In the manual of the ADI-2 DAC in chapter 31.16 Block Diagram you can see, how the two sockets are connected. Both are sourced from the only one digital to analog converter. There is no possibility to adjust the volume separately (except the Hi/Lo setting for the 1/4" TRS).

If you look at the block diagram in the ADI-2 Pro manual (17.2 Preamp and following, 34.6 Balanced Phones Mode) you can see the difference. Phones 1/2 and phones 3/4 have complete separated digital to analog converter and signal processing. Therefore each socket has own setting (volume, parametric EQ, loudness, crossfeed, ...), even balanced phones can be used.

As long as you don't plug an EMI into the 1/4" jack, nothing will break, but one of the two headphones will either be too quiet or too loud, and there's nothing you can do about it. Look to the manual: Technical Specifications / Analog Outputs, the difference between the sockets is fixed to 3+7 = 10 db or 3+22 = 25 dB!

I am not 100% sure whether the two phones output of ADI-2 DAC FS can operate at the same time.
But even if, then you should also be aware of, that the 2nd output is only for IEM headphones.

If you want to have the flexibility to use two normal headphones or if you would like to be able to compare two different headphones, then you need two separate headphone outputs like the ADI-2 Pro FS R BE and ADI-2/4 Pro SE offer.
Then you need two outputs with independent volume control that can stay plugged in, which allows you to make quick headphone changes by only changing them on your head. For such quick A/B comparisons, there is not time for plugging the cables additionally.

Those two models with two D/A converters for supporting two normal headphones also give you the ability to use balanced phones if you wish, where only the ADI-2/4 Pro has the new Pentaconn plug for balanced headphones.

Thanks rawac for such informative details. I think I will take some time to read the manual, but not sure I can understand technique details. But I know 10 dB is already ten times difference and 20 is 20x. I am glad to know that to get the shared listening, I just need to get a 3.5mm one-to-two headphone jack, that’s great! I guess it will double the load, do you think the ADI-2 DAC can still afford to provide it?

And, do you know when I have a headphone (Currently I have a Sennheiser IE 500 Pro and HD 660S), what’s the key factor to determine which jack into which I should plug my headphone? I think it should not be the size of the plug because people can always convert it using adapters. I think it should base on the ohms of the headphones but I don't know any more exact rule.

It’s a pity that ADI-2 DAC does not provide balanced output, because I think the next headphone I will buy is the IE 900. Do you think the balanced output is something that can make a big difference?

Sorry for so many questions..

Plug the HD-660S (104 dB SPL /1V) into the 1/4” output and the IE 500 (126 dB SPL /1V) into the 1/8” IEM output.
This will give you about the same loudness on both ‘phones.

ADI-2 DAC’s 1/4” is 25 dB louder than the 1/8”.
IE 500 is 22 dB louder than 660S.
That’s just a nominal difference of 3 dB, not considered differences in sound.

BTW: ADI-2 doesn’t care much about load impedance by the ‘phones.
You can plug 2 or more ‘phones (e.g. 10 pairs of HD-660S) into one output, using y-adapters, no problem.

KaiS, Many thanks for the calculation!

You searched in the wrong place. +22 dBu = 0.775 V + 22 dB ≝ 0.775×10^(22/20) (V) = 0.775×10^1.1 (V) ≈ 0.775×12.589 (V) ≈ 9.7565 V

To be exact, dBu relates to the Voltage that creates 1 mW into 600 Ohm.
That’s about 0.774596692 Volts, or round 0.775 V.

Decibel for linear physical quantities (like e.g. Voltage “U”), is calculated:
x dB = 20 * log10 (U1 / U2), where U2 can be a reference value.

Decibel for squared physical quantities (like e.g. Power “P”), is calculated:
x dB = 10 * log10 (P1 / P2), where P2 can be a reference value.
(Power is a squared physical quantity, as it’s calculation incorporates a squared physical base quantity: P = U² / R)

The reason for the dB calculation difference:
For a given level difference you can use the either Voltage or Power based dB calculation, and get the same result.

E.g. if you double the Voltage you get 4 times the Power, both calculate to +6.0206 dB.

Thanks for the detailed explaination! Yest, I got it.

My understanding is like the following.   The standard definition of decible (dB) is dB = 10 * log(Value/reference) which gives a logrithm ratio of some value related to a reference value.  No one would just change the constant scaler 10 to 20 for no reason.  But when the value you are concerning is power and you indeed don't really want to use power, then you can use voltage and the assumption is that the impedence is a constant.  So you plugin the formula and get:  dB = 10 * log( (U^2/R) / (U_0^2/R) ) = 10 * log( (U/U_0)^2 ) = 2 * 10 * log( U/U_0 ) = 20 log( (U/U_0). In the previous case, U_0 is 0.775 V. In this sense, I think the dBu is not a dB for voltage, it's actually a dB for voltage squared (which is proportional to power), that's the reason why we need divide it by 20 instead of 10 to the the logirithm ratio.

