[ab initio]

  Equations are nice an all, but Tyll has nice webpage on how planar magnetics work.  I sense that practicality of how both work will be more insightful.
 
http://www.innerfidelity.com/content/how-planar-magnetic-headphones-work
 

 
I'm quite familiar with Tyll's page and his and others articles on orthodynamic headphones. They do a nice job about doing a show-and-tell of the inside of an orthodynamic headphone and showing how the arrangement of magnets and etched traces differ from the classing driver layout. It's also nice how they explain what issues these types of design changes attempt to address over the classing dynamic driver design.
 
Aside from that, the article does nothing to address how "damping factor" affects the driver's fidelity. There's a quick snippet that gives some wishy-washy discussion (without any references) that claims orthos suffer from a lack of damping (from what i can tell, based mostly on the fact that modders focus a lot of attention on adding damping materials to the headphones---which is about all modders can do to headphones in the first place). The claim is in direct contradiction with one of the articles linked at the end of his post (the wisdom audio article) which claims that orthos naturally have very tight control over the motion of the diaphragm (This article that is nothing more than some advertising doesn't have any sources cited for these claims either). Who to believe?
 
It really doesn't matter. I performed the relevant analysis above, all we need are typical headphone parameters for orthos and we can quantitatively assess the importance of damping factor.
 

  ^_^. ab initio you killed us here.
 
about the dynamic vs planar, there are simply too many reasons for differences to just try and deduce a behavior from the dynamic drivers IMO.
the magnetic field is different and if the principles obviously remain the same, as I said, the fact that the magnetic field can be the same in all positions with a planar has to impact the capacity for control. but it needs more power to work the same, so that means more opposing force... it's hard to get a clear view of the entire system(at least for me).
 
the weight and flexibility of the moving part is different too, so the mechanical damping might not be the same as dynamic drivers(I say might but I'm pretty sure it's not).
 
also the air flow has to be massively impacted by the magnets on each side that are obstructing the path. I'm not sure you can assume the electrical damping to always be of same importance be it to get the membrane at a precise position, or to make the driver stop.
I'm sorry I can't follow you on the formulas, my last years at school where about optic and photography ^_^.
maybe someone can kidnap Tyll, I remember a video where he had broken opened a planar. and he often mentioned talking a great deal with the audeze boys.

 
In my analysis above, I'm not trying to capture higher-order effects that depend on details like whether I have a dynamic driver with a suspended speaker cone with a coil centered around a magnet stator vs having an array of magnets with traces etched on a membrane. I was addressing the operating principal of these drivers and their most basic level by reducing them to an approximate mass-spring-damper system and corresponding electrical equivalent.
 
The result of the analysis is that the speaker system is defined by it's mechanical properties, M, C, and K, and its electrical properties, Zeb, and the electro-mechanical coupling which is given by B*l. These are the properties that describe the system whether the speaker is "dynamic" or "orthodynamic". The fundamental equations are exactly the same. This is why "damping factor" applies to orthodynamics in the exact same way as it applies to dynamic drivers. This is what I've demonstrated in my previous post. My analogy here is that of classic bicycles and recumbent bicycles. Both operate by exactly the same principle: you get on and pedal, turn the steering wheel to steer and squeeze the brakes to stop; however, the layout is quite a bit different and results in some key performance parameters being a bit different (like aerodynamic drag). Yes, orthodynamics are layed out differently than dynamic drivers and some of the values of key parameters are different, but its the same underlying physical principle at play and the governing equations are identical.
 
 
What everybody else here seems to be arguing is that between typical "dynamic" and "orthodynamic" drivers, there are differences in the constants that define the system. This is certainly true. M, C, K, B*l, Zeb, etc. are all different between the two different types of designs and even among the various individual drivers within each class. I am not arguing against this at all! These are exactly the types of parameters I'm asking the community here if they can give me typical values for the each of the types of drivers. The difference in these parameters between dynamic and orthodynamic drivers doesn't change the fact that the equation governing how "damping factor" influences the driver is exactly the same. What the differences  in these parameters will do is change how the impact of varying damping factor compares in size to the other forces at play in the transducer's response. I've given you guys the fundamental equations, all you guys have to do is tell me what the typical values for each type of driver are and we can calculate it directly.
 
