Approximating Headphone Volume Output (dB)

[tinyman392]

I originally posted this on another forum (iFans) here: http://www.ifans.com/forums/showthread.php?t=364522
 
I do think that it is an important concept to get down for a headphone forum, so I am going to repost here as well.  Some disclaimers before I begin, I have double checked the units and they do add up to dB when multiplied out (stage by stage), so that part is accurate.  However, I still have to warn that these results may not be 100% accurate and will only give you a ballpark estimate at your actual listening levels.  The best way to get the actual number would to get a dummy head and an SPL meter (kind of expensive).  This method requires a calculator, and a little understanding of high school physics.
 
On the other forum I did include a quick and dirty command line program (Windows only, sorry Mac and Linux users) that would calculate it for you granted you inputted correct information.  I don't know what the rules are like here concerning executables, so I won't upload it here.  It is uploaded on the other forum if you need it (it comes with source code to be compiled for other machines; use a C-based compiler).
 
Again, no guarantees of 100% accuracy, but it will get decent numbers to give you a good estimate.  Let me know if you have any questions.  This guide also assumes that your volume bar increases voltage at a parabolic rate (what I found most devices to do: phones, iPods, laptops, etc).
 
Quote: Tinyman392

I don't know how many of you actually pay attention to this. However, a large volume output can really damage your hearing, even further, anything over ~75-85 dB can cause your ears to tune out low and high frequencies. This in turn can actually decrease the quality of your music (Inner Fidelity). Anything over 90 dB can damage your hearing if exposed too long.

Ideal listening levels should be around 60-80 dB. I generally listen myself around 70 dB (calculated). So how exactly do we calculate this (estimated) listening level? Well it's simple, first we need some information though:

1. Maximum voltage output of your device (search google for this), the iPod Touch 4G puts out a maximum of 1.110 v (60 mW max or 30 mW per channel).
2. Headphone specs (specifically the impedance and the sensitivity ratings) - Find it on the box of your headphones, or online through the manufacturer's website (or even the reseller's site sometimes)

Disclaimer: this guide will not get you exact numbers. It will merely give you a ballpark estimate of how loud you are actually listening. To get actual numbers, obtain a mic, dummy head and test your own headphones.

So let's start by calculating the actual voltage that comes out of your headphone jack. When you push up on your volume, you actually don't increase in a linear, straight manner. Instead, it's squared. To get the voltage your increasing use this formula below:

Actual Voltage (volts) = Max Output Voltage (volts) * (volume level fraction)^2*

The volume level fraction is actually the amount of the total volume you are outputting. So if you listen at half the max volume of the iPod, your volume level fraction is 1/2 = .5. If you listen to it at 30%, then the fraction would be 30/100 = 3/10 = .3. If you listen at 3/8, then your fraction would be 3/8 = .375. It's important that you keep this a fraction under 1.

The next thing we need is actually the current going through your headphones. The current is generated using the formula below for current through a parallel circuit.

Current = Voltage * (1/R1 + 1/R2 + 1/R2 + 1/R4...); where R1, R2, R3, etc are resistors. Your headphones have two of these both rated the same. So we can simplify that whole statement to:

Current (amps) = Voltage (volts) * (2/R); Where R = Impedance rating (ohms) of the headphones. Thusly, we get the equation:

Current (amps) = Voltage (volts) * 2 / Impedance (ohms)*

With the voltage and current we can calculate power generated by your device at a given volume level. This is why we calculated the current and voltage. So use the equation below:

Power = Current (amps) * Voltage (volts)*

Or if you want the long one:

Power = (Voltage (volts) * 2 / Impedance (ohms)) * (Max Output Voltage (volts) * (volume level fraction)^2)

Now to get the actual sound pressure level (SPL) coming out of your headphones (note this is in Pa), we can use the formula below:

SPL (Pa) = Sensitivity * Power*

Or the long one:

SPL (Pa) = Sensitivity * (Voltage (volts) * 2 / Impedance (ohms)) * (Max Output Voltage (volts) * (volume level fraction)^2)

Now, this number is going to range anywhere from 0 to 15 (maybe higher if you have special headphones). We need to convert this number into one we can use. To convert the SPL (Pa) to SPL (dB) to get the decibel rating we want you can do one of two things:

Use this converter*

Or you can use this formula:

SPL (dB) = 20 log(SPL (Pa) / (2*10^-5)) / log(10) (dB)
*We are stating that the log has a base of ten when we divide by log(10). On any given calculator, the log button has a default base of 10 (check this for your self), that is, log(10) = 1. Ln is the natural log which has a base of e. In the equation above, log can be replaced with ln, but you'll have to divide by ln(10) still since ln(10) is not 1.

