Why isn't "loud" just "loud" regarding headphones?

I’ve been trying to figure out what is the difference between voltage and wattage when it comes to headphone amplifiers.

At first, I thought that more watts = more better, disregarding all other factors.

Then I read that it also depends on the voltage that is being sent out. Then I got confused. I know about ohms law, but not how it is supposed to affect headphones and how they are driven.

If one of you has the time to dumb it down for me, even if only explained partially, I’d be grateful :slight_smile:

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with planars you want as much current (amps) as you get get usually and it really comes into play with less sensitive planars. same goes for dynamics at higher ohm usually but not nearly as drastically as overall wattage usually in my experience. if you aren’t using headphones known to be particularly demanding i woudn’t worry too much about power, honestly.

Usually for piezoelectric, the pressure/displacement is tied to the voltage. Now a dynamic driver isn’t a piezoelectric, but usually in electrodynamics diaphragm displacement is tied to voltage, although this may not be true to every driver design. As @Pokrog said, planar drivers are a good example of devices where current plays a more important role.

If you’re asking why some manufacturers put voltages alongside power, or sometimes just the output voltage, than it’s more complicated. Basically this comes down to design philosophy, a.k.a how you measured and designed your amplifier. Also, usually people don’t want to show results/products in a bad light, so they may use a different measurement to “look better than competition”.

The difficulty is that 10W can be 100V at 0.1A, or 1V at 10A, or anything in between.

So the amplifier is either voltage limited (= current spare but can’t “force” it through) or current limited (= current so high it can not get the voltage up).