Speaker sensitivity

[Drew in the desert]

Just curious about speaker sensitivity ratings. I think I have an idea, but how is sensitivity measured?

[Ray Garrison]

Not very well.

First, there are two terms that get all crossed up in advertising and marketing. These terms are "efficiency" and "sensitivity". They get used interchangeable, but really mean two different things.

Efficiency is a measure of how good a job a speaker does of turning electrical power provided by an amplifier into acoustical power. A 100% efficient speaker fed one electical watt of power (say 0.35 amps at 2.83 volts, using an imaginary speaker with a pure 8 ohm impedence) would produce one acoustic watt of sound. (One acoustic watt is *VERY* loud, by the way.) There is no such thing as a 100% efficient speaker. The most efficient horns are, uh, I think, on the order of 10% efficient.

Sensitivity is a measure of how loud a sound a speaker will produce given some specific voltage input. It is usually specified as something like "95dB / 2.83 Volt / 1 Meter" which means if you put 2.83 volts into the speaker, and measure the sound pressure level 1 meter away from the speaker, the SPL should be 95dB.

Everybody used to print specs like "95dB / 1 watt / 1 meter". This is not really an accurate measure of either efficiency or sensitivity. Saying the output is "95dB" doesn't really tell us how much total acoustic power we're producing, because we don't know anything about the environment in which that sound is radiating. For some reason manufacturers are starting to print sensitivity specs more often. Problem with this is, saying "95dB / 2.83 volts / 1 meter" doesn't tell us how much power the speaker is sucking out of the amp to produce that 95dB level. If the speaker is actually a pure 8 ohm load, the speaker is amp is pushing 1 watt into the speaker. However, if the speaker's impedence falls to, saw, .8 ohm (Apogee Scintilla) the speaker is actually sucking 10 watts out of the amp. Conversely, if we have a 16ohm Lowther here, the amp is putting out a half a watt. Same spec, three different speakers. One would play at 104dB with a little bitty 4 watt tube amp, another would need 80 watts of solid state power (dumping 10 amps of current) to play at the same level.

All of which goes to say that comparing specs on efficiency or sensitivity from one manufacturer against another is pretty pointless, in anything other than the most general, broad terms.

Within a given manufacturer's line, assuming they use the same rules for measuring all their speakers, you can use these specs to evaluate how much louder the next bigger speaker in a given line might play with the same equipment.

[MrMcGoo]

Ray,

Good explanation!

Most folks do not look at the speaker's impedance when looking at sensitivity. RF-7s go down to about 3 ohms in the bass frequencies and present closer to a 6 ohm load than 8 ohms. Hence the RF-7 is not as easy to run as the 102 db/1w/1m suggests.

Bill

[tubeglow]

OK, here I go again....Ray, I agree with some of what you say but I don't think you quite went far enough. I agree that 2.83v @ 1 mmeter is the standard for effeciency but...you neglected to mention that the standard requires a DB standard such as 20-20,000hz +/- 3db. Or in the case of the Khorns (35-17,0000hz+/-5db). (This is a quote from a Klipsch brochure of the the early 80's). So the published response would be ....104db @ 35-17,000Hz+/-5db.Since it is a given that it takes more *** to drive low frequencies to a loud level than mids or highs, the FR with a given +/- db range is somewhat important.

[Ray Garrison]

tubeglow -

That's kind of my point. Getting any meaningful info out of a sensitivity or efficiency spec requires much more info than any manufacturer provides. You need to know (at a minimum) the complex impedence of the speaker, the characteristics of the room in which the speaker was measured, the nature of the signal that was used to drive the speaker, the details of the positioning of the microphone and something about the pickup characteristics of the microphone, whether we're measuring narrow band or wide band response, whether the measuring was done in a time windowed fashion or not (did the reverbrant field of the room increase the apparent output of the speaker?) and about a bazillion other things.

The old Audio magazine used to print a plot showing how the complex impedence (reactance and resistance) of the speaker under test varied with frequency. Here's their take on the Klipschorn:

As you can see, it's not a simple question to say "How efficient is my speaker?"

[dougdrake]

And wasn't it typically stated that the sensitivity was "x" at 1000hz?

[WMcD]

There is quite a bit going on.

At least the microphone is placed one meter or about 39 inch away. Or more distant and some adjustment of the numbers are done.

Actually, the big problem is the difference between "voltage sensitivty" at 2.82 volts. Versus actually figuring out at every frequency how much power is being delivered and SPL output in terms of input power (true efficency). The variation in impedance is the problem.

