[ADUHF]

All recordings have the same maximum sustained peak level. That is the maximum comfortable volume you set it to with your amp....

...I might not be explaining that clearly. But I'm trying.

You're doin fine bro. I guess sustained peaks are different than average sustained levels. Hence the confusion.

Another way to look at it would be peak comfortable listening volume. For the loudest sustained peak level, that would be about 80dB. Add 10dB for a transient peak here and there and you are at 90dB. If an amp can produced 90dB, you should be fine. Anything over that would be overkill.

I'm not sure how this fits into the equation, but a figure I've often run across on sites like Gearslutz is around 85 dB. I'm not sure if that's for mixing or mastering though. And whether that includes the headroom that you all are referring to. This would be for loudspeaker monitors in a studio btw, rather than headphones.

I would think that the approximate range used for 0 dBFS for mastering might be more relevant to this question than the range for mixing though.

 

[ADUHF]

If you want to get technical, you should add the headphone's impedance (R_L = load impedance) to the amp's output impedance (R_S = source impedance), which would be important for an amp with very high output impedance, like an OTL.
P = V²/(R_L+R_S)

The above formula seems to imply that an amp with higher impedance (say 50 ohms) needs a higher voltage to achieve the same power as an amp with lower impedance (say 1 ohm). Does that seem correct?

I would've thought it was the other way around. But I'm not really sure.

 

[VNandor]

I'm using it in the sense that it is used in recording studios.

Another way to look at it would be peak comfortable listening volume. For the loudest sustained peak level, that would be about 80dB. Add 10dB for a transient peak here and there and you are at 90dB. If an amp can produced 90dB, you should be fine. Anything over that would be overkill.

I think that clears it up. So you define headroom as the difference between the transient peak level and the sustained peak level while I used it as the difference between the average level and the transient peak level. I could ask how "sustained" we are talking about exactly but ultimately it doesn't matter. At least I understand where you are and your numbers are coming from.

 

[VNandor]

The above formula seems to imply that an amp with higher impedance (say 50 ohms) needs a higher voltage to achieve the same power as an amp with lower impedance (say 1 ohm). Does that seem correct?

I would have thought that it was the other way around. But I'm not really sure.

Let's say you feed some arbitrary signal to the amplifier and you measure the output voltage with a multimeter. You adjust the gain until you read 1V. If you repeated the measurement but this time you used a load (for example you connected a headphone to the amp) the voltage would actually drop a little bit and you would have to increase the gain a bit to read the same voltage on the multimeter. If the amplifier has a high output impedance the drop could be considerable, if the output impedance was low, the drop would be negligible. Ultimately the ratio between the output impedance and the headphone's impedance is what actually matters.

This is because the output impedance of the amplifier and the impedance of the headphones are creating a voltage divider with each other. If the output impedance is too high, plenty of voltage is going to drop on that instead of on the headphones. The higher impedance a headphone has, the less likely this is going to be a problem.

As a sidenote, the output impedance isn't literally a resistor slapped to the output of the amp, it's just an abstraction to help with modelling how the amplifier would work with different loads.

Oh, to also answer the question, all in all, the headphones would take the same voltage but the amplifier would have to be set to a higher gain to achieve that same voltage.

 

[megabigeye]

120dB is the threshold of pain and into the range where you incur serious hearing damage from short exposure. It's like laying down on the sidewalk next to a jackhammer. If you play Dark Side of the Moon at 120dB you'll be deaf before you get to the second track on the album. Most people listen to music below 80dB (80dB isn't pleasant). That is a much more reasonable target. The difference between 80 and 120dB is massive. I'd suggest arbitrarily picking 80dB. That will give you plenty of leeway.

