This question occurred to me recently and honestly I'm still not completely sure I really understand the answer. It may be fun to hear what others would think and maybe learn something. I suppose it's quite simple in the end but I'm just stratching my head around that for a while.
Suppose we have a woofer with a moving mass m and force factor Bl. Current through its voice coil will cause force on a cone F = Bl * I, which will give it an acceleration a = F/m = (Bl * I) / m.
Now increase both m and Bl twice, i.e. make the cone twice as heavy and the magnet twice as stronger. With the same current, the acceleration will stay the same, as will the velocity and excursion. BUT, kinetic energy of the cone will double, because it is twice as heavy. Twice as much energy must have been taken from somewhere. From where exactly? How this really works?
It seems that it leeds to a slight difference in electrical impedance but...
Suppose we have a woofer with a moving mass m and force factor Bl. Current through its voice coil will cause force on a cone F = Bl * I, which will give it an acceleration a = F/m = (Bl * I) / m.
Now increase both m and Bl twice, i.e. make the cone twice as heavy and the magnet twice as stronger. With the same current, the acceleration will stay the same, as will the velocity and excursion. BUT, kinetic energy of the cone will double, because it is twice as heavy. Twice as much energy must have been taken from somewhere. From where exactly? How this really works?
It seems that it leeds to a slight difference in electrical impedance but...
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It is a transducer, it takes some kind of energy (electrical) and turns it into another (mechanical).
And it´s not very efficient, only a small part of the electrical energy is turned into mechanical work.
A higher BL speaker motor will be more efficient, "transducing" more energy taken from amplifier.
Since speaker becomes more efficient, not acoustically since we are not changing piston movement so no change in SPL (useful work) but only moving just cone (more of it but at same speed / distance / acceleration) , useless work, transducer absorbs same energy as before, just doing more mechanical work (and wasting less).
Say instead of absorbing 1W and turning 1% of it into mechanical work, it still absorbs 1W but now turns 2% of it into mechanical work.
That it does not profit us is not the question asked.
And it´s not very efficient, only a small part of the electrical energy is turned into mechanical work.
A higher BL speaker motor will be more efficient, "transducing" more energy taken from amplifier.
Since speaker becomes more efficient, not acoustically since we are not changing piston movement so no change in SPL (useful work) but only moving just cone (more of it but at same speed / distance / acceleration) , useless work, transducer absorbs same energy as before, just doing more mechanical work (and wasting less).
Say instead of absorbing 1W and turning 1% of it into mechanical work, it still absorbs 1W but now turns 2% of it into mechanical work.
That it does not profit us is not the question asked.
🙄 Let's say you move doubled mass with the same acceleration and velocity by hands. Will the task be easier in this case?
Acoustic output is the same....but you have more mechanical force transferred into the driver frame/enclosure.
Thanks.
So the kinetic energy of the cone actually comes from the magnetic field of the magnet? And when the cone stops, is returned back? That would also be my initial reasoning but I can't think of it more rigorously. (We could think of a very high Cms not to step into the picture or consider the situation high above the resonance.)
OK. For the electrical-to-mechanical efficiency to increase, the electrical impedance must change (for the same 1W of input power the current must be different otherwise the heat generated would also be the same). How does that actually work in this case?
Energy doesn't come from the magnet it comes from the mains via the amplifier and is the product of the current through the voice coil and the voltage across the driver terminals. The current through the voice coil is the voltage provided by the amplifier minus the back emf created by the voice coil moving through the magnet field divided by the electrical impedance of the voice coil.
If you want to examine efficiency then the back emf, Blv, needs to be determined. V is the velocity of the cone and needs to be expressed in terms of known mechanical and acoustical quantities such as suspension stiffness and damping, cone mass, acoustical impedance. This sort of thing is present in most text books on acoustics but is also programmed in speaker design software and may well be included in the write up for such software.
If you want to examine efficiency then the back emf, Blv, needs to be determined. V is the velocity of the cone and needs to be expressed in terms of known mechanical and acoustical quantities such as suspension stiffness and damping, cone mass, acoustical impedance. This sort of thing is present in most text books on acoustics but is also programmed in speaker design software and may well be included in the write up for such software.
