Help to understand "current drive"

Oh, please, come on! Perfection? Did I ever mention perfection and that component qualities or characteristics are perfect, whatever is meant by the latter. However, it is extremely desirable that a speaker magnet assembly provides a reliable magnetic field, and one requirement of that field, is constancy of intensity. Another such requirement, but in the case of resistors, is their material must have a 'linear' characteristic. Yes, for real materials one may get other coefficients for the polynomial representing voltage against current, but the most prominent, the coefficient of the linear term, must be far bigger than the rest. Pedantically, it would not be linear, but for real applications, it can satisfy most situations.
You really should read some of the references in this thread, to get an idea how far from perfect everything is.
 
Last I checked E = I * R, so E and I are linked. At least as long as we're not talking about superconductive materials.

He's (probably) fundamentally correct in that the voltage does not enter into the equations directly, but the current does. The equations probably start with F = B*I*L, where F is the magnetic force, B is the magnetic flux density, I is the current, and L is the length of the conductor (think single-turn coil). Look, ma! No voltage in the equation!! Voltage no matter! Obviously the maths explode once you move to multi-turn coils and take the homogeneity of the magnetic field into account, but I bet the fancy equations just solve F = B*I*L in a small unit area/volume and add it up at the end (otherwise known as integration).
But if E = I*R, then I = E/R, so the voltage does enter into the equations as soon as you take even the DC resistance of the voice coil into account. And the current becomes complex once you account for the inductance of the voice coil.

Fundamentally, there should not be any difference between current drive and voltage drive. An ideal current source forcing 1 A through an 8 Ω speaker will impose a voltage of 8 V across the speaker. Similarly, an ideal voltage source imposing 8 V across that same 8 Ω speaker will force 1 A of current through the speaker. The two scenarios are equivalent.
The only way I can see the two sources being different is how they handle the back-EMF. There's no way to avoid the back-EMF. If you force the speaker cone forward (either by voltage or by current) and reduce the current/voltage to zero instantaneously, the suspension on the speaker cone will move the cone back to the equilibrium position and this will generate a current in the speaker voice coil. An ideal voltage source will shunt this back-EMF current to ground, thus, keeping its output voltage constant. An ideal current source will change its output current by an equal and opposite amount to the back-EMF, thereby cancelling the back-EMF current and allowing zero change in the voltage across the voice coil. Again, in the ideal case, there is no difference between the two. Thus, if there is any difference between current and voltage drive in audio speaker applications, I bet they're due to the non-idealities of the amplifiers and not due to the different operating modes.

I encourage those who are curious about the differences between voltage and current drive to look into Thevenin and Norton equivalent circuits.

Tom
The current drive only have advantage when combined with the non-ideal parameter of the speaker.
If the speaker is an ideal resistor within the whole frequency range, then the voltage drive and current drive are exactly the same.That can not be the reality.
 
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Yeah, simplified, current through voice coil makes the cone assembly move and acoustic sound.

With voltage amplifier the current depends on impedance in the circuit voice coil is part of, and when impedance is non-linear the current is non-linear and acoustic output is non-linear.

What makes the impedance non-linear is stuff in driver motor, heating of voice coil, hysteresis, varying inductance by excursion, perhaps more, and all these are inherent in the driver. Dilute their effect on current by increasing stable impedance in the circuit.
 
For frequencies far from resonance, where the loudspeaker impedance is dominated by the voice coil resistance and self-inductance, it's quite bizarre:

We build voltage output amplifiers that are almost perfectly linear over a wide signal level range (levels well above their noise floor and still below clipping). To achieve that, we use lots of negative feedback, so the distortion is determined by the feedback network. By using low-temperature-coefficient resistors sized such that the signal hardly causes any self-heating, and possibly series connections of unit resistors, the distortion is kept very small indeed.

Then we let the 4000 ppm/K-temperature-coefficient resistance of a voice coil that heats up quite a lot and its non-linear self-inductance convert the voltage into a current that drives the loudspeaker.
 
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frugal-phile™
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Well, there's the air (spring) also if you use a closed box.

The box can certainly provide significant damping. The whole idea of aperiodic boxes revolves around that. ESA suggests low Qm driver in a heavily damped (with cotton) sealed box. I prefer other approaches.

Speaker (driver(s)+box), amplifier (and what connect them), is a system that has to be considered as a whole.

dave
 
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Neurochrome.com
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In fact, this story also comes around to the lack of damping with current drive, only the mech losses give damping. That is also why you see that resonance peak in the freq response - it is not damped.
So... Random thought: Presumably the manufacturers of drivers measure their drivers with amps that have voltage outputs. What if the driver characteristics are significantly different when used with a current-output amp? Then all the enclosure math will be off.... No?

Tom
 
Yes, on driver resonance impedance has peak and voltage amp causes less current in the circuit and the peak doesn't appear as such in acoustic response. Current amp doesn't mind about the impedance, it makes sure there is current as requested and basically frequency response follows impedance.

