[cspirou]

Lately I have been thinking about what makes certain headphones work with certain amplifiers. There is a well known rule of thumb that the impedance of a headphone should be at least 8x the output impedance of an amplifier to have sufficient damping. So high impedance headphones like the HD-650 or the Tesla T1 should work with just about any amp as long as they have sufficient power.
 
However it appears that a major exception to this rule are planar magnetic headphones like those from Audeze and HifiMAN. Just look at these impedance curves for some of these headphones.
 
 

 
They are all completely FLAT! This means that the impedance is purely resistive. Dynamic headphones have a frequency dependent impedance which typically peaks around the 60Hz-100Hz. Having a low output impedance is important for dynamic headphones because the different impedances will affect the voltage output at different frequencies, leading to distortion. Planar magnetics affect the output voltage uniformly due to the fixed impedance, which means that the result should just be a shift in voltage by a factor x. This appears to be a major reason why planar magnetics are so clear to begin with.
 
Based on this reasoning it seems that you can completely ignore the 8:1 impedance ratio when looking for a headphone amp for planar magnetics and just focus on other factors like power output. In fact you might even prefer a 1:1 ratio since power output maximizes when the output impedance and headphone impedance are equal, something people tended to do with tube amps due to the lower power output. Probably also means that these are ideal for professional uses since you can plug these into a studio console or a laptop without having to worry about differences in sound coloration.
 
If I had the equipment I would measure a set of planar magnetic headphones with a variable output impedance to see the differences in frequency response and distortion. However maybe some of the more technical minded readers out there with a bit of spare time would be willing to do this experiment. Or just plug in a pair of Audeze headphones in different amps and see if the result is worse compared to Grados in the same amps.

 

[castleofargh]

there is another thread where the planar/ortho vs impedance is discussed (with lots of theories).
 
but to answer to your post specifically, you can find dynamic headphones with an almost flat impedance response over frequencies. and I don't know of any evidence suggesting that those dynamic headphones are better than some with fluctuating impedance response.
I wouldn't go there to find the reason why planars are good.
 
 
about the 1/10 ratio, even if it ends up having little impact on the headphone's side, I wouldn't dismiss the fact that the amp might not behave as well when presented with a low impedance load(low relative to its own output). the results will vary with the amp, but I know I wouldn't take a chance unless I have exhaustive measurement data for the amp I'm looking to buy?

 

[cjl]

Based on this reasoning it seems that you can completely ignore the 8:1 impedance ratio when looking for a headphone amp for planar magnetics and just focus on other factors like power output. In fact you might even prefer a 1:1 ratio since power output maximizes when the output impedance and headphone impedance are equal, something people tended to do with tube amps due to the lower power output.

This isn't really true. If you have a fixed source impedance and vary the load impedance, you're right that you'll get maximum power when the source and load impedance are identical. With a fixed load impedance though (the headphones), you get peak power when the source impedance is zero (a pure voltage source).

 

[SilverEars]

  This isn't really true. If you have a fixed source impedance and vary the load impedance, you're right that you'll get maximum power when the source and load impedance are identical. With a fixed load impedance though (the headphones), you get peak power when the source impedance is zero (a pure voltage source).

This is maximum efficiency not power transfer.  The impedance of the source and load must be equal for maximum power transfer.  
 

 

[cjl]

  This is maximum efficiency not power transfer.  The impedance of the source and load must be equal for maximum power transfer.  
 

Once again, no. For a fixed voltage supply and load impedance, the maximum power applied to the load occurs at zero source impedance. Suppose you have a 10V supply voltage and a 10 ohm load, for example. With a zero ohm source, your total impedance is 10 ohms, you'll have 1 amp of current, and 10 watts of power supplied to the load. If the source is 10 ohms, on the other hand, you'll have a total impedance of 20 ohms, half an amp of current, and only 2.5 watts supplied to the load.
 
