FiiO E17 "ALPEN" - First Impression + Final Thought - Page 328

E07K is not a unit that should be better than E17. I no longer have the E17 but the E07K doesn't have the current for some harder to drive headphones at my preferred listening volumes.

It depends on what you want. 

For me, I noticed that the E07K exhibits noise on my non sensitive nor low impendence IEM's which may be a problem for IEM users

I'm not totally impressed w/the E17 and my high impedance or inefficient phones -- DT880 & K501 respectively -- not quite enough gain even on +12db in every situation (and with a 1volt source) and I feel like making myself a little more deaf ... but boy does it shake my head when I drive my Focal Spirit One with it!  And the sound-stage and prat I get from my Marshal Majors with it was a surprise and eye opener.
 

how?

E17: Output power: 250mW @ 16 Ohm / 215mW @ 32Ohm / 29mW @ 300 ohm

E07: Output power: 250mw(16Ω); 36mw(300Ω)

supposed to be the same/better in theory, btw what headphones are you talking about?

Those numbers are incorrect. The E17 numbers are correct but the E07K numbers written in the manual show it as having 20-30% more ouput wattage than the E17. FiiO told me that that was incorrect. They will have an update out soon for it.

And kep in mind that I said current. Not power. They are different things. In all actuality, very high impedence 600+Ohm headphones actually require less power BUT more current. 

they are the same thing:

P=I^2*Rload

you want more current to drive your load(headphones) so it basically means more power

And kep in mind that I said current. Not power. They are different things. In all actuality, very high impedence 600+Ohm headphones actually require less power BUT more current. 

Do you know Ohm's law?  Because it dictates the opposite of what you just stated.  V (voltage) = I (current) * R (resistance)....and... P (power) = V (voltage) * I (current) = I * (I * R) using variable replacement replacing V with (I * R).

Then further the law of electon flow is voltage (potential) is provided, and current is drawn from the source based on however much is required & permitted given the resistance/impedance; with the exception that if the source is not capable of providing the current at the given voltage (i.e. source does not have the power capability), the voltage will be drawn down to the level where voltage * current = the maximum power that the source can provide.  So with V held constant, the higher the impedance for the same voltage, the less current is drawn. 

What high impedance headphones require categorically is higher voltage to drive them, and less current will be drawn, for the same milliwatt power required input to them to produce the same decibel level out.  And if a high impedance headphone requires high current, that implies [is a derivative way of saying] it requires high power.

Now, apply that linear equation to headphone's audio output level, which is a function of power, not just voltage or current alone.  As stated just before, for a 32 ohm version of a headphone to produce the same Db level as the 600ohm version of the same headphone, they require the same power input to them.  Let's imagine this headphone needs 100mw to produce a certain decibel level whatever that may be (I have no idea of what the real power required to produce that sound level, but using that as an example).  We know our Rs, and we know our P.  Our equivalent equation in terms of current and resistance cited on the 1st line of this post becomes:

So where I1* V1 = 100mw (variable #1 representing the 600ohm headphone) and I2 * V2 = 100mw (variable #2 representing the 32 ohm headphone), using the variable replacement we have I1 * I1 * 600 = I2 * I2 * 32.  And that should clearly be self evident that to make both sides equal, i1, the current necessary to drive the 600 ohm load, is far less than I2, the current necessary to drive the 32 ohm load.

Do you know Ohm's law?  Because it dictates the opposite of what you just stated.  V (voltage) = I (current) * R (resistance)....and... P (power) = V (voltage) * I (current) = I * (I * R) using variable replacement replacing V with (I * R).

Then further the law of electon flow is voltage (potential) is provided, and current is drawn from the source based on however much is required & permitted given the resistance/impedance; with the exception that if the source is not capable of providing the current at the given voltage (i.e. source does not have the power capability), the voltage will be drawn down to the level where voltage * current = the maximum power that the source can provide.  So with V held constant, the higher the impedance for the same voltage, the less current is drawn. 

What high impedance headphones require categorically is higher voltage to drive them, and less current will be drawn, for the same milliwatt power required input to them to produce the same decibel level out.  And if a high impedance headphone requires high current, that implies [is a derivative way of saying] it requires high power.

Now, apply that linear equation to headphone's audio output level, which is a function of power, not just voltage or current alone.  As stated just before, for a 32 ohm version of a headphone to produce the same Db level as the 600ohm version of the same headphone, they require the same power input to them.  Let's imagine this headphone needs 100mw to produce a certain decibel level whatever that may be (I have no idea of what the real power required to produce that sound level, but using that as an example).  We know our Rs, and we know our P.  Our equivalent equation in terms of current and resistance cited on the 1st line of this post becomes:

So where I1* V1 = 100mw (variable #1 representing the 600ohm headphone) and I2 * V2 = 100mw (variable #2 representing the 32 ohm headphone), using the variable replacement we have I1 * I1 * 600 = I2 * I2 * 32.  And that should clearly be self evident that to make both sides equal, i1, the current necessary to drive the 600 ohm load, is far less than I2, the current necessary to drive the 32 ohm load.

exactly what I thought when Bowie was complaining about the current, that post should be stuck on to the headphone impedance thread