What determines how sensitive a headphone is?
Jan 4, 2010 at 2:42 PM Post #31 of 39
Quote:

Originally Posted by plonter /img/forum/go_quote.gif

......just tell me if i am correct in the next assumption:
given two headphones with the same SPL level, the more the impedance, the harder it will be to drive?




"If your cans are low impedance but are very sensitive (the case of Grados and most IEMs) then despite their asking more current from the source, they still manage to sound very loud because they need very little power to do so."

So using my collection as an example, the Grados and the Senns are the same low impedance (32ohms) but their sensitivity means they do not need a huge amount of power to go loud. The Goldrings have a higher impedance, but they are more sensitive than the Grados, and they need more power to go loud. So two headphones with the same SPL the more impedance, the more power is needed -

Grado SR80 32ohms SPL 98
Goldring NS1000 300ohms SPL 101 (with NR on)
Senn PX200 32ohms SPL 115.


Hence, if put all three into an ipod the Senns are fine, the grados are a bit strangled and the Goldrings sound terrible when compared to putting all three into an MF X-CANV8P. That is because the ipods output is milli watts compared to the X-CANS 1.3 watts.

I hope I have got this right and picked out the salient points!
 
Jan 4, 2010 at 3:03 PM Post #32 of 39
Quote:

Originally Posted by plonter
......just tell me if i am correct in the next assumption:
given two headphones with the same SPL level, the more the impedance, the harder it will be to drive?



Well that depends on the amp. If you've got something that's able to supply copious amounts of current but hasn't the greatest voltage swing then it'll drive the lower impedance cans much better. However if it's an OTL Tube amp then it'll drive the higher impedance cans more easily because OTLs aren't good amping low impedance headphones, but they do nicely with high impedance loads that need a decent voltage swing.

This happens in the real world. My Mini3 does a much better job of driving my ER4s than my Darkvoice THA332. However when it comes to my HD600s the Darkvoice does beautifully and the Mini3 struggles a bit.

Basically if you have two headphones with the same efficiency but different impedences they're just looking for a different balance of current and voltage to get the same amount of power out of an amp.
 
Jan 4, 2010 at 3:39 PM Post #33 of 39
Quote:

Originally Posted by Menisk /img/forum/go_quote.gif
Well that depends on the amp. If you've got something that's able to supply copious amounts of current but hasn't the greatest voltage swing then it'll drive the lower impedance cans much better. However if it's an OTL Tube amp then it'll drive the higher impedance cans more easily because OTLs aren't good amping low impedance headphones, but they do nicely with high impedance loads that need a decent voltage swing.



Menisk, how do you measure voltage swing and copious amounts of current? Is one watts and the other amps?

From earlier in this thread "Ohms is related to sensitivity in that watts is related to ohms (and sensitivity is given in watts). However in its truest sense sensitivity is related to efficiency". The use of apprently different terms for the same thing is confusing.

I am hoping this is an idiots guide for idiots like me whose brain does not work when it comes to science stuff!
 
Jan 4, 2010 at 3:56 PM Post #34 of 39
Quote:

Originally Posted by Prog Rock Man /img/forum/go_quote.gif
Menisk, how do you measure voltage swing and copious amounts of current? Is one watts and the other amps?

From earlier in this thread "Ohms is related to sensitivity in that watts is related to ohms (and sensitivity is given in watts). However in its truest sense sensitivity is related to efficiency". The use of apprently different terms for the same thing is confusing.

I am hoping this is an idiots guide for idiots like me whose brain does not work when it comes to science stuff!



Voltage Swing is referring to the amount of potential difference (Voltage) that an amp can put across a driver. We'll have a look at my Mini3 for example. My Mini3 runs off a 9V battery. This 9V is split into 4.5V available for each channel. The voltage swing for my Mini3 is therefore +/- 4.5V because that's the largest potential difference it can put across a driver. If you try to turn the volume up anymore it'll clip the top of the wave off.