In fact, if you even look at the wikipedia page from which that figure originates, you'll find the following quote:
 
"A note of caution is in order here. This last statement, as written, implies to many people that for a given load, the source resistance must be set equal to the load resistance for maximum power transfer. However, this equation only applies if the source resistance cannot be adjusted, e.g., with antennas (see the first line in the proof stating "fixed source resistance"). For any given load resistance a source resistance of zero is the way to transfer maximum power to the load. As an example, a 100 volt source with an internal resistance of 10 ohms connected to a 10 ohm load will deliver 250 watts to that load. Make the source resistance zero ohms and the load power jumps to 1000 watts."

 

[SilverEars]

Gotcha.  It makes better sense now. So the source impedance at 0 is when it's at maximum efficiency and transfer.  

 

[cspirou]

This isn't really true. If you have a fixed source impedance and vary the load impedance, you're right that you'll get maximum power when the source and load impedance are identical. With a fixed load impedance though (the headphones), you get peak power when the source impedance is zero (a pure voltage source).

Thanks for the correction. For tube amps you get maximum power when you use the taps that match impedance but that is really more a case of changing load impedance to match the high tube impedance rather than the other way around.

But that was a small part of a bigger point I was making in terms of damping and quality.

On another note, am I the only one that thinks this forum should be called Sci-Fi instead of Sound Science?

 

[castleofargh]

On another note, am I the only one that thinks this forum should be called Sci-Fi instead of Sound Science?

but what would you call the cable section then?

 

 

[KamijoIsMyHero]

Supernatural

 

[Joe Skubinski]

It's about damn time! Absolute control of a transducer is what a near perfect amplifier should provide.

Once again, no. For a fixed voltage supply and load impedance, the maximum power applied to the load occurs at zero source impedance. Suppose you have a 10V supply voltage and a 10 ohm load, for example. With a zero ohm source, your total impedance is 10 ohms, you'll have 1 amp of current, and 10 watts of power supplied to the load. If the source is 10 ohms, on the other hand, you'll have a total impedance of 20 ohms, half an amp of current, and only 2.5 watts supplied to the load.

In fact, if you even look at the wikipedia page from which that figure originates, you'll find the following quote:

"A note of caution is in order here. This last statement, as written, implies to many people that for a given load, the source resistance must be set equal to the load resistance for maximum power transfer. However, this equation only applies if the source resistance cannot be adjusted, e.g., with antennas (see the first line in the proof stating "fixed source resistance"). For any given load resistance a source resistance of zero is the way to transfer maximum power to the load. As an example, a 100 volt source with an internal resistance of 10 ohms connected to a 10 ohm load will deliver 250 watts to that load. Make the source resistance zero ohms and the load power jumps to 1000 watts."

 

[Greenears]

  Once again, no. For a fixed voltage supply and load impedance, the maximum power applied to the load occurs at zero source impedance. Suppose you have a 10V supply voltage anNd a 10 ohm load, for example. With a zero ohm source, your total impedance is 10 ohms, you'll have 1 amp of current, and 10 watts of power supplied to the load. If the source is 10 ohms, on the other hand, you'll have a total impedance of 20 ohms, half an amp of current, and only 2.5 watts supplied to the load.
 
In fact, if you even look at the wikipedia page from which that figure originates, you'll find the following quote:
 
"A note of caution is in order here. This last statement, as written, implies to many people that for a given load, the source resistance must be set equal to the load resistance for maximum power transfer. However, this equation only applies if the source resistance cannot be adjusted, e.g., with antennas (see the first line in the proof stating "fixed source resistance"). For any given load resistance a source resistance of zero is the way to transfer maximum power to the load. As an example, a 100 volt source with an internal resistance of 10 ohms connected to a 10 ohm load will deliver 250 watts to that load. Make the source resistance zero ohms and the load power jumps to 1000 watts."

General idea, but not quite.  In the real world no source has zero impedance - this is akin to a perpetual motion machine.  If you invent such a thing and try to patent it, the patent office will invoke their never-used right to inspect your invention.  Try it  So given whatever small output resistance the source has, the maximum possible power to the load is to match impedance. ELE101, or maybe 201.  Can't remember that far back.  Also this is for linear circuits, and nothing is linear when it begins to melt.  Don't try this at home, or at least more than 6 ft from a fire extinguisher. 
 
