Sennheiser HD 490 Pro Headphone Review

[IAtaman]

Every amplifier is V driven and try to compensate for Impedance rise by increasing V

They do not increase V to compensate impedance rise, they decrease current to compensate impedance rise to keep the Voltage constant. They are, after all, constant voltage sources ideally. What you are describing is constant current amplifier, and most audio amps are not current amps.

 

[ZolaIII]

They do not increase V to compensate impedance rise, they decrease current to compensate impedance rise to keep the Voltage constant. They are, after all, constant voltage sources ideally. What you are describing is constant current amplifier, and most audio amps are not current amps.

Either can be limiting factor and as always it's V/A = W but hire V is limiting factor! 500 mV vs 5800 mV regarding how much power it gives to 300 Ohms load and it's only a little bit less then 25 dB SPL difference. So 300 Ohms rated with 116 dB/V sensitivity headphones will be driven the same with 500 mV one as 92 db/V one with 5.8 V one. 32~16 Ohms rated one will get 4~8 x more mW from same V (of course max V on stages that can provide more of it will be limited on such to protect it regarding max A/V) and that's +6~9 dB SPL. You won't drive headphones so loud but again headroom of min +6 to 9 dB for peeks need to be counted in. Main reason to use optimal amplifier stage to sensitivity is so that it can retain sufficient performance on lower SPL output levels (65 dB and under) and of course sufficient power (performance is not an Isue today) on high SPL levels (90+ but try to restrain yourself for the sake of your hearing health).

 

[solderdude]

I measured the HD490Pro and got these results (producer pads)

Power rating: 0.3W
Max. Voltage: 6V (17Vpp)
Max. current: 50mA
Max. S.P.L. 125dB
Impedance: 120Ω (measured)
Efficiency: 100dB @ 1mW
Sensitivity: 109dB @ 1V

Not surprisingly about the same as HD560S (same driver)
Impedance: 121 Ω (measured)
Efficiency: 99 dB @ 1mW
Sensitivity: 108dB @ 1V (the data sheet mentions 110dB/V)

This means on a E.U. phone one can reach 103dB peaks (peaks in the entire spectrum combined) so about 93dB average.
Note that sensitivity is 2dB higher at low frequencies.

 

[ZolaIII]

I measured the HD490Pro and got these results (producer pads)

Power rating: 0.3W
Max. Voltage: 6V (17Vpp)
Max. current: 50mA
Max. S.P.L. 125dB
Impedance: 120Ω (measured)
Efficiency: 100dB @ 1mW
Sensitivity: 109dB @ 1V

Not surprisingly about the same as HD560S (same driver)
Impedance: 121 Ω (measured)
Efficiency: 99 dB @ 1mW
Sensitivity: 108dB @ 1V (the data sheet mentions 110dB/V)

This means on a E.U. phone one can reach 103dB peaks (peaks in the entire spectrum combined) so about 93dB average.
Note that sensitivity is 2dB higher at low frequencies.

And in many ways that's optimal as crosstalk will be high, it will be less self impedance of amplifier dependant (then low impedance one's), 500 mV (EU smartphone/dongle restricted) stage can drive it more than sufficient and 1, 2 V one to a +6, 12 dB more SPL and you really won't nead ever more than that, and SNR will be better (again than on low impedance one's). Now please take in account how really bad most of interface headphone outputs are (on all metrics; SINAD, power, SNR) and how this will primarily be used with such and you see whole logic behind it (never the less same logic applies and on all other source). Meaning mid sensitivity (about 110 dB/V) and mid impedance (100~120 Ohms) is best way to go anyway regarding different possible sources.

 

[staticV3]

True and hence the reason any such technical talk needs to be about sensitivity, not efficiency. I don't care how much power comes out the back of the headphone even though that gets included in efficiency. We care about how loud the headphone sound is as it is perceived by our ear. Not total energy.

And how loud the headphone gets is dictated by its sensitivity, not by how much power it allows the Amp to output.

Voltage is routinely a barrier to SPL as evidenced by anemic power you get out of Sennheiser HD6XX class of headphones. Replace them with a 32 ohm headphone and the job becomes simpler.

Only because those 32Ω headphones usually also have higher sensitivity.

Switch the 600Ω headphone to a 32Ω headphone with identical sensitivity and it will become harder to drive (more current, same SPL)

By the same token, if you raise the bass impedance by yet another 40%, it makes the voltage demands even worse.