I think the reason that a lot of discussion on Head-Fi goes in circles is because too much of it is based purely on arbitrary theory. I'm not criticizing you for that, it's the way things seem to work around here. Everyone talks about specs as arbitrary numbers on a page, and they don't translate them into real world listening with human ears. A lot of "experts" on yootoob make it worse by claiming to be able to hear numbers that only bats can hear. So everyone just grabs a worst case number out of the air and cites that. Then the next guy picks a slightly larger number "just to be safe" and uses that. The next person takes that number and makes it a little bigger "just to be safe"... and so on. Pretty soon you're at the threshold of pain and it isn't at all safe! There's no anchor in reality to give context to those numbers.

Yes, 120dB is the threshold of pain, and I think hearing damage is incurred in something like 7 seconds. That's why I picked it as my "arbitrary" number that shouldn't be exceeded.
If I understand what @VNandor was saying about their music (was that this thread? Now I can't remember), I don't know if it's that unreasonable, though-- if you're rocking out at an average of 90dB and your music has transient peaks of +30dB, you'd want your amp to be able to handle that. That sounds hellishly loud to me, but it also doesn't sound entirely out of the question, either.

I agree with your second point. There's a lot of arbitrariness and opinion-as-fact and "just-to-be-safe" on Head-Fi. I try to be as accurate as I can and point out where I'm being arbitrary. I realize that my 120dB was a worst case and that others will probably run with it. "Well... 125dB, just to be safe..." "125dB? Better make it more like 150dB, just in case..." "150dB?! Only if you're not a real man and can't handle 200dB!"
On the other hand, I think that a lot of Head-Fi'ers would see a lower, more reasonable number like 100dB or 110dB and think it's not high enough and infer that I don't know what I'm talking about.

The above formula seems to imply that an amp with higher impedance (say 50 ohms) needs a higher voltage to achieve the same power as an amp with lower impedance (say 1 ohm). Does that seem correct?

I would've thought it was the other way around. But I'm not really sure.

Yes, I think you're right.
The easiest, most succinct way to understand it (at least for me) is to plug in different numbers for the variables.

I believe the Bottlehead Crack as an output impedance of 120Ω in stock form, so let's use that using @skhan007's headphones with 97dB @1mW and 300Ω. We already know it takes 501mW to reach 120dB, so:
0.501W = V²/(300Ω+120Ω) = V²/420Ω
V = √(0.501W*420Ω)
V = 14.5V

So it requires 2.2V more than an ideal amp with 0Ω output impedance.

 

[skhan007]

Yes, 120dB is the threshold of pain, and I think hearing damage is incurred in something like 7 seconds. That's why I picked it as my "arbitrary" number that shouldn't be exceeded.
If I understand what @VNandor was saying about their music (was that this thread? Now I can't remember), I don't know if it's that unreasonable, though-- if you're rocking out at an average of 90dB and your music has transient peaks of +30dB, you'd want your amp to be able to handle that. That sounds hellishly loud to me, but it also doesn't sound entirely out of the question, either.

I agree with your second point. There's a lot of arbitrariness and opinion-as-fact and "just-to-be-safe" on Head-Fi. I try to be as accurate as I can and point out where I'm being arbitrary. I realize that my 120dB was a worst case and that others will probably run with it. "Well... 125dB, just to be safe..." "125dB? Better make it more like 150dB, just in case..." "150dB?! Only if you're not a real man and can't handle 200dB!"
On the other hand, I think that a lot of Head-Fi'ers would see a lower, more reasonable number like 100dB or 110dB and think it's not high enough and infer that I don't know what I'm talking about.


Yes, I think you're right.
The easiest, most succinct way to understand it (at least for me) is to plug in different numbers for the variables.

I believe the Bottlehead Crack as an output impedance of 120Ω in stock form, so let's use that using @skhan007's headphones with 97dB @1mW and 300Ω. We already know it takes 501mW to reach 120dB, so:
0.501W = V²/(300Ω+120Ω) = V²/420Ω
V = √(0.501W*420Ω)
V = 14.5V

So it requires 2.2V more than an ideal amp with 0Ω output impedance.

