Can E9 drive LCD2 well
Dec 12, 2011 at 3:14 PM Post #18 of 43
In a nutshell, if you're able to hit a good enough volume on your headphones without cranking it to 90%+ on everything, you're probably good as far as amps are concerned. 
 
Dec 12, 2011 at 3:43 PM Post #20 of 43
The LCD-2 will work well with the E9. I have listened and compared many inexpensive amps with the LCD-2 and an expensive amp I used to have and they sound extremely close. Tube amps will allow you to fine tune the sound a little more. 
 
Dec 12, 2011 at 4:38 PM Post #23 of 43
From what I've learnt on the forum, if you have a good headphone with high impedance, it's necessary to have an amp, otherwise, the sound from headphone will be muddy and unclear. Amp will make a huge difference.
 
Dec 12, 2011 at 4:43 PM Post #24 of 43
This is actually a bit of misconception. Higher impedance implies a less efficient headphone, but doesn't necessarily mean it is inefficient. The efficiency of a headphone is calculated from a combination of the impedance and also the sensitivity of dB/mW or dB/V. 
 
For example of low impedance headphones that are difficult to drive but have relatively low impedance ratings, look at everything Hifiman makes. They are some of the most difficult headphones to drive, but have impedance ratings in the 30s. 
Quote:
From what I've learnt on the forum, if you have a good headphone with high impedance, it's necessary to have an amp, otherwise, the sound from headphone will be muddy and unclear. Amp will make a huge difference.



 
 
Dec 12, 2011 at 4:50 PM Post #25 of 43


It will probably sound horrible through your laptop since laptop soundcards are generally atrocious. It may sound alright through the Ipad2
 
 

well do they give like 80 or 85 % of it capacity? or less ? if it less then it wont make sense to me to spend 1k on this i would rather get a hooker for 1 year nonstop 
 

From what I've learnt on the forum, if you have a good headphone with high impedance, it's necessary to have an amp, otherwise, the sound from headphone will be muddy and unclear. Amp will make a huge difference.

 
umm try it with you ipod or what ever you have without amp and tell us if it worth to spend 1k$
 
 
Dec 12, 2011 at 5:03 PM Post #26 of 43
Roughly speaking, most portable sources have .6V of output. Most computer outputs have around 1-1.2V. The Fiio E5, E6, and E7 all have 1.2-1.8V depending on the impedance of the headphones. The Fiio E9 can reach up to around 6-7V. To reach 110dB SPL with the LCD-2, you need closer to 2V but not everyone needs that kind of volume from their headphones. That should put it relatively in perspective. 
 
Dec 12, 2011 at 5:08 PM Post #28 of 43

 
Quote:
This is actually a bit of misconception. Higher impedance implies a less efficient headphone, but doesn't necessarily mean it is inefficient. The efficiency of a headphone is calculated from a combination of the impedance and also the sensitivity of dB/mW or dB/V. 
 
For example of low impedance headphones that are difficult to drive but have relatively low impedance ratings, look at everything Hifiman makes. They are some of the most difficult headphones to drive, but have impedance ratings in the 30s. 


 


well lesson failed
 
 
Dec 12, 2011 at 5:18 PM Post #30 of 43

Sure. 
Power in watts is calculated by taking (Voltage * Voltage) / Impedance. So, for an iPod with 0.6V of power, we have (0.6V * 0.6V) / 50 ohm = 0.0072W (7.2 mW) for power into an LCD-2. 
 
We can use the power calculation to calculate the volume from the headphones. The LCD-2 has a sensitivity of 91dB/mW. dB is calculated from dB = sensitivity in dB/mW + 10 * log(power in mW). In the iPod with LCD-2's case we have 91 + (10 * log(7.2)) = 99.6dB.
 
For comparison against the E9, which I suspect has about 6V of output into 50ohms, we have (6 * 6) / 50 = .72W (720 mW). 
dB = 91 + 10 * log(720) = 119dB, way louder than you would ever want to listen to, really.
 
I guess I was wrong when I said 9V for the LCD-2, I don't know why that number stuck out in my head because it was completely off. It's much closer to 2.2V.
Quote:
Could you show us a formula how you calculate this, so we can use it for future reference.



 
 

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