[kwkarth]

Quote:

The only reason any of this is matters is you were trying to convince people they didn't have to worry about the Asgard transient as it was long and slow. And I still maintain that's not true. A brief fast transient will result in less driver excursion (and also less thermal dissipation).
 
As I said I can take my bench power supply and ever so slowly and gently ramp up DC into any headphone until they're damaged. The driver will still bottom out despite the gentle ramp. But I can probably apply that same "lethal voltage" to the same headphones for 1 mS and they'll easily survive.
 
Next time I'm at one of the shows with the headphone vendors I will make it a point ask them about turn on/off transients and what sort of damage they've seen, etc. We already have an answer from AKG that even a few tenths of a volt can be dangerous.
 
The issue with the calculator that Joe and I have been trying to get across is SPL is proportional the the voltage applied. So just because SPL is involved doesn't render the whole thing irrelevant as you keep suggesting. I was only offering it as evidence excursion decreases with increasing frequency. The calculator is assuming flat frequency response. So 1 volt at 5000 hz is 105 dB SPL and 1 volt at 50 hz is still 105 dB SPL. But the 50 hz example will result in far more excursion for the same voltage than 5000 hz. I wasn't suggesting trying to use the calculator to provide the exact mm of excursion.

This is just crackers.  You still don't get it do you?  

 
 

 

[kwkarth]

Quote:

But if you don't know anything about the driver design and only know the voltage applied, you won't know the driver excursion either.
 
My reasoning was thus:
1. a driver with flat frequency response would give the same dB SPL at the same volatage no matter what frequency it is applied at and the driver excursion would be as given in the calculator if the enclosure assumptions are met (no mention of the "subwoofer" on the page, I agree it's a speaker-centric page, however open headphones should be effectively equivalent to infinite baffle)
2. when a frequency response chart if measured like this

It is done my driving the headphones with tones of different frequencies of the "same voltage"
http://www.headphone.com/learning-center/about-headphone-measurements.php
 
3. Therefore, when you drive the K701 with a tone at 1kHz and another tone at 10Hz and you reach, say, 30dB SPL with the 1kHz tone and 20dB SPL (less 10dB) with the 10Hz tone, you know that you have driven them with equal voltage although you don't know what that voltage is--because 10dB is the difference between 1kHz and 10Hz on the graph.  (not strictly accurate because this graph is measured through a dummy head;  if the graph were measured through a simple acoustic coupling this should be accurate)
 
Knowing the frequency response you can convert from dB SPL to voltage and vice versa while solving for the driver excursion relative to different frequencies;  without the frequency response you wouldn't be able to solve for the driver excursion no matter whether you start with dB or volts.

I'm not sure I'm following your line of reasoning correctly, what are you trying to say?  Where are you going with this?
 
 

 

[Joe Bloggs]

Quote:

 
The issue with the calculator that Joe and I have been trying to get across is SPL is proportional the the voltage applied. So just because SPL is involved doesn't render the whole thing irrelevant as you keep suggesting. I was only offering it as evidence excursion decreases with increasing frequency. The calculator is assuming flat frequency response. So 1 volt at 5000 hz is 105 dB SPL and 1 volt at 50 hz is still 105 dB SPL. But the 50 hz example will result in far more excursion for the same voltage than 5000 hz. I wasn't suggesting trying to use the calculator to provide the exact mm of excursion.
 

Or if you solve for dB instead of excursion in the calculator, you find that for the same excursion 500Hz creates exactly 40 more dBs than 50Hz.  Ditto 1000Hz vs 100Hz.  So you could conceivably have drivers that respond with the same excursion to the same voltage applied at whatever frequency if these drivers' frequency response peak at 20kHz, are down 40dB at 2000Hz, 80dB at 200Hz, 120dB at 20Hz

 

[kwkarth]

Quote:

Or if you solve for dB instead of excursion in the calculator, you find that for the same excursion 500Hz creates exactly 40 more dBs than 50Hz.  Ditto 1000Hz vs 100Hz.  So you could conceivably have drivers that respond with the same excursion to the same voltage applied at whatever frequency if these drivers' frequency response peak at 20kHz, are down 40dB at 2000Hz, 80dB at 200Hz, 120dB at 20Hz

Yes, and your point?  The acoustic coupling of transducer to air becomes less and less efficient as frequency decreases because the wavelength gets longer and longer.  So to maintain the same SPL, as you lower the frequency, driver excursion or diameter must increase to maintain SPL.  There is no application of this to the headphone issues we're discussing here.
 
