RDO Question: Why doesn't Qms affect SPL?

[bwbass]

I need some help getting my mind around how resistively damped operation works in dual voice coil drivers.

I understand that when just one coil of a DVC speaker is driven, the other coil can be shorted with or without a resistor to create electromechanical drag and change the Qts of the driver. According to Adire's White Paper on RDO [pdf], it's the Rms, and consequently Qms that change as resistance changes, but Qes stays the same.

And since sensitivity is calculated from Fs, Vas, and Qes, and none of those parameters are related to Qms, SPL should not change as the amount of resistive damping is increased.

My question is, how is this possible? My understanding is that Rms/Qms represent mechanical losses in the driver, with mechanical energy being lost as low-level heat dissipated by the suspension components (and in this case, resistance in the secondary voice coil?). If these losses (Rms/Qms) increase, how can the driver have the same acoustic output with the same electrical input?

It sounds like something for nothing to me...? Help!

[DanWiggins]

Think of Qms as "mud". Efficiency as horsepower. Driving your car through a lot of mud will result in slowing you down -drag and all. Of course, the horsepower doesn't change, your actual car velocity does.

Likewise with a speaker. You do have a greater resistance to motion (lower Qms), but still have the same ability to turn electrical input power to acoustic output power.

Dan Wiggins
Adire Audio

[bwbass]

Okay, but in that analogy when you put the same amount of power to the wheels, less work (movement of the car) gets done in the same amount of time because some of the power was spent rearranging the mud.

That's less efficient, isn't it?

Is the problem with my understanding that efficiency of the motor does not always correlate directly to acoustic output because of mechanical and electrical damping?

[5th element]

Think of it in another way if you put a speaker in a free air situation (on an open baffle) and apply a signal, providing the baffle is wide enough and loudspeaker resonance low enough, the spl will remain constant in line with the midrange, as you descend in frequency.

Now if you put a box behind this driver its excursion is greatly reduced but the SPL generated is exactly the same assuming a flat alignment is chosen.

Putting the box there in a way accomplishes the same role of the second voicecoil being wired up to a resistor or just shorted. Instead of altering the box size you can change the value of resistor to achieve the same thing.

Thats how I understand it but it could be wrong im not an expert like dan.

[kelticwizard]

Originally posted by 5th element
Think of it in another way if you put a speaker in a free air situation (on an open baffle) and apply a signal, providing the baffle is wide enough and loudspeaker resonance low enough, the spl will remain constant in line with the midrange, as you descend in frequency.

Now if you put a box behind this driver its excursion is greatly reduced but the SPL generated is exactly the same assuming a flat alignment is chosen.

I am not understanding that statement at all.

In both Infinite Baffle and Closed Box situations, the SPL is caused by the cone motion-no reinforcement from anywhere else.

In what way can a Closed Box cause a decrease in cone motion while maintaining SPL?

I don't understand. In both cases, the SPL is generated by the cone excursion,and the back wave is blocked from cancelling. The cone excursion and SPL should stay the same, down to the point where the box causes a rolloff in both cone excursion and SPL simultaneously.

[Ron E]

Originally posted by bwbass
And since sensitivity is calculated from Fs, Vas, and Qes, and none of those parameters are related to Qms, SPL should not change as the amount of resistive damping is increased. My question is, how is this possible? My understanding is that Rms/Qms represent mechanical losses in the driver, with mechanical energy being lost as low-level heat dissipated by the suspension components (and in this case, resistance in the secondary voice coil?). If these losses (Rms/Qms) increase, how can the driver have the same acoustic output with the same electrical input?

You do lose output from shorting the coil - you lose output at and near resonance as the losses are increased. Plot it out in software.

Rms, Qts, Qms, Qes are defined only at resonance and do not affect the output at frequencies well above resonance where the sensitivity equation holds true. The "true" sensitivity equation (the one that gives you insight into the physics) uses fundamental parameters:

no=K*Bl^2*Sd^2/(Mms^2*Re)
dB=112+10*log(no)

To briefly explain things to a general audience, at resonance, the system is "loss-controlled". Below resonance, it is compliance controlled and above resonance where the sensitivity equation holds, the system is mass-controlled..... Both (sufficiently) above and below resonance, damping doesn't matter.

[bwbass]

Originally posted by Ron E


You do lose output from shorting the coil - you lose output at and near resonance as the losses are increased. Plot it out in software.

Rms, Qts, Qms, Qes are defined only at resonance and do not affect the output at frequencies well above resonance where the sensitivity equation holds true. The "true" sensitivity equation (the one that gives you insight into the physics) uses fundamental parameters:

no=K*Bl^2*Sd^2/(Mms^2*Re)
dB=112+10*log(no)

To briefly explain things to a general audience, at resonance, the system is "loss-controlled". Below resonance, it is compliance controlled and above resonance where the sensitivity equation holds, the system is mass-controlled..... Both (sufficiently) above and below resonance, damping doesn't matter.

Your "general audience" thanks you! This makes perfect sense.

Now the real question I've been trying to unravel:

Ascentant Audio's Atlas drivers have one 4 ohm coil, intended to be driven, and one 2 ohm coil, intended for RDO control only. The specs on Ascendant's site (www.ascendantaudio.com) show Qes changing as the resistance across the secondary coil changes, along with Qts, Bl, and sensitivity.

Would everybody agree here that it's more likely Qms and Qts changing, and that Bl and sensitivity would remain constant?

I'm inclined to believe this is the case, since Bl imust be measured when the driver is at rest, when there would be no motion for the shorted coil to damp.

[catapult]

Yes, Ascendant's specs appear to be in error. To check it, calculate backward from their Q values to get Bl and you will find it changes depending on whether the second coil is shorted or open. That shouldn't happen. That said, everyone says they are great drivers; just take the published T/S parameters with a grain of salt.

[bwbass]

Originally posted by catapult
Yes, Ascendant's specs appear to be in error. To check it, calculate backward from their Q values to get Bl and you will find it changes depending on whether the second coil is shorted or open. That shouldn't happen. That said, everyone says they are great drivers; just take the published T/S parameters with a grain of salt.

Yes, lots of people have bought these drivers and built boxes from these specs, and I've only heard great things about them.

[catapult]

All you really need to build a sub are Fs, Qts and Vas. I imagine those values are correct on the Ascendant web site. It looks like he worked backward from a known Qts and changed Qes to make it fit when he should have changed Qms.

FWIW, I modified Dan Wiggins' RDO spreadsheet to work with the Atlas drivers and came up with about the same values for Qts as Chad did but I got there a different (more correct) way by changing Qms instead of Qes.