I was thinking iron pole center of the voice coil, like with inductors it boosts inductance. Leave it air core and inductanse drops so you can wind more to get higher Re with same inductance. Inductance needs to be certain to meet certain high frequency response. Leaving core out would reduce magnetic field strength though.
I have no idea if this would work, just trying to reason with little information nuggets 🙂
Last edited:
Oh right, you mean it in a way like any iron core raises inductance in a inductor.
Problem is that we need that iron pole center to move our pisten forward and backwards.
The problem is not having an inductance.
This can be compensated with a RC filters parallel te the speaker (to some extend).
The main problem is any change of the value of this inductance.
Even with the best speakers out here, with the best linear Le(x) and not having any cone excursion.
You only need one nasty standing wave in the cabinet or other issues, that reflects back into the voice coil's impedance, and you're screwed.
Yeah but if the circuit imoedance is high Le(x) doesn't make difference in acoustic output 😉 thats the idea I tried to think over, have high impedance driver whose Le is tailored to give flat frequency response in a high impedance circuit. external RC network, zobel, would make low impedance path for backEMF negating the idea some. Perhaps it's necessary, I don't know how to design drivers. There aren't many examples for high circuit imoedamce, if any, is there? 32 or 64 ohm drivers? 🙂
Well just academic thought experiment with very little knowledge. I guess this kind of things aren't even necessary as low enough distortion performace can be achieved without, by system design and with very low distortion drivers if needed.
Well just academic thought experiment with very little knowledge. I guess this kind of things aren't even necessary as low enough distortion performace can be achieved without, by system design and with very low distortion drivers if needed.
Last edited:
That still doesn't matter, it's all about the CHANGE in impedance.
Not the absolute value of the impedance.
I guess one could argue about the fact that the relative change might be less for higher impedances.
I don't know if that is true, but than we also have to talk about an order of magnitude more, ideally more.
So sitting around the 80 ohm or 800 ohm.
I forgot on top of my head what it also does to the BL (for the same amount of SPL), so maybe someone else can quickly answer that.
Yes there are 64 ohm speakers;
https://www.wavecor.com/html/64_ohm_fullranges.html
I have seen other custom OEM 32 ohm and 64 ohm woofers as well.
Not the absolute value of the impedance.
I guess one could argue about the fact that the relative change might be less for higher impedances.
I don't know if that is true, but than we also have to talk about an order of magnitude more, ideally more.
So sitting around the 80 ohm or 800 ohm.
I forgot on top of my head what it also does to the BL (for the same amount of SPL), so maybe someone else can quickly answer that.
Yes there are 64 ohm speakers;
https://www.wavecor.com/html/64_ohm_fullranges.html
I have seen other custom OEM 32 ohm and 64 ohm woofers as well.
This is the key thing: Constant current makes constant acoustic output. Impedance varying with excursion doesn't make any difference to acoustic output when current stays constant. Absolute value of driver impedance doesn't matter either, because current is not defined by it but by the constant current source.
Unlike with constant voltage amplifier, where impedance varying with excursion changes current, which means acoustic output changes with excursion. Here driver impedance defines current, defines acoustic output.
Only current makes force in motor and moves the cone, makes acoustic output.
You can think that constant current source speaks directly to acoustic domain. But constant voltage source must use the driver impedance to translate the voltage into current, which makes some error.
Unlike with constant voltage amplifier, where impedance varying with excursion changes current, which means acoustic output changes with excursion. Here driver impedance defines current, defines acoustic output.
Only current makes force in motor and moves the cone, makes acoustic output.
You can think that constant current source speaks directly to acoustic domain. But constant voltage source must use the driver impedance to translate the voltage into current, which makes some error.
Last edited:
No. The motor issue that is behind the nonlinearity is not fixed by constant current alone.
No, you just literally described a constant voltage source.
Just look at my simulations again, or start doing them yourself.
Otherwise we keep on talking in circles.
You're clearly misreading a few things. (or we are just fully miscommunicating )
I am not talking about cone excursion itself, I am talking what things like Le do when there is cone excursion.
Or rather the stability of the Le.
Point is, as soon as the impedance changes for whatever reason, it will have an direct effect on frequency response itself.
(instead of causing distortion)
They both make errors.
- With voltage non-linear behavior resulting in distortion (voltage and current are out of phase and not equal in shape)
- With current that results in non-linear behavior in the frequency response, since (like you said) it's directly proportional to it.
^ Think about it for a second, I think you are mixing up the voltage drive and current drive for some reason.
