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Approximating Headphone Volume Output (dB)

Discussion in 'Sound Science' started by tinyman392, Dec 21, 2011.
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  1. 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
     
     
  2. itchyblood
    what's the max voltage output of a fiio e9?  I just find mW and ohms for that value.
     
  3. Head Injury
    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.
     
  4. 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).
     
  5. tinyman392


    Quote:


    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. 
     
  6. Head Injury
    Quote:

    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.
     
  7. itchyblood
    Okay Thanks guys!
     
  8. jcx
  9. 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.
     
    Capture.png
     
    (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
     
    Capture.png
     
    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.
     
     
     
  10. 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
     
     
  11. 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..[​IMG]
    any suggestion, please?
     
  12. 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.
     
  13. 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:

    while

    khaos974 assumes
    Quote:

    where power = voltage^2 / impedience.

    Which one is correct ?  Or does it depend on choice of earphone ?
     
  14. 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.
     
  15. arisu
    Quote:
     
    Thanks for the link.
     
    Look like tinyman392's formula for actual output voltage might also be wrong then.
    Quote:
     
    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.
     
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