 

 

[Joe Bloggs]

Quote:

I'm not sure I'm following your line of reasoning correctly, what are you trying to say?  Where are you going with this?
 
 

 
Simply put, that it's possible to get accurate relative numbers for excursion from that piston excursion calculator simply by disregarding the unrealistic flat frequency response assumption and inserting realistic frequency response into the calculation.  Example (for how many times?)
 
Suppose you know your drivers to be 1 inch in diameter and have a sensitivity of 100dB/V @ 1kHz.  Plugging 1,1,1000,100 into the calculator gives you 1.1mm Xmax.  Now you know that your drivers are moving 1.1mm @ 1V @ 1kHz, because 100dB->1V at 1kHz.
 
Now suppose you know that the frequency response of the drivers are such that at 4Hz they are 50dB down from 1kHz.  Therefore, their sensitivity is 50dB/V @ 4Hz.  Plugging 1,1,4,50 into the calculator gives you 218mm Xmax.  Now you know that your drivers are moving 218mm@1V@4Hz, because 50dB->1V at 1kHz.
 
Now if this were true of headphone drivers driving them with a 1V 4Hz signal would make them jump right out of the phones.  The one big problem I can think of is that the dB SPL is measured at 1m away, which is not how you measure headphones.  If headphones, even big open air ones, could make a sound as loud as 100dB from 1m away, you'd be seriously worried about their health.
 
So I think the above calculation method could give you an accurate picture of the RELATIVE amount of excursion at different frequencies but not the absolute amount.

 

[Joe Bloggs]

Quote:

Yes, and your point?  The acoustic coupling of transducer to air becomes less and less efficient as frequency decreases because the wavelength gets longer and longer.  So to maintain the same SPL, as you lower the frequency, driver excursion or diameter must increase to maintain SPL.  There is no application of this to the headphone issues we're discussing here.
 
 

 
Are you saying that the calculations don't work because frequency response is to be measured coupled to the headphones?  I guess you have a point there--is there a formula for how head coupling affects the excursion / SPL relationship?
 
And K702s are open headphones?  They are open on both sides when not coupled to anything as in the driver wrinkling video.

 

[nwavguy]

Quote:

 
Now if this were true of headphone drivers driving them with a 1V 4Hz signal would make them jump right out of the phones.  The one big problem I can think of is that the dB SPL is measured at 1m away, which is not how you measure headphones.  If headphones, even big open air ones, could make a sound as loud as 100dB from 1m away, you'd be seriously worried about their health.
 
So I think the above calculation method could give you an accurate picture of the RELATIVE amount of excursion at different frequencies but not the absolute amount.

Thanks Joe, I appreciate your help and confirmation here. I had pretty much given up trying to get my point across. But not to get too far off topic, as I keep saying, all this only matters to point out big fat long duration "thumps", or whatever you want to call the Asgard transient, create more excursion, and hence a higher risk of damage than the more typical short fast "clicks" even if they're the same voltage. There's nothing "crackers" about that.
 

 

[kwkarth]

Quote:

Simply put, that it's possible to get accurate relative numbers for excursion from that piston excursion calculator simply by disregarding the unrealistic flat frequency response assumption and inserting realistic frequency response into the calculation.  Example (for how many times?)
 
Suppose you know your drivers to be 1 inch in diameter and have a sensitivity of 100dB/V @ 1kHz.  Plugging 1,1,1000,100 into the calculator gives you 1.1mm Xmax.  Now you know that your drivers are moving 1.1mm @ 1V @ 1kHz, because 100dB->1V at 1kHz.
 