Let's try thought experiment, which hopefully clears any possible confusion:
Input 100Hz tone and 1000Hz tone, there is ten cycles of 1000Hz tone while one 100Hz cycle goes through, or 5 cycles on single half cycle. Now lets assume acoustic frequency response is flat so 1000Hz must make much less excursion, so 100Hz dominates the voice coil position in gap, the x. Now lets pause time for a second to a point where 100Hz tone is at such phase that the cone is at extreme and x equals xmax. Now, according to Le(x) impedance at 1000Hz is different than previous cycle of 1000Hz, 100Hz tone modulates Le at 1000Hz. But does it modulate sound in acoustic domain?
If you have constant current source, high circuit impedance to keep the current constant, the Le(x) would be perhaps 0.01% of total circuit impedance and have about no effect on current in the circuit, no effect on acoustic output. The 1000Hz would not modulate as you are keeping current constant at 1000Hz no matter what the 100Hz does. Le(x) distortion does not make into acoustic domain here. Other distortion mechanisms would like Bl(x) or Sd(x), but lets keep it simple and use Le(x) only.
If you have constant voltage, low circuit impedance to keep constant voltage, the Le(x) would vary impedance perhaps 10% and modulate current, which would modulate the acoustic output. Now 100Hz affects acoustic output at 1000Hz.
You must know, that current is what makes cone move, which equals acoustic output. Cone excursion is inherently linked to acoustic output, it makes the acoustic output. Le(x) would need to make appearance in current in order to make appearance in acoustic domain. If you have constant current source which ensures current is truly constant the varying impedance is compensated by the source, eg. diluted by the high circuit impedance, which means Le(x) does not vary current but voltage, and does not appear in acoustic domain.
Here simple simulation to demonstrate. As source impedance is increased the frequency response changes because the driver impedance does not equalize the acoustic output anymore, current in the circuit is dominated by the source and not by the driver. Emulating Le(x) at xmax just use series inductor and toggle it on or off. As circuit impedance increases, as we move from voltage control to current control, Le(x) makes less and less effect on acoustic output.
Pardon the legend is not in order for some reason. Current, and acoustic output, drops as I'm keeping voltage constant while increasing circuit impedance so the legend is from smallest to highest from top to bottom. Important bit is that on the highest circuit impedance case, where current is most constant, the Le(x) makes no difference. You see the effect diminish as the circuit impedance is increased.
Here is the test setup, just use enclosure tool to generate frequency response and impedance, or use measured frequency response and impedance if you have one.
Let's try thought experiment, which hopefully clears any possible confusion:
Input 100Hz tone and 1000Hz tone, there is ten cycles of 1000Hz tone while one 100Hz cycle goes through, or 5 cycles on single half cycle. Now lets assume acoustic frequency response is flat so 1000Hz must make much less excursion, so 100Hz dominates the voice coil position in gap, the x. Now lets pause time for a second to a point where 100Hz tone is at such phase that the cone is at extreme and x equals xmax. Now, according to Le(x) impedance at 1000Hz is different than previous cycle of 1000Hz, 100Hz tone modulates Le at 1000Hz. But does it modulate sound in acoustic domain?
If you have constant current source, high circuit impedance to keep the current constant, the Le(x) would be perhaps 0.01% of total circuit impedance and have about no effect on current in the circuit, no effect on acoustic output. The 1000Hz would not modulate as you are keeping current constant at 1000Hz no matter what the 100Hz does. Le(x) distortion does not make into acoustic domain here. Other distortion mechanisms would like Bl(x) or Sd(x), but lets keep it simple and use Le(x) only.
If you have constant voltage, low circuit impedance to keep constant voltage, the Le(x) would vary impedance perhaps 10% and modulate current, which would modulate the acoustic output. Now 100Hz affects acoustic output at 1000Hz.
You must know, that current is what makes cone move, which equals acoustic output. Cone excursion is inherently linked to acoustic output, it makes the acoustic output. Le(x) would need to make appearance in current in order to make appearance in acoustic domain. If you have constant current source which ensures current is truly constant the varying impedance is compensated by the source, eg. diluted by the high circuit impedance, which means Le(x) does not vary current but voltage, and does not appear in acoustic domain.
Here simple simulation to demonstrate. As source impedance is increased the frequency response changes because the driver impedance does not equalize the acoustic output anymore, current in the circuit is dominated by the source and not by the driver. Emulating Le(x) at xmax just use series inductor and toggle it on or off. As circuit impedance increases, as we move from voltage control to current control, Le(x) makes less and less effect on acoustic output.
