Sreten is right (as usual). If the frequency is lowered, the cone moves more. If the box is a closed box, the cone amplitude will be approximately proportional to 1/f² down to the cutoff frequency of the subwoofer. Below that frequency the amplitude of the cone will be approximately constant, but the sound level will decrease, approximately proportional to f².
Hi,
I am kind of new to this speaker lark, but that does not sound right.
I imagine a DC of 1Volt will cause the speaker Voice Coil to move to a new position. Let's suppose for a moment it moves 5mm.
Now apply a -1Vdc to the speaker and it will move the other way 5mm( provided the motor is symetrical).
Now apply a very low frequency square wave of 2Vpp and the speaker will move 10mmpp i.e. +-5mm.
Now apply a very low frequency sine wave of 2Vpp and again the speaker cone will move +-5mm (10mmpp).
I imagine that this speaker will continue to move 10mmpp as the frequency is increased to audio frequencies and continue to do so through some of the audio range.
A big exception is the range around resonance when the movement will increase substantially.
At high frequency the movement will decrease from the standard 10mmpp.
What is the hf displacement reduction frequency? I suspect it is related to energy absorbed by accelerating the cone and the inductance of the VC. Any predictions?
I am kind of new to this speaker lark, but that does not sound right.
I imagine a DC of 1Volt will cause the speaker Voice Coil to move to a new position. Let's suppose for a moment it moves 5mm.
Now apply a -1Vdc to the speaker and it will move the other way 5mm( provided the motor is symetrical).
Now apply a very low frequency square wave of 2Vpp and the speaker will move 10mmpp i.e. +-5mm.
Now apply a very low frequency sine wave of 2Vpp and again the speaker cone will move +-5mm (10mmpp).
I imagine that this speaker will continue to move 10mmpp as the frequency is increased to audio frequencies and continue to do so through some of the audio range.
A big exception is the range around resonance when the movement will increase substantially.
At high frequency the movement will decrease from the standard 10mmpp.
What is the hf displacement reduction frequency? I suspect it is related to energy absorbed by accelerating the cone and the inductance of the VC. Any predictions?
AndrewT said:
Svante and sreten know what they are speaking of.
At DC, the only thing that matters is stiffness, at ~3x resonance and up all that matters is mass, neglecting inductance. At resonance, the contributions of stiffness and mass are equal, and damping dominates.
1/f^2 is correct - every time you halve frequency, excursion increases by 4 for the same SPL. The reason sealed boxes don't double with each halving below resonance is because the 12dB/oct rolloff reduces excursion requirements.
A speaker that had constant excursion with frequency would have a 12dB/oct rising response....
The excursion formula depends on whether the speaker is enclosed or not and what sort of enclosure there is. In free air, it acts the same as if it were in a very large box - like 10000 cubic feet or so so you can enter that into a program that calculates escursion.
I have WinISD beta but it does not model the excursion of the driver. I just downloaded the pro version but it is asking for parameters of the driver that I don't have.
Is there a formula for excursion vs frequency for a ported enclosure? It is tuned to 14 Hertz. I know that around the tuning frequency the cone of the driver won't be moving much and most of the SPL will be coming from the port. So there probably is a complex relationship between the two, but for higher frequencies I would imagine there is a formula.
Is there a formula for excursion vs frequency for a ported enclosure? It is tuned to 14 Hertz. I know that around the tuning frequency the cone of the driver won't be moving much and most of the SPL will be coming from the port. So there probably is a complex relationship between the two, but for higher frequencies I would imagine there is a formula.
Maybe this graph helps of a butterworth aligned closed box helps?
The black curve is the response of the system, the red is the cone amplitude. Above the cutoff frequency (50 Hz) the cone amplitude is tilted by -12 dB/oct but the response is flat. Below the same frequency the response tilts by 12 dB/oct and the cone amplitude is flat.
It is at the low frequencies that the intiution works best.
An externally hosted image should be here but it was not working when we last tested it.
The black curve is the response of the system, the red is the cone amplitude. Above the cutoff frequency (50 Hz) the cone amplitude is tilted by -12 dB/oct but the response is flat. Below the same frequency the response tilts by 12 dB/oct and the cone amplitude is flat.
It is at the low frequencies that the intiution works best.
Nichol1997 said:
It may be indirect, but go to
http://www.diysubwoofers.org/prt/ported3.htm
The SPLmax is the SPL at Xmax+15% assuming that the suspension is linear. Excursion and SPL will actually be slightly less than predicted due to nonlinearities.
In the simple formulas on that page, you can use Fb=0 to simulate a sealed box.
Have fun!
Cal Weldon said:
Just the musings of an idiot here (me, not Cal), but could it be that there is a direct relationship between wave length and excursion for a given SPL?