So we get the final equation as:

SPL (dB) = 20log(SPL (Pa) / (2*10^-5)) dB*

Plugging in what SPL (Pa) is, you get the long equation:

SPL (dB) = 20 log(Sensitivity * (Voltage (volts) * 2 / Impedance (ohms)) * (Max Output Voltage (volts) * (volume level fraction)^2) / (2*10^-5))

^^If you do this equation, don't type it in wrong

And double check your work).

Please note you'll get more accurate results (since you'll make less human errors) if you just calculate each thing and right down the number it produces and plug those into the small equations (which are marked with a star, *, at the end of them.. The big equations scare me (especially the last one)... If you are using the bigger equations, just skip to the final equation; plug and chug it all down and hope you didn't type anything in wrong

And there you have it, the actual listening level you are listening at (approximated). Now let's see how scary your numbers are... Happy (safe) listening everybody.
 
Sources used: http://www.innerfidelity.com/content/loud-music-sucks

 

 

[itchyblood]

what's the max voltage output of a fiio e9?  I just find mW and ohms for that value.

 

[Head Injury]

Quote:

what's the max voltage output of a fiio e9?  I just find mW and ohms for that value.

You can determine it from the ohms and mW. V = sqrt(R*W), R being ohms and W being mW/1000.
 
I'll save you the trouble and say it's about 7 Vrms max, less into low impedances.
 
Ah, on the subject, that's something the OP is missing. Output impedance can reduce the voltage into low impedances. Some amps are also current limited, so will get reduced power into low impedances.

 

[tinyman392]

Based on the 2V RMS conversion to peek voltage output, V RMS = V Peak / sqrt(2); the peak voltage is 2.8284 volts.  Please do note that this amp may not do do power output in a squared fashion like I have above.  It may actually do linear, but I"m not sure.  If it does do linear, don't square your (% of volume) when you do the actual voltage calculation...  I do know there are some amps that do this, I think FIIO might be one of those amps (don't take my word for it though). 
 
Another way to get actual voltage output is to get an aux cable and connect it to a volt meter (one end goes to grounding contact while the other will go to one of the other contacts for either left or right).

 

[tinyman392]

Quote:

You can determine it from the ohms and mW. V = sqrt(R*W), R being ohms and W being mW/1000.
 
I'll save you the trouble and say it's about 7 Vrms max, less into low impedances.
 
Ah, on the subject, that's something the OP is missing. Output impedance can reduce the voltage into low impedances. Some amps are also current limited, so will get reduced power into low impedances.

The post was mainly intended for portable devices as it came from an iPod forum, so I didn't modify it for anything.  That MaxVoltage*(%volume)^2 is actually a really special function that is unique to most laptops and mobile devices, as this is normally how they are hooked up to deliver voltage outputs.  That multiplier may have to be changed for different amps and dacs.  As you increase the volume on different devices, the actual increase in voltage may not follow that same parabolic increase I saw on my iPod.  Instead, it could be linear, or some other function of the volume output. If you know the actual voltage output (EG you measured it), you can change that whole equation there to just the actual voltage.  Otherwise, there will be some sort of relationship to the %volume and actual voltage output (with respect to the max voltage output): linear, squared, cubed, etc.
 
But you are correct, this wasn't made with amps and other things in mind, for that I apologize. 

 

[Head Injury]

Quote:

Based on the 2V RMS conversion to peek voltage output, V RMS = V Peak / sqrt(2); the peak voltage is 2.8284 volts.  Please do note that this amp may not do do power output in a squared fashion like I have above.  It may actually do linear, but I"m not sure.  If it does do linear, don't square your (% of volume) when you do the actual voltage calculation...  I do know there are some amps that do this, I think FIIO might be one of those amps (don't take my word for it though). 
 
Another way to get actual voltage output is to get an aux cable and connect it to a volt meter (one end goes to grounding contact while the other will go to one of the other contacts for either left or right).

Where are you getting 2 Vrms for the E9? That's the line out, not headphone out. Use the power output at 600 ohms.
 