Pardon the legend is not in order for some reason. Current, and acoustic output, drops as I'm keeping voltage constant while increasing circuit impedance so the legend is from smallest to highest from top to bottom. Important bit is that on the highest circuit impedance case, where current is most constant, the Le(x) makes no difference. You see the effect diminish as the circuit impedance is increased.
Here is the test setup, just use enclosure tool to generate frequency response and impedance, or use measured frequency response and impedance if you have one.
Last edited:
I think you are not considering that Le must have effect in excursion in order be audible in acoustic domain, so it must affect current in the circuit in order to have effect on excursion. Perhaps excursion word is confusing here, let's use force instead.
Le(x) must modulate current in order to modulate force to the cone, any electrical signal must affect current in order to appear in acoustic domain. If there is no change in current, it does not affect force on the cone and does not appear in acoustic domain. If current is modulated, force is not modulated, acoustic output is modulated. This is key thing to understand. Unlike with voltage amplifier, where circuit current is dominated by driver impedance, here Le makes directly into acoustic domain by modulating impedance, modulating current from our constant voltage, modulating current modulates force to move the cone.
It is important to do thought experiment with it. It all happens simultaneously. As our input makes current into the circuit the cone is sent moving, which simultaneously in real time varies the impedance, which affects driver impedance in realtime, which changes the acoustic output in realtime if the driver impedance dominates circuit impedance, has (meaningful) effect on current in the circuit.
Last edited:
You are right, the non-linearity stays, it is physical property of the driver and varies Le(x) no matter what the circuit impedance is. But acoustic distortion is fixed: The constant current source, high circuit impedance, changes effect of the Le(x) non-linearity from modulating current to modulating voltage.
Z=U/I, and as Z varies with excursion
=> if you have I constant then U varies with Z.
=> if you have U constant then I varies with Z.
Voltage over voice coil has no effect on acoustic output, only current does, so all we care is current through voice coil and how stable is that. How does it depend on excursion? As per above
=> In constant current case, current through voice coil does not depend on excursion as current is held constant.
=> In constant voltage case, current through voice coil does depend excursion as current is determined by driver impedance, and varies with it.
In constant current case Le(x) can vary the voltage all day long and have no effect on acoustic output.
Last edited:
Yes.
Think about it, the Le is not resistance of the coil or anything like that, it is electromagnetic property that opposes change in current in conductor due to magnetic field appearing around the conductor causing opposing current, making the initial current appear lower. Impedance is literally relationship ratio of voltage to current. Now, if this current has no meaning to total current that flows through the voice coil, as the current is regulated to be constant outside of the driver, it does nothing to how motor engages, it does not affect force made by the motor and motor engages evenly, if I interpret your words correctly.
High circuit impedance prevents the Le make change in current. This literally means Le has no effect in acoustic response with ideal current source. With real world current source there would be some, but less and less the higher the circuit impedance.
Think about it, the Le is not resistance of the coil or anything like that, it is electromagnetic property that opposes change in current in conductor due to magnetic field appearing around the conductor causing opposing current, making the initial current appear lower. Impedance is literally relationship ratio of voltage to current. Now, if this current has no meaning to total current that flows through the voice coil, as the current is regulated to be constant outside of the driver, it does nothing to how motor engages, it does not affect force made by the motor and motor engages evenly, if I interpret your words correctly.
High circuit impedance prevents the Le make change in current. This literally means Le has no effect in acoustic response with ideal current source. With real world current source there would be some, but less and less the higher the circuit impedance.
Last edited:
I've been trying to bring up that driver datasheet frequency response is like that because it was measured with constant voltage. The driver was designed to be used with low circuit impedance to give ~flat acoustic frequency response. Impedance of the driver itself works as "passive filter", manipulating circuit impedance and manipulating acoustic output.
Now, if you hook up this kind of driver to high impedance circuit where the driver impedance does not EQ the current anymore, thus not EQ:ing the acoustic output, the response looks wrong. But thats the normal response of the driver with constant force applied at all frequencies! The impedance of the driver reduces force to the cone from voltage amplifier, where impedance is high. Le drops highs and motional impedance peak reduces acoustic output at driver resonance.
Now, if you raise circuit impedance you take this EQ away, current is not affected by the driver so it resonates freely at driver resonance, and highs peak because wavelength gets shorter compared to volume displacement and constant force makes more output.