And German speakers can read the original webpage: https://sengpielaudio.com/Rechner-db-volt.htm 

In German, this is called “Ein Streit um Kaisers Bart” (a dispute over the emperor's beard).  In literature, you will also find that the u stands for “unloaded.” Some also say that it stands for “unit” as in VU. This would be a topic for the Mythbusters.

https://en.wikipedia.org/wiki/Decibel

The already mentioned Eberhard Sengpiel (born 1940, died 2014, German sound engineer, musician, lecturer at Berlin University, https://en.wikipedia.org/wiki/Eberhard_Sengpiel or https://de.wikipedia.org/wiki/Eberhard_Sengpiel) mentions both: https://sengpielaudio.com/ElektrSpanndB … worten.pdf

The basic definition is much more often referred to as power. I don't think it's that important. The main thing is to understand the difference and use both formulas correctly. Before the power amplifier, it's more about voltage, and after, it's more about power. At the beginning, I was looking at the power in the case of "0 dB are 1 Bell = 1E1 = 10 = factor 10". Because the headphone output is after the power amplifier. I should have included the word "power" in my posting.

The audio community does things a little different. It's a set of unique units and jargon.
I'm sure someone created dbu's to make things easier to manipulate. But it's only a factor of .775.
I find myself constantly grabbing my calculator to convert voltage and power levels from dbu.

The block diagram is correct. The DAC output is +19 dBu (not labeled). Also a block diagram is a simplified version for a better understanding of the signal flow. It might not show the exact levels as they exist inside the unit.

Thanks KaiS. I was aware of the 1 mW into 600 ohms communications standard...but I never connected to Vu.

Interesting I.R.E. proceeding article also. I had to chuckle how they determined distortions based on average people's perceptions using equipment that generates lots of distortion based on today's standards. But It makes sense at the time there was a need to establish some kind of standard.

It's better to stop these speculative and misleading exercises. dBu is a unit of voltage, by its very definition.

Sometimes it helps to be old enough to have had to work with a slide rule. It's the same formula, just phrased differently. I try an other way to explain.

It's about the relationship between linear and all logarithmic scales.

First, the multiplication x · y
log(x · y) = log(x) + log(y)
A multiplication on the linear scale becomes an addition on the logarithmic scale.

Then the exponential function xEy (or x^y)
log( xEy ) = y · log(x)
An exponential function on the linear scale becomes a multiplication on the logarithmic scale.

So if
dB = 10 ln (P/P[0]) = 10 ln (V² · R/V[0]² · R)
we can cancel out R and combine the identical exponents
dB = 10 ln (V/V[0])²
Then bring the exponent to the front
dB = 10 · 2  ln (V/V[0]) = 20 ln (V/V[0])

If you take physics very seriously, the unit Volt is also eliminated from the fraction, as is the resistance in Ohms, and dBu is dimensionless.
9.756671941 V = 0.775 V · 10E(22 dBu / 20)
Not only must the values on the left and right sides of the equal sign be equal, but so must the units. If the factor 10E(22 dBu / 20) were Volts, the result on the left would be V² = V · V.

For non-German readers: Dezibel is the German writing for decibel and we use U instead of V in physic formulas.

@rawac I know you wanted to say that dB{x} is intended to tell ratios, which is dimentionless. That is true and is the key to the understanding.

To make it very clear of my point:

u1 = 5   dBu = 0.775V * 10^(5/20) = 1.3782 V
u2 = 10 dBu = 0.775V * 10^(10/20) = 2.4508 V

u2 is 5 dBu higher than u1.

Assumping a same impedency, u1 and u2 produce powers p1 and p2.  What's the decibel power difference of p1 and p2?

p2 is higher than p1 with 10 * log(u2^2 / u1^2) = 10 * log(2.4508^2 / 1.3783^2) = 5 dBm.

In this way, the logorithm difference of powers and voltages matches perfectly.  And, the real point is, per my assumption, otherwise I cannot think of another cause for this math manipulation of using 20 instead of 10, that when a people see from some specs that there is 5 dBu increasing in voltage he can immediately conclude that there is 5 dBm increasing in power!  -- That's all for power actually, and I fully agree with and leant from KaiS that the reason behind this is simply because voltage are much easier to be measured than powers.

Exactly, the 600 Ohm is a good insight!

Yes, it’s my mistake by saying dBu are ratios. You are absolutely correct in they are volts. I may be just wanting to emphasise that those kinds of quantity are measured based on ratios.

But for the other part. No. I learned from school decibels are always 10xlog things; I also checked internet that this is true. My original intention of posting the question was driven by that I didn’t believe that in foundation math units, one thing could have different
definitions.  20x is not the first hand definition and there is only one definition which is 10x. 20x is just the same definition applying on squared values like V^2; and the squares move down and timed to the 10 becoming 20 - that’s a result of the calculation based on the single definition and the purpose of using V^2 is to make sure the  dBu difference will equal to the corresponding dBm difference. Just because of the used quantities are indeed V^2’s, so what the resulting dBu compared are essentially powers - a single dBu has two rules, indicating a volts quantity and telling you the power difference at the same time.

I just found the following WiKi which states exactly what I understood including the reasoning for how the 20 came from:

https://en.wikipedia.org/wiki/Decibel

As an engineer I might see things more from the practical side:

10 Volt is 10 times 1 Volt, no matter of load impedance, which actually might even vary due to nonlinear resistance (think light bulb or semiconductor as load).

No defined load impedance = no defined power..

20 dBV is another expression for 10 V , so 0 dBV is 1 Volt, still no power in sight – so why bother?!

Same for dBu (0.775 V) - no power relationship included.

BTW: schoolbook often contain outdated or over-complicated information, even more so if school was multiple decades ago.

I remember (and still have) some of my old physics and electronics engineering formula and table books - most of which don’t contain much useful information from today’s point of view.
The best formula collections I have are the ones I created by myself.

Yes, I agree that practically we should focus on voltages here, seeing the V^2 behind is more a theoretical view.