Originally Posted by superjawes
Efficiency actually tells you how hard a headphone is to drive. Low efficiency means you need more power input for the same power output from a more efficient model.
Impedance really tells you what kind of power you need for your headphones. A 300 Ohm headphone wants more voltage and less current (and the impedance will resist the current anyway). A 32 Ohm headphone will draw more current due to less resistance, but the voltage will be lower. If you have an ideal amplifier (one that looks like an ideal voltage source), the total power (number of Watts) deliverd to each headphone model will be the same (assuming the same efficiency, too). The only thing that changes is the ratio of voltage to current.
Now, I think your response to JohnBal really comes down to John using the wrong terminology. He didn't so much mean that a higher impedance driver is easier to drive, but that the driver will pair better with the amp. This is because amplifiers are not ideal sources. The output impedance is indicative of this. The pairing isn't about efficiency, but electrical dampening effects. It does also change how power is delivered. Output impedance is part of a model. We can simplify the internals of the amp to get a good idea of how it will work by reducing it down to an ideal voltage source in series with a resistor, and the value of that resistor is the output impedance. The headphones complete the circuit, allowing current to flow out of the voltage source, through the output resistor (output impedance), and through the input resistor (headphone impedance).
Two important points. 1) This circuit acts as a voltage divider. You want a relatively low output impedance on the amp so that most of the voltage from the source makes it into the headphones. Otherwise you're limiting how much power can get from the source to the driver. 2) The total impedance will resist the current. This also limits the power draw by limiting the current. The example I always point to is the max power of Schiit's Valhalla, since they show max power for 50, 300, and 600 Ohm headphones.
This is by no means a complete explanation. I've basically ignored the frequency response of these things, and headphones/speakers, being reactive devices, will have impedances that vary with frequency. But I hope it helps at least a little.
This got a little long winded, but it all comes down to this: Efficiency
determines how easy it is to drive a pair of speakers or headphones. Impedance determines what kind of power you need (higher voltage or higher current).
Well, yes and no. You're confusing the issue. What you're referring to is generally quoted as how much sound is produced per watt (or milliwatt), at a given frequency, as measured by a sound pressure level meter, a set distance from the drivers. This value does not necessarily correlate with impedance. In other words, they are two different specifications.
As an example, let's look at the three different impedances offered by Beyerdynamic, for their DT 770 model:
Beyerdynamic DT770 / 250 ohm: 96 db @ 1mW / 500 Hz.
Beyerdynamic DT770 / 80 ohms: 96 db @ 1mW / 500 Hz
Beyerdynamic DT770 / 32 ohms: 96 db @ 1mW / 500 Hz
You will note that the manufacturer of this headphone lists the sensitivity as exactly the same, for three different impedances. So the sound level produced by each of these cans is exactly the same, given a 1mW input at 500 Hz. Using your terminology, they are equally efficient. Of course, they don't specify what sort of amplifier they used to achieve that result. They also ignored the effect of frequency (though I give them props for including the frequency used in this measurement).
Using the points in your discussion, if you drive any of these headphones with an amplifier that is capable of providing the necessary voltage and current, then theoretically you ought to get the same sort of result. Without resorting to a lengthy explanation on ohms law and calculating impedance using imaginary numbers, let's just postulate that you get the same result, if and only if your amplifier can indeed match the right voltage and current to the load presented to it. I assert that the load for each of those headphones is different, notwithstanding their equal efficiency ratings. Each requires a different voltage and current to achieve the same power, and thus the same sound level output. But those are two different things.
So back to the original question: is there any advantage to higher impedance headphones? Let me modify my previous answer thusly:
If you can hear the difference, AND if you have an amplifier that can produce the necessary voltage and current needed, then higher impedance headphones are better.
If you can't hear the difference, AND/OR you do not have an amplifier that can produce the necessary voltage and current needed, then skip the higher impedance cans. Go with what you can drive with an ipod alone, and leave the arguments to people who write long messages late at night on mysterious internet forums...