Now, if you hook up this kind of driver to high impedance circuit where the driver impedance does not EQ the current anymore, thus not EQ:ing the acoustic output, the response looks wrong. But thats the normal response of the driver with constant force applied at all frequencies! The impedance of the driver reduces force to the cone from voltage amplifier, where impedance is high. Le drops highs and motional impedance peak reduces acoustic output at driver resonance.
Now, if you raise circuit impedance you take this EQ away, current is not affected by the driver so it resonates freely at driver resonance, and highs peak because wavelength gets shorter compared to volume displacement and constant force makes more output.
Last edited:
The opposing current you seem to speak of works with varying strength and cannot be free of distortion. Constant current can not fully restore cone movement.
An accelerometer on the cone would be one way to give you the information you're looking for.
An accelerometer on the cone would be one way to give you the information you're looking for.
Yeah, you are right about the distortion mechanism, opposing current varies which makes the total current vary, which is total current through voice coil, the cause for distortion!
But don't confuse accellerometer to this yet, it is not about restoring anything, it is about preventing Le altering the current, prevent Le affecting the cone movement, preventing Le(x) showing up in acoustic domain. This is the distortion reduction we see when increasing circuit impedance. There is no counter force but prevention of distortion force, or how you want to imagine it.
This is fundamental for understanding what it is all about: shift your perspective of thinking about it from "correcting something with counter measure" to "preventing something happening in the first place".
There is no counter force or anything, there is just prevention of the change that Le would make to the force by modulating current in the circuit.
To help thinking: All of it happens simultaneously, current is same on any part of the circuit at any given moment, this is why the external impedance between driver terminals affects what happens inside the driver, how the driver parameter modulates the circuit current, or not! Voice coil is part of the circuit, thus current through voice coil is limited by the circuit impedance, impedance in series with the voice coil.
In other words if we allow Le to modulate circuit current, we allow the modulation to appear in acoustic domain, distortion. If we prevent Le from modulating circuit current, we prevent modulation for that part in acoustic domain, prevent that distortion mechanism appear in acoustic domain.
But don't confuse accellerometer to this yet, it is not about restoring anything, it is about preventing Le altering the current, prevent Le affecting the cone movement, preventing Le(x) showing up in acoustic domain. This is the distortion reduction we see when increasing circuit impedance. There is no counter force but prevention of distortion force, or how you want to imagine it.
This is fundamental for understanding what it is all about: shift your perspective of thinking about it from "correcting something with counter measure" to "preventing something happening in the first place".
There is no counter force or anything, there is just prevention of the change that Le would make to the force by modulating current in the circuit.
To help thinking: All of it happens simultaneously, current is same on any part of the circuit at any given moment, this is why the external impedance between driver terminals affects what happens inside the driver, how the driver parameter modulates the circuit current, or not! Voice coil is part of the circuit, thus current through voice coil is limited by the circuit impedance, impedance in series with the voice coil.
In other words if we allow Le to modulate circuit current, we allow the modulation to appear in acoustic domain, distortion. If we prevent Le from modulating circuit current, we prevent modulation for that part in acoustic domain, prevent that distortion mechanism appear in acoustic domain.
Last edited:
https://rmsacoustics.nl/activecontrol.html
They combine feedback and constant current.
The upside is low current
The downside is extrem voltage amplitude on the amp
But result is low distortion
This may seem irelevant in the passive component discussion, byt maybe it isn't.
Let's say you have a midcone with resonance like measured in this thread. (Thank you, tmuikku)
If you put a (tuned to resonance) RLC filter in series with the mid, the amp will see a high impedance at reconance. The woofer will also se an high impedance towards the amp at resonance. That equates to a current amp at the resonance of the RLC filter. So we have two effects:
Let's say you have a midcone with resonance like measured in this thread. (Thank you, tmuikku)
If you put a (tuned to resonance) RLC filter in series with the mid, the amp will see a high impedance at reconance. The woofer will also se an high impedance towards the amp at resonance. That equates to a current amp at the resonance of the RLC filter. So we have two effects:
- The signal is attunated by the RLC cirquit so the resonanse of the cone is less excited.
- The speaker element sees a current source at it high frequency cone resonance and may have lower distortion because of that. (Only around RLC resonance)
Last edited:
Example of RLC series filter use on active filter system and RLC + magnesium driver mid: https://vcllabs.com/project04/
Look for: "Harmonic distortion improvement of the midrange transducer" in the long article
Last edited:
Yes, which works in theory.
In practice, even with a high-pass filter, there will be some cone excursion happening. Depending on how well the driver is designed, this will still shift the Le.
Which for current drive immediately means the freq resp will change.