It seems to be the conversion from mechanical energy to acoustical energy which the most inefficient.
a = (Bl * i) / m = (2Bl * i) / 2m
For the same current the movement of the cone will be allways exactly the same no matter how heavy it will be, provided that Bl gets raised in the same ratio. But the work done to accelerate the cone will be higher and higher as the cone gets heavier and heavier. What actually does this work? And as the cone slows down and stops, where does this kinetic energy go?
It seems that the impedance will get (slightly) more and more capacitive. But what's really going on?
For the same current the movement of the cone will be allways exactly the same no matter how heavy it will be, provided that Bl gets raised in the same ratio. But the work done to accelerate the cone will be higher and higher as the cone gets heavier and heavier. What actually does this work? And as the cone slows down and stops, where does this kinetic energy go?
It seems that the impedance will get (slightly) more and more capacitive. But what's really going on?
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The radiated acoustic output in watts, W, in an old copy of Kinsler and Frey is:
W = (Bl)^2 * R_r * E^2 / (Z_m^2 * Z_I^2)
where R_r is the radiation resistance, E the voltage at the terminals, Z_m the mechanical impedance (acoustic impedance plus mechanical impedance of cone/coil assembly) and Z_I the electrical impedance at the terminals.
Don't know if that answers you question but the above quantities although functions of frequency are known/measurable so you can plot the output against frequency.
PS Do you understand the concept of impedance?
W = (Bl)^2 * R_r * E^2 / (Z_m^2 * Z_I^2)
where R_r is the radiation resistance, E the voltage at the terminals, Z_m the mechanical impedance (acoustic impedance plus mechanical impedance of cone/coil assembly) and Z_I the electrical impedance at the terminals.
Don't know if that answers you question but the above quantities although functions of frequency are known/measurable so you can plot the output against frequency.
PS Do you understand the concept of impedance?
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Does this document help?
https://adireaudio.com/wp-content/uploads/2018/12/Adire-Audio-Woofer-Speed-by-Dan-Wiggins.pdf
https://adireaudio.com/wp-content/uploads/2018/12/Adire-Audio-Woofer-Speed-by-Dan-Wiggins.pdf
Not at all under the conditions expressed in the example.
Mechanical force comes from B*L*I
If you double mass and double BL as stated, you do NOT double I ; what for?
I don't know. Try to explain and I will tell you if I get it 🙂
- You can think of the driver in a vacuum. It doesn't chage anything.
Not really, no. It just states what I state, without any deeper explanation. That's what I'm missing.
I don't quite follow what is not clear to you. You only have 2 differences (increased moving mass and motor strength) in 2 hypothetical drivers.
Fs=30Hz
Re=5
Vas=100 lit (50 lit); Vas reduction increases mass twice
Qes=0,4 (0,2); Qes reduction increases Bl twice
Qms=4
Sd=350cm2
Put these into a simulator.
Fs=30Hz
Re=5
Vas=100 lit (50 lit); Vas reduction increases mass twice
Qes=0,4 (0,2); Qes reduction increases Bl twice
Qms=4
Sd=350cm2
Put these into a simulator.
Who does know? Is it only the energy transfered through the amplifier (in and out) or (at least partly) the static magnetic field?
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Your question was already answered by Andy in post #7. Since the second driver's Bl is 2x the first one, it is also 2x as much efficient in energy transfer. That is the reason its spl is the same as with first one even though the second one has 2x more moving mass.
Yep.
In simple terms: if you doubled BL, you either doubled the magnet strength (B) or the length of the voice coil wire (L), or a combination of the two. To get the same current with the doubled BL, you need twice the voltage (this is very obvious if you doubled the wire length L). This means you doubled the electrical power going into the driver. This is were you double the energy.
No.
To double BL there are 2 "pure" ways and any intermediate combination of them, none of which requires drive voltage or power increase at all as you assume.
1) if you double B you double efficiency , no increased current or power at all.
2) if you double L and at the same time double section you keep resistance , you keep I as before, and efficiency doubles.
Only problem is voice coil mass doubles ... no big deal since the situation described *requires* moving mass doubling anyway.
Again no increased current, power or voltage at all.
3) or any combination of the above (partly increasing B, partly increasing L in any proportion)
4) while the end of your premise negates efficiency increase 😕
If you double energy consumed or power applied you have not increased efficiency, you are going against what we are discussing here. 😕
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