Again:
No. Just raising the impedance doesn't increase the voltage demands. Only lowering the sensitivity does.

We are talking about inability to create power. High impedance headphones are enemy of low voltage source amplifiers. I gave you example of this. A dongle with 1 volt output would not be strained but simply not produce much loudness at high impedance loads.

And I explained that power is not the primary force that produces SPL.

After all, some headphones need just 53μW to produce 94dB SPL (AKG K371) while others need 35mW (DCA Aeon RT Closed).
Same SPL, over 600x as much power consumed.

Therefore, you cannot think of high impedance alone as an attribute impeding power, and therefore loudness.

Because again: Power is not our primary source of loudness. Voltage is.

Higher impedance at equal sensitivity means the headphone needs less power to produce the same SPL. Not more voltage to compensate for the drop in power.

 

[amirm]

Switch the 600Ω headphone to a 32Ω headphone with identical sensitivity and it will become harder to drive (more current, same SPL)

Common fallacy in these arguments. There is no such thing as "switch from 600 to 32 ohm." A lot of things will change in that transformation.

 

[MayaTlab]

This is another common confusion I see in headphone/speaker forums, if you know the harmonic distortion levels of headphone or speaker, you automatically know the IMD levels too. Both harmonic and inharmonic distortion comes from same reason, the driver as a piston is not in precise position its supposed to be.

As the driver goes further, the restoring force pulling it back is higher so the error in position is bigger. Now if you have headphone with 1% THD at 100hz and lets say 0.01% THD at 7200hz, becose the driver excursion is much higher at the 100hz, the poor 7200hz that would otherwise be clean will be riding on top of the 100hz driver motion and pushed into the highly non-linear region, thus making it highly distorted too. And since this is time variable, depending on at what phase of the 100hz cycle the driver is at, it will create time variable distortion of the 7400hz sine, thus creating inter modulation tones

Its like waterfall plot and freqency/phase response graphs. If you know the frequency and phase response of a headphone, you can predict exactly how will fast will it decay across the spectrum in waterfall plot.

I get that there's a relationship there, but so that I understand you well, are you suggesting that IMD can be directly predicted by harmonic distortion measurements ? My understanding is that you rather need simulations.

And again, are we expecting it to be significant factor for the HD 490 Pro ? I can understand that this could raise concerns, but the HD 490 Pro measures really nicely compared to other open over-ears in terms of harmonic distortion to the point where, in that regard, I'm not even certain that we're facing an audible phenomenon at common listening levels, even when EQed to a Harman-ish target.

 

[amirm]

And how loud the headphone gets is dictated by its sensitivity, not by how much power it allows the Amp to output.

You can't be serious. By that notion I can have an amplifier with 0.0001 milliwatts of power and have any SPL I want to have?

Given the same drive level, higher sensitivity will produce higher SPL. Lower the power and SPL will drop.

 

[staticV3]

Common fallacy in these arguments. There is no such thing as "switch from 600 to 32 ohm." A lot of things will change in that transformation.

But they don't change on the 490 Pro.

Once frequencies approach its impedance peak, nothing changes but the impedance.

Same sensitivity, just higher impedance ->the headphone gets momentarily easier to drive.

You can't be serious. By that notion I can have an amplifier with 0.0001 milliwatts of power and have any SPL I want to have?

Of course there are limitations, but different headphones/IEMs/loudspeakers having different efficiency is a fact you cannot deny.

As explained above: you can switch from an Aeon RT Closed to an AKG K371, reduce your Amp's output power by 99.85%, and still get the same SPL.

Power doesn't make loudness.

 

[ZolaIII]

@staticV3 nope higher sensitivity countered with higher impedance so the SPL remains about the same. If you make only pasive headphones in order to resolve possible issues with different sources (to possible extent) going with mid sensitivity and impedance just like Sennheiser did and it's about average for all measured so far regarding mV (as V again translate transparent across the impedance loads). There is no need for higher impedance as you would gain very little to nothing (regarding SNR and crosstalk) and you lose on both with low impedance even if you gain regarding mW power in output and SPL to it. High sensitivity low impedance are hardest to drive properly and not regarding A but in the sense of preserving (multitone) SINAD above SPL level on lower SPL levels (50~65 dB) again to SPL sensitivity at 1 V (as 0 dB). And this can easily fall in audible range (both regarding SNR and SINAD).

 

[staticV3]

@staticV3 nope higher sensitivity countered with higher impedance so the SPL remains about the same.

