What determines how sensitive a headphone is? - Page 3

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.

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.

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

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.