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Driver comparison (1 Viewer)

Apr 24, 2003
I have a question. I've been looking at drivers, and something puzzles me.

I look at this woofer: http://www.partsexpress.com/pe/showd...ID=16239&DID=7

and I look at the Tumult: http://www.adireaudio.com/diy_audio/...ire/tumult.htm

The Peerless has about 3.5dB more efficiency at 1W than the Tumult and a lower resonance (and is smaller), while the Tumult has higher power and excursion.

Shouldn't the Peerless reach lower frequencies than the Tumult? And wouldn't they play comparably loud at max power? Could someone explain to me the technical reasons why the Peerless has a higher efficiency with a lower resonance in a smaller woofer?

Michael R Price

Jul 22, 2001
Jason, you ask a valid question and it is true that the Peerless woofer can probably match the Tumult's output at higher frequencies. But once we get under around 50Hz, the story changes.

Firstly the Peerless woofer has a much lower Q, so it will roll off sooner for any given box size (and be less efficient at lower frequencies than the Tumult). Also, clean output at low frequencies (say, under 40Hz) tends to be limited by excursion capability much more than by amplifier power. The Tumult displaces 4[?] times as much air linearly, so its output capability at lower frequencies is much greater than the Peerless, provided there is enough amplifier power to reach near the excursion limit. I think there are other technical reasons, but other more knowledgeable folk will chime in soon.
Apr 24, 2003
Thanks. What about why the higher efficiency of a smaller driver with a smaller cone? I still don't understand that.

Justin Ward

Supporting Actor
Jun 6, 2002
My understanding is that when drivers are designed compromises must be made. For example a Tumult has huge excursion likes small boxes but has low sensitivity. The Tempest has good sensitivity, good excursion but requires a large box. I believe that when dealing with excursion, efficiency and box size, when two of the attributes are locked the other is basically determined.

Maybe the experts can explain why this is so.
Oct 1, 2002
Very good questions, and very good answers thus far. I just want to add that although the Peerless does have a better efficiency at 1 watt, once the sub starts moving that changes. My guess is that the Peerless unit has a parabolic BL curve, so it's efficiency will go down as it loses BL with excursion. The Tumult keeps the same BL throughout most of it's excursion, so it will stay roughly the same.

It also looks like they used a standard linear spider. This means that their Kms curve is most likely parabolic as well, meaning Fs, Qts, and Vas will all change considerably with excursion. This will cause higher distortion in many ways including effecting frequency response and box alignment.

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