The headphones A has an impedance of 32 Ohms.

The headphones B has an impedance of 250 Ohms.

Both of the headphones are similar by size.

Is it right that the headphones B will play quieter by 250/32=7.8, if we talk about the sound pressure level?

7.8 times is approximately 9 dB.


4 Answers 4


The only spec that matters is sensitivity. For normal speakers, sensitivity is usually listed as dB Per Watt at 1 meter. For headphones, it is usually listed as db per milliwatt at "ear distance".

There are lots of things that can affect sensitivity, and impedance is just one of those.

What I can say is that at the same "volume setting", one of these headphones will consume about 7.8 times more wattage than the other. But how that impacts volume is completely up in the air because we don't know the sensitivity.

  • \$\begingroup\$ Yes, sensitivity will absolutely 100% tell you how load a headphone will be at a given power, and to know the relative power between two different sets of cans you need to know their impedances and the output impedance of the amp driving them. You're right -- you need to know the whole chain. \$\endgroup\$ Mar 22, 2013 at 18:29

If all else were equal, then yes, in theory. However, you really can't make that assumption because there are so many other variables regarding how the two different headphones might be designed. There can be significant variation in the mass of the vibrating element, its area, how far it can travel, the magnetic coupling, the strength of the magnetic bias field, how the pressure from the back side of the vibrating element is handled, how well the front side pressure is coupled into your ear, etc, etc.

You really can't make any meaingful assumptions about loudness from just the impedance.

  • \$\begingroup\$ Not quite. If all else were equal, except for the impedance, You would calculate power through each headphone for the given source impedance to find which is louder. Chances are that the source impedance is closer to 32 ohms than 250 ohms, and the low impedance headphone would be louder, but it isn't a sure thing \$\endgroup\$ Mar 22, 2013 at 17:26
  • \$\begingroup\$ @Scott: Source impedance is usually very low. In other words, you can think of the amplifier as being a voltage source, so the power is inversely proportional to the headphone impedance. Of course all else is not even close to equal, so this really doesn't matter anyway. \$\endgroup\$ Mar 22, 2013 at 18:28

Headphone B may be more sensitive than headphone A - it has more turns on its magnets and this can make it more sensitive to signals and make it louder in some cases.

EDIT - I originally made a mistake in the calcs below so I'm editing this bit...

An amplifier's headphone socket usually has resistors in series with the output to restrict the power taken by headphones and if the resistor is (say) 100 ohms, the voltage across A would be much lower than B. Taking only impedance into account this would mean power into B is about 0.5dB more than A

Please note that the above are just simple examples showing how wrong it might be to assume A is louder than B

  • \$\begingroup\$ Weird. I would consider putting a resistor in series on the amp outputs to be really stupid. It will interact with the variation of headphone impedance over frequency, and drastically influence the resulting frequency response of the headphone. A quick Google search shows that many HP amps do this, however, including ones from well respected designers. I will have to talk with some of them to see what their opinion is about this. \$\endgroup\$
    – user3624
    Mar 22, 2013 at 17:27
  • \$\begingroup\$ Pro audio headphone amps typically have about 50 ohms of output impedance, usually more than 10 and less than 100 \$\endgroup\$ Mar 22, 2013 at 18:21
  • \$\begingroup\$ @Scott: The circuits I have seen have nearly zero output impedance, as long as the power level is within their capabiliy. There may be some resistance in series as protection, but that resistor is usually inside the feedback loop, so the active impedance is basically 0. \$\endgroup\$ Mar 22, 2013 at 18:30
  • \$\begingroup\$ Tascam (TEAC) output impedence is 94 ohms (and the 250 ohm headphone would push more power) (tascam.com/content/downloads/products/138/e_mh-8_om_va.pdf). Presonus (ubiquitous in sound studios worldwide) is 51 ohms presonus.com/products/HP4/techspecs. My ART headphone amp is 12 ohms. It's actually difficult to find a pro level headphone amp that's less than 10 (though Furman makes one at 0.5). Maybe pro audio headphone amps are high output impedance to keep audio engineers from going deaf or something ;) , but they're usually not zero. \$\endgroup\$ Mar 22, 2013 at 18:55
  • \$\begingroup\$ @OlinLathrop That's what I thought too, until I looked at various circuits (proprietary and ones on the 'net). A surprising number actually have a real 50 to 200 ohms output impedance. \$\endgroup\$
    – user3624
    Mar 22, 2013 at 20:32

If \$R_1 \$is the output impedance of the amplifier, and \$R_2\$ the resistance of the headphones, power through the headphones is \$ I^2 R_2 \$, and \$I=\frac{V}{R_1+R_2}\$. If you run through the exercise of maximizing the power (taking the derivative with respect to R2, setting it equal to zero, yada yada yada), you'll find that power is maximized when \$ R_2=R_1. \$

Thus, the answer to your question is "it depends on what's driving the headphones". If the amp has a 250 ohm output resistance, the voltage across the low impedance headphones will be too low to push real power, and the high impedance headphones win for power.


simulate this circuit – Schematic created using CircuitLab


Since some (i.e., me) are making assumptions about the headphones and saying we can't make assumptions about the driver, and others are making assumptions about the driver and saying we can't make assumptions about the headphones, and others correctly point out that we're missing a parameter, I'm removing all assumptions and offering a real example using cans and amps that I can buy or borrow and have them in a few days.

For an amp, Presonus HP4, output impedance 51 ohms, input impedance 10K.

For headphone A, let's go with Grados- a fine set of phones, but a little bit uncomfortable to wear for long stretches. RS1i SPL/1mW 98 Impedance 32 ohms

Heaphone B is Sennheisers Sound pressure level (SPL) (/mW assumed!) 102 Impedance 64 ohms

Lets just make the input to the amp be a 0.2 Volt sine wave at 5kHz, coming from a low impedance source), and amplification will be 0 dB

Power at the phones will be I * Vphones, where I = Vin/(Rphones+Ramp), and Vphones = Vin*Rphones/(Rphones+Ramp).

Thus, power=(Vin^2)Rphones/((Rphones+Ramp)^2), and "loudness" = powersensitivity

The "loudness" of the 32 ohm Grados will be 18.2 dB.
The "loudness" of the 64 Ohm Sennheisers will be 19.7 dB.

If you use a different amp, with an output impedance of 0.01 ohm, then:

The "loudness" of the 32 ohm Grados will be 122.4 dB.
The "loudness" of the 64 Ohm Sennheisers will be 63.7 dB.

  • \$\begingroup\$ Audio power amplifiers typically have a very low output impedance (a fraction of an ohm)! \$\endgroup\$ Mar 22, 2013 at 17:28
  • \$\begingroup\$ Yes. but the point is that the power through each headphone can be calculated--and I've driven plenty of pairs of headphones with things that are not audio amps. \$\endgroup\$ Mar 22, 2013 at 17:43
  • \$\begingroup\$ Curious about just which part of the answer someone thinks is incorrect enough to merit a downvote? If R2=0, then the power ratio is 8.9dB, and if R2 does not equal 0, you still get the right solution. \$\endgroup\$ Mar 22, 2013 at 17:49
  • \$\begingroup\$ @Leon-- the headphone amp I'm holding right now is 12 Ohm, the Behringer UCA202 is 50ohm, and IEC 61938 (silly though it may be) recommends 120 Ohm output impedance. \$\endgroup\$ Mar 22, 2013 at 17:58
  • \$\begingroup\$ Those look like load impedances, not output impedances. Do you know the difference? \$\endgroup\$ Mar 22, 2013 at 21:41

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