An amplifier is also limited in the amount of current that it is able to supply. So the same thing happens, if you try to draw too much current it'll clip the top of the waveform because it can't pull anymore.

So if I have an amp that has a large voltage swing but isn't able to supply a great deal of current, like my Darkvoice, it's going to do reasonably well with a pair of 300ohm Senn HD600s, but it's going to struggle with a pair of 27ohm ER4-Ps because they'll require more current and it's lacking in that department

If I've got an amp designed with current output in mind like my Mini3 it'll do poorly with my Senns because they'll want more than 4.5V sometimes and so the waveform will get clipped however it'll be able to supply more than enough current to my 27ohm ER4-Ps and so it'll drive them much better than my Darkvoice.

When you get told an amp is a 1W amp they'll likely tell you that it's 1W at a certain impedance however if you change the impedance the amplifiers max output changes because you're changing the balance of current vs voltage to attain that power output. You'll find that power amp is capable of outputting nearly twice the power into 4ohm speakers than into 8ohm speakers because power amps are designed to be optimized for current output because loudspeakers have low impedance. My Darkvoice will be able to put much much more power into my Senns than into my ER4-Ps because when I use my ER4-Ps I shift the balance towards current which is where my OTL Darkvoice struggles.
 
Jan 4, 2010 at 8:10 PM Post #35 of 39
Thanks Menisk, examples makes it easier to understand, which is why I refer back to my kit.

Thanks as well to boomana who has provided more details with another thread relating to the sensitivity of different headphones;

http://www.head-fi.org/forums/f4/db-...lly-go-168037/
 
Jan 11, 2010 at 7:25 AM Post #36 of 39
I think that the majority of the confusion on this topic is caused by the fact that the majority of sensitivity ratings are given in dB/mW. by the virtue that not all headphones have equal impedance, it will require different voltages to achieve the same amount of power(mW).

This is important because what amplifiers produce is voltage, actually voltage gain. The 1/4 in connector on your amplifier provides a difference in potential across the connections made. Thats it. What this does then in produce a current. The current is a resultant. Amplifiers do not "push" current through anything. The current is the result of the voltage divided by the impedance in ohms. Thus you can multiply the current(in amps, which is a measurement of coulombs per second, a coulomb is a measurement of charge) by the voltage to get the power, in mW. Hence you can see the problem here. Even if two headphones have equivalent sensitivity in dB/mW, for a given setting of the volume knob, they can produce different SPL's based on the fact that each one is getting a different amount of power because of the difference in impedance in the headphones.

for example:
Say you have a headphone amp that with the volume knob at 9 O'Clock produces 1 volt at the jack. And we have two headphones, one with a impedance of 150 ohms and a sensitivity of 100dB/mW, and the second with a impedance of 300 ohms and a sensitivity of 100dB/mW.

the first headphone:
we take the 1volt and divide by the 150 ohms to get .00667 amps.
Then to get power, which is what our sensitivity is measured in, we multiply amps by volts, or 1 x .00667 and we get 6.66 mW.

for the second headphone:
we take the 1 volt divide by 300 ohms to get .00333 amps.
then multiply to get 3.33 mW

so you can see that the first headphone will be louder becuase it is getting twice as much power even though it is receiving the same voltage and has the same sensitivity rating. This is why it is so hard to compare the loudness of headphones based on sensitivity ratings measured in dB/mW.

But if headphones all headphones were measured in dB/V, this would be negated because for any given position of the volume knob on your amp your are getting a certain amount of voltage, which doesnt change with different headphones.

but if we give another example, same amplifier providing 1 volt, but with two different headphones. The first headphone with a sensitivity of 99dB/mW and impedance of 30 ohms, and the second headphone with a sensitivity of 96dB/mW and impedance of 60 ohms.

first headphone:
1 volt divided by 30 ohms is .033 amps, meaning you are getting 33mW

second headphone:
1 volt divided by 60 ohms is .066 amps, meaning you are getting 66mW

Now we can see the second headphone is using exactly twice the power of the first headphone. And because 3dB's is a doubling of sound pressure, it takes twice the power to get 3 extra decibels of sound. So this shows that both the headphones will be producing the same volume, because there is a 3dB difference between the two, with twice the power going to the headphone with 3dB lower impedance.

so this shows that even with two totally different headphones, you can get equal volume for a given setting of volume on your amp.

