Originally Posted by nikongod
There should be an equation to explain this similar to:
V=I*R or P=V*R
(P=V*R ? It's either P=V*I or using ohm's law: I^2*R or V^2/R)
There is, since it works like a simple voltage divider: Vout = Vin * Zload / (Zload + Zout) so 8 Vrms with 100 ohms output impedance and 600 ohms load (see my previous example) results in 6.857 Vrms --> 78 mW into the headphones.
(Btw, if we take a look at the impedance curve of the headphone we can calculate the influence of the output impedance with the same formula. Let's say the headphone has 700 ohms at 100 Hz and 600 ohms at 1 kHz. --> 20*log10(7 / 6.857) = 0.18 dB bass "boost")
There is no law relating output power an output impedance.
Actually if there was no relation you could not calculate the output impedance using math like this:
Zout = (Zload * (Vnoload - Vload)) / Vload
But I guess what you're trying to say is that one cannot reason output power from output impedance or vice versa, which of course is right. Though it's still less likely that a headphone amp with high output impedance will deliver lots of power into low impedance loads.
While you could write an equation to estimate the maximum output power of a given amplifier into a given load if you define unloaded voltage, max output ucrrent, load impedance, and output impedance there is no reliable way to estimate the maximum output power of an amplifier by looking at just its output impedance and the load impedance.
No but some manufacturers provide specs with a high and a low impedance load. You can use both values to interpolate and even estimate the power outside that range. Extrapolating output power into higher impedance loads should be fairly accurate.
I do agree that at some point the realities of the situation do create limits, but you can not in any way say that since an amplifier has a 1ohm output impedance it can put out a bunch of power - which is the correlation that the OP looks like it was leaning towards. The info for output voltage or current are not given - only output impedance and load.
The op has specified the output power into a 32 ohms load and close to zero output impedance. That's enough information to calculate e.g. the min. power the amp can output into loads ranging from 32 up to thousands of ohms.
Maximum power transfer does not apply to almost any audio application.
I know, I was just saying that there's a well defined relationship between output power and output impedance. As firev1 wrote, impedance/voltage bridging is the preferred configuration for headphones but it's also a matter of choice as you wrote. I'm sure some headphones sound better with higher output impedance.Edited by xnor - 2/15/12 at 1:47pm