Guessing that the layout of the magnets makes dynamic drivers susceptible to damping factor issues where orthodynamics are immune is no more helpful to this discussion than guessing that conductor material makes copper cables susceptible to transmission line effects where silver cables are immune is to cable debates. (and I know how most of you feel about that! )
 
I haven't said anything in my arguments to claim that orthodynamic drivers rely on electrical damping for proper function. I'm saying that the governing equations for ortho and regular dynamic drivers are exactly the same to zeroth order. I'm trying to fix the common misconception that somehow orthodynamic drivers are inherently immune to damping factor issues because they have purely resistive impedance. As far as damping is concerned, there is zero relationship between the frequency-dependent impedance character of a driver and whether or not is it susceptible to damping factor issues.
 
Cheers

 

[davidsh]

Quote: ab initio

 ...there is a tendency for the diaphragm to overshoot and ring at frequencies below mechanical resonance

Can you please elaborate on this part? What about at higher frequencies, and what if the driver is over-damped to begin with?
 
I do appreciate your contributions by the way.

 

[Steve Eddy]

I haven't said anything in my arguments to claim that orthodynamic drivers rely on electrical damping for proper function.

Oh really? You said it was just as relevant to orthos as it was dynamics.

I'm trying to fix the common misconception that somehow orthodynamic drivers are inherently immune to damping factor issues because they have purely resistive impedance.

Look, if the impedance is purely resistive, then by definition there is no mechanical resonance manifesting itself in the electrical domain. That being the case, the whole notion of damping and damping factor are moot. Instead of damping and damping factor issues, the amplifier's output impedance just represents simple resistive loss.

As far as damping is concerned, there is zero relationship between the frequency-dependent impedance character of a driver and whether or not is it susceptible to damping factor issues.

Nonsense. If the load is purely resistive, damping is moot.

Here's the equivalent circuit.



se

 

[davidsh]

I really don't know what to believe. I don't quite see why current would be induced when the diaphragm/driver is moving because it is manipulated by the electrical signal of the amp. Wouldn't dismiss it though. I am kind of lost, honestly.

 

[SilverEars]

If you look at HE-6's impedance graph, it's pretty straight forward.  It's resistive as it doesn't vary with frequency.  Look at HD800 graph below.
Here some cable measurements to compare with the HE-6 impedance graph which are flat.
 

 
Here is HD800's impedance graph
 

 

[Bill-P]

I really don't know what to believe. I don't quite see why current would be induced when the diaphragm/driver is moving because it is manipulated by the electrical signal of the amp. Wouldn't dismiss it though. I am kind of lost, honestly.

 
Because the electrical signal of the amp is actually as a result of how much "resistance" (or "impedance") the diaphragm/driver is presenting.
 
Basically, I think you can look at it this way: as soon as you connect a headphone to the amplifier, the headphone is just asking the amplifier to send it something. The amplifier then looks at the headphone and sends the appropriate signal. The drivers/diaphragms then "dance" to the signal sent by the amplifier. It's a two-way relationship.
 
That's why at the same position of the volume knob in the same amplifier, headphones with different impedance and sensitivity produce sounds at different volumes most of the time. I'm saying most of the time because a headphone with higher impedance but also higher sensitivity may output the same sound pressure as a headphone with lower impedance but also lower sensitivity. Also that assumes the load presented (how much the headphone is requesting of the amplifier) does not exceed the amplifier's limits.

 

[ab initio]

Quote: ab initio

 ...there is a tendency for the diaphragm to overshoot and ring at frequencies below mechanical resonance

Can you please elaborate on this part? What about at higher frequencies, and what if the driver is over-damped to begin with?

