Current drive for speakers

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Sorry to bring up this old topic, but I've been thinking (again, my apologies). Does current drive really help out with power compression? It doesn't make speakers behave like they do with voltage drive, but that doesn't necessarily mean that current drive is perfect. We know that joule heating causes the voice coil to heat up, which means that unless the conductor is made of constantan or a similar material, the resistance will go up. This means that current drive does the opposite that voltage drive does - it'll put more power into a hot voice coil.

This should also apply to inductance effects. It should be safe to say that varying inductance (impedance) causes distortion by modulating the voltage. As before, instead of fixing the power to the speaker, current drive will do the opposite thing that voltage drive does.

So why did Hawksford measure a decrease in distortion by using current drive?

http://www.essex.ac.uk/dces/researc...J12 Distortion reduction MC current drive.pdf

EDIT: You can use constantan for voice coils. You just lose about 10dB of efficiency compared to copper.
 
I apologize for the lack of a more rigorous answer, but my intuition is to model a driver as a zero resistance coil in series with a resistor. Further, current through the coil is proportional to electromagnetic force.

In the case of constant voltage, as the resistor heats, the current and EMF decrease. In a constant current situation, EMF is constant. The resistor does receive additional heating.

Similar logic works for the inductive component of the voice coil.
The 20 DB reduction makes sense to me. I assume that other mechanical non-linear systems such as surrounds prevent a more dramatic reduction in distortion.

Doug
 
454Casull said:

I'm still stuck at this point. Are you saying that acoustic power is derived from the coil current ONLY and has nothing whatsoever to do with the electrical power through the coil?
It's possible to work it out from a few simple equations:

i * Bl = f
f / m = a
v = int(a)dt
f * v = P

therefore

P = i * Bl * int(i * Bl / m)dt
(do please point out if I've made a mistake!)

where
i = current through voice coil
Bl = force factor
f = force on cone
m = moving mass
a = acceleration
P = instantaneous power

Note that there is no relationship to the electrical power wasted in the voice coil resistance.

EDIT: Oops, calculus.
 
454Casull said:

So it is around Fs that the efficiency of transfer is higher?
I suppose that depends on how well damped the speaker is, but yes, in general every resonant system is more efficient at the resonant frequency. It's easy to get a feel for this by imagining a pendulum; it requires just a little bit of pushing to make it swing at its natural frequency, but to make it swing, say 20% faster is much harder because you constantly have to force it to change direction when it doesn't want to.
 
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