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

post #4906 of 6535
Quote:
Originally Posted by m8o View Post

But I quoted McNuggetsPie and was replying to him ... I know you know it.  smily_headphones1.gif  And yes, I understand the sentiment of the 1st sentence.
Quote:
Originally Posted by kalbee View Post

The E07k's functions and versatility caught up a bit to the E17 but still the internals are not all the same. E17 is still more versatile (in terms of variety of supported inputs) AFAIK.
The amp section is not the same either.
The panda will do this very often tongue.gif He scans too fast and presses reply without re-reading.

In reply to both of you it was on purpose. I was agreeing to your post and thus quoted it to add to it.

As Head-Fis Official Panda, that is my final say on this matter ....
Lol smily_headphones1.gif
post #4907 of 6535
Quote:
Originally Posted by bowei006 View Post

I love to help but some very basic questions that are the basis of a product do get old after a while. So yes.

Time to go read another thread? rolleyes.gif
post #4908 of 6535
Quote:
Originally Posted by Chris J View Post


Time to go read another thread? rolleyes.gif

Nah, I enjoy this thread

 

 

700

post #4909 of 6535

Using the E17's to drive Hifiman HE500s lol. They do a surprisingly good job at +12 gain, and 30 vol.

post #4910 of 6535

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.
 


Edited by m8o - 12/7/12 at 9:56pm
post #4911 of 6535
Quote:
Originally Posted by bowei006 View Post

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

 

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?

post #4912 of 6535
Quote:
Originally Posted by m8o View Post

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.
 

 

you need our Mont Blanc(E12) which can push out almost 1.4W into a 32 ohms loader . a Monster Portable headphone amp consider the size of it.

post #4913 of 6535
I'm really impressed by this little machine. Baring in mind the cost and quality of the unit/accessories, I think it's up there with "bang for the buck" big hitters. I wouldn't say it's night and day from the E7 but it's very noticeable a pleasant improvement and it is defo a keeper. I've sold most of my cans now so am just running PSB's through it, it's not a particularly hard can to drive so can't comment on that. I will test with my friends 650's when he's back from the big apple. What I can comment on is pretty much what has already been said. Soundstage, presentation, separation, depth, all extended and improved and considerably so. It's obviously going to lack the "Ummph" that my LD SE has but I can harldy put that in my pocket now can I. A resounding YES from me for this ;little box of delights:D
post #4914 of 6535
Quote:
Originally Posted by KamijoIsMyHero View Post

 

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. 

post #4915 of 6535
Quote:
Originally Posted by bowei006 View Post

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

post #4916 of 6535
Quote:
Originally Posted by KamijoIsMyHero View Post

they are the same thing:

 

P=I^2*Rload

 

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

Power (Watts) = Voltage(Volts) x Current (amperes)

 

So while it is not a fully direct thing they of course do hold a very similar meaning. I think I should have said Voltage instead of Current but all is well.

 

http://www.head-fi.org/a/headphone-impedance

 

 

 

Quote:

Pros and cons of high impedance headphones:

  1. Headphone impedance is usually increased by thinner wire and most importantly more turns of wire in the voice coil. More turns or loops creates a larger field (area of magnetic influence). In layman's terms more magnetic force for the coil to move the diaphragm.  Thinner wire usually works out to a lighter, more responsive diaphragm. Depending on the headphone design, this may lead to more accurate response.
  2. The displacement (amount of movement) of the diaphragm (the part that vibrates to produce sound) can be better controlled via a more accurate flux (magnetic field to pull and push the diaphragm).
  3. Difficult to drive for small headphone amps with low output voltage and low gain.
  4. Most high impedance headphones need an amplifier with higher voltage gain and higher output voltage, e.g. the 600 ohm Beyer DT770/880/990 series. 
  5. Allow Solid State Op Amps to work more efficiently with less distortion. Have a look at Op Amp data sheets and a graph of distortion vs. output impedance for most audio Op-Amps and you'll get the idea. This is a very complex subject, but most Op Amps are designed to output (typically) up to 10 Volts into 600 ohm loads or higher.

 

Pros and Cons of low impedance headphones:

  1.  Headphone impedance is usually decreased by thicker wire and less turns of wire in the voice coil. The magnetic field is built up by more current.
  2. Easier for small and/or portable headphone amps to drive. For example: an iPod or MP3 player headphone jack. Many small and/or portable headphone amplifiers are designed to output a volt or two into low impedance, high efficiency headphones, e.g. Grado headphones.
  3. Low impedance, low efficiency headphones usually sound better when driven by a desktop amplifier, e.g. Audeze LCD-2 or AKG K70X.
  4. Low impedance headphones usually sound better when driven by a solid state or a transformer coupled vacuum tube amplifier.  Low impedance headphone do not usually work well with Output Transformerless vacuum tube amps.

 

post #4917 of 6535
Quote:
Originally Posted by bowei006 View Post

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.


Edited by m8o - 12/8/12 at 5:02pm
post #4918 of 6535
Quote:
Originally Posted by m8o View Post

 

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.

I will guess that the 10 seconds it took me to write my responses before resulted in a confusion.

 

My point above was basically what your last sentence stated but without all the math.

post #4919 of 6535
Quote:
Originally Posted by m8o View Post

 

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

post #4920 of 6535
Quote:
Originally Posted by KamijoIsMyHero View Post

 

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

Sorry about that. Going back and re reading my lesson on impedance's and what not it seems I confused a few terms.

 

Here is my friend's explanation of it if anyone is curious:

 

 

Quote:Borisu

Say you have 2 headphones, one with 32Ω impedance and the other with 300Ω impedance.

Assume both require 1mW for listening levels (which is quite high) and constant impedance for all frequencies.

The the 32Ω and 300Ω phones will need (0.18V, 5.6mA) and (0.55V, 0.18mA) respectively.

So the question is if the amp performs better outputting more current or higher voltage swing.

For most cases amplifier tend to perform better outputting less current (to do with the amp's output impedance.)

 

^ BTW, the only thing I've really said up there is that most amps perform better with high load impedances, but somehow adding some statistics and terminology will make me sound more like an expert thus more believable. Its basically mostly BS that isn't reverent to the topic.

THAT'S HOW MARKETING WORKS! 

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