New capacitive motional feedback method

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Hi all,

After years of hard meditation, failures, retries, despair, totally destroyed loudspeaker drivers and even some success I decided to make webpages about the new capacitive motional feedback method. Pages can be found here:

www.servospeaker.com

Pages are not too pretty (web design is not exatly my strong part) :), but I hope the content is understandable...

Warning: This system is potentially dangerous to your drivers, and maybe even yourself as there are some high-voltage parts involved (no current needed though) so take care of yourself (and your drivers)...

Comments are welcome!
 
Congratulations on the work you have put into this project. The idea of a cylindrical capacitive sensor works very well.
It is very linear and you can use various methods to measure the capacitance, including radio frequencies such as used in some condenser mics.
The ibasic idea however is not novel so I would be doubly careful with your patent application.

Andrew
 
heater said:

Now I guess you have a switch some where that enables/disables the motioan feedback.
So my question is how does it sound when you throw the switch to the ON position ?
Can you hear a big change ?

I would say that the bass sounds generally tighter. Also when playing constant, say 20 Hz note with a lot of deflection, the sound is more - should I say "convincing", since the THD is lower. You just don't hear so much of those harmonics. Also the long term compression effect due to voice coil heating is reduced. This effect I haven't measured yet, but it was actually very noticeable when I tried to get constant output for distortion measurements and the output level (without feedback) just kept dropping when adjusting the volume up.

Well, of course this was only my own opinion that may be a bit biased. ;)

At lower levels effect is not so noticeable, since much of the speaker THD comes from spider/surround/motor nonlinearities that are proportional to cone deflection.

Puggie: I don't see any problem with SEAS L22. It seems to have the same voice coil diameter as L18RNX. Some drivers however may not have enough "depth" between top of the voice coil former and pole piece to reliable attach the outer cylinder of the capacitor. I dont know if this is the case with L22's. I have done some experimenting with L26RFX/P and atleast there is plenty of room (and 51mm voice coil).
 
Very nice! I was looking to create and patent a slightly different motional feedback system. I like yours, as it is sensitve to small fluctuations of the driver cone, wheras most are only concerned with the gross level changes at high levels of deflection.

The reality of the human ear, is that it is the miniscule differences in peak levels -they are the 'intelligence' of how we actually hear. This means that your system has far more fidelity, to that 'ear' that 'we' actually use to hear things. Ie, measurements are important-- but their weighting/design is even more critical.
 
KBK said:
I was looking to create and patent a slightly different motional feedback system.

What was the operating principle in your system? Was it also capacitive?

I actually first played with simple plate capacitor with other plate connected straight to the cone and other to the basket (so that the distance between them changed with cone deflection). Didn't really work as the capacitance was unlinear and very small causing noisy signal. And cone breakup caused chaotic sounds coming out of the system (this proto was with paper cone). :)
 
(JPK) Interesting paper (your PDF). Very nice. I have a couple of questions. You assume the capacitance varies linearly with displacement. What about edge effects. I would expect these to be more significant at high excursions. Perhaps the deviation form linearity is significantly less that the inherent woofer nonlinearity?

Second, your formula 2. You have Uout = R dQ/Dt, which is simply Uout = RI, since I = dQ/dt. Then you write Uout = R Ucap dC/dt. I assume Ucap is the voltage across the cap.

However, dQ/dt = CdU/dt + UdC/dt. So shouldn't IR = RCdU/dt + RUdC/dt?

I haven't worked though the details so perhaps I missed something?
 
john k... said:
(JPK) Interesting paper (your PDF). Very nice. I have a couple of questions. You assume the capacitance varies linearly with displacement. What about edge effects. I would expect these to be more significant at high excursions. Perhaps the deviation form linearity is significantly less that the inherent woofer nonlinearity?
Do you mean by edge effects that the capacitor is working in nonlinear region? This can be eliminated by making the linear region (length of the capacitor cylinders+ some spare) longer than the xmax of the driver.

Second, your formula 2. You have Uout = R dQ/Dt, which is simply Uout = RI, since I = dQ/dt. Then you write Uout = R Ucap dC/dt. I assume Ucap is the voltage across the cap.
That Uout = R dQ/Dt was there instead of Uout = RI to make it more evident that we are interested in capacitance change. Voltage over the capacitor is kept constant (connected to virtual ground as Charles states) but the capacitance is changing, thus generating current. And yes, Ucap is voltage across the cap.

