Okay, then let's disregard that for a second and say that I'm wrong. The O2 doesn't have any "roll off" that is significant.

This just occurred to me. Going back to the same thing about voltage drop earlier, assuming the following:

[V out] is the output voltage from the amp.

[Z out] is the output impedance of the amp.

[V load] is the load voltage delivered to the headphone.

[Z load] is the impedance of the headphone.

Then:

[V load] = [V out] * [Z load] / ([Z load] + [Z out])

Obviously, this means the higher [Z load] is, the more [V load] will be closer to [V out], thus less voltage drop, or less attenuation at this peak.

However, going by this equation:

[Power] = ([V load])^2 / [Z load]

It looks to me like the higher [Z load] is, the more [Power] will drop.

Let's put some numbers in.

Assuming [Z out] = 50 Ohm, [Z load] = 250 Ohm, and [V out] is 2.7v, then:

[V load] = (2.7v) * (250 Ohm) / (250 Ohm + 50 Ohm) = 2.25v

[Power] = (2.25v)^2 / 250 = 0.02025W = 20.25mW

Now, assuming [Z out] = 50 Ohm, [Z load] = 350 Ohm, and [V out] is the same 2.7v, basically just [Z load] increased, then:

[V load] = (2.7v) * (350 Ohm) / (350 Ohm + 50 Ohm) = 2.3625v (obviously less voltage drop)

[Power] = (2.3625v)^2 / 350 = 0.015946875 = 15.95mW

Oops... so power output actually suffered. By almost 5mW.

Let's drop [Z out] to an ideal 0 Ohm now, so:

[V load] = (2.7v) * (250 Ohm) / (250 Ohm + 0 Ohm) = 2.7v (no drop in voltage)

[Power] = (2.7v)^2 / 250 = 0.02916W = 29.16mW

And do the same for 350 Ohm:

[V load] = (2.7v) * (350 Ohm) / (350 Ohm + 0 Ohm) = 2.7v (no drop in voltage)

[Power] = (2.7v)^2 / 350 = 0.02083W = 20.83mW

Oops... now it's a 8mW drop.

It looks to me like the impedance peaks actually cause power to suffer more for higher impedance headphones. So... what am I missing in this picture?

Edited by Bill-P - 10/10/12 at 2:08pm