Could we double check the math? I’m getting 200 mW using the formula on the last page. I think you had 93 listed vs 97, for the 120-97 exponent.

Another question: since 120 dB is crazy loud, if I substitute 85 dB in the formula (which produces a negative value), are the results still accurate?

 

[megabigeye]

Could we double check the math? I’m getting 200 mW using the formula on the last page. I think you had 93 listed vs 97, for the 120-97 exponent.

Another question: since 120 dB is crazy loud, if I substitute 85 dB in the formula (which produces a negative value), are the results still accurate?

D'oh!

You're right. This is why I'm not a mathematician. I'll spend an hour pouring over grammar and making sure my language is just so, but I screw up the math almost every time.
With the right math I got 199.5mW and 7.7V

And yes, it still works with negatives in the exponent. It'll show you how little power headphones actually need under normal (i.e., not mind-bogglingly loud) listening conditions.
P_T = 10^(85-97)/10 = 0.063mW That's 63µW! Microwatts, not milliwatts!
V = √(0.000063W*300Ω) = 0.14V The idea that you'd need more than 100 times that voltage is... Misguided.

 

[skhan007]

No worries! Great information regardless. Yes, I too came up with 0.063, and was curious if that made sense. I guess these 300 Ohm ZMF’s can work under a range of mW values. Now, how people feel about the sound in lower mW settings is up for interpretation.

Lastly, I find it absolutely amazing that you chose the Bottlehead Crack as your example. You’ll never guess what amp I ordered just this morning!! Talk about a wild coincidence!

 

[sander99]

I adjusted the following quote to the corrected 199.5mW, but I round it to 200mW:

I believe the Bottlehead Crack as an output impedance of 120Ω in stock form, so let's use that using @skhan007's headphones with 97dB @1mW and 300Ω. We already know it takes 200mW to reach 120dB, so:
0.200W = V2/(300Ω+120Ω) = V2/420Ω
V = √(0.200W*420Ω)
V = 9.16V

You are forgetting one thing here: This way the amp delivers 0.200W in the combination of headphone and output impedance, and only a part of that power ends up in the headphones, and the rest is "burned" inside the amp, so you actually need an even higher voltage!
Let me try it differently, first let's go back to the result of your first calculation (I corrected this quote to 7.75V - added one decimal for precision):

V = 7.75V

Now to be precise we want the amp to deliver this voltage over it's output connector. As @VNandor already said the output impedance and the load impedance act as a voltage devider.
From that we can infer that the amp with 120Ω output impedance internally has to deliver (300+120)/300 * 7.75V = 10.85V
Doublecheck: 10.85V / 420Ω gives 0.0258A, the headphones get P = I^2 * R = 0.0258^2 * 300 = 0.200W (that was 0.199692 so rounded 0.200)
That looks correct!
(And "burned" inside the amp is P = I^2 * R = 0.0258^2 * 120 = 0.0799W).

 

[megabigeye]

No worries! Great information regardless. Yes, I too came up with 0.063, and was curious if that made sense. I guess these 300 Ohm ZMF’s can work under a range of mW values. Now, how people feel about the sound in lower mW settings is up for interpretation.

Ah! Reading this I was struck with the thought that part of the reason people might think a headphone is "under driven" / "not driven properly" / whatever has to do with the equal loudness curve. Remember the "loudness" button on old receivers? That adds bass and treble for lower volume listening since our hearing is less acute at those frequencies.

As an aside, I wish some developer would create a dynamic equal-loudness app based on our hearing. I can't imagine it would be that hard (says the guy with absolutely no coding or psychoacoustics knowledge).

Lastly, I find it absolutely amazing that you chose the Bottlehead Crack as your example. You’ll never guess what amp I ordered just this morning!! Talk about a wild coincidence!

Nice! I've always been curious, but I can never justify it to myself since I have headphones that wouldn't work well with it. I have the Quickie + Quicksand combo, which was a ton of fun to put together and that I've enjoyed a lot (but that makes a ton of noise and that I'm much to lazy to fix/mod). I've thought about getting the SEX, but I don't know that I want that much voltage just hanging around my apartment... Also, I'm lazy and cheap.

