Sennheiser HD 490 Pro Headphone Review

[IdleTalk]

I hope this post could provide some constructive insight.
The picture above is the impedance response graph of Beyerdynamic T1 1st gen.
You can see that it has about 1200Ω peaks at 80Hz and nominal impedance of 600Ω at 1kHz and above.

The table below is from my personal measurement of sine waves with a simple multimeter.

Frequency	Voltage	Current
20Hz	1.857V	2.30mA
40Hz	1.899V	1.70mA
80Hz	1.912V	1.48mA
160Hz	1.913V	1.82mA
320Hz	1.913V	2.30mA
640Hz	1.912V	2.70mA
1280Hz	1.904V	2.84mA

Power consumption at...
80Hz: 1.912V * 1.48mA = 2.83mW
1280Hz: 1.904V * 2.84mA = 5.41mW
At 80Hz, where the impedance peak is, the power to produce (almost) same SPL of sine wave is halved.

You can read impedance from the graph but you can also calculate it from the measured voltage and current using the Ohm's Law (R=V/A).
The calculated impedance at...
80Hz: 1.912V / 1.48mA = 1292Ω
1280Hz: 1.904V / 2.84mA = 670Ω
There is about 10% margin (which is not small) because of unprofessional equipments and possible sample discrepancies but you should be able to see the logic.

I encourage all of you to do similar experiments for yourself.
Proper reasoning with a simple equipment can be a lot more "scientific" than cargo-worshipping a particle accelerator at CERN.

 

[PuX]

They have a lot of "advantages" over HD400Pro, which is the proper comparison in my opinion - easy to swap, washable pads; 2 tuning options with included pads; included DAW plug-in; cable that can be connected to the right or left ear cup and has built-in coil to reduce microphonic effects. These are the ones I can think of. They might not matter to you and me, but then again these are not built for you and me. These are built to be used in professional or home studios.

I think it is important to understand these are not SUVs, these are trucks built for people who intend to use them as trucks.

the real question is why can't Sennheiser build a model with all of those advantages at a reasonable price, i.e. 120% of the price of HD 560 S.

 

[IdleTalk]

the real question is why can't Sennheiser build a model with all of those advantages at a reasonable price, i.e. 120% of the price of HD 560 S.

Well, 560s can be cheap because it shares many parts with other 5xx line headphones.
But for HD490pro only few parts shares same molds.
One of them are ear cup yokes which can be seen in Momentum 4 & Accentum. But even then Sennheiser did not reuse the headband of those outdoor headphones because they are not apt for long-time studio use. I find it a crazy luxury design choice.

 

[IAtaman]

the real question is why can't Sennheiser build a model with all of those advantages at a reasonable price, i.e. 120% of the price of HD 560 S.

As I tried to explain, I don't think the target audience for this product is the casual audio hobbyist who does not want to spend more than $200 on headphones. Given Sennys experience in marketing products to audio professionals, I think their price is very likely to be reasonable. And again, it is value offered by the product dictates its price, not the cost of manufacturing it. So Senny can most likely manufacture a good headphone they can sell for 120% of HD400 Pro. The reason they don't is because they think they can get more for it.

 

[amirm]

"Impedance is middle of the road but rises in bass to over 200 ohm (where you need most power)"

No, please. An impedance peak is where least power is needed.

With such a high impedance, you are voltage limited. At that impedance, you get far less power out of your source if it doesn't have high voltage drive.

The easiest to drive headphones are the ones that are low impedance and have high sensitivity. Then even a dongle with low output voltage can drive them. Boost the impedance up high and the same source will struggle to deliver power.

See this example out of many:

You have just 10 milliwatts. Same dongle now into much lower impedance:

Now you have six times the power. The HD490 falls in the former category and hence requires more power than that dongle can provide.

 

[IdleTalk]

With such a high impedance, you are voltage limited. At that impedance, you get far less power out of your source if it doesn't have high voltage drive.

