[milkpowder]

Quote:

Originally Posted by PeterR Ok. Sensitivity is output per 1 V. As we know the impedance, we can calculate that 1 V gives us 31.25 mW. Now, if 1 mW gives us 100 dB(SPL), 31.25 mW will give us 31.25 times that amount obviously. Our hearing doesn't work very linear and covers an enormous range, so we use a logarithmic scale to specify levels. One Bel (the "B" in dB) is the base 10 logarithm of the ratio of two figures. If P2 is 10xP1, the difference is 1 B. More commonly used is the "deci"-Bel, which means 1/10 of a Bel. 1 B = 10 dB. The nice thing with a logarithmic scale is that you can multiply by adding. A power ratio of 31.25:1 is log(31.25)=1.5 B = 15 dB. So 1V into 32 Ohms means 15 dB more power than 1mW. If 1mW gives us 100 dB(SPL), 15 dB more input power gives us also 15 dB more output, so 1V in means 115 dB(SPL) out.

Excellent. Can you check to see whether I've got this right?

I'm trying to convert the 105db/V sensitivity of the 62ohm K701 into db/mW efficiency.

1. P=1/62ohm=16.13mW
2. Log(16.13)=12.07dB
3. As 1V at 62ohm delivers 12.07db less power than at 1mW
4. so efficiency is 105-12.07dB/mW = 92.93dB/mW

If 1V to 1mW, then you subtract the dB to dB/V to get dB/mW.
If 1mW to 1V, then you add the dB to dB/mW to get dB/V. (Step 1 and 2 are the same. Reverse step 3 and 4)

 

[PeterR]

Quote:

Originally Posted by milkpowder Excellent. Can you check to see whether I've got this right? I'm trying to convert the 105db/V sensitivity of the 62ohm K701 into db/mW efficiency. 1. P=1/62ohm=16.13mW 2. Log(16.13)=12.07dB 3. As 1V at 62ohm delivers 12.07db less power than at 1mW 4. so efficiency is 105-12.07dB/mW = 92.93dB/mW If 1V to 1mW, then you subtract the dB to dB/V to get dB/mW. If 1mW to 1V, then you add the dB to dB/mW to get dB/V.

Exactly.

 

[ThirtySixBelow]

I did a little calculation on two headphones. The HD595 seems to be regarded as a headphone that doesn't need an amp, and the HD600 does.

HD595
50 Ohm
104dB sensitivity
calculated ~90dB/mW

HD600
300 Ohm
97dB sensitivity
calculated ~91dB/mW

That looks pretty darn close to me in efficiency assuming my calculations match. It seems like the HD600 will need more voltage for it's impedence rating but then why do the sensitivities and efficiencies come out so close? Why would the HD600 need an amp to drive it but the HD595 doesn't? Either I miscalculated, or my logic is totally off.

 

[lini]

Generally, I'd rather recommend to forget those manufacturer specs - especially if you compare over longer time periods, because then different measurement methods & systems will make it even worse. If you still want to compare: Don't rely on the name of the spec, rather look at the units - especially regarding the efficiency/sensitivity confusion, to which PeterR has already hinted. If it's dB/(m)W, then it's efficiency - i.e. sound pressure per power; if it's dB/V, then it's sensitivity in its strict sense, i.e. sound pressure per voltage. Especially in the last few years, manufacturers seem to make more and more a mess of these specs - hence you might want to look further, if you find nonsense like this spec currently on the German Sennheiser website for the HD 650: "Schalldruckpegel bei 1kHz ... 103 dB (1 Vrms), at 1 mW". Now, what's that? So you open the product flyer pdf link to find a "Schalldruckpegel ... 115 dB bei 1 kHz/1 Veff", which apparently is a proper sensitivity spec, from which you can conclude that the other spec is meant to be efficiency and the "(1 Vrms)" in it either hasn't been edited away or is supposed to mean that they have measured the efficiency with a voltage of 1 V and the resulting current instead of applying 1 mW...

