Need help! really confused with amp/dac choice for dt990 250 ohm pro
Jan 11, 2015 at 10:04 PM Thread Starter Post #1 of 38
ConWood

ConWood

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Soooooo... I looked all over and I'm getting various opinions which make me more and more confused.
 
The amps/dacs I am looking to buy to help power the dt990 250 ohm pro's is the fiio e10k. I currently have a xonar dg sound card in my pc that I don't mind bypassing because I don't use the dolby headphones on it really at all. I initially put it in there as a small upgrade from on-board sound when I build my pc. As for the fiio e10k some people are saying they are not good to power the headphones and other say they work beautifully. So I am just a little confused.
 
My question basically is: Does the fiio e10k power the Beyer DT990 250 ohm pro headphones and the reach a reasonable volume while still sounding great? OORRRR... Should I just get an amp and use my xonar dg as my source/dac? and what amp would be a good suggestion? budget ~130ish.
 
Thanks for any help you give to this audio noob! 
 
Jan 11, 2015 at 10:37 PM Post #2 of 38
Normally I recommend getting an amp with enough power to reach 110 dB. This allows you to crank the average volume up to 90 dB and still have room for the +20 dB peaks some dynamic albums might have. Most people will use less, in certain situations you might use more, but it's a good overall volume to aim for in most cases.
 
Beyerdynamic lists the DT990 Pro as having a sensitivity of 96 dB/mW. So we want to figure out how much power we need for 14 more decibels. It just so happen that if you divide the change in decibels by 10, then do 10 to that power, you get the change in power. In this case that's 10 ^ ( 14 / 10 ), or just about 25. So now we know we need an output of 25 mW for 110 dB with the DT990.
 
Now, how powerful is the E10K? This will depend on the headphone's impedance and the amp's voltage output. Sometimes the amp's spec page will publish the max power output at various impedances, in which case you can just use that. The E10K only tells us the power at 32 ohms which isn't much help, because the DT990 has an impedance of 250 ohms. But it does tell us the max Vp-p (peak-to-peak voltage) of 7.39 Vp-p,  which we can use. First, though, we need to find the Vrms (root mean square voltage), which is more useful for audio. It's easier than it sounds; for a sine wave like our audio signal, you halve Vp-p to get the amplitude, then divide that by the square root of 2. In this case that's ( 7.39 / 2 ) / sqrt(2). The E10K is capable of up to 2.6 Vrms in a best case scenario.
 
What's that mean for power? Converting to power is pretty easy too, you just square your Vrms and divide by the headphone's impedance, which gives you a value in full Watts. Then multiply by 1000 to get that same value in mW. Then the E10K should ideally be capable of ( 2.6 ^ 2 ) / 250 * 1000 = 27 mW at 250 ohms.
 
Now, most amps can't drive their max voltage into every impedance, it can fall off due to output impedance on the amp or a limited amount of current. That's a story for another post, though. The E10K should be pretty capable of 25 mW into 250 ohms, it's not a difficult load. But, I've seen DT880s measured as needing up to 3 Vrms for 110 dB as opposed to the 2.6 we tested here, meaning official specs were not accurate. Still, remember that's only if you need volumes of 110 dB. As volume goes down power goes down rapidly (about half the power per 3 dB down).
 
I think you'll be just fine with the E10K. If you feel safer with something else, look for something capable of 36 mW at 250 ohms, or 3 Vrms, using the equations I posted if needed. The Schiit Fulla is an option.
 
I hope this mess helps someone.
 
Jan 11, 2015 at 11:08 PM Post #3 of 38
You, good sir, are wonderful and beyond enlightening. I appreciate the math and explanation! With this information I am pretty sure I will be going with the e10k after I get my classes for next semester paid for. Stupid engineering fees on top of college prices suck... 
 
Once again thank you for everything it really helped me out!
 
Feb 25, 2017 at 12:16 AM Post #4 of 38
head injury said:
  Normally I recommend getting an amp with enough power to reach 110 dB. This allows you to crank the average volume up to 90 dB and still have room for the +20 dB peaks some dynamic albums might have. Most people will use less, in certain situations you might use more, but it's a good overall volume to aim for in most cases.
 
Beyerdynamic lists the DT990 Pro as having a sensitivity of 96 dB/mW. So we want to figure out how much power we need for 14 more decibels. It just so happen that if you divide the change in decibels by 10, then do 10 to that power, you get the change in power. In this case that's 10 ^ ( 14 / 10 ), or just about 25. So now we know we need an output of 25 mW for 110 dB with the DT990.
 