Now where this gets tricky with amp pairing is in the current. Notice i said that current is a resultant. Which is entirely true, you dont set current, you set voltage. but what if the amount of current that is needed for a given voltage set by the volume knob of your amp is greater than that which can be provided by the power supply? what happens here is that we get voltage sag. Which means that the voltage at the connector will drop.

But even this isnt the end of the story. because our headphones transducers arent perfect they have whats called a impedance versus frequency graph. So this means that different amounts of voltage are required for different frequencies to produce equivalent sound pressure levels. So what can happen because of this is that for certain frequencies the amp can run out of the necessary current, but for others be ok. So you will get voltage drops for some frequencies and not others changing the sound of the headphones from what it is supposed to. this is where i guess that some of amp pairing would come in.

I tried not to ramble to terribly much with this post, but im not the most deft writer :p also i took several liberties in my discussion, that the electrical engineers will cringe at, such as impedance equaling resistance, which i know is not true, but i felt that for the discussion, my points could be made more easily this way, and a higher number of people would understand them. also if you see any glaring errors, please say so
smily_headphones1.gif
 
Feb 25, 2010 at 6:12 PM Post #37 of 39
[size=small]This is an interesting thread. There seems to be a lot of misconceptions regarding headphone sensitivity, efficiency, impedance and the voltage and current needed to drive the headphone without clipping and distortion. [/size]

[size=small]The confusion is compounded by the fact that manufactures are inconsistent in the manner that they communicate specifications. Sometimes headphone sensitivity is listed in dB (SPL) * when 1 milliwatt of power is applied to the headphones (this is sometimes written as dB/mW). Sometimes headphone sensitivity is listed in dB (SPL) when 1 Vrms of voltage is applied to the headphones (this is sometimes written as dB/V). For example, on their website, Sennheiser gives the following specifications for their HD 650: [/size]

[size=small]Nominal impedance: 300 ohms[/size]
[size=small]Sound pressure level (SPL): 103 dB (1 Vrms).[/size]

[size=small]Sensitivities in dB/mW and dB/V are not always directly comparable to each other. To compare dB/V to dB/mW, we need to know the nominal impedance of the headphones. By my calculations, 103 dB/V (at 1 Vrms into 300 ohms) is equivalent to 97.8 dB/mW (at 1 mW). [/size]

[size=small]And sometimes headphone sensitivity is listed in dB (SPL) WITH NO REFERENCE TO AMOUNT OF POWER OR VOLTAGE APPLIED. For example, on their website, Sennheiser gives the following specifications for their RS-180: [/size]

[size=small]Impedance: 32 ohms[/size]
[size=small]Sound pressure level (SPL): 106 dB[/size]

[size=small]This type specification is not very useful. The specification 106 dB, by itself, says nothing about the sensitivity of the headphones. We are left to assume that they mean 106 dB (at 1 Vrms into 32 ohms). This assumption seems reasonable since Sennheiser list other headphones this way. Under that assumption, the RS 180 sensitivity is 91.1 dB/mW (at 1 mW). [/size]


[size=small]* SPL is an acronym for Sound Pressure Level[/size]

[size=small]==== Extra Credit =============[/size]
[size=small]For the geeks out there, the equation to convert dB/V to dB/mW is [/size]

[size=small]dB/mW = dB/V + 10*log(R*P/V^2)[/size]

[size=small]where [/size]
[size=small]dB/V is the sensitivity in dB (SPL) at 1 Vrms of voltage into impedance R, 103 dB/V into 300 ohms in this case[/size]
[size=small]R is the nominal impedance, 300 ohms in this case[/size]
[size=small]V is the reference voltage, 1 Vrms in this case[/size]
[size=small]P is the reference power, 0.001 watt in this case[/size]
[size=small]dB/mW is the sensitivity in dB (SPL) at 0.001 watt of power (that is 1 milliwatt)[/size]
[size=small]log is the logarithm base 10[/size]

[size=small]In the above example we have:[/size]

[size=small]dB/mW = 103+10*log(300*0.001/1^2) = 98.7[/size]
 
Mar 3, 2010 at 4:02 AM Post #38 of 39
Impedance also has a phase component. If at some frequency, the phase is close to 90 degree, it will be very hard to drive, unless the sensitivity is very high at that frequency.