However, dQ/dt = CdU/dt + UdC/dt. So shouldn't IR = RCdU/dt + RUdC/dt?
Sorry, I don't get this... :confused:

BTW: Charles, what is ICTA?
 
Sorry, I don't get this... :confused:

BTW: Charles, what is ICTA? [/B][/QUOTE]



What I was getting at is that the charge on a capacitor is equal to the voltage across it times the capacitance, Q = C V. Current is the time rate of change of Q, I = dQ/dt. This would be the current flowing into your current to voltage converter. If C were constant I = C dV/dt, what we typically see if physics books. However, if the capacitance is not constant, I = d(CV)/dt = CdV/dt + VdC/dt. I guess I don't understand Charles' comment about a virtual short because if there is charge on the cap then there must be a voltage across it. If the two plates on the cap are at the same potential (virtual short) then Q must be zero for all time and there would be no current. I guess what is important is the relative magnitude of the two contributions.

As for edge effect what I meant was that when the inner and outer plates of the cap are aligned, the effect of the finite length of the plates on the capacitance will be different than when, for example, the woofer moves outwards and the plates over lap. This would introduce nonlinear behavior. Again, it probably becomes an issue of the relative magnitude of the different contributions.

As I said, I think you have a good idea here. I'm just wondering about the magnitude of the effects I mentioned.
 
Nuutti said:


What was the operating principle in your system? Was it also capacitive?

I actually first played with simple plate capacitor with other plate connected straight to the cone and other to the basket (so that the distance between them changed with cone deflection). Didn't really work as the capacitance was unlinear and very small causing noisy signal. And cone breakup caused chaotic sounds coming out of the system (this proto was with paper cone). :)


As far as I can tell, no-one has 'done' my version yet, so I have to keep my mouth shut on that one. No hints :)

But I do like yours, very nice.
 
Always the jokester that John K : http://www.musicanddesign.com/icta.html
Hope he hasn't forgot

Maybe he is not THAT John K. :confused:

Regarding the MFB: There is certainly a voltage accross the capacitor: A constant (and high) voltage is connected to one plate of the cap. The other plate is connected to a virtual short - the summing node of an inverting op-amp stage. So there is no voltage change across the cap. Only a capacitance change is taking place, leading to charge- and discharge- current that is proportional to dC/dt.

I think Nuuti has definitely done his homework.

Regards

Charles
 
john k... said:
What I was getting at is that the charge on a capacitor is equal to the voltage across it times the capacitance, Q = C V. Current is the time rate of change of Q, I = dQ/dt. This would be the current flowing into your current to voltage converter. If C were constant I = C dV/dt, what we typically see if physics books. However, if the capacitance is not constant, I = d(CV)/dt = CdV/dt + VdC/dt. I guess I don't understand Charles' comment about a virtual short because if there is charge on the cap then there must be a voltage across it. If the two plates on the cap are at the same potential (virtual short) then Q must be zero for all time and there would be no current. I guess what is important is the relative magnitude of the two contributions.

Now i get it. :) In the system in the Voltage across the cap is kept constant. Other end is connected to hv (1200V in prototypes) and other to virtual ground (short?) at the negative input of op-amp. So formula I = CdV/dt + VdC/dt is reduced to I = VdC/dt.

Now, when I looked at the doc, it really isn't stated that the voltage over the cap is kept constant (outer cylinder connected to hv)... Should fix that.

As for edge effect what I meant was that when the inner and outer plates of the cap are aligned, the effect of the finite length of the plates on the capacitance will be different than when, for example, the woofer moves outwards and the plates over lap. This would introduce nonlinear behavior. Again, it probably becomes an issue of the relative magnitude of the different contributions.
The inner cylinder doesn't "see" the length of the outer because the end (farther away from the pole piece) of it is separated by grounded shield. So the finite length doesn't cause nonlinearity as long as the other end of the outer cylinder is kept well over that shield (and other end over the inner cylinder's bottom edge).
... Well that was quite chaotic explanation. Maybe a picture would make it more understandable...
 

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