[Me x3]

I was wondering about this problem:
I have some different headphones here with some different specs.

AKG K44
32 Ohms
115db/V

Marshall Major
32 Ohms
115db/mW

Yamaha HPH-200
48 Ohms
Yamaha only says SPL: 98.5dB
(I thought: ok! Nobody talks about efficiency then those are db/V - but then I realize that most people on the internet say this are 98.5dB/mW, what seems correct because Yamaha's are easier to drive than Akg's. I mean with the same power they seem to give more volume, but not more than Marshall Majors)

My question is: How much db/mW is the efficiency of Akg's K44 then?

First I thought

V = I . R

Then

1V / 32 Ohms = I = 0.03125 A
 
Power = V . I =  0.03125 W = 31.25mW

Ok! I have now the efficiency: 115db/31.25mW

But then I realized that here we are not talking about a continuum voltage.
In fact the imput when sensitivity tests are made is an alternate wave of 1kHz as far as I know.

Then: How much db/mW is the efficiency of Akg's K44?

Thanks! And greetings from Argentina.

 

[skamp]

115 dB/V - (10 * log10(1000 / 32 Ω)) = 100 dB/mW

Don't ask me why, I just know it's correct. I derived that formula off a blog that I'm not allowed to link to because the author is a former head-fi member who was banned.