Now suppose you know that the frequency response of the drivers are such that at 4Hz they are 50dB down from 1kHz.  Therefore, their sensitivity is 50dB/V @ 4Hz.  Plugging 1,1,4,50 into the calculator gives you 218mm Xmax.  Now you know that your drivers are moving 218mm@1V@4Hz, because 50dB->1V at 1kHz.
 
Now if this were true of headphone drivers driving them with a 1V 4Hz signal would make them jump right out of the phones.  The one big problem I can think of is that the dB SPL is measured at 1m away, which is not how you measure headphones.  If headphones, even big open air ones, could make a sound as loud as 100dB from 1m away, you'd be seriously worried about their health.
 
So I think the above calculation method could give you an accurate picture of the RELATIVE amount of excursion at different frequencies but not the absolute amount.

Sorry, this is not a relevant or valid approach.

 
 

 

[Joe Bloggs]

Thanks for the succint and informative explanation.

 

[nwavguy]

Quote:

 
Sorry, this is not a relevant or valid approach.
 

Well it's two against one. Anyone else want to join in? Anyone had headphones that only bottomed out on deep bass notes? There's a reason for that.
 
 

 

[Joe Bloggs]

I think if we obtained a frequency response of a pair of headphones as measured with an uncoupled mic spaced some distance away from the phones instead of coupled directly to the phones, the approach I outlined above would be very valid.  Anyway phones removed from the head do have no bass but is it 50dB's worth of no bass?
 
edit: the replies can't be coming in much faster if this were on instant messaging, eh?

 

[nwavguy]

In the name of science I just put my ultra valuable very high-end Sennheiser HD 201s to the test on my arbitrary wave form generator which can do a single cycle "burst" of a sine wave at any frequency down to 0.000001 hertz (really!). Sure enough, they bottom out (make ugly noises) far easier as you go down in frequency. So I'm satisfied even if kwkarth isn't. Case closed.

 

[Shike]

To increase SPL the driver MUST move closer to xmax.  To produce bass frequencies of equal SPL as others, xmax or driver size must be increased over other frequencies.  Since we can't increase size the driver must be pushed and pulled further, meaning higher chance for bottoming out.
 
I think we're all on the same page here ^
 
However,
 
I'm a bit confused by the calculator too.  If the headphones are theoretically flat, and mostly a resistive load (some are) . . . then the power applied is going to be the same regardless of frequency.  At that point it becomes the FR curve that determines what's going to bottom out.  If the raw FR shows them being flat the bass frequencies will, and even in some cases where it's not due to the amount of xmax necessary to achieve those lower frequencies.  Basically, it isn't necessarily getting louder just being at the same voltage though, but it may induce more mechanical stress due to the inherent FR curve.
 
Then again, frequency is also a wave . . . the faster the wave the higher it goes.  As such a faster/higher frequencies seems to help prevent bottoming out inherently, I think?
 
 
. . . I think I'm going to take a break, this theoretical stuff drives me up the wall.

 

[maverickronin]

Quote:

Well it's two against one. Anyone else want to join in? Anyone had headphones that only bottomed out on deep bass notes? There's a reason for that.

My HD650s did that in a big way once when I was testing their bass extension with a sine wave generator.

 

[jcx]

dynamic headphones are similar to, but a little different from loudspeakers radiating into a room
 
you can see a bass frequency mass-spring resonance in some headphone's electrical impedance graph - at frequencies higher than the peak the diaphragm motion is "mass controlled" the force provided by voice coil current x magnet field is integrated to frequency dependent velocity, and a 2nd time for displacement so the x_max falls quickly as frequency increases - there are complicating issues with back EMF, acoustic radiation impedance vs frequency, diaphragm break up that conspire to give relatively "flat" acoustic frequency response over most of the audio range
 
below resonance the diaphragm surround/suspension spring controls the motion turning the voice coil force into a displacement of the spring - for really low frequencies this is basically a constant x_max determined by the force from the current from amp output voltage/coil DC resistance x the magnet field strength pushing against the "spring" - the volume of the air trapped by (some styles) earcups also adds to the spring stiffness - you may see more motion with the driver in open air
 
diaphragm displacement is typically max at the bass mass-spring resonance frequency for constant amplitude V sine wave drive, usually somewhat less, to flat for frequencies below resonance