Yeah, if cone moves 5mm it will make Le shift no matter what the circuit impedance is, or how the cone moved.
You could vibrate the cone with a stick and the Le would shift per Le(x), but shifting of Le does not affect the acoustic output, unless you short driver terminals with low impedance and let backEMF current flow! Now, the shift in Le appears as current and will apply force to the cone making distortion! Leave the circuit open, and there is no current in the circuit, no force on the cone other than your stick.
Le is not some property that is acoustic domain and always in effect, it is mechanism in electric domain that makes force through opposing current, which flows strong depending if circuit impedance is low, or flows poorly if impedance is high.
This is more difficult subject than I ever thought, very very hard to explain so that it was easy to understand 😀
Perhaps this?:
Think there is voltage amplifier inside the driver that makes independent current to the circuit, independently of what ever power amplifier you have. Back EMF is voltage induced to voice coil due to the coil moving in a magnetic field. What ever pushed the cone to motion activates the back EMF voltage amplifier which does it's own thing on top of what your actual power amplifier is doing. Now, what ever current flows in the coil makes force to the cone and measures in acoustic domain. If you prevent back EMF making current into the circuit you measure less effects of back EMF in acoustic domain.
I have abstracted actual power amplifier away from this example purposely, assume what ever amplifier and replace with a resistor equivalent to its output impedance (Norton / Thevenin equivalent) and the cone is moved by a long stick instead. Hopefully it's now easy to imagine what the effect of back EMF is? As the stick moves the cone the voice coil induces voltage into the coil, which makes current flow over the impedance as its load. As the cone moves, the load varies per Le(x), so the current varies. As the voice coil is in the same circuit, the current makes force to the cone in addition to what the stick does, and as it varies per Le(x) you'd measure distortion. But, if you disconnect the circuit, or put very high impedance there in series, current from the back EMF is much less, and there is about no extra force acting on the cone, apart from the stick. Now you would not measure any distortion because there is no other force affecting the cone movement than the stick itself.
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Yes, of course 🙂
But to go back why I started this constant current side track thought experiment story, (which has been long enough now) is to get an idea what it will bring us compared to the passive solutions.
Speaking from a very general sense: a lot of trouble and hurdles.
These can be solved and fixed obviously.
But it's most definitely not a one size fits all approach!!
It's not hard to understand why a constant voltage source is so much more practical from a general point of view.
Besides, quite some of those drivers with copper bells and ironless whistles will have pretty good distortion to begin with.
So it kinda defeats the purpose.
Except for your odd little break-up peak.
No, definitely not. The reason for using current drive is exactly to get rid off Le's effec on the voice-voil current. So it does neither cause a lowpass pole anymore nor does Le(x) have an effect anymore.
Regards
Charles
I have done some calculations earlier and found the voltage swing needed at Fs, for constant current amps, to be very large.
50 ohm vs 5 ohm. That is 10 times the voltage if the current is constant.
I know it is done in 70V systems but they are not very high output. So maybe som health and safety issues for high current output, constant current amps?
One would probably want some current (and voltage?) limiting capability in the amplifier to prevent hitting amplifier limits, and also prevent burning voice coil. You could pour current through the coil until it snaps open. Before that its resistance would rise and voltage would rise even higher on the amp?
For same reason the voltage peaks up, there is peak in frequency response at Fs which needs to be EQd down. EQ the frequency response down before amplifier, which would reduce current at Fs, and voltage, and flatten the frequency response 😉 If you did this with passive network between amplifier and driver it would equalize the frequency response but leave current and voltage from amplifier as it was. With constant voltage amplifier this EQ happens naturally, the drivers electric damping that makes the impedance peak, lowers current in the circuit equalizing the frequency response.
For same reason the voltage peaks up, there is peak in frequency response at Fs which needs to be EQd down. EQ the frequency response down before amplifier, which would reduce current at Fs, and voltage, and flatten the frequency response 😉 If you did this with passive network between amplifier and driver it would equalize the frequency response but leave current and voltage from amplifier as it was. With constant voltage amplifier this EQ happens naturally, the drivers electric damping that makes the impedance peak, lowers current in the circuit equalizing the frequency response.