The easiest to drive headphones are the ones that are low impedance and have high sensitivity. Then even a dongle with low output voltage can drive them. Boost the impedance up high and the same source will struggle to deliver power.

See this example out of many:
[pic]

You have just 10 milliwatts. Same dongle now into much lower impedance:
[pic]

Now you have six times the power. The HD490 falls in the former category and hence requires more power than that dongle can provide.

Sir, you are confusing [different headphones with different nominal impedance] and [different impedance at different frequency of a same headphone].

This error, which became very obvious thanks to your reply, has sustained at least 3 years since I pointed out it in your T1V2 review.

When you feed sweep or white noise signal into a headphone, you are giving the same voltage though different frequency ranges into a same headphone. If you get same SPL at the impedance peak, you get same SPL with lower power (same voltage * lower current).
Please read my post #101 and take some time to think about it.

 

[staticV3]

even a dongle with low output voltage can drive them. Boost the impedance up high and the same source will struggle to deliver power.

Yes, but the 490 Pro doesn't need much power at bass frequencies.

Its impedance peak in that region makes it more efficient in that frequency band, alleviating the Amp's current requirements.

You're thinking about this from the wrong point of view.

"An impedance resonance means the Amp cannot supply as much power at that resonance point, making the headphone harder to drive"

That would perhaps be true if the headphone was primarily driven by current, but in reality it's a voltage driven device.

Current just follows suit.

After all, your frequency response sweeps were captured at constant voltage, not constant current or power.

So in reality, while playing the sweep (or music for that matter), current will dip near the impedance resonance point with no detriment to the headphone's response or slam or any of that.

The impedance resonance is simply the point of greatest headphone efficiency, not of greatest Amplifier strain.

 

[ZolaIII]

Sir, you are confusing [different headphones with different nominal impedance] and [different impedance at different frequency of a same headphone].

This error, which became very obvious thanks to your reply, has sustained at least 3 years since I pointed out it in your T1V2 review.

When you feed sweep or white noise signal into a headphone, you are giving the same voltage though different frequency ranges into a same headphone. If you get same SPL at the impedance peak, you get same SPL with lower power (same voltage * lower current).
Please read my post #101 and take some time to think about it.

Let's just say both of you are right. First it does need to give required voltage and be able to sustain it as it won't go down on fast real time loads and secondly the resistance rise of 2x and current (A) with it won't be huge compared with much lower impedance and much higher rise in speakers with much higher Voltage of course. When conditions are harsh where V reaches the limit and stays at it while Amperage rises and so do temperature, leaking and diffusion it's a problem but we are talking about power amplifiers soft and hard cliping (voltage collapse).
As in your little table V is very close as it is to 2 V already logical is to drive it with something that can provide 2x more voltage (being balanced or unbalanced stage that gives more V). This way you will be in safe area regarding cliping, best performance delta for such one that provides such V as 2x V is +6 dB. Low V stages (0.5/1 V) simply won't be able to drive it on such SPL and different stages do perform different on different impedance loads, usually anaemic one's are horrible on such loads. So you pick something like Apple EU USB dongle to drive 15~32 dB 110+ dB SPL/V headphones/earphones, 100+ will pass with 2V and below that you go more. It's much more simple with dB SPL/V then per mW so I hope we will have it like that once. Would be good to take a lower base SPL from 94 to 90 dB and required +9 dB peak headroom to drive them as even 94 dB is too extreme (even 90 is as as reference in room mono white noise [-20 dB peek] calibration point is 83~85 dB).

 

[amirm]

When you feed sweep or white noise signal into a headphone, you are giving the same voltage though different frequency ranges into a same headphone.

So? I am using a very powerful/high-voltage headphone amp to drive it. A voltage limited source may not be to maintain that constant SPL.

 

[amirm]

Yes, but the 490 Pro doesn't need much power at bass frequencies.

Its impedance peak in that region makes it more efficient in that frequency band, alleviating the Amp's current requirements.

