I'm sure SH, for one, will have an answer for this.
Is there a method for a guy with normal audio components to produce a 1 watt signal so I can duplicate the industries standard method of measuring speaker efficiency.
Contacting the manufacturer is not an option, as I made em myself and I don't know!!
UNLESS: I do know the published sensitivity of each driver in the speaker, so would the result be an average of all the drivers?
Which brings up another question.
At what frequency does a manufacturer measure the sensitivity? If he measures at 40hz with a 10" driver in a 3-way, and that drivers sensitivity is 90Db, he will get a different result than measuring at 10Khz if the tweeters sensitivity is 93Db.
That said, he must be averaging right ? or does the passive crossover mess with this?

To measure sensitivity, a manufacturer (I'm going to outline the procedure used while I worked for Altec Lansing) sets up a loudspeaker inside an anechoic test chamber. An anechoic chamber is basically a room that has large wedges of sound absorbing material on it's walls so that there are absolutely no reflections of the sound. It simulates being outdoors in free space, like being suspended 100' in the air with nothing around you.

A microphone is placed one meter in front of the speaker and the chamber is sealed. Then the speaker is fed a constant voltage signal of 2.83 volts (assuming an 8 ohm speaker) that sweeps from 20Hz to 20,000Hz. As it does, the output of the speaker is recorded on graph paper that has it's vertical scale calibrated in decibels sound pressure level, and the horizontal scale calibrated in frequency. (I'm sure this is done by computer now.)

The resulting curve is basically a frequency response curve of the speaker, but they can tell the absolute ouput of the speaker in db sound pressure level verses frequency. There is usually a frequency that is standardized as where the sensitivity is taken, but sometimes the entire frequency spectrum is looked at and 'eyeball' averaged for sensitivity.

In other instances, pink noise is used to measure sensivity, using the same physical setup.

You can approximate what they are doing at home by using a source of pink noise from a test DVD, or the internal noise generator of the 950 for that matter. This noise is usually filtered to center around 1,000Hz, but it will give a reasonable indication of sensitivity. Feed the speaker with this signal and set the voltage at the input termainals of the speaker to be 2.83 volts AC, measured on a fairly good meter (a multimeter will do).

Using the RadioShack sound level meter, set it for "C" weighting, and "slow" response, and place it 1 meter in front of the speaker, on axis with the center of the cabinet, with the microphone facing the ceiling.

The reading in decibels is approximately the sensitivity of the speaker for one watt, at one meter.

For best results, the room should be pretty dead, acoustically.

"Then the speaker is fed a constant voltage signal of 2.83 volts (assuming an 8 ohm speaker)"

Thanks SH, your explanation was very interesting.
How important is the accuracy of the voltage signal and are you ballparking an 8 ohm speaker?
My actual measured impedances are:
Front L&R =3.5 ohms
Center = 4 ohms
Surrounds = 7 ohms

I think I can find the published figure for the latter two in their respective documentation.

There are two basic schools of thought on this one. The most previlent (and probably correct) is that you compute the input voltage based on the approximate average impedence with a preference to a standard figure.

So, for a speaker with a 'real' impedence of 7 ohms most folks would use ~2.83 volts which is 1 watt at 8 ohms.

A speaker with 3-5 ohms would probably get tested as a 4 ohm load, etc.

Some outfits just use ~2.83 (2.82843) volts since speakers are not, strictly speaking ( ) power referenced devices. They are designed to deliver response linear to voltage, not power. They draw variable power depending on all sorts of variables, including signal frequency and other things like what the moving parts are doing at the instant.

It's OK, since amps are (mostly) designed to approximate perfect voltage sources.

For a 4 ohm system, 2 volts into 4 ohms gives 0.5 amp, so (0.5 x 2) = 1 watt.

Are you sure you're not measuring DC resistance?

Impedance is not easy to measure with really basic instruments. Generally an 8 ohm impedance speaker will measure about 3.5 ohms DC resistance, give or take.

The formula for watts is:

driving voltage squared, devided by the impedance in ohms=watts.

You want to drive the speaker with 1 watt.

You should try to get close to the correct drive voltage, but as I stressed in my original post, this will only give you an approximation of what the factory can measure.

[This message has been edited by soundhound (edited January 28, 2003).]

I decided to try soundhounds test for measuring efficiency and it seems to work very well using the test tone of the 950. I coundn't get exactly 2.83 VAC because the signal varied about +/- .1 VAC but the sound meter showed 88db compared to the factory spec of 86db.
While I was at it I decided to try and measure a slight hum that comes from the speakers if I put my ear to them.
The readings were:
1.8 mV 950 off, 770 off
3.5 mV 950 off, 770 on
3.9 mV 950 on, 770 on
Going by soundhounds formula the amp itself is producing .5db on it's own which I assume is do to power cords and cable TV lines which do have a lot of physical contact behind my unit. Like I said this hum is only noticable with your ear to the speaker.

I did have a question on wattage being output by the amp. If my speakers have a nominal impedance of 6 ohm and a minimum of 3.6 ohm how would you ever figure the wattage not knowing the impedance at any given moment? I played some music and used my meter to capture the highest voltage reading which was 9.63 VAC. That works out to 15.5 watts @ 6 ohm and 25.8 watts at the 3.6 ohm. Is there another way to get the wattage without guessing what the the impedance is at any given moment? I assume the amperage output from the 770 is dependent on the load the speakers is imposing on the amp. Correct or not?




[This message has been edited by Keta (edited January 30, 2003).]

You might try to plug in 6 ohms into the formula to find the voltage required for 1 watt. This might give you a measurement closer to spec from the factory. The difference in impedance across the speakers frequency band is the reason that the factory measures efficiency with a tone that is swept across the whole audio range. With this information, they can arrive at an 'average' efficiency for the speaker.

The lower the impedance that an amplifier drives, the more the current will be. The amplifier puts out a constant voltage (within certain impedance limits), and the load determines the current, which in turn can be used to calculate the watts dissipated. Voltage times current=watts. These are all ohms law calculations where you can calculate either the voltage, the current, or the resistance (impedance).

You could use a watt meter, but in reality speakers are voltage referenced devices, so worrying about real watts isn't to worthwhile IMO.