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Yeah maybe safety would be an issue if lots of power was needed. If you want current drive you likely want very high efficiency speaker as well, and in hifi application not much power would be needed.
Tubes have high output voltage and low current don't they? They are also dangerous for that 😀 And have step down transformer to interface with low impedance speaker circuit. Why not leave it high impedance output with less step down in the transformer?
From the other post:Say I use a current amp.
As i understand the cone accelleration is constant vs frequency
Force factor is Newton/Ampere is constant for a woofer.
So if F = ma, a is proportional to current from amp and the same over all frequencies from Fbox and up?
Why is it not linear response from fbox to say fbox x 4 ?
Hi,
please read this article: https://www.edn.com/loudspeaker-operation-the-superiority-of-current-drive-over-voltage-drive/
It's quite emotional text, but if you just ignore that part and focus on the scientific content, the impedance. it's golden for understanfing the stuff 🙂
please read this article: https://www.edn.com/loudspeaker-operation-the-superiority-of-current-drive-over-voltage-drive/
It's quite emotional text, but if you just ignore that part and focus on the scientific content, the impedance. it's golden for understanfing the stuff 🙂
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Then I have to get back to this. And the consept of constant I. Around Fresonance Z = Rc + RLC where RLC is electrical transform of the physical properties of the speaker element. And RLC part is caused by a resonance in back EMF.
At resonance where R can be 10 times Rc, is it still fruitful to talk about a constant current? When current from Back EMF is 10 times the current of the amp?
There is no, or very little, current from backEMF if you have constant current from amplifier!🙂 the back EMF is voltage, and when circuit impedamce is very high, with ideal current source it would be infinite, there is no current from backEMF! 🙂
Both the power amplifier and backEMF voltage source inside the motor operate on same circuit, both would make current in the circuit and trough voice coil and affect acoustic output when circuit impedance is low, because low impedance means high current for given voltage. All and any current through voice coil makes acoustic sound. If you increase circuit impedance then current from the backEMF voltage source reduces. It's all in the article.
Basically, velocity of cone is same for given excursion no matter how you set the excursion, with voltage or current power amplifier or without any amplifier vibrating the cone with a stick. This means back EMF voltage would be same on any of these cases. If circuit is open (high impedance), the back EMF voltage would not make any current flowing and there is no force in the motor from backEMF, no acoustic distortion. If circuit is shorted (very low impedance), back EMF voltage makes maximal current in the circuit, which makes force in the motor and now you'd measure acoustic distortion.
When the back EMF can make current flow in the circuit, you'd feel extra force in your hands if you used the stick method. You'd measure reduction in acoustic output if you used a voltage source as it means low circuit impedance must be low. Using current source means circuit impedance must be high and you would observe no change at all.
Basically, velocity of cone is same for given excursion no matter how you set the excursion, with voltage or current power amplifier or without any amplifier vibrating the cone with a stick. This means back EMF voltage would be same on any of these cases. If circuit is open (high impedance), the back EMF voltage would not make any current flowing and there is no force in the motor from backEMF, no acoustic distortion. If circuit is shorted (very low impedance), back EMF voltage makes maximal current in the circuit, which makes force in the motor and now you'd measure acoustic distortion.
When the back EMF can make current flow in the circuit, you'd feel extra force in your hands if you used the stick method. You'd measure reduction in acoustic output if you used a voltage source as it means low circuit impedance must be low. Using current source means circuit impedance must be high and you would observe no change at all.
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OK I have to contemplate on that😉 I asumed it was some kind of feedback curcuit path to ground that the back EMF voltage would cenerate a current through. Maybe I mixed it up with sensing resistor to measure EMF.
I also se the author of the article says in the comments:
I also se the author of the article says in the comments:
Yeah that all is also deducible with info in the article. The particular wording goes to specifics how amplifiers would change output impedance. To understand circuit impedance and how the driver operates use ideal amplifier, see Norton and Thevenin theorem and how to do circuit analysis. This is basis to understanding the stuff. Basically power amplifier can be abstracted away completely, replaced by resistor equal to it's output impedance. Then analyze what current the backEMF voltage source can make.