 
Here I think the terseness of my statement makes it both confusing and misleading. For a rigorous discussion of forced, damped oscillators, I  recommend "Fundamentals of vibrations" by Leonard Meirovitch. 
 
Consider the classing forced mass-spring-damper system (as described in previous posts) where F(t) = M*x'' + C*x' + K*x. Rewriting in terms of the undamped natural frequency w0 = sqrt(K/M) and the damping factor z = C/ ( 2*sqrt(K*M)), gives the an expression for the one-degree-of-freedom system:
 
x'' + 2*w0*z * x' + w0^2 * x = F/M [ * ]
 
Notice that when both the acceleration and velocity terms (the first and second terms in eqn [ * ]) are very small (namely, for frequencies much smaller than the natural frequency) then the force acts as a displacement controller:
 
w0^2 * x = F/M 
rearrange ->
x = F/(M*w0^2 ) [ ** ]
 
Here, we can see that the displacement is a given by the force, scaled by the factor 1/ (M * w0^2). In other words, the displacement of the system follows directly from the force input. This is the low frequency limit and is illustrated by Fig. 1 as w -> 0 that the amplitude of the response (called "|G(iw)|") -> 1. This figure was copied from Meirovitch.
 

Figure 1. 
 
There is a neat feature here in Figure 1 that shows that for sufficiently underdamped systems (namely, z <  0.7071 ), below the natural frequency the magnitude of the response is always greater than 1. That means that if you supply a sinusoidal forcing on any system with a damping ratio less than 0.7071 and below that system's natural frequency, then the displacement amplitude will always be greater than | F/(M*w0^2 ) |. Thus, if we think of the forcing as a displacement controller, then the amplitude of the response in an underdamped system with .7071 will always overshoot the intended amplitude at frequencies below the natural frequency.
 
There's more to this story than simple harmonic excitation. Consider a the step response of this system (or the square wave excitation which is like the step response when the square wave frequency is much lower than the natural frequency). First, let's look at the critically damped case where the overshoot is zero (i.e., z = 1). Fig 2 shows the scenario where the natural frequency is twice the square wave frequency. Here, the oscillator displacement just about has enough time to reach its full displacement before the square wave switches polarity. Perhaps more interesting is the case where the excitation frequency is far below the oscillator's natural frequency (Fig 3). Here, you can see how the displacement rapidly follows the force input without overshoot.  
 
 

Figure 2. Critically damped oscillator subject to square wave pulse train. Square wave pulse train has frequency 1/2 of the oscillator's natural frequency.
 

 
Figure 3. Critically damped oscillator subject to square wave pulse train. Square wave pulse train has frequency 0.05 times the oscillator's natural frequency.
 
 
When the system is underdamped (z < 1), the system will always overshoot the step response. You can see examples in Figs. 4a--c.
 
(a)
(b)
(c)
Figure 4. A slightly underdamped oscillator always overshoots the step response (or square wave train's with sufficiently slow frequency) (a) shows the over shoot for z = .8 and the natural frequency of the system 20 times the square wave frequency. (b) shows  the over shoot and ringing for an underdamped (z = 0.5) oscillator with natural frequency 20 times the square wave frequency. (c) shows the overshoot for a slightly underdamped "slow" oscillator with frequency only twice the square wave frequency (z = 0.8).
 
When the force frequency is much faster than the natural frequency of the oscillator, then the system never has enough time to "catch up" with the force signal. Consequently, the response amplitude asymptotically approaches 0 as the force frequency becomes increasingly larger than the natural frequency (w/w0 -> infinity). The behavior is clearly shown in the amplitude response plot (Figure 1). Here, it is informative to think of the square wave as a Fourier series (sum of sine waves) and recall that the fundamental has the same frequency as the square wave frequency with decreasingly important higher harmonics at higher frequencies.
 