Now, how powerful is the E10K? This will depend on the headphone's impedance and the amp's voltage output. Sometimes the amp's spec page will publish the max power output at various impedances, in which case you can just use that. The E10K only tells us the power at 32 ohms which isn't much help, because the DT990 has an impedance of 250 ohms. But it does tell us the max Vp-p (peak-to-peak voltage) of 7.39 Vp-p,  which we can use. First, though, we need to find the Vrms (root mean square voltage), which is more useful for audio. It's easier than it sounds; for a sine wave like our audio signal, you have Vp-p to get the amplitude, then divide that by the square root of 2. In this case that's ( 7.39 / 2 ) / sqrt(2). The E10K is capable of up to 2.6 Vrms in a best case scenario.
 
What's that mean for power? Converting to power is pretty easy too, you just square your Vrms and divide by the headphone's impedance, which gives you a value in full Watts. Then multiply by 1000 to get that same value in mW. Then the E10K should ideally be capable of ( 2.6 ^ 2 ) / 250 * 1000 = 27 mW at 250 ohms.
 
Now, most amps can't drive their max voltage into every impedance, it can fall off due to output impedance on the amp or a limited amount of current. That's a story for another post, though. The E10K should be pretty capable of 25 mW into 250 ohms, it's not a difficult load. But, I've seen DT880s measured as needing up to 3 Vrms for 110 dB as opposed to the 2.6 we tested here, meaning official specs were not accurate. Still, remember that's only if you need volumes of 110 dB. As volume goes down power goes down rapidly (about half the power per 3 dB down).
 
I think you'll be just fine with the E10K. If you feel safer with something else, look for something capable of 36 mW at 250 ohms, or 3 Vrms, using the equations I posted if needed. The Schiit Fulla is an option.
 
I hope this mess helps someone.
Click to expand...

man this formula applies for any headphone?
10 ^ ( 14 / 10 )
/10 from: It just so happen that if you divide the change in decibels by 10, right?
10 ^ from: then do 10 to that power, right?
14 from: 110-96 db, right?
what is this ^ means so you can have 25 as a result?
 
( 7.39 / 2 ) / sqrt(2)
7.39 is base from the spec of e10k?
/2 from where?
what is sqrt(2)? from where? what is the number so you have the result 2.6?
 
( 2.6 ^ 2 ) / 250 * 1000
2.6 vrm
^2 from?
/250 from headphone impedance
1000 to make it mw
 
what i can't understand is what is this (^) means? and what is sqrt(2)? 
 
Feb 25, 2017 at 12:28 AM Post #5 of 38
benn beckman said:
  man this formula applies for any headphone?
10 ^ ( 14 / 10 )
/10 from: It just so happen that if you divide the change in decibels by 10, right?
10 ^ from: then do 10 to that power, right?
14 from: 110-96 db, right?
what is this ^ means so you can have 25 as a result?
 
( 7.39 / 2 ) / sqrt(2)
7.39 is base from the spec of e10k?
/2 from where?
what is sqrt(2)? from where? what is the number so you have the result 2.6?
 
( 2.6 ^ 2 ) / 250 * 1000
2.6 vrm
^2 from?
/250 from headphone impedance
1000 to make it mw
 
what i can't understand is what is this (^) means? and what is sqrt(2)? 
Click to expand...

 
^ means to the power of. so for example 2^2 is 2 to the power of 2.
sqrt is square root.
 
Feb 25, 2017 at 12:56 AM Post #6 of 38
dulalala said:
   
^ means to the power of. so for example 2^2 is 2 to the power of 2.
sqrt is square root.
Click to expand...

still i can't understand, if 2*2 means to power 2, then 10^1.4 is 10 to power 1.4??
i never know that square root before, what is that mean? is it a number or what? or is it exist in scientific calculator so i just put the number there and the result come?
man please help me, this formula can help me to choose what kind of headphone or amplifier if i know that 2 missing link.
please
 
Feb 25, 2017 at 1:04 AM Post #7 of 38
√  this is the symbol for square root. 
For power it would be 101.4 in this case.
I'm sorry if this is a bit rude but I'm either assuming you know nothing about maths, or English isn't your first language. In any case, I think it would be much easier for you to ask for assistance rather than trying to figure it out yourself. If you have no background knowledge in science or math you may accidentally make a wrong calculation and end up buying a product that would not fit the specification you require.
 