It goes into details when you take account how actual amplifier works and how it's operation is affected by backEMF for example. Effects in acoustic domain are ~same with ideal or with real world amplifier. It is quite likely there is additional distortion from the amplifier, and so on.
The section you have underlined is something I've been trying to lift up in this thread as well. About all drivers we can buy off the shelf are designed for low circuit impedance, for voltage amplifiers. This means that their acoustic frequency response depends on their own impedance affecting circuit current in order to equalize the acoustic output, making it flat. This also means that any variance with excursion of the driver impedance also shows in acoustic output. And, as Esa refers above their acoustic frequency response would not be flat in high circuit impedance, because their effect on current is reduced, so the EQ effect of driver impedance does not work anymore.
One could design driver/speaker for current drive, that has flat frequency response with high circuit impedance, whose own impedance is not needed to do EQ, and show no distortion (in this regard). Such driver would likely have bad frequency response with voltage amplifier.
It goes into details when you take account how actual amplifier works and how it's operation is affected by backEMF for example. Effects in acoustic domain are ~same with ideal or with real world amplifier. It is quite likely there is additional distortion from the amplifier, and so on.
The section you have underlined is something I've been trying to lift up in this thread as well. About all drivers we can buy off the shelf are designed for low circuit impedance, for voltage amplifiers. This means that their acoustic frequency response depends on their own impedance affecting circuit current in order to equalize the acoustic output, making it flat. This also means that any variance with excursion of the driver impedance also shows in acoustic output. And, as Esa refers above their acoustic frequency response would not be flat in high circuit impedance, because their effect on current is reduced, so the EQ effect of driver impedance does not work anymore.
One could design driver/speaker for current drive, that has flat frequency response with high circuit impedance, whose own impedance is not needed to do EQ, and show no distortion (in this regard). Such driver would likely have bad frequency response with voltage amplifier.
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I know current drive work in theory, And I have done my calculations 10 times. But still some room for confusion;-) It is the practical implementation that is the problem. So still the amp stability af current amps and big voltage swing represents a large hurdle.
The backEMF manifests as the higher impedance. Normally a fixed Voltage drive would pass lower current through the higher impedance. The current source will put more current through than that, because it also puts a higher Voltage drive across the driver.
Back to what I was saying before. It must be doing something because the current source will now be putting more Voltage across the driver to achieve this. The backEMF results in the higher impedance. The higher impedance increases the drive Voltage from a current source.
Yeah and that part is very true, voltage must go high if you want to have certain current over high impedance. To remedy you can reduce current and impedance.
You could reduce impedance for the amplifier by flattening the load impedance with shunts, but this would allow the backEMF current flow again 😉 So instead you could design a speaker such that mechanical damping reduces the driver resonance impedance peak. You could use a driver with low Le.
Demand for current is determined by system sensitivity and how loud you want it to be. If you need 80db + peaks at listening position, calculate back how much that is at 1meter. Then figure out what current you need over what impedance to keep voltage swing sane to make real world system. Just increase system sensitivity by making the speaker big enough, so that you need very little current (and voltage swing) to reach SPL target.
You could reduce impedance for the amplifier by flattening the load impedance with shunts, but this would allow the backEMF current flow again 😉 So instead you could design a speaker such that mechanical damping reduces the driver resonance impedance peak. You could use a driver with low Le.
Demand for current is determined by system sensitivity and how loud you want it to be. If you need 80db + peaks at listening position, calculate back how much that is at 1meter. Then figure out what current you need over what impedance to keep voltage swing sane to make real world system. Just increase system sensitivity by making the speaker big enough, so that you need very little current (and voltage swing) to reach SPL target.
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Yeah it requires more voltage swing from amplifier, but has no effect on acoustic output in a sense that it has with voltage drive.
I also se now that EQing Fbox is not that easy because of the high Q and "floating" F box regards to temperature, speaker element variance ++
But then as I understand now, we are not talking about existing speaker elements.
But still i am intrigued by the practical solution by https://rmsacoustics.nl/audiodesign.html
But then as I understand now, we are not talking about existing speaker elements.
But still i am intrigued by the practical solution by https://rmsacoustics.nl/audiodesign.html