Eh, sorry if I come across as a know-it-all. I tried to spout off my response between my kid bossing me around (geez, kid, can't you see I'm
busy talking to strangers
online?!) and my dog complaining that she needs to go out NOW (again,
busy!), and, well, I probably end up seeming more like an asshat than I want. Not that I'm not an asshat. I just don't want to
seem like an asshat. I try to make up for it with cute gifs.
But seriously, sorry if your blood pressure went up when you saw my post. I really didn't mean it.
Yes, when impedance goes up, then required current goes down and voltage goes up,
assuming that the required power (efficiency) stays the same. But if the efficiency changes, then it's not as simple.
Power needed for a given SPL can be calculated like this:
P
t = 10
t-e/10
P
t = power required to reach target SPL "t", t = target SPL (dB), e = headphone efficiency (dB/mW)
For the Mk2 to reach 100dBSPL: P
Mk2 100dB = 10
100-94/10 ≈ 3.98mW
For the Mk1 to reach 100dBSPL: P
Mk1 100dB = 10
100-102/10 ≈ 0.63mW
Using
this calculator, the sensitivity of the Mk2 is 109.2dB/1V, and the Mk1 is 108dB/1V.
You can either use a similar equation to the one above for 100dB: V
t = 10
t-s/20, where s = sensitivity (dB/1V)
V
Mk2 100dB = 10
100-109.2/20 ≈ 0.35V
V
Mk1 100dB = 10
100-108/20 ≈ 0.4V
Or use Ohm's law (Hi, Georg!):
P = VI, I = V/Z, so P = V
2/Z, V = √(PZ)
V
Mk2 100dB = √(0.00398*30) ≈ 0.35V
V
Mk1 100dB = √(0.00063*250) ≈ 0.4V
Mk2 needs 6.3x as much power as Mk1.
Mk2 needs 7/8 as much voltage as Mk1.
Despite the low impedance of the new drivers, but because of their lower efficiency, they require nearly the same amount of voltage as the older higher Z drivers.
It's made even more confusing that Beyer's own copy says that the new headphones are for devices that can't provide enough voltage the old ones. "If only my phone had 1/8 more voltage! I could listen 1.2 decibels louder!! Grrr!!!"
I'll stop now before I find another stupid typo I have to fix. And hopefully before you've become
completely apoplectic.