I'm trying to get a better feel of how exactly the drivers in headphones work.

Specifically, I want to know if it's possible to calculate how much electricity being sent to the drivers in a pair of headphones or speakers is turned into waste heat vs actual motion of the diaphragm.

For example, in a pair of ATH-M50s, the drivers have an impedance of 38 ohms with a maximum input power of 1600 mW at 1 kHz.

I don't know what else would be needed for the calculation, but could anyone give me a general sense of how efficient such drivers would be with their power?

Also, as a side question, what happens to the energy in the vibrating diaphragms? Is most of it dissipated as sound, or is it converted into heat as well through the flexing of the material?

  • 1
    \$\begingroup\$ I imagine there's a standard head defined somewhere that models the human outer ear canal and sound pressure level is read at a certain depth - probably ear-drum position. That should give a measure of output power. Input power would be calculated by measuring voltage and current. Efficiency, \$ \eta = \frac {P_{OUT}}{P_{IN}} \$. A frequency sweep or multiple test points across the audio spectrum would enable an average to be calculated. \$\endgroup\$
    – Transistor
    Commented Jul 29, 2016 at 1:33
  • \$\begingroup\$ The standard head is VERY IMPORTANT because sound wavelengths are long compared to the headphone/ear distances. \$\endgroup\$
    – Whit3rd
    Commented Jul 29, 2016 at 3:58
  • \$\begingroup\$ @transistor Thanks for helping, would you be able to explain more if I provided measurements for a specific pair of headphones? Here's a good set measured in a controlled environment with good equipment. \$\endgroup\$ Commented Jul 29, 2016 at 4:45
  • \$\begingroup\$ My comment was based on my interest in audio over many decades but I think @jbarlow has answered better than I could. He is right that all speakers have low efficiency. \$\endgroup\$
    – Transistor
    Commented Jul 29, 2016 at 8:11
  • \$\begingroup\$ Obligatory note of caution here : you really don't WANT headphones to be efficient, especially if you're planning to put 1.6W into them (and if you are - don't. Just don't...) Headphones can go dangerously loud without the usual cues that the same SPL from speakers would give you. I don't just mean that kick in your gut from the bass and vibrating bones, but windows rattling, cups walking off tables, etc. 90dB sounds moderate, 115dB sounds a bit loud, and you can easily go louder - and damage your hearing - without any clues. See canford.co.uk/HEADPHONE-LIMITERS \$\endgroup\$
    – user16324
    Commented Jul 29, 2016 at 11:16

2 Answers 2


Calculating headphone sensitivity

The essential parameter is the sensitivity of the driver, usually expressed in sound pressure level (dB) per watt of electrical (dB SPL/W), as measured by test equipment at 1 m. Typically this parameter is frequency sensitive.

In the case of the headphones in your comment, they give the specification "90 dB SPL at 0.13 mW". Like most specifications in the audio electronics industry, they omit details of test conditions. They don't give us the frequency range or distance at which this measurement was taken. Typically this parameter is given with a recording device at 1 m, but for reasons I'll explain, it's quite clear that was not the case here.

For now, let's assume they measured this at 4.5 cm, allowing 1 cm between the driver and outer ear, and 3.5 cm for the average depth of the ear canal in adults. Let's call it 90 dB SPL at 0.13 mW, 0.045 m, probably at 1 kHz but we'll ignore the effects of frequency.

Sound attenuates according to an inverse square law. We can calculate the attenuation factor in dB as 20 log(0.045/1) = –27 dB. Decibels are logarithms, so adding/subtracting them is multiplication/division of the real quantity.

Subtraction our attenuation of –27 dB, we now know that: $$ 90\space\mathrm{dB}\space\mathrm{SPL} @ 0.13\space\mathrm{mW},\space0.045\space\mathrm{m} =63\space\mathrm{dB}\space\mathrm{SPL} @ 0.13\space\mathrm{mW},\space1.0\space\mathrm{m} $$

0.13 mW converts to –9.9 dBm or –39.9 dBW. To eliminate the "0.13 mW" component of this non-standard sensitivity, we subtract this negative factor, or rather add it, thus giving:

$$ 63\space\mathrm{dB}\space\mathrm{SPL} @ 0.13\space\mathrm{mW},\space1.0\space\mathrm{m} =103\space\mathrm{dB}\space\mathrm{SPL/W}\space@\space1.0\space\mathrm{m} $$

A standardized sensitivity of 103 dB SPL/W at 1 m converts to an efficiency of 12%. This sensitivity and efficiency is reasonable for a high end set of cans and probably typical for earbuds. Acoustic devices are, shall we say, resoundingly inefficient. Remember that when a loudmouth politician is yammering on about something.