None of this applies to audio.  You don't want to be taxing anything to the point of maximum power output.

 

[miceblue]

If you have a fixed source impedance and vary the load impedance, you're right that you'll get maximum power when the source and load impedance are identical.

Is that true?

 

[cjl]

  General idea, but not quite.  In the real world no source has zero impedance - this is akin to a perpetual motion machine.  If you invent such a thing and try to patent it, the patent office will invoke their never-used right to inspect your invention.  Try it  So given whatever small output resistance the source has, the maximum possible power to the load is to match impedance. ELE101, or maybe 201.  Can't remember that far back.  Also this is for linear circuits, and nothing is linear when it begins to melt.  Don't try this at home, or at least more than 6 ft from a fire extinguisher. 

Still no, unless you have a variable load impedance.
 
If you have an output impedance of 0 ohms, a load of 10 ohms, and a 10 volt signal, you'll have 1 amp of current and 10 watts delivered to the load
If you have an output impedance of 0.1 ohms (realistic, near-zero) and everything else the same, you'll have 0.9901 amps of current and 9.803 watts delivered to the load
If you have an output impedance of 10 ohms (and everything else the same again), you'll have 0.5 amps of current and 2.5 watts delivered to the load.
 
The near-zero output impedance still gives you much, much more power than the output impedance matched to the load. It's only if you vary the load (rather than the output impedance) that you get your result:
 
If you have an output impedance of 0.1 ohms, a load of 0.1 ohms, and a 10 volt signal, you'll have 50 amps of current (if your amp can deliver it, of course) and 250 watts delivered to the load. This is of course far, far more than the amount delivered in the second scenario above, and is the most power you can deliver with an output impedance of 0.1 ohms and a 10 volt source. It isn't terribly relevant to headphones or audio though. In every case where you have a fixed load impedance (the transducer), you will deliver more power with a lower output impedance for the same peak voltage swing. If you have a 5 ohm output and a 3 ohm output, the 3 ohm will always deliver more power to the headphones for the same voltage output. A 1 ohm would be better still. A 50 ohm would be worse. This applies whether you're driving 4 ohm speakers, or whether you're driving 600 ohm headphones.
 
 
(I'm starting to feel like a stuck record here. Please, do the math before trying to disagree with me)

 

[cjl]

Is that true?

Yep. As I said above, if you have a fixed source impedance and voltage available, and you vary the load, the peak power delivered to the load occurs when the load impedance matches the source impedance.

 

[Greenears]

  Still no, unless you have a variable load impedance.
 
If you have an output impedance of 0 ohms, a load of 10 ohms, and a 10 volt signal, you'll have 1 amp of current and 10 watts delivered to the load
If you have an output impedance of 0.1 ohms (realistic, near-zero) and everything else the same, you'll have 0.9901 amps of current and 9.803 watts delivered to the load
If you have an output impedance of 10 ohms (and everything else the same again), you'll have 0.5 amps of current and 2.5 watts delivered to the load.
 
.....
 
(I'm starting to feel like a stuck record here. Please, do the math before trying to disagree with me)

Sigh.  I have done the math, and passed the course.  I have the book on my shelf.  They really test you on this stuff, I'm not making it up.
 
0 ohm source doesn't exist.  
0.1 ohm source delivers maximum power to a 0.1 ohm load (if it doesn't melt first)
10 ohm source delivers maximum power to a 10 ohm load
 
Just try the 10 ohm source into a 9 or 11 ohm load.  You'll see the power to the load is less than 10 ohm load.  So that is your maximum.
 
The mistake you are making is for 0.1 ohms you are plugging in only 10 ohm and not 0.1 ohm load, and 0.1 yields 250 W of load power or something crazy which is much higher than 9.803W.  
 
Hope that helps you out.