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
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.
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.
10-01-12: Rower30
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.
I had indicated that the average SPL of the recordings I referred to were "perhaps in the low 70's." I doubt that anyone would want to play them at an average level of 85 db, with peaks of 115-120 db. That is simply too loud. As a point of reference, as noted in this thread 8 hours is the limit of permissible continuous exposure to 85 db, beyond which hearing damage can be expected to occur.
Some of that precious little energy is wasted as heat as you say, but the majority is imaginary in vector.
What do you base that on?

That would say that a speaker whose impedance has small phase angles across the audible frequency range would approach 100% efficiency. My understanding is that most speakers do not have as much as 10% efficiency. You cited a figure of 5%. And it is rare for a speaker to have phase angles that exceed or even approach 45 degrees across broad parts of the spectrum (although that can occur across narrow ranges of frequencies). Meaning that their impedance is mostly resistive. That can be seen in the measurements that are provided by John Atkinson in Stereophile's speaker reviews.

As an additional point of reference, note in this Wikipedia writeup that the acoustic power radiated by a jackhammer is all of about 1 watt! If the 50 or 100 electrical watts or thereabouts that may be sent into a speaker at times were reduced to that kind of acoustic power level (or less) primarily as a result of non-resistive impedance characteristics (as opposed to conversion to heat), it would say that the impedance would have to be almost entirely either inductive or capacitive (i.e., having phase angles approaching +90 or -90 degrees) across nearly the entire frequency range. Which, frankly, is nonsensical, as well as being completely inconsistent with JA's measurements.

Kijanki, LOL :-)

Regards,
-- Al