 

[megabigeye]

I adjusted the following quote to the corrected 199.5mW, but I round it to 200mW:

You are forgetting one thing here: This way the amp delivers 0.200W in the combination of headphone and output impedance, and only a part of that power ends up in the headphones, and the rest is "burned" inside the amp, so you actually need an even higher voltage!
Let me try it differently, first let's go back to the result of your first calculation (I corrected this quote to 7.75V - added one decimal for precision):

Now to be precise we want the amp to deliver this voltage over it's output connector. As @VNandor already said the output impedance and the load impedance act as a voltage devider.
From that we can infer that the amp with 120Ω output impedance internally has to deliver (300+120)/300 * 7.75V = 10.85V
Doublecheck: 10.85V / 420Ω gives 0.0258A, the headphones get P = I^2 * R = 0.0258^2 * 300 = 0.200W (that was 0.199692 so rounded 0.200)
That looks correct!
(And "burned" inside the amp is P = I^2 * R = 0.0258^2 * 120 = 0.0799W).

Thanks for the clarification!

I forgot about it being a voltage divider. This makes sense.

 

[ADUHF]

I'm not sure how this fits into the equation, but a figure I've often run across on sites like Gearslutz is around 85 dB. I'm not sure if that's for mixing or mastering though. And whether that includes the headroom that you all are referring to. This would be for loudspeaker monitors in a studio btw, rather than headphones.

I would think that the approximate range used for 0 dBFS for mastering might be more relevant to this question than the range for mixing though.

Did a little poking around, and it looks like 85 dB is a fairly common reference level used in both mixing/mastering and also in home theaters. The 85 dB reference listening level is usually set with an approximately -20 dBFS signal though. Which means 0 dBFS should be 20 dB above that, at approximately 105 dB. Low frequencies may excurse a bit higher in volume though, up to 115 dB. Due to greater dispersion, and room gain, I guess.

So it looks like 105 to 115 dB is a pretty good range for the maximum clip point on an amp. (Why audiophiles need an extra +5 dB on top of that, I dunno. But I assume they're just rounding up to 120 for the sake of simplicity. )

The 85 dB reference level is supposedly fairly loud though. So most folks actually listen at about -10 dB (or even lower) below that. Which would put any 0 dBFS transients more in the 95 to 105 dB range.

 

[ADUHF]

Each +10dB requires 10x power, so to determine how much power you need:
P_T = 10^(T-S)/10
Where P_T is target power, T is target volume, and S is headphone sensitivity.
Since headphone measurements are in milliwatts, this equation spits out milliwatts.

Using 120dB as T with your headphones:
P_T = 10^(120-93)/10
P_T = 10^2.7
P_T = 501mW

Using the equations above, I get P = 10^[(115-97)/10] = 10^1.8 = 63 milliwatts. Which is well below the ADI-2's ~300mW clipping power for a 300 ohm load (according to Amir's review).

So it looks like the ADI-2 does indeed have sufficient headroom to reach reference listening levels with a 300-ohm headphone, with power efficiency in the 97db/mW range. The Sennheiser HD800S is supposedly even more efficient at 100 dB/mW, so it should have enough headroom as well. Let's confirm it though, just for grins...

P= 10^[(115-100)/10] = 10^1.5 = 31.6mW

Looks pretty good.

It would be nice to confirm Amir's max power ratings for the ADI-2 with another independent source though.

 

[ADUHF]

NM

 

[Kammerat Rebekka]

^^There are quite a few ‘headphone calculators’ out there that do the maths for you.
http://www.digizoid.com/headphones-power.html
Then again I’ve tried several amps that on paper seemed like a poor match for my HE500, yet in a blindtest up against a kilobuck monster amp, I could detect no differences outside of cranking the volume beyond what I find pleasing.