Feb 25, 2017 at 4:17 AM Post #8 of 38
dulalala said:
  √  this is the symbol for square root. 
For power it would be 101.4 in this case.
I'm sorry if this is a bit rude but I'm either assuming you know nothing about maths, or English isn't your first language. In any case, I think it would be much easier for you to ask for assistance rather than trying to figure it out yourself. If you have no background knowledge in science or math you may accidentally make a wrong calculation and end up buying a product that would not fit the specification you require.
Click to expand...

it's okay mate, yes actually my english is bad, i'm still learning. i just don't know the symbol on english, if you explain me like this i can understand. i'm not good and math either, but this is the basic, if i know what you mean in my langunge this is easy. this formula will help me to buy the next headphone than matching with an amp i have or vice versa. thank you mate.
once again thank you very much mate
 
Feb 25, 2017 at 4:29 AM Post #9 of 38
benn beckman said:
  it's okay mate, yes actually my english is bad, i'm still learning. i just don't know the symbol on english, if you explain me like this i can understand. i'm not good and math either, but this is the basic, if i know what you mean in my langunge this is easy. this formula will help me to buy the next headphone than matching with an amp i have or vice versa. thank you mate.
once again thank you very much mate
Click to expand...

There is this website which maybe able to help. If you have the specifications, you can input it in and it will calculate the required info for you. I hope it helps! (http://www.digizoid.com/headphones-power.html)  
 
Feb 25, 2017 at 6:59 AM Post #12 of 38
benn beckman said:
  mate what other indicator that i can use for calculate how much power that an amp can produce if the vp-p unknown?
i'm curious mate, there is no vp-p on creative sound blaster e5 specs. only output impedance 2.2 ohm and thd <0.005%
can i use those specs to know how much power they can produce @250 ohm?
Click to expand...

Unfortunately without the proper specifications you cannot find out how much power they output at 250 ohms. You would need to find out it's output power at a specific impedance in order to calculate the power at different impedance. I have tried searching for the specs myself and I was not able to find it either. 
 
Feb 25, 2017 at 7:28 AM Post #13 of 38
dulalala said:
  Unfortunately without the proper specifications you cannot find out how much power they output at 250 ohms. You would need to find out it's output power at a specific impedance in order to calculate the power at different impedance. I have tried searching for the specs myself and I was not able to find it either. 
Click to expand...

oh ok can you explain me what is RMS?
and i'm trying to figure this:
 V = IR
P = (V^2 / R) = VI
Click to expand...

what is
P
R
IR
VI
means?
Here is how I see it.

There are voltage swing limitations, and current drive limitations.

USB provides a 5V / 500mA supply, and that's it. So...

EDIT: Note USB provides 5V to GND, which the Fulla-2 converts to +/-5V using a "high-current dual-polarity switching regulator, with inductor filtering and local regulation". This means the internal devices are going to see 10 Vpp, 5 Vp, and +/-5 V rails .

1) Maximum Power, 300 ohms: 40mW RMS per channel

Means the system is voltage swing limited to V = sqrt(R*P) = sqrt(300 * 0.04) = 3.46 Vrms (Remember, USB is limited to 5 V. Since 3.46 Vrms is about 4.89 V peak you are kind of close to what the port can give you anyway).

The device may be capable of giving you **** loads of current, but if it can only swing 3.46 Vrms, it will only give you 11.5 mA rms. Unless you do something funky, you cannot give the load more than the supply is giving you: 5 V. And to that subtract all non-idealities and realities that Jason mentioned. I'm surprised how close the device can swing from the rails!

2) Maximum Power, 16 Ohms: 550mW RMS per channel

Means the system at this point is current drive limited to I = sqrt(P/R) = sqrt(0.55 / 16) = 185 mA rms. (Remember, USB is limited to 500 mA so with 2 channels the 185 mA x 2 = 370 mA rms or ~500 mA which is kind of what the port can give you anyway)

The device may be capable of giving you a swing of 3.46 Vrms as implied by 1), but since it can only give you 185 mA rms per channel, you are only going to get 2.96 V rms.

Now, can we not be voltage limited into 300 ohms, and not current starved into 16 ohms? Yes, proly get another more powerful amp, and get voltage limited into another impedance and current starve into another impedance. May need to not use the USB port for power, and there goes portability.

As it is, IMO the drive limitations of the Fulla 2.0 are pretty competitive in it's class and should be able to drive most cans out there using a single simple portable solution. YMMV.

BTW, voltage swing limitations and current drive limitations may be (slightly) different depending on load impedance as well.  
Click to expand...


 ​
Click to expand...

10 Vpp, 5 Vp, and +/-5 V rails.
is Vp = 1/2 Vpp ?
 
what is voltage limited swing? i mean what is the different with current drive limited?
 
Remember, USB is limited to 500 mA so with 2 channels the 185 mA x 2 = 370 mA rms or ~500 mA which is kind of what the port can give you anyway)
is 5V = 500mA?
 
how much Vrms to V?
how much Vrms to mA?
 
