Why Do We Use Big Drivers For Low-Frequency Material?

It’s easy to say that we have to move more air, but there’s more to it.

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The opinions expressed are mine only. These opinions do not necessarily reflect anybody else’s opinions. I do not own, operate, manage, or represent any band, venue, or company that I talk about, unless explicitly noted.

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There’s a certain intuitiveness to the idea that a subwoofer driver (especially one that radiates directly, as opposed to being horn-loaded) is big. Or rather, that the subwoofer driver has a large diaphragm relative to a high-frequency driver. If you want a low-frequency noise, and you want it loud, it just makes sense that you need something big for making it. Bears, for instance, have lower voices than doormice.

For the average person in entertainment, I don’t really find satisfaction with the standard explanation for why we use big drivers to produce LF information. We say things like, “Ya gotta move more air, dude!” and then move on. Sure we have to move more air, but WHY? It doesn’t help that the topic seems to be avoided by many sites that talk about sound. My guess is that it probably has something to do with the physics being more hairy than a lot of audio humans are ready for. It’s the kind of material that you’d expect to find in acoustical engineering classes, as opposed to a live-sound engineering course. (Acoustical engineering is “classical” engineering, whereas being a sound engineer for entertainment emphasizes equipment operation.)

As such, even folks like me end up “feeling around in the dark” in regards to the question. We know that there’s more to it all, but how it works is tough to piece together.

This article is all about me trying to piece it together. A big “thank you” is due to Jerry McNutt, an honest-to-goodness Product Design Manager at Eminence Loudspeakers. Two years ago(!), he was kind enough to answer some of my questions about this topic, and I’ve been chewing on those answers sporadically since then.


Please be aware that this is a “best attempt.” My conclusions may not be exactly correct, but I don’t have an easy way to really verify them. Treat this all as food for thought seasoned with at least one grain of metaphorical salt.

Sound Intensity vs. Frequency

Intensity is a measure of power over area, or watts applied per square meter at the observation point. Most of us don’t think of sound level in terms of intensity as defined by physics. We’re used to dB SPL. Conversions are definitely possible, but that’s not the point here. The point is that intensity does relate to frequency, and greater intensity means that something is perceived as being louder.

If you want to actually calculate intensity of sound with real units, there’s a fair bit of math involved in figuring out how to do so. The end result of all that figuring still looks a bit intimidating to those of us used to moving no more than three terms around. According to the physics.info site:

I = 2π^2ρƒ^2v∆x^2max

But…if all that’s desired is to make comparisons regarding how intensity varies with frequency, everything that isn’t “ƒ” can be set to a value of 1:

I (abstract comparison) = ƒ^2

If we start with good ol’ 1 kHz as a reference point, the abstract comparison intensity is 1000^2, or 1,000,000. If we go down an octave, the frequency is 500 Hz. Five-hundred squared is 250,000.

In other words, if everything else but frequency is held constant, then going down an octave means the sound intensity drops by a factor of four.

To really drive this home, let’s consider the frequencies of 60 Hz and 6000 Hz. We would generally expect the low side to be produced by a big ol’ subwoofer, and the high side to be in compression-driver territory.

I (abstract comparison) = ƒ^2 = 6000^2 = 36,000,000

I (abstract comparison) = ƒ^2 = 60^2 = 3,600

36,000,000 / 3,600 = 10,000

In terms of power, a factor of 10,000:1 is jaw-dropping. Pushing an itty-bitty compression driver with one watt is common. Pushing one with 10,000 watts, well…

Two vs. Four

From the above, I think you can get an idea of the importance of “moving more air” to keep everything manageable. We have to do something to counteract the intensity drop from lower frequency. It’s actually a multi-factor problem, of course, because real-life tends to be that way. We can move more air by making a driver undergo longer excursion (forward/ back movement), but there’s only so much that’s doable. Closely related to that is more drive power. That’s good, but again, there’s only so much that’s reasonable. If we’re going to shove more air molecules around, we need to also have more diaphragm area.

One of the best tidbits I got from my conversation with Mr. McNutt was in regards to the advantage of using a squared term instead of a linear term. Doubling a driver’s excursion (the linear term) certainly gets you something, but doubling the driver radius (the squared term) gets you much more.

For the sake of argument, let’s simplify a loudspeaker driver’s diaphragm into being a piston that pushes hydraulic fluid around. We’ll conveniently use a driver that starts with 1 mm of excursion, because it will make the math easier. My guess is that most compression drivers can handle rather less excursion than that, but this is just an example. The radius will be 25.4 mm (that’s like a 2″ diameter compression driver, if you want to visualize it).

Displacement Volume = Area * Excursion

Displacement Volume = (pi*25.4mm^2) * 1mm = 2027 mm^3

If we double the linear term to 2 mm of excursion, the displacement doubles to 4054 mm^3. Nice, but if we double the squared term and leave the excursion alone:

Displacement Volume = (pi*50.8mm^2) * 1mm = 8107 mm^3

That’s a fourfold increase in the amount of fluid the piston moved. When it comes to loudspeakers, making a small driver have a very long excursion is impractical, but making a driver with a larger surface area is commonplace. So, if we consider an 18″ diameter subwoofer (228.6 mm radius) that can handle an excursion of 8 mm:

Displacement Volume = (pi*228.6mm^2) * 8mm = 1313386 mm^3

That’s 648 times more displacement, gotten mostly by making the driver bigger.

I can’t say exactly how all this works out with real drivers, real air, and the real equation for intensity. However, even with rough approximations it seems pretty clear that it’s much easier to move a lot more air if you have a big diaphragm available. The squared term is very important in getting the necessary results.