AndrewT said:
Hmmm.......
No, I read it and answered the question in terms of the conditions given.
Seems to me a complete fluke that your total misunderstanding of the
situation gives you a get out clause for a sealed box below resonance
rolling off at 12dB/octave. Nothing to do with the original question.
🙂/sreten.
Hi Sreten,
Sorry, you are due an apology for my misunderstanding and using that to misinterpret your answer.
But I still read your answer as constant SPL and not constant voltage.
I wonder if I am confusing power with energy?
Any thoughts?
Sorry, you are due an apology for my misunderstanding and using that to misinterpret your answer.
But I still read your answer as constant SPL and not constant voltage.
I wonder if I am confusing power with energy?
Any thoughts?
AndrewT said:
Hi,
that's because in terms of the question I didn't really care whether
its an "idealised" driver with flat response or a real driver EQed to
have a flat response in the frequency range of interest.
I took amplitude constant to mean either of the above, an area of flat response.
Energy = power x time, so I don't follow your question.
🙂/sreten.
This is an interesting thread. Sort of a "basics of drivers".
To me the original question isn't clear - maybe that's why the answers haven't been.
Amplitude of what? Sound pressure? That is what we might assume. But could it be the "amplitude" of the voltage, the current, the power? Nichol obviously isn't implying movement amplitude, because he asks if that changes.
Since I'm an amp guy, I tend to think of amplitude in terms of voltage. If I have an amplifier with a gain of 10, I want the output voltage to be exactly 10X the input voltage no matter what the frequency (within the BW).
So when I read Nichol’s question - I think voltage. If a constant voltage is swept across the voice coil, will cone excursion change with frequency? That seems to be what Andrew is getting at with his DC questions.
If the cone excursion at a given voltage does change with frequency - why does it change? I suspect the answer may be hidden in the nest of formulas posted above, but I can't see it.
(My apologies to Nichol if that is not what he was asking - it's just what I read when I see the question.)
To me the original question isn't clear - maybe that's why the answers haven't been.
Nichol1997 said:
Amplitude of what? Sound pressure? That is what we might assume. But could it be the "amplitude" of the voltage, the current, the power? Nichol obviously isn't implying movement amplitude, because he asks if that changes.
Since I'm an amp guy, I tend to think of amplitude in terms of voltage. If I have an amplifier with a gain of 10, I want the output voltage to be exactly 10X the input voltage no matter what the frequency (within the BW).
So when I read Nichol’s question - I think voltage. If a constant voltage is swept across the voice coil, will cone excursion change with frequency? That seems to be what Andrew is getting at with his DC questions.
If the cone excursion at a given voltage does change with frequency - why does it change? I suspect the answer may be hidden in the nest of formulas posted above, but I can't see it.
(My apologies to Nichol if that is not what he was asking - it's just what I read when I see the question.)
Hi,
there is a simple model you can apply to a 2nd order alignment,
sealed box, open baffle, tweeters, or a reflex box below port frequency.
(note a reflex box uses the driver Vas not the effective box stiffness)
The equivalent circuit of 2nd order is a LC (+some R) circuit.
L=mass, C= stiffness, R=damping 2nd order high pass.
Below the resonant frequency the river is in the stiffness region
where C dominates. As described by AT the excursion remains
constant, consequently output falls at 12/dB octave. Above the
resonant frequency is the mass region where L dominates, the
output remains constant, the cone velocity is constant, so this
means the excursion much reduce with increased frequency,
or increase with reduced frequency. If it stayed the same
output would rise at 12dB/octave. the other way of looking at
it is Sv's Fsquared or 1/Fsquared relationships for excursion.
To ask the right question you almost need to know the answer
already. So to answer a simplistic question you have to make
assumptions as to what is meant. My assumption is the basic
behaviour above box resonance was being enquired about.
🙂/sreten.
there is a simple model you can apply to a 2nd order alignment,
sealed box, open baffle, tweeters, or a reflex box below port frequency.
(note a reflex box uses the driver Vas not the effective box stiffness)
The equivalent circuit of 2nd order is a LC (+some R) circuit.
L=mass, C= stiffness, R=damping 2nd order high pass.
Below the resonant frequency the river is in the stiffness region
where C dominates. As described by AT the excursion remains
constant, consequently output falls at 12/dB octave. Above the
resonant frequency is the mass region where L dominates, the
output remains constant, the cone velocity is constant, so this
means the excursion much reduce with increased frequency,
or increase with reduced frequency. If it stayed the same
output would rise at 12dB/octave. the other way of looking at
it is Sv's Fsquared or 1/Fsquared relationships for excursion.
To ask the right question you almost need to know the answer
already. So to answer a simplistic question you have to make
assumptions as to what is meant. My assumption is the basic
behaviour above box resonance was being enquired about.
🙂/sreten.
So as frequency increases, so must cone speed. The cone has to “get there and back” faster.
That increase in speed means increased acceleration.
Higher acceleration needs more power.
But as the voltage is constant, there isn't more power applied.
Thus excursion must decrease.
Is that it?