As Cal points out, a DC signal just pushes the cone until it hits some sort of physical restriction. I suppose there is no specific SPL per se in a DC thud.
And a sine wave pushes out to its apex, then reverses until its nadir. So, if the cone were to move at a constant speed (constant voltage?), a low frequency wave would have more time, and thus distance, to travel before reversing direction than a higher frequency.
But, is this relationship linear with frequency for a given voltage? Squared for a given SPL?
max
yes, all three of you forgot to read the question.sreten said:
That is almost how a speaker works in a linear fashion (+- small non linearites). Apply 1mV and it moves 1unit. Apply 10mV and it moves 10units. Apply 1V and it moves 1000units. Apply a battery (1v5) and it moves 1500units. Try it!!!Cal Weldon said:
That is exactly the relationship that Sreten & Co. were quoting.
not quite, it doesn't hit a stop. The force generated by the coil reacting against the magnetic field BALANCES the force from the surround and spider. increase the drive force (more current) and the suspension has to displace farther to generate more restoring force to maintain the balance (for every force there is an equal and opposite reaction, some quote by a long dead Englishmen, Yeah I know sometimes the English get it right).
a low frequency at constant voltage drive moves at low speed and a high frequency moves at high speed. Now change it to constant power input (nearly constant SPL) and the SPEED of cone movement is substantially constant for constant power (that is my understanding of constant SPL, am I wrong?).
Good question, Clarify it for all of us?
Nichol1997 said:
Actually I missed this very clear question in the very first post...
The complex formulation of the cone amplitude in a closed box is:
x(w) = K / (1 + jw/(Qtc*wc) + (jw/wc)²)
Neglecting the phase, by taking the absolute value this turns into:
|x(w)| = K / sqrt( (1-(w/wc)²)² + (w/(Qtc*wc))² )
...where wc is 2*pi*fc, fc is the resonant frequency in the box, Qtc is the Q of the driver mounted in the box.
It can be seen that at low frequencies this turns into
|x(wlow)|=K
ie independent of frequency, and for hight frequencies:
|x(whigh)|=K / (w/wc)²
ie sloping by -12dB/octave (due to the 1/w² relation).
Hope I got it right...
the power of a voltage sine wave doesn't depend on its frequency, at a given constant amplitude. unless the load the wave is applied to is frequency dependant, of course.
constant SPL is constant power (power of the electric signal)(with a flat response speaker), and constant power is constant amplitude (amplitude of the electric signal) (with the same flat response speaker, wich also has a constant impedance). in order to fullfill all these conditions, this speaker has to increase its excursion to continue dlivering constant SPL, as frequency decreases.
having said all that, this is wrong:
"a low frequency at constant voltage drive moves at low speed and a high frequency moves at high speed"
Hi,
sounds like my thinking is **** for elbow. (damned dictionary won't let me use a collocquial phrase).
Is that right?
sounds like my thinking is **** for elbow. (damned dictionary won't let me use a collocquial phrase).
seems to be saying that the constant amplitude I have been arguing ONLY applies below resonance.
Is that right?
I think this needs clarification:
At low frequencies, below resonance fs, this is true:
-The movement is spring controlled ie x=F*Cms *
-The cone excursion is proportional to the applied voltage.
-The cone velocity (speed) is iproportional to f.
-The cone acceleration is proportional to f².
At higher frequencies, above the resonance fs, this is true:
-The movement is mass controlled ie a=F/Mms **
-The cone acceleration is proportional to the applied voltage.
-The cone velocity (speed) is proportional to 1/f.
-The cone excursion is proportional to 1/f².
*Cms is the compliance of the driver, x is the excursion, F is the driving force (infinite baffle assumed)
** Mms is the moving mass of the driver, a is cone acceleration.
At low frequencies, below resonance fs, this is true:
-The movement is spring controlled ie x=F*Cms *
-The cone excursion is proportional to the applied voltage.
-The cone velocity (speed) is iproportional to f.
-The cone acceleration is proportional to f².
At higher frequencies, above the resonance fs, this is true:
-The movement is mass controlled ie a=F/Mms **
-The cone acceleration is proportional to the applied voltage.
-The cone velocity (speed) is proportional to 1/f.
-The cone excursion is proportional to 1/f².
*Cms is the compliance of the driver, x is the excursion, F is the driving force (infinite baffle assumed)
** Mms is the moving mass of the driver, a is cone acceleration.
AndrewT said:
Sealed box is principally identical, it is just that the resonace frequency is slightly shifted upwards.
Bass reflex box is similar except around the helmholtz (box) resonance where the cone amplitude has a local minimum.
Tranmission line is similar to the bass-reflex box, in turn.
So, for the understanding, they are pretty much the same, at least a bit away from the system resonances.
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