Let's start with some basics. The power being put into a load (speaker too) is P= V time V divided by R. (My keyboard is not working on the virgule.)

You may have seen P = volts times amps. True. But by Ohm's law, A = V divided by R. So it is the same thing.

You assume that R is 8 ohms. To make P = 1 watt, you have to find some V times V to equal 8. The magic number is the square root of 8 which is 2.82.

Therefore we can use the figure of 2.82 volts as a kinda standard. We say, if there was an 8 ohm pure resistive load, there would be 1 watt being put into the speaker. This is why specs will say something like "1 watt at one meter (2.82 volts)".

No speaker is a true 8 ohm resistive load. The Nyquest plot presented above shows how very much it is not. Some of it shows the capacitive and inductive effects which change over frequency. The Y axis. Resistance is on the X.

We also see the effect of the autotransformer in the midrange crossover where the load goes way up in value. Follow around the ring and you'll see that in frequencies where the midrange is acting, the resistance is very high, up to about 40 ohms.

OTOH, we can see the very low impedance of the woofer down to 4 ohms, and then also the tweeter at about 8 ohms above 6 kHz. (I can't see it while writing; bear with any minor inaccuracy.)

If you look at the graph for the actual R, it is actually never at 8 ohms. (Take a point on the circle and draw down to the R axis (X) to find R.)

This means that at every point in the frequency response, the speaker is NOT absorbing exactly 1 watt. The mid is absorbing much less. The woofer at 4 ohms is absorbing more. You have to crank the math.

The mid in the K-Horn is more efficent or IOW more sensitive (too loud). The autotransformer works to present less of a load and thus cut down power delivered into it. It is a step down transformer giving less voltage to the K-33 driver.

But what a headache in figuring out electrical power input.

This headache is why the standard is to just assume 8 ohms, and then say that 2.82 volts is going deliver 1 watt.

Now if you look at this math, you will see that a 4 ohm woofer load has a bit of a cheat factor. We're assuming an 8 ohm load connected to our amp. But the 4 ohm woofer is soaking up twice the current, and therefore twice the power.

Is this "fair". (The question mark doesn't work either.) It is if the amp can deliver the same 2.82 volts when heavily loaded. Most can. Some shut down.

Gil

[Griffinator]

Great explanation, William.

[tubeglow]

Ray,

Great post....thanks for the the stuff fom Audio mag. BTW, nice letter to Stereophile. The more information that people have regarding hi-fi, the better off we all are. I was just trying to make the point that the more information a company gives us, the better we are equiped to make a judgement on the stuff that they are trying to sell us. Even then, there is no substitution for hearing stuff with your own ears. Mark

[Trey Cannon]

FYI: We do all thoes test in the chamber using pink noise w/ a 6db crest factor, filtered. We use an RMS Volt meter to set the voltage to as close to 2.83V as we can. Using a B & K measuring amp set to slow on the responce, we look at the output from the speaker at 1 meter. From there we do some math and produce a number.

When doing this test I take many data points, to make sure the result is as right as it can be.

Ray and Gill, very good job explaining this....kudos

[WMcD]

I made a bit of mistatement. At some points there is an 8 ohm R in the load, plus some reactance. And at some points there is is pure resistive load. This is where the curve passes through the X (or R) axis.

The J number is a wonderfully way to convert a near impossible calculus problem into algebra. Don't worry about it.

Al K. and I had a bit of a disagreement a few years ago on the proper name for this type of graph. Contrary to Al's thoughts, I'm quite sure it is a Nyquest plot and not a Smith chart. But I couldn't come up with the correct name at the time.

The plot describes the complex impedance seen at the input to the crossover. We'd have to figure out whether it is A or AA or AK, by the date.

It is interesting to just follow the parts of the curve which correspond to the bass, midrange, and tweeter. We know the crossover points are at about 400 Hz (bass to mid) and 6000 Hz (mid to tweeter).

We can see that between 400 and 6000 Hz there is an increase in impedance. That is the effect of the auto transformer. It comes back down for the tweeter.

- - - -

I use an LMS (DOS) system for testing the home made K-Horn. The testing system will put out a continuous tone at 1000 Hz in a set up mode. I measure the voltage to the speaker and adjust the volume control on the amp for 2.82 volts on an RMS voltmeter.

In fact this is very loud and I use 0.282 volts. By some math, this is 20 dB down from the reference. But it is easy to visually add 20 dB to any plotted curve.