Andrew's point is that the question was about a driver "held in free air" and you keep talking about boxes. If it's the same thing, say so. 😉
That increase in speed means increased acceleration.
Higher acceleration needs more power.
But as the voltage is constant, there isn't more power applied.
Thus excursion must decrease.
Is that it?
sreten said:
Andrew's point is that the question was about a driver "held in free air" and you keep talking about boxes. If it's the same thing, say so. 😉
Thank you everyone, I understand the relationship between cone movement vs frequency now. But, now I have a different question.
Here is my setup, I have an 18" subwoofer in a ported enclosure that is tuned to appoximately 14 Hertz with 500 Watts running to it. THe driver has an Xmax of 27 mm and a power handling of 800 watts. I am trying to understand why I can't get the cone to move peak to peak at higher frequencies. By higher frequencies I mean anything from 40 to 90 Hertz. I have a test tone cd and the subwoofer will really start to move when I play anything less than 30 Hertz. I understand the concept of Hertz. For instance, I know that at 60 Hertz the cone has to move back and forth 60 times in one second.
My question is, why can't I get the cone to move peak to peak with higher frequencies? One would think that by just giving it more power it will move more. On that thought, am I limited by the amount of power the voice coil can handle?
Here is my setup, I have an 18" subwoofer in a ported enclosure that is tuned to appoximately 14 Hertz with 500 Watts running to it. THe driver has an Xmax of 27 mm and a power handling of 800 watts. I am trying to understand why I can't get the cone to move peak to peak at higher frequencies. By higher frequencies I mean anything from 40 to 90 Hertz. I have a test tone cd and the subwoofer will really start to move when I play anything less than 30 Hertz. I understand the concept of Hertz. For instance, I know that at 60 Hertz the cone has to move back and forth 60 times in one second.
My question is, why can't I get the cone to move peak to peak with higher frequencies? One would think that by just giving it more power it will move more. On that thought, am I limited by the amount of power the voice coil can handle?
WHat are the T/S parameters?
Model it and you will see that the excursion drops very fast. It reduces by 4 for each octave you increase, which means that you would need 16 times the power to reach the same excursion., then 16*16 or 256 times the power the octave above that.
Model it and you will see that the excursion drops very fast. It reduces by 4 for each octave you increase, which means that you would need 16 times the power to reach the same excursion., then 16*16 or 256 times the power the octave above that.
Ron E said:
The driver is a Ascendant Audio Avalanche 18.
Okay, so the only thing that is limiting the driver from moving peak to peak at 60 Hertz is the amount of power I have and the amount of power the voice coil can handle, correct?
I thought there might be some kind of physical constraint like a cone can only move so fast.
I think it gets back to Newton's second law again, which is what Ron was talking about above
At 60 Hz the cone has to move twice as fast as it does at 30 Hz. Right? Soooo, you are going to need twice as much power to move it the same distance twice as fast.
You can throw more and more power at it, but there comes a point when the voice coil starts to heat up and you get power compression. You either run out of power or the coil burns out.
There may be increased resistance from the cone suspension with increased speed - I don't know about that. Ron probably does.
At 60 Hz the cone has to move twice as fast as it does at 30 Hz. Right? Soooo, you are going to need twice as much power to move it the same distance twice as fast.
You can throw more and more power at it, but there comes a point when the voice coil starts to heat up and you get power compression. You either run out of power or the coil burns out.
There may be increased resistance from the cone suspension with increased speed - I don't know about that. Ron probably does.
Nichol1997 said:
The 800 watts that the driver can handle is limited by the heat in the voice coil. If you feed the driver more than 800 watts, there is a risk that the glue holding the coil together is damaged due to the heat. But there is another limit too, and that is xmax for the driver, if the cone moves more than xmax permits, the sound is distorted*.
Now, since the cone excursion varies with frequency, these two factors limit the maximum output level in different frequency ranges. The black curve is the response of the driver. The thin gray curve is the maximul output level of the system. The red curve (which is partially hidden under the thick gray curve) shows the margin from the response until xmax is reached. The black dashed cuve shows the same for the power dissipation in Re.
The resulting margin, for which neither of Xmax or Pmax is exceeded is the thick gray curve. If the thick gray curve is added to the response, this results in the maximum output level, the thin gray curve.
So, it is understandable that you cannot reach xmax for higher frequencies, since the maximum input there is limited by (available and sustainable) power and not xmax.
An externally hosted image should be here but it was not working when we last tested it.
*There are of course gray zones here, but for the sake of understanding this is a good model.
Hi,
Or in other words if you could xmax it above 50Hz you would then be
very likely to have overexcursion / distortion / endstop thumping lower
down the frequency range, especially with a port frequency of 14Hz.
Also note Svante curves are typical for a sealed box, vented looks different.
🙂/sreten.
Or in other words if you could xmax it above 50Hz you would then be
very likely to have overexcursion / distortion / endstop thumping lower
down the frequency range, especially with a port frequency of 14Hz.
Also note Svante curves are typical for a sealed box, vented looks different.
🙂/sreten.
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