynamic range...sooner than later you can't enjoy the music.

...This statement is self-contradictory. An increase from 1 watt to 2 watts IS a 3db increase (as is an increase from 10 watts to 20 watts), and requires twice as much amplifier power (as the second sentence indicates), not an amp that is ten times bigger....
You're right, it should be twice, not ten. Every 3dB is twice the power.

The "real" part of the impedance component is the entire vector sum of the X-over and the driver both. The entire speaker is measured, not just a part of it. Most energy going into a speaker never makes a peep. "Real" power vectors are not reflected, that's why you want them to be large (but they aren't)...they do work. Some of that precious little energy is wasted as heat as you say, but the majority is imaginary in vector.

And no, you can't remove imaginary components of an impedance curve by physicaly moving your physicaly.