Yes, i remembered that plannars are exception, they require some voltage. Thats why i added the 2nd sentence.
Though other Hps with such impedance usually work.
I want to make sure people aren't buying 400i without an amping strategy (or any low efficiency headphone for that matter). They are going to need significant power, and not all amps are going to work. I dont mean to go on a rant, but i think i am going to anyway...
Its not voltage that low impedance headphones need, its current. For example you will not be able to use OTL amps like the bottlehead crack and some of woo audio's offerings. They have plenty of voltage, but cannot supply the current needed to properly power low impedance headphones. Per Ohms law, if an amp has a set max output voltage (which all do - this is is your output rail voltage), higher impedance headphones need less current than low impedance headphones. 15V/32ohm=0.47A, 0.47A*15V=7W. Whereas 15V/300=0.05A, 0.05A*15=0.75W. Most amps are going to be able to do 0.75W, but most are NOT going to run 7W without letting go. This is an extreme example as this would be at clipping, but you get the idea.
Lets look at a more reasonable case, comparing the HE400i to a 300ohm headphone like the HD600 using a Schiit Magni 2 which can do 1.2W into a 32ohm load, with 15V rails. Lets say 0.2W out from the amp:
0.2W/15V=0.0133A. output current max from the Magni
0.0133A^2*32ohm=5.7mW swing +- with the hifimans
whereas that same current through lets say an HD600 would be: 0.0133A^2*300ohm=50.1mW.
Keep in mind though that as i said previously, damping factor is important to consider. Damping ratio in audio can be thought of as a metric used to describe the ability of the source (the amp in this case) to properly control the movement of the headphone transducer. High damping ratio means the headphone is overdamped, where a damping ratio of 1 would mean it was critically damped(amp Zout=HP impedance), and less than 1 would mean it was underdamped (amp Zout > HP impedance). If you think about this in terms of a control system, an underdamped system would ring before settling on its new position. We dont want this. Headphones are designed to sound their best with high damping ratios (for example, nobody is making amps with 600ohm outputs for Beyers. Low output impedance is a figure of merit.)
The downside(?) of a really high damping factor is that it is not optimal for power transfer from the amp to the headphone. This means that your 250mW e17, assuming it has an output impedance of 1ohm (not listed in the specs) will not be able to transfer all of that power to the load.
p_loss=10log((1*32)/(1+32)^2)=-15.32dB = 0.03W or 30mW
For the sennheisers:
p_loss=10log((1*300)/(1+300)^2)=-24.8dB = 0.0033W or 3.3mW
This puts us at -24.3mW for the same drive power with the hifimans. Obviously this is impossible, since the headphone is not generating any power, so we pick a new output power, knowing that it must be above 30mW to ensure we wont clip. Lets say we want to see +-31mW at the headphone:
sqrt(0.031W/15V)=0.04546A
Now check how much power we need at the amp for this output current:
0.045A*15V=0.6819W. This over half of the magnis total output power, and much more than the Fiio e17 is capable of. Note that the output power has more than tripled!
This is quantified in terms of output sound volume by the sensitivity rating of the headphone. Remember our rating of 93dB/mW? This means that 1mW will get us 93dB of sound pressure level (SPL). doubling power to 2mW gets us +3dB, and halving power to 0.5W gets us -3dB.
We know it takes 682mW to get to 93dB on the hifimans with the e17. Now this is probably more volume than you will need. I measured my present volume setting on my main rig at 55dB with a crappy phone decibel meter, but I probably sometimes listen at closer to 65dB. So lets calculate the power needed to get 65dB. We know 1mW gives us 93dB, and halving that power gives us -3dB, so 0.5mW would result in a volume of 90dB. 0.25mW would result in 87dB etc.
93/10log(0.001)=65/10log(x), x=0.000051mW
How much current to achieve 30.000051mW?
sqrt(0.030000051W/15V)=0.0447A
how much power is this from the e17?
0.0447A*15V=0.6705W. Still not gonna happen. Keep in mind i am assuming the fiio is using 15V rails. This is likely because it is probably using opamps for the amplification stage, and that is a common supply voltage for such devices. A smaller rail voltage would serve lower impedance headphones better, as they could get more current! (ohms law)
So those 250mW dont really give us any room to play with on low impedance headphones with low sensitivity. Even if you are able to get up to the listening levels you desire, you wont have much headroom. And you wont be able to play dynamic music without clipping. On the otherhand though, the loss does not really hurt the sennheisers, since as seen above they could pull 50mW with the same max output current!
This should demonstrate that low impedance/low efficiency headpones are going to need an amp, and that any amp will not do if you want to drive them properly. You can repeat these calculations for any amp that you know the levels of the voltage rails, and the output impedance. Now keep in mind this calculation is conservative, and does not account for voltage drop from the output stage of the amp. In reality you are going to have more like ~+-14.6V. There are other aspects such as frequency response and the transfer function of the amp, as well as non-uniform headphone impedances over frequency that i am not accounting for. Also, i am in no way affiliated with Schiit, Hifiman, Sennheiser, or Fiio. I also apologize if this came off as aggressive, but i havent seen these calculations done anywhere and i am hoping it can be helpful to people. If anyone has any questions, feel free to PM me. I think i have derailed the OP enough already.