To belabor the point. Let's just assume I'm using 2.82 volts. It you look at the load presented by the speaker at 1000 Hz, it is not 8 ohms. Therefore it is not getting 1 watt of electrical power. In fact, much less. Again the assumption is that we're working with 8 ohms of R.

The rest of the test is to let the system run a frequency response curve. The mike is 1 meter away.

With 2.82 volts, the frequency response curve from 40 to 15000 Hz is bouncing around a bit. However, the average value is 104 dB in that frequency range.

A result like this allows Klipsch to say that the "efficency" is 104 dB at one watt at one meter.

BTW, the vintage Sony amp does not have a problem with a 4 ohm load. It does put out 2.82 volts. Err. Maybe a smidge less where the load drops to 4 ohms, but nothing to worry about.

Now you know what I know.

Gil

[kenratboy]

Trey:

I love having mods that do more than ban people and edit posts to embarass forum members

[Andre_Roy]

Thanks to all of you , im a newbie learning , from all you cool people,

Thanks again EH

[DrWho]

one thing i've wondered a lot about...

why do you say the amplifier is working harder on a 4ohm load versus an 8ohm load? It seems to me that with less resistance, less voltage would be needed to obtain the same current. Is it because the amplifier is going to put out the same voltage independent of the load and thus the current increases? I know I'm directly applying the concepts of direct currents, but isn't it practically the same concept with alternating currents? or is there some math/physics here that I'm overlooking?

an extension to this question (or at least i think it's related)...what is the correlation between the frequency response and how the sensitivity and impedance together change with frequency? It would seem to me that the 3ohm lows would be significantly louder than the 40ohm mids, especially considering that the 3ohms would also be getting that much more wattage (too lazy to do the math). does the sensitivity decrease with a decreasing impedance? (if so, then why?) are there more factors involved?

Feel free to hit me with all the complicated math...worst case scenario, i still have my calc and diff EQ textbooks handy for reference.

[WMcD]

This is difficult to describe without diagrams and a firm grasp of electrical engineering. And, no, it is not a matter here of a.c. or d.c. But d.c. analogies are more easy to explain.

One important issue is whether the amplifier is a true voltage source. That is to say, can it put out the same voltage despite the resistive load. This depends on how true it is to the voltage source anology.

- - -

For example. Let's take a big Die Hard battery from Sears at full charge. That is a pretty good voltage source. It sits at about 13 volts. If you put a little light bulb across it (high resistance and little current), or then the big load the electrical starting motor which cranks over the gasoline engine, (low resistance and high current), the output voltage remain the same. The engine cranks. Battery voltage remains high. You may have a gauge on the dashboard which shows it.

But what happens when the same battery is not charged up totally and is not acting like a voltage source. To some extent, our amps act like half charged automotive batteries.

Lets see what happens with a half charged automotive battery.

The dome light works well enough (a high resistance, little current demand). Gee, this looks like the automotive battery is turning out constanstant voltage. In fact we can turn on an additional load like the radio.

But when we try to crank the starter, and there is a demand for current, the voltage sags. The dome light goes out. The battery sags in voltage because there is too much of a load. The starter motor doesn't crank. Time to find someone with jumper cables.

Much the same goes on with our amplifiers. They are not true voltage sources over all load situations.

It is quite correct that they can maintain voltage when loaded wtih the dome light, and then the additional load of the radio. But if you try the starter, or more lights, voltage sags.

- - - -

You are pretty much on the correct thought about the issue of impedance where the action of the auto transformer (it could be a normal transformer). The K-55 is much more efficent than the rest of the motors and horns in the system. Therefore it make sense that we can use a step down transformer to knock down voltage being put into it.

That is all for tonight.

Best to all,

Gil

[Marvel]

I know my speakers are very sensitive. It takes almost nothing at all to insult them.

Marvel

[dougdrake]

I think the best analogy I heard was from Ray Garrison. I'm recalling this as best I can, but...

He likened the phenomenon of lower resistance causing stress for an amp to a water pump and hose situation. If you have a water pump capable of pushing out x gallons/minute, if you connect it to a proper sized hose it runs along just fine, the motor doesn't overheat from going too fast or from pumping the well dry.

However, if you hook said water pump to a 4' storm drain, which provides virtually no resistance to the pump, the pump may burn out by over-revving and by running dry.

As a runner, my analogy is trying to run down a steep hill - you can't control it and you can end up at a high rate of speed resulting in a very nasty spill. But, on a generally flat surface, you can keep it all good.

[DrWho]

yay it makes perfect sense now, thanks a bunch