Why are you talking about current? Weak sources like dongles are limited by output voltage. Many for example max out at 1 volt. Power = V^2/R. Increase R and you will have less power.

As to sensitivity, it is in the frequency response:

At 20 Hz, we are 4 dB lower than at 425 Hz which means that you need fair bit more power to drive the headphone. At 70 to 80 Hz where impedance peak exists, the sensitivity is the same at 425 Hz where 94 dBSPL reference exists. So it is not more sensitive in bass.

 

[staticV3]

Why are you talking about current? Weak sources like dongles are limited by output voltage. Many for example max out at 1 volt. Power = V^2/R. Increase R and you will have less power.

Sure. Doesn't matter.

Power is not directly what drives a headphone to produce SPL.

Instead, we feed a headphone a given voltage, and it in turn draws current according to its impedance response.

Let's say we take a weak 1V dongle to drive the 490 Pro.

If we were to create a copy of the 490 Pro with identical characteristics, except that it had a flat impedance response without that resonance, then it would make that copy a less efficient, harder to drive, more current hungry headphone.
Same frequency response, same sensitivity, more current drawn at the same SPL.

The 490 Pro's impedance peak makes it an easier to drive load for the weak dongle. Not the other way around.

As to sensitivity, it is in the frequency response:

At 20 Hz, we are 4 dB lower than at 425 Hz which means that you need fair bit more power to drive the headphone.

No. The 4dB bass roll-off does not by itself mean that you need more power to drive the headphone.

If the impedance response of the headphone were flat, and you tried to drive the headphone to the same SPL at 20Hz, as you were at 425Hz, then to compensate for the 4dB roll-off, indeed more power would be required (~2.5x).

At 70 to 80 Hz where impedance peak exists, the sensitivity is the same at 425 Hz where 94 dBSPL reference exists. So it is not more sensitive in bass.

Efficiency≠Sensitivity.

At 70-80Hz, the 490 Pro is about as sensitive as it is at 425Hz (equal voltage -> equal SPL).

At 70-80Hz however, the 490 Pro is more efficient than it is at 425Hz (equal voltage at higher impedance -> lower current -> less power -> nonetheless equal SPL).

 

[IdleTalk]

So? I am using a very powerful/high-voltage headphone amp to drive it. A voltage limited source may not be to maintain that constant SPL.

There isn't any relevance to another source/headphone that may or may not fail to do something.
What we are talking about is a fixed setup where the same headphone (what you measure) is driven by the same source/amp (what you use).

The impedance varies in the same headphone & the same source/amp at different frequency.
That is what the impedance response means and you measured it yourself.

P.S. Your very powerful/high-voltage headphone amp, which could easily put enough power to very high nominal impedance of 600Ω (of T1 for example), does not have to put that amount of power to 1200Ω at the impedance peak, because the headphone is more efficient there.

 

[IdleTalk]

Let's just say both of you are right. First it does need to give required voltage and be able to sustain it as it won't go down on fast real time loads and secondly the resistance rise of 2x and current (A) with it won't be huge compared with much lower impedance and much higher rise in speakers with much higher Voltage of course. When conditions are harsh where V reaches the limit and stays at it while Amperage rises and so do temperature, leaking and diffusion it's a problem but we are talking about power amplifiers soft and hard cliping (voltage collapse).
As in your little table V is very close as it is to 2 V already logical is to drive it with something that can provide 2x more voltage (being balanced or unbalanced stage that gives more V). This way you will be in safe area regarding cliping, best performance delta for such one that provides such V as 2x V is +6 dB. Low V stages (0.5/1 V) simply won't be able to drive it on such SPL and different stages do perform different on different impedance loads, usually anaemic one's are horrible on such loads. So you pick something like Apple EU USB dongle to drive 15~32 dB 110+ dB SPL/V headphones/earphones, 100+ will pass with 2V and below that you go more. It's much more simple with dB SPL/V then per mW so I hope we will have it like that once. Would be good to take a lower base SPL from 94 to 90 dB and required +9 dB peak headroom to drive them as even 94 dB is too extreme (even 90 is as as reference in room mono white noise [-20 dB peek] calibration point is 83~85 dB).

