Originally Posted by Dark_wizzie
Can you do a for dummies version of your HD800 calculation? How did you get 2.6vRMS, is that the average case or the worst case scenario? So if we want to calculate the worst case scenario for HD800/O2 instead we should look at minimum impedance because that is hardest to drive? Then that would be 350 impedance which is a larger number than the 300 ohms listed on the headphones itself.
So you believe O2 will easily drive HD800, T1, LCD2/3.
And, higher impedance = requires more voltage, less current. Lower impedance = less voltage, more current. So in a way, having either a very high impedance or very low impedance can be an issue for an amp.
So, with Sennheiser headphones, it's pretty easy. They specify their sensitivity at 1V RMS, so you don't actually have to worry about their impedance at all (unless you want to calculate power required or current). The HD800 are rated at 102dB at 1VRMS. From there, you can calculate the voltage needed for 110dB as shown here:
1) How much gain do we need compared to the reference? We want 110dB, reference specified is 102dB, so we want 8dB more SPL
2) For 8dB gain, 8 = 20*log(V2/V1)
3) V1 = 1 volt (specified reference level)
4) Rearranging the equation, 8/20 = log(V2)
5) 10^(8/20) = 10^log(V2) = V2
6) V2= 2.512
If you want to calculate the current or power requirements, you can then look up the impedance at the desired frequency, and use P = V^2/R and V = I*R to figure them out. If you want the worst case power and current, use the minimum impedance to do these calculations.
If you had a pair that specified in dB/mW, you would have an extra step involved. Let's try it with the HE-6:
1) How much gain do we need? Desired = 110, reference = 83.5, gain = 26.5dB
2) 26.5 = 10*log(P2/P1) <----- Note the change in the leading constant - for voltage, use 20*log(V2/V1), for power, use 10*log(P2/P1)
3) P1 = 0.001W (1mW)
4) 26.5/10 = log(P2/0.001)
5)10^(2.65) = 10^log(1000*P2) = 1000*P2
6) 447 = 1000*P2 ---> P2 = 0.447W
7) P = V^2/R, R = 50, P = 0.447
8) V = sqrt(0.447*50)
9) V = 4.73
(and yes, I believe the O2 will easily drive the T1, HD800, and any of the LCD series)
As for your last statement, that is true - either very high or very low impedance can mean that a given amp will be unable to drive the headphones, and in either case, a very low sensitivity will increase the difficulty of driving the headphones.
Edited by cjl - 5/9/14 at 9:54am