Higher sensitivity combined with higher impedance would boost SPL per Vrms and massively boost SPL per mW.

 

[solderdude]

Funny... Amir and staticV3 are arguing the exact same point to each other but from a different perspective.

Basically amplifiers are voltage and current limited (staticV3 sometimes posts nice plots about this aspect).
Drivers can have different impedances, different impedance curves (over a frequency range) and different sensitivity.
Effectivity is determined by the sensitivity and impedance (which can differ per frequency band).

Most amplifiers have increased distortion in lower impedances. So... with the same sensitivity (but different effectivity) the higher impedance headphones are usually in the advantage. Also these will draw less power which, when playing really loud, could improve service life of a battery fed device on 1 charge.

When the output is voltage limited only a higher sensitivity headphone can play louder. Often these are also low impedance so there's that.

 

[ZolaIII]

Higher sensitivity combined with higher impedance would boost SPL per Vrms and massively boost SPL per mW.

Sensitivity as it is, impedance x2 = W/2 = - 3 dB. Now read solderdude note and take a look at impedance plots. You you need to hit it harder.

 

[staticV3]

Sensitivity as it is, impedance x2 = W/2 = - 3 dB.

If sensitivity and voltage stay the same and you increase the impedance by 2x, then output power drops by 50% but loudness stays the same.

Same sensitivity + same voltage = same SPL.

W/2 = -3 dB is only true if sensitivity and impedance stay the same.

 

[IdleTalk]

Every amplifier is V driven and try to compensate for Impedance rise by increasing V but (...)

A voltage driven amp does not compensate for impedance rise by increasing voltage.
It feeds same voltage into all frequency range and lets current drop when impedance rises.
Please remind voltage is constant unless you touch the volume knob.
Thats what voltage drive means.

Despite current drop (=power drop, because V is constant) at the impedance peak, SPL remains almost same because the efficiency (dB/mw) is higher there. You can experiment this for yourself with your multimeter, voltage driven amp (output Z = 0), and sine wave generator on internet.

What you describe is current drive, not voltage drive.
A current driven amp feeds same current into all frequency range.
I order to do that, a current driven amp tries to compensate for impedance rise by increasing voltage.
As a result a current drive amp boost bass, because impedance peak of dynamic driver headphones are typically in bass region.

 

[solderdude]

Perhaps add that a voltage drive amp turns a bit more towards acting like a current drive amp when the output resistance is increased.
Some (AV)Receivers and integrated amps with headphone out derived from the speaker output can have high output impedances ranging from 50 to 300ohm !

For HD490PRO the tonality will slightly change with output resistances > 50Ω. Most audio interfaces, they will likely be connected to, will be lower than that.

0.2Ω, 10Ω, 32Ω, 120Ω

The higher impedance will help in this case too.

These HD490PRO are not hifi headphones but intended for pro usage. These headphones, however, are more than good enough to also be used as a hifi headphone which not all pro-headphones are really suited for.

 

[ZolaIII]

If sensitivity and voltage stay the same and you increase the impedance by 2x, then output power drops by 50% but loudness stays the same.

Same sensitivity + same voltage = same SPL.

W/2 = -3 dB is only true if sensitivity and impedance stay the same.

Now read first sentence in mine and hit yourself again.

 

[staticV3]

Now read first sentence in mine and hit yourself again.

Which one?

@staticV3 nope higher sensitivity countered with higher impedance so the SPL remains about the same.

On a regular voltage drive Amp, increasing the headphone sensitivity and impedance will boost SPL.
It will not remain the same.

Sensitivity as it is, impedance x2 = W/2 = - 3 dB.

On a regular voltage drive Amp, leaving the headphone sensitivity as-is but doubling the impedance will result in W/2.
But it won't reduce volume by 3dB.
In fact, it will increase volume by improving damping factor.

 

[ZolaIII]

@IdleTalk it's not either or, it's combined and not fixed but influenced also by temperature. Much more complex than all of us are trying to represent it and what's even more funny is that we are arguing about essentially same thing interpreting it only partially (initial comment how both of you are right). I give up on this!

 

[ZolaIII]

Which one?


On a regular voltage drive Amp, increasing the headphone sensitivity and impedance will boost SPL.
It will not remain the same.


On a regular voltage drive Amp, leaving the headphone sensitivity as-is but doubling the impedance will result in W/2.
But it won't reduce volume by 3dB.
In fact, it will increase volume by improving damping factor.

No 130# (I Shell Be Free - Bob Dylan).
Have a nice time.