One can "EQ the impedance" with conjugate network so that where driver has its resonance the shunt network flows the extra current and driver current reduces, frequency response remains similar as with voltage amp. Or mechanically/acoustically dampen/tune the box to reduce resonance and impedance. Or with DSP/active EQ. As you suspect its different, but its not impossible task, just something that needs to be considered, its system level issue in a way. If speakers are already bought and not suitable for current drive as such, and can't do DSP because for what ever reason its better to stick with voltage amp :)

Key is to think what the pros and cons are for both, and if its worth it. As its DIY forum one could make hybrid that takes pros from both voltage and current drive, use voltage amp and do impedance manipulation. Or what ever rocks the boat :)
 
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So... Random thought: Presumably the manufacturers of drivers measure their drivers with amps that have voltage outputs. What if the driver characteristics are significantly different when used with a current-output amp? Then all the enclosure math will be off.... No?

Tom
The quality factor of the main resonance will be much higher, but loudspeaker manufacturers usually already measure and specify that. The Thiele and Small parameter Qms is the quality factor with current drive.

Besides, the treble increases. You can see how much by multiplying the frequency response under voltage drive with the impedance curve, both of which are normally specified. After all, under small-signal conditions, the voltage is just the impedance times the current.

Normally there are no data in a typical loudspeaker datasheet that show the effect of current drive on distortion.
 
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So... Random thought: Presumably the manufacturers of drivers measure their drivers with amps that have voltage outputs. What if the driver characteristics are significantly different when used with a current-output amp? Then all the enclosure math will be off.... No?

Tom
The basic measurement to obtain T/S data is the impedance sweep, two of them (free-air and free-air with added mass, or free-air and in known closed cabinent).
Ideally, it wouldn't matter how that sweep is made wrt drive impedance, the driver's parameters do not change by how it's driven, only the complete system parameters change.

High impedance drive naturally leads to (unduly) high excursion around resonance, especially when the goal to establish the T/S params at moderate to high signal levels which then could lead to wrong T/S params as the operating point was unsuitable, violating the linearized small-signal regime T/S params live in by definition.

The point is, even when staying halfway in the linear range of operation the T/S-params are still signal-dependent, the biggest factor often being VC heating, and ambient temperature. For example, sweep direction can have an influence as the VC heats up during the sweep. Pro manufacturers always do a preconditioning for several hours to break in and heat up the driver.

IMHO, T/S params are way overrated as to what accuracy to expect from them (better than 10% is stellar precision here). Whenever you see a datasheet with stuff like Qes=0.3421 or Fs=43.2Hz you know something is fishy.
 
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I am going to sum up the messages that have arrived me:

- benefit for current drive strongly depends on the drivers that were used,
the cheaper the driver - the higher the benefit

- benefit of current drive depends on the output impedance of the current amp,
the higher the output impedance - the higher the benefit

- current drive can get problematic around fs, more critical at higher output impedance

- the benefit of current drive (if any) shows up in lower intermodulation (and that is very welcome to me!)

- to really compare current drive with voltage drive individual equalizing is a must

- the interaction between amplifier and speaker is hardly understood, even specialists dont agree about current drive

- the current amps to build are all based on Esas ideas how to do it, none of them is really low THD (when tested with resistive load)

- my question for help on building such an amp (detail: output impedance) seemes to be unheared...

- my question to the musical timing when the current is in phase with the controlling voltage (and the amps
output voltage is therefore phaseshifted) is still not answered...
Good summary.

The sad part is this:
"the interaction between amplifier and speaker is hardly understood, even specialists dont agree about current drive".
A result of the reductionist divide and conquer approach so common in science and engineering this day and age. I can tell you that I really have had extremely tiring discussions with speaker as well as electronic engineers who had extreme difficulties to grasp the concept of local degenerative feedback happening in the driver depending on source impedance.

As for output impedance, once you are at or above 5x the driver's impedance in the region of interest it close enough to "pure" current drive. Any typical constant current circuit will do (sense-resistor in ground leg is the easiest but a Howland current pump is also quite simple and the speaker can be grounded or used with a bridged topology -- the second amp simply inverting the signal of the first, tapping off the signal before the sense resistor).

As for phase, voltage and current are connected by the impedance as usual. "Musical timing" (whatever that may mean) is only a function of the transfer function (frequency response of SPL magnitude and phase) -- however, note the "fine-print" difference disclaimer I made earlier.
 
IMHO, T/S params are way overrated as to what accuracy to expect from them (better than 10% is stellar precision here). Whenever you see a datasheet with stuff like Qes=0.3421 or Fs=43.2Hz you know something is fishy.
If manufacturer measured random batch of 10 or 100 or 1000 drivers, with the usual 15% spread of TS parameters, why is fishy to calculate and publish in the datasheet a median value (like Qes=0.3421)?
Common sense should be applied: expect up to +/- 15% diference from the median value of published values Qes=0.3421 or Fs=43.2 Hz. Or measure the damn drivers...
 
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I had a look at Joe Rasmussens current amp. He has measured an output impedance of about 250 Ohm. Can this be calculated somehow?
I don't know Joe Rasmussen's amplifier, but for a chip amplifier with single-sided output and feedback via a current sensing resistor: look up the open loop gain in the datasheet, add 1, multiply by the current sensing resistance. If there is an extra resistor or an RC network shunting the loudspeaker for stability reasons, calculate the parallel value of (open loop gain + 1) Rsense in parallel with whatever is connected in parallel with the loudspeaker.