TSB: To confuse you even more, checking the specs on the German Sennheiser site, they state an efficiency of 97 dB/mW (tech specs on web page) and a sensitiviy of 114 dB/V for the HD 600 (product flyer pdf) - for the HD 595 they only publish a sensitivity of 112 dB/V (both on the web page and in the pdf)...

Greetings from Hannover!

Manfred / lini

 

[ThirtySixBelow]

Yes, I am now thoroughly confused, but I guess I'll just ask if I get a headphone if it will be good with an amp or not

One thing I am curious about though is what effect the Ohmic rating has on the sound of the speaker, if any. I notice things like subwoofers are about 2-8 ohms and then there are headphones at 300 ohms which leads me to believe it either has something to do with the power necissary to drive something like a 12" subwoofer vs a tiny headphone driver, or it has something to do with the frequency of the sound needed. If two sets of speakers with different impedences have corresponding power to make them equal in volume, will there be any sound differences?

 

[coloboxp]

Not exactly. Let me try to explain to the best of my knowledge how this would work, with examples.

Sensitivity is the volume of sound produced by a headphone per unit of electrical energy, while impedance is the amount of current a device will draw from the amplifier in the first place. This is why higher impedance is easier to drive-- it's taking less energy from the amp.

e.g., running off its 9V battery, a Xenos OHA-REP will deliver 185mW to a 32 ohm headphone, but only 35mW to a 300 ohm one. (according to the Xenos website, I chose the Xenos because they give very full specifications)

So let's figure the maximum sound pressure level (SPL) for three headphones from an OHA-REP, two with an impedance of 32 but different sensitivities, and one with an impedance of 300 but about the same sensitivity as one of the 32's. We'll use j-curve's handy excel sheets.

Grado SR line (32;98db/mW)-- 121dB

Senn HD580/600 (300;97)-- 112dB

AKG K26p (32;110)-- 133dB

now for fun, let's throw in a K701 (120;93). The Xenos does 75mW into 120.-- 112dB

In general terms, 10dB difference equals about ten times the real pressure and twice the perceived volume. Our three different figures here (112, 121, 133) equal a maximum of really rather loud, extremely loud, and unbearably loud. Of course, you'd likely be losing sound quality to play at full volume, so you could say that the Xenos with a decently strong source in (a whole other ball of wax, that, here let's just say something stronger than an iPod) will be just enough for your K701 or HD600, plenty for your Grado, and overkill for your K26p.

Also notice that the K701's lower impedance in relation to the Senn makes up for its lower sesitivity to produce the same maximum SPL. Interesting, no? The high-impedance phones do have some advantages, though: little or no sound floor hiss, and you can plug in more than one in parallel.

by any chance anyone got a copy of that excel sheets?, the link is broken

 

[pecotunkA]

I am not an expert on headphones, but since there seems to be some complex explanations here I thought I would give my two cents.

I understood higher impedance (typically what is being referred to as "power hungry"), as simply designed from beginning to end for the stronger strength of the signal.
in other words the only difference is to think of them as "designed for a higher output current".

therefore the first thing I think of is a lower noise floor, since the specified electrical signal for optimal performance needs to be stronger. lower noise floor can at times be a characteristic of "high-end".
next I would think arguably "more detail" if perceived or not is another question. which would be another high-end characteristic.
how I come to the second conclusion is the same concept as a lower noise floor. if the electrical signal is stronger, think of a waveform of a signal (music) pumping through low impedance pair of headphones, except for high impedance think of the waveform as "stretched" vertically because of the stronger electrical signal (current). so in the stretched high impedance waveform, there will be more data in between where it was previously compressed for low impedance.
- you could think of it/compare it to image upscaling. except with image upscaling the resulting data point is "a calculated estimated interpolation" while what shows itself in the waveform with high impedance headphones would be actual genuine data.

therefore, the sensitivity spec is the more accurate way to judge perceive loudness. hope this is easy to understand and is somewhat accurate.
edit - I changed the word "voltage" to "current", because I am pretty sure impedance affects current more. for example at the end of a long USB cable, there are more current losses than % voltage losses.