I was curious about how driver excursion in a headphone corresponds to the volume that music is playing at.
For this question, let's use my current headphone as an example. It's the Audio Technica MSR7NC, a noise cancelling variant of the popular MSR7's.
Here are its measurements and specifications when the noise cancelling is off…
and here are its measurements and specifications when the noise cancelling is on.
So when I plug my headphones in to recharge them, I typically leave on the noise cancelling. According to the first chart I provided, the noise cancelling provides up to 15 decibels of noise cancellation in some frequencies, and a bit less in others.
Given this, for simplification, let's say that this results in the headphones "playing" some frequencies at 15 decibels for the 18+ hours that I'm not using them.
Then, let's say that I listen to the headphones for a few hours at normal listening levels, say 65-70 decibels.
I understand that decibels are logarithmic, so 65 decibels should be 10,000 times louder than 15 decibels, but I find it hard to believe that the drivers in the headphones are doing 10,000 times more work.
Does anyone have a way to estimate the excursion made by headphone drivers based on their diameter (45mm) and their volume of 15 decibels, at least relative to their excursion at 50 decibels louder? Finally, I want to be able to measure the total excursion of my headphone drivers after a given amount of time (like hours) at a given decibel. How many hours of active noise cancelling is equal to a few hours of normal volume listening in terms of driver excursion?
I want to know this because I don't want to switch my headphone's noise cancelling on and off all the time as the switch seems a little flimsy, but I also want to know what I'd be putting the drivers through this way (I charge them right next to my pc which is computing 24/7).
Thanks!
Best Answer
Acoustic power (measured in Watt) or sound instensity (measured in Watt/m^2) can be compared logarithmically via the bel (B) although the 1/10th of the unit is commonly used, the decibel (dB). So 65 dB (6.5 B) is [6.5 - 1.5 = 5] 10^5 times the acoustic power and sound intensity, respectively, of 15 dB (1.5 B), i.e. 100,000 times as high. Since work is power * time, indeed, the drivers are doing 100,000 times as much work in a given time at 65 dB vs 15 dB.
However, the sound pressure level (SPL) has a square-root-relation to power, i.e. 65 dB is \$ \sqrt{100,000} \$ ~ 316 times the SPL of 15 dB. Now, driver excursion happens to be proportional to SPL at a fixed frequency / sound signature/frequency spectrum. So it is ~ 316 times smaller at 15 dB than at 65 dB. Here is a convenient tool to directly calculate/estimate driver excursion from driver size, SPL and frequency: http://www.baudline.com/erik/bass/xmaxer.htm
As for subjective human perception, loudness is used and can be estimated from SPL via Steven's power law (https://en.wikipedia.org/wiki/Stevens%27s_power_law): For acoustic stimuli it states that the perceived loudness has a cubic-root-of-the-square, i.e. \$ x^ {\frac 2 3} \$ -relation to SPL. So for the 65 dB vs 15 dB-case it is: \$ (\sqrt{10^5})^{ \frac 2 3} \approx 316^{\frac 2 3} \approx 46 \$. So we perceive 65 dB as about 46 times as loud as 15 dB. You may have heard the rule of thumb that a 10 dB rise in SPL is perceived as twice as loud which is at least in the same ball park (\$ 2^5 = 32 \$).
Edit: Thanks very much to Transistor for converting all the crude mathematical content into the nice format it is now in (and thereby also teaching me how to)!