V = sqrt(600 x 0.08) = sqrt(48) = 6.9 Vrms
 
And at 16 ohms:
 
V = sqrt(16 x 1) = 4 Vrms
 
You can't ignore max current and output impedance even if you ignore amps. Portable players will be current limited into low impedance loads, and some will have impedances greater than 1 ohm. Using max voltage makes for a good estimate, but the only way to know for sure is to measure it.

 

[itchyblood]

Okay Thanks guys!

 

[jcx]

the Headwize site is archived, try the library:
 
http://gilmore2.chem.northwestern.edu/library.htm
 
 
specifically for hearing safety:
 
http://gilmore2.chem.northwestern.edu/articles/hearing_art.htm

 

[khaos974]

I usually a different formula where the listening level is set by listening to regular music.
 
dB(SPL) = S + 20 log ( Va / Vref )
 
where:
 
- dB(SPL) is the average listening volume for a track.
- S is the sensitivity of the headphones expressed in dB/Vref, Vref is often 1V and the specification is dB/V
- Vref is the reference voltage used in the dB/V spec, or the calculated one from the dB/mW spec ( Vref = sqrt (0.001 W * Zload)), Zload is the impedance of the headphones
- Va is the 'average' voltage at the headphone output corresponding to to the listening volume and the specified track.
 
Va = Vmax * 10 ^ ( (Gvc + Gtrack) / 20 ))            (1)
 
where:
 
- Vmax is the max Vrms output of the headphone out, usually 1 or 2 Vrms for on board computer outs
- Gvc is the gain in dB set with the Windows volume control (2)
- Gtrack is the level of your track compared to a 0 dBFS sine wave (3)
 
 
(1)The idea comes from 20 log ( Va / Vmax ) = Gvc + Gtrack, that is to say that you need to subtract the gain of the volume control and the gain of the track to find the voltage output compared to the max voltage
 
(2) Control Panel > Sound > Properties > Level > right-slick on the number and select dB. Gvc is - 15 dB in this case.
 

 
(3) Gtrack is - 14 dB  in this case, the Dynamic Range Meter plugin for foobar is available here: http://www.jokhan.demon.nl/DynamicRange/index.htm
 

 
NB: The above calculation takes into account the average level of the track, not the peaks, I find it a more accurate approximation of the average listening volume, It also assumes a 0 ohm output impedance, otherwise replace Vmax by Vmax * ( Zload / (Zload +Zout) ), where Zout is the output impedance and Zload the impedance of the headphones.
 
 

 

[geetar7]

Check-out this online calculator:
 
http://www.headphone-amplifier.com/calculator.htm
 
It took me a while to find this through google searches.  I have not verified their math or assumptions.  For example, comparing headphones must use the same units for effectivity or be referenced to the same impedance - tricky.
 
A search for this link here on head-fi yielded several results.
 
David
 

 

[Atzki]

Thank you very much for your post. It was REALLY what we need for our thesis project. My group is developing an android application that would estimate the headphone volume output among other things.

My only problem is that I can't search for the device's (samsung mobile) maximum voltage output. The results just pertain to its battery (is that it? i kinda doubt it). So I really can't follow through..

any suggestion, please?

 

[Atzki]

Ok. I've already satisfied my concern above. So anyway, as I've stated, I'm doing my thesis project and I would really like to use this as a reference of my formula but I think I need to prove that this is really credible--I AM convinced it is, only, I don't know how to explain that and defend this against the panelists. i would REALLY appreciate suggestions given asap. PLEASE.

 

[arisu]

The formulas given by tinyman392 and by khaos974 seem to give different results.

From what I can understand, the difference comes from the following:

tinyman392 assumes
Quote:

SPL (Pa) = sensitivity(Pa/W) * power

while

khaos974 assumes
Quote:

SPL (Pa) = sensitivity(Pa/V) * voltage

where power = voltage^2 / impedience.

Which one is correct ?  Or does it depend on choice of earphone ?

 

[xnor]

Sensitivity can be specified in both dB SPL/V or dB SPL/mW. It doesn't matter which you use as long as you don't mix up the formulas.
 
Tinyman's formulas seem to be wrong.. also see this.

 

[arisu]

Quote:

Tinyman's formulas seem to be wrong.. also see this.

 
Thanks for the link.
 
Look like tinyman392's formula for actual output voltage might also be wrong then.
Quote:

Actual Voltage (volts) = Max Output Voltage (volts) * (volume level fraction)^2

 
Is ipod's volume level fraction linear to actual output level in dB as described in this ?  If yes, then we would expect actual voltage to be exponential to volume level fraction instead.