Feb 25, 2017 at 7:53 AM Post #14 of 38
benn beckman said:
  oh ok can you explain me what is RMS?
and i'm trying to figure this:
what is
P
R
IR
VI
means?
10 Vpp, 5 Vp, and +/-5 V rails.
is Vp = 1/2 Vpp ?
 
what is voltage limited swing? i mean what is the different with current drive limited?
 
Remember, USB is limited to 500 mA so with 2 channels the 185 mA x 2 = 370 mA rms or ~500 mA which is kind of what the port can give you anyway)
is 5V = 500mA?
 
how much Vrms to V?
how much Vrms to mA?
Click to expand...

RMS - root mean square
P - power
R - resistance
IR - I is current and R is resistance so IR is current x resistance. V=IR is the formula.
VI - V is voltage and I is current so VI is voltage x current. P=VI is the formula.
Vpp - is voltage peak to peak. So basically the difference from the highest voltage value to the lowest value which is also just the highest voltage peak x2. So 2Vp=Vpp. 
Voltage Swing - is the highest voltage peak to peak possible without clipping.
Current Drive Limited - Limiting the incoming current.
"is 5V = 500mA?" - I have no clue what you mean by this but it is dependent on resistance. The base formula is V=IR.
Vrms - It is the root mean square of the voltage, meaning it is the average/mean of the voltage squared, then square rooted. So formula would be √ (ΣV2/n).
 
Feb 25, 2017 at 1:55 PM Post #15 of 38
dulalala said:
  RMS - root mean square
P - power
R - resistance
IR - I is current and R is resistance so IR is current x resistance. V=IR is the formula.
VI - V is voltage and I is current so VI is voltage x current. P=VI is the formula.
Vpp - is voltage peak to peak. So basically the difference from the highest voltage value to the lowest value which is also just the highest voltage peak x2. So 2Vp=Vpp. 
Voltage Swing - is the highest voltage peak to peak possible without clipping.
Current Drive Limited - Limiting the incoming current.
"is 5V = 500mA?" - I have no clue what you mean by this but it is dependent on resistance. The base formula is V=IR.
Vrms - It is the root mean square of the voltage, meaning it is the average/mean of the voltage squared, then square rooted. So formula would be √ (ΣV2/n).
Click to expand...

ok 
V=IR
P=VI
R=v/I
I=V/R or P/V depend what we know
 
2Vp=Vpp 
i understand
 
Voltage swing yes i understand
Current drive limited means the limit of minimum current right?
 
5V=500mA from Remember, USB is limited to 500 mA 
but i forgot this USB provides a 5V / 500mA supply
so its not means 5V = 500mA but. USB can provide 5V or 500mA right? i mean its in different unit? am i wrong?
 
so in reality on headphone spec list usually on spec list we know R is impendance and what is V and the missing link and still remain mistery to me is I on what unit we described current as?
if R in ohm, Sensitivity in db, what is V and I, and how i look those two on headphone or amp spec list?
i mean on this thread http://www.head-fi.org/t/708158/fiio-e07k-enough-to-power-the-250-ohm-dt-990-pro-dt990-or-akg-550#post_13292996
if i apply this formula 10^(110-headphone sensitivity/10) to find the power we need then akg k702 will be easy to drive
akg k702 have 105db, so 10^(5/10)=3.16mW but he says that akg is hard to drive since low impendance need more power. i'm confuse because like fidelio x2 is 32ohm and it easy to drive.
so what is the missing link?
 
edited
after i read calculator that you gave me carefully so current unit is mA right? Vrms is unit for V and P unit is mW right?
so it can be like this 
V=(Vpp/2)/sqrt2
P=(V^2)/R *1000
 
to reach 110 on beyer dt990 pro need V: 2.51 Vrms  I: 10.04 mA  P: 25.2 mW
to reach 110 on akg k702 need V: 1.78Vrms  I:28.71 mA  P:51.1mW
 
a new question is what is the different SPL/mW and SPL/V on that calculator? since mW is P and V is in different unit? V unit is Vrms right?
i use SPL/mW for beyer, and the result is correct to this formula 10^(14/10)
but i use SPL/V because in akg website it write 105db SPL/V
 
second from where is 10^ and /10 on this formula 10^(14/10) which what i know only 14 from (110-96) but i don't know 10^ and /10 come from, it can be apply on SPL/mW right? what about the formula for SPL/V?
 
3.46 Vrms is about 4.89 V
3.46 Vrms=4.89V how is the formula?
 
3.46 Vrms, it will only give you 11.5 mA rms
3.46 Vrms=11.5 mA rms what is the formula?
 
185 mA rms per channel, you are only going to get 2.96 V rms.
185 mA rms=2.96 Vrms ?
 
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