The efficiency is so low, perhaps the industry prefers to not talk about it. Many loudspeakers can be around 80-90 dB SPL/W @ 1 m, with efficiency below 1%. Sensitivity is the parameter of interest, because our perception of loudness is roughly logarithmic, and it allows for a simple equation of speaker sensitivity (dB) + amplifier gain (dB) = loudness (dB). From there, you determine how much speaker and amplifier you need to hit a desired loudness.

Calculating efficiency from sensitivity

The conversion is simply:

Efficiency = 10^((Sensitivity in dB – 112)/10)

The dB SPL scale is a relative scale, with 0 dB SPL defined as 20 µPa of sound pressure or equivalently, 1 pW of acoustic power.

How do we know they didn't measure 90 dB SPL @ 0.13 mW, 1.0m? That converts to 130 dB SPL/W, or about 63% efficiency. The most sensitive drivers are only about 20% efficiency.


Poor impedance matching between will reduce the power transfer from amplifier to headphones. As usual, ideally the output impedance of the amplifier is equal to input impedance of the driver.

Impedance is frequency sensitive as well.

  • \$\begingroup\$ Hi @jbarlow, thanks a lot for your informative answer, I really appreciate it when my curiosity is lent such expertise. However, I also wanted to know, since acoustic devices are so inefficient, where does the rest of the energy go? I assume a fair amount of it is converted into heat, but do the voice coils mostly heat up, or is the vibrating diaphragm responsible for dissipating more of the energy? Should I be concerned about voice coil heating effecting the life of my high end headphones? I've heard horror stories about blowing out high end audio gear this way. Once again, thank you. \$\endgroup\$ Commented Jul 31, 2016 at 0:46
  • \$\begingroup\$ You're welcome. While some energy is stored mechanically or magnetically, it all becomes heat eventually. sound.westhost.com/articles/speaker-failure.html suggests that loudspeakers fail because of amplifier clipping, which exerts maximum mechanical stress on the surround while pushing as much current as possible through the voice coil. Voice coils can reach quite high temperatures. The good news is if you drop the input power by just 3 dB you have decrease power by 50%, and save your hearing too. Headphones seem more like to fail from being dropped or damage to the external wiring. \$\endgroup\$
    – jbarlow
    Commented Jul 31, 2016 at 6:41
  • \$\begingroup\$ Okay, so would it be safe to say that given the 12% efficiency that means that 12 percent of the energy is being turned into mechanical energy in the diaphragm and the rest is heating up the coil? Or does the diaphragm also contribute significantly the energy loss? I'm not sure if the 12% efficiency signifies the amount of energy being turned into pure sound or the energy that goes to actually exerting physical motion on the diaphragm. \$\endgroup\$ Commented Jul 31, 2016 at 8:17
  • \$\begingroup\$ 12% is the acoustic power, or the power outside the speaker that is moving air around. The rest of is expended inside the speaker. I don't know how much is in the voice versus diaphragm, but I suspect most of it is in the voice coil. The diaphragm is damped, so it does not store much energy. The voice coil is pushing the diaphragm to closely replicate the input voltage, and precise control means more energy expended. Fans also push air around and make noise, but can be much more efficient because they're not trying to replicate a complex waveform. \$\endgroup\$
    – jbarlow
    Commented Jul 31, 2016 at 9:43

For accurate measurements you need to have precise values for both the input and the output of the headphones. An anechoic chamber would be needed in order to avoid reflections. You need to apply a sine signal from a signal generator and measure the electrical power that goes into the headphones. Using the measurement microphone and an acoustic analyzer you should be able to calculate the output power and then get the Output / Input ratio. The result will vary across the frequency range of the headphones, so you should sweep the frequency.


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