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Looking at the data sheet of the MAX4410 headphone driver, lower impedance headphones will get more power.

Is this true in general for laptop headphone drivers? especially with regards to 8-32 ohm impedance ?

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  • \$\begingroup\$ I just stumpled upon the tag:amplifier tag and your wording: with "laptop headphone drivers", do you refer to the drivers in the headphone (where "driver" would be the right word) or to the amplifier in the laptop (where you should say "amplifier")? \$\endgroup\$ Jun 17, 2014 at 11:41

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Headphone impedance and anything related is quite complex, since you are normally interested about things like "sound" and "loudness", which are influenced by a lot more than only this single factor. I recommend reading this article on head-fi.org, which sums it up very well.

To get to your question: The output of your headphone jack is controlled by voltage (if you change your computers volume level, you are actually changing the voltage at the headphone jack). So the power your laptop delivers is inversely proportional to the impedance: \$S = U^2/Z\$. That is true if you compare different headphones at the same voltage of your laptop. But normally, you want to compare them at the same loudness, since this is what you as a user normally care about: How loud is the sound that reaches your ears? So this is the point where the headphone´s efficiency comes into play and makes any calculations vain.

A little example: Imagine you have to pair of headphones, one having a low impedance and a high efficiency, the other having a high impedance and a low efficiency. If you take your low-impedance-pairs,they will consume a lot more energy at a set voltage from your source (e.g. laptop) than the high impedance phones. However, you will probably never need to set the same voltage level because of their higher efficiency. They might sound at 30% voltage as loud as the others at 50% or 80%, and then the difference in impedance can´t be compared that easily.

You should also consider, that low-impedance loads usually put larger stress on their amplifier. In the cited Head-Fi.org article, there is a rule of thumb:

As a rule of thumb, the load impedance (headphone) should be at least eight times higher than the amplifier output impedance.

Of course, there are a lot more things to considered when buying headphones (like sound quality). I recommend thorougly testing anything your are planning to buy. And, to sum it up: If headphones have a low impedance, they consume more power at the same output level of their input device. That doesn´t say anything about loudness, though.

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  • \$\begingroup\$ So assuming similar sensitivity, low impedance creates stronger volume(leaving aside "sound") ? \$\endgroup\$ Jun 17, 2014 at 10:51
  • \$\begingroup\$ Your answer would be more clear if you avoided the ambiguous "output level" and instead stuck to "voltage" and "volume". \$\endgroup\$
    – Phil Frost
    Jun 17, 2014 at 11:21
  • \$\begingroup\$ @PhilFrost thanks for that suggestion. Sometimes I am struggling to put my thoughts into (english) words... I will edit that in. \$\endgroup\$ Jun 17, 2014 at 11:28
  • \$\begingroup\$ @hulkingtickets: Exactly. But your assumption (similar sensitivity) will rarely be met by actual headphones. However, in most cases headphones with an impedance >50 Ohm are rarely suited for mobile use. \$\endgroup\$ Jun 17, 2014 at 11:38
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Looking at the headphones as a simple resistor, and the sound source as a simple voltage, it is quite plain using Ohms Law that yes, that is true.

If the headphones are 8Ω and the sound source is 1.2V (for example), the current would be \$I=\frac{V}{R}\$, which is 150mA, and the power (\$P=\frac{V^2}{R}\$) would be 180mW.

If the headphones were 4Ω the results of the same formulae would be 300mA and 360mW.

So yes, the lower the impedance the higher (in basic terms) the power, but also the more strain on the amplifier.

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  • \$\begingroup\$ What about the output impedance of the headphone amp? And how much is it in general ? \$\endgroup\$ Jun 17, 2014 at 10:52

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