Please remind that voltage is constant.
When the signal is flat across the frequency range (like the test tone), same voltage is fed across whole frequency range.

You seem to be thinking that the voltage should rise at impedance peak, but that does not happen with a voltage driven amp (output impedance = 0) Such happens only with current driven amp (output impedance = ∞).

With a current driven amp with very high output impedance, voltage rises at impedance peak and causes SPL rise.
That's the logic behind the bass boost with high output impedance amps.

However what we are talking about here is the case of voltage driven amp, where constant voltage is fed and flat SPL response is expected.

 

[staticV3]

@amirm
Just think of your loudpeaker reviews.

When measuring a speaker's impedance response, it's the dips in impedance which tell us how hard a speaker is to drive, since those are areas at which the speaker will draw especially high current from the Amp:


We don't care if the impedance peak right next to it goes up to 8, 10, or 50Ω, because peaks in the impedance response do not bother an Amp (on the contrary: they relieve the Amp).

And no, those peaks are not points in the spectrum where the speaker needs especially high voltage to reach the measured response.
After all, the response was measured at a constant 2.83V:


So: Same voltage + higher impedance inevitably means less current->less power->higher speaker efficiency & less strain on the Amp.

Exactly the same concept applies to headphones.

 

[lewdish]

Its closer to my preference curve, think most major audiophile manufactures prefer to recess the 1-2khz region a little more these days compared to where harman starts their gain

 

[johny_2000]

Its closer to my preference curve, think most major audiophile manufactures prefer to recess the 1-2khz region a little more these days compared to where harman starts their gain

In my opinion, this is better. This means that I don’t have to dampen this range with EQs, as in previous years of headphone models.

Perhaps the trends in frequency response were caused by changes in the art of music, which with each passing decade moves further and further from vocals and wind instruments to electronic instruments and effects. Music becomes simpler and more compressed. So there is less need for high fidelity and resolution in the midrange of music for the listener.

The same Harman curve simply systematizes the preferences of most listeners. During the time from the beginning of this research to the present day, about a dozen years have passed, music has changed, and preferences in listening to it have also changed. It's time to update the curve.

 

[amirm]

Efficiency≠Sensitivity.

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.

 

[amirm]

We don't care if the impedance peak right next to it goes up to 8, 10, or 50Ω, because peaks in the impedance response do not bother an Amp (on the contrary: they relieve the Amp).

You can't compare headphones to speakers. Headphones have a massive impedance range from 8 ohm to over 600 ohm. That causes current to be an issue at one end of the spectrum and voltage at the other. 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. By the same token, if you raise the bass impedance by yet another 40%, it makes the voltage demands even worse.

 

[amirm]

So: Same voltage + higher impedance inevitably means less current->less power->higher speaker efficiency & less strain on the Amp.

We are not talking about strain. 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.

 

[ZolaIII]

@IdleTalk 500 mV into 300 Ohms give 0.9 mW and into 16 Ohms it's 16 mW.
5800 mV into 300 Ohms give 113.5 mW.
Ix2=W/2=-3 dB. Every amplifier is V driven and try to compensate for Impedance rise by increasing V but doesn't have time to get it down on Impedance fall and then current (A) increase and if it's more than it can give it clips (driven at or close to it's max V) and as it continues to do so it over heat's and as it does it leaks more and A rises even more and now it can't keep V anymore and it hard clips (voltage collapse if safety doesn't engage and shot it down to prevent demage prior to it). This is much more problem with power amplifiers where both V and A will be much higher than on headphone one, principle is the same. Solution you don't use amp which doesn't have enough power (V determined) to drive the thing sufficiently in the first place (and with a headroom counted in).