Theoretically, if the phase is exactly 90 degree, even if the amp is capable of outputting infinite voltage and infinite current, the power output will still be zero.

So if the phase is high at some important frequency, the amp needs to be able to output much higher voltage and current, than what is suggested by the numbers of impedance and sensitivity.
 
Mar 29, 2014 at 5:23 PM Post #39 of 39
[size=small]This is an interesting thread. There seems to be a lot of misconceptions regarding headphone sensitivity, efficiency, impedance and the voltage and current needed to drive the headphone without clipping and distortion. [/size]

[size=small]The confusion is compounded by the fact that manufactures are inconsistent in the manner that they communicate specifications. Sometimes headphone sensitivity is listed in dB (SPL) * when 1 milliwatt of power is applied to the headphones (this is sometimes written as dB/mW). Sometimes headphone sensitivity is listed in dB (SPL) when 1 Vrms of voltage is applied to the headphones (this is sometimes written as dB/V). For example, on their website, Sennheiser gives the following specifications for their HD 650: [/size]

[size=small]Nominal impedance: 300 ohms[/size]
[size=small]Sound pressure level (SPL): 103 dB (1 Vrms).[/size]

[size=small]Sensitivities in dB/mW and dB/V are not always directly comparable to each other. To compare dB/V to dB/mW, we need to know the nominal impedance of the headphones. By my calculations, 103 dB/V (at 1 Vrms into 300 ohms) is equivalent to 97.8 dB/mW (at 1 mW). [/size]

[size=small]And sometimes headphone sensitivity is listed in dB (SPL) WITH NO REFERENCE TO AMOUNT OF POWER OR VOLTAGE APPLIED. For example, on their website, Sennheiser gives the following specifications for their RS-180: [/size]

[size=small]Impedance: 32 ohms[/size]
[size=small]Sound pressure level (SPL): 106 dB[/size]

[size=small]This type specification is not very useful. The specification 106 dB, by itself, says nothing about the sensitivity of the headphones. We are left to assume that they mean 106 dB (at 1 Vrms into 32 ohms). This assumption seems reasonable since Sennheiser list other headphones this way. Under that assumption, the RS 180 sensitivity is 91.1 dB/mW (at 1 mW). [/size]


[size=small]* SPL is an acronym for Sound Pressure Level[/size]

[size=small]==== Extra Credit =============[/size]
[size=small]For the geeks out there, the equation to convert dB/V to dB/mW is [/size]

[size=small]dB/mW = dB/V + 10*log(R*P/V^2)[/size]

[size=small]where [/size]
[size=small]dB/V is the sensitivity in dB (SPL) at 1 Vrms of voltage into impedance R, 103 dB/V into 300 ohms in this case[/size]
[size=small]R is the nominal impedance, 300 ohms in this case[/size]
[size=small]V is the reference voltage, 1 Vrms in this case[/size]
[size=small]P is the reference power, 0.001 watt in this case[/size]
[size=small]dB/mW is the sensitivity in dB (SPL) at 0.001 watt of power (that is 1 milliwatt)[/size]
[size=small]log is the logarithm base 10[/size]

[size=small]In the above example we have:[/size]

[size=small]dB/mW = 103+10*log(300*0.001/1^2) = 98.7[/size]

 
Sorry to bump such an old thread, but my math shows
 
103+10*log(300*0.001/1^2) = 97.8
 
please let me know if I got that wrong. I'm trying to compare a few headphones and this formula is very useful. 
 

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