Lastly, consider the over-damped system (z > 1). This system will never overshoot. The higher the damping ratio, the slower the system is to respond to changes in the forcing signal.  For any periodic forcing, the amplitude response of an overdamped system is strictly less than unity, regardless of the frequency (e.g., See Figure 1). Figure 5 shows how the response of the system is less than the forcing input for an overdamped system. To show how over damping increases the time it takes for the system to follow the input forcing, Figure 6 compares the response of 3 systems with damping ratios of 1, 1.5, and 10. Recall the critically damped system (z= 1) is the least-damped a system can be without overshoot. Consequently, this system is the fastest to follow an input forcing signal without ringing. Increasing the damping increases the time it takes for the system to relax to the forcing input.
 

 
Figure 5. Over damping reduces the amplitude of the system response.
 

 
Figure 6. Comparison of overdamped systems.
 
Cheers

 

[Steve Eddy]

Please, somebody put this thread out of its misery.

se

 

[MacedonianHero]

Please, somebody put this thread out of its misery.

se

^^ I'm with you Steve. 

 

[ab initio]

 

 

Oh really? You said it was just as relevant to orthos as it was dynamics.

That's correct. The physics apply just the same. Please re-read what I said and don't turn this into a straw man where you purposely misrepresent what I'm saying .

I'm trying to fix the common misconception that somehow orthodynamic drivers are inherently immune to damping factor issues because they have purely resistive impedance.

Firstly, I am incorrect in this statement. It should have read ".... because they appear to have purely resistive impedance."
 

Look, if the impedance is purely resistive, then by definition there is no mechanical resonance manifesting itself in the electrical domain.

By virtue of being a coupled electro-mechanical system, the system will be governed as I described in my post above. You should be able to take the fundamentals that I outlined in my previous posts and combine that with the "equivalent circuit" representation for a *dynamic driver and show this to be fundamentally true.

That being the case, the whole notion of damping and damping factor are moot. Instead of damping and damping factor issues, the amplifier's output impedance just represents simple resistive loss.

 

As far as damping is concerned, there is zero relationship between the frequency-dependent impedance character of a driver and whether or not is it susceptible to damping factor issues.

 

Nonsense. If the load is purely resistive, damping is moot.

I stand by my statement. It is accurate and supported by the physics of the system. Whether or not a pair of real headphones are sensitive to damping factor in an audible way and at audible frequencies depends on the relevant parameters (i.e., the Theile small parameters that I've been asking for). I haven't asserted anything stronger than that.  Let me remind you what I'm saying to make sure we are clear about what I am arguing and what I am not arguing :

  ....
 
Here, I have tried to demonstrate that there is nothing fundamental about the electrical characteristics of any type of moving coil transducer (e.g., dynamic headphones, orthodynamic headphones, woofers, etc.) that renders the concept of electrical damping (which depends on the damping factor) not applicable. 
 
... I think I have clearly and fully elaborated on the position I took at the beginning of this discussion while trying to address some of the confusion regarding "damping factor" and whether or not planardynamic headphones are theoretically immune to it.
 
...
All I'm trying to do is to address what appears to be a lack of fundamental understanding on the mechanics of "dynamic" type headphones. 
 
Cheers
 

Here's the equivalent circuit.

se

No. This is a simplified circuit.  The AES paper I linked to in a previous post shows the equivalent circuit for a speaker:

This paper provides us with a framework for discussing the electromechanics of a speaker using common terminology. Using the terminology in the paper, can you you show how this circuit simplifies to your equivalent circuit and how that relates to the mechanical damping of the physical driver? This would be a great help towards helping us gain the physical insight as to how damping relates to the physical operation of the speaker and how it manifests it self in the electrical characteristics of a speaker.
 
My previous posts have already laid out the groundwork for the mechanics of a dynamic driver, and I tried to relate how the mechanics couple to the electrical circuits via the product of the magnetic field strength and voice coil length B*l. Recall that the induced EMF in the driver circuit is given by V = B*l*u, where u is the velocity of the speaker's diaphragm (pushing the voice coils through the magnetic field) and that the force on the diaphragm is given by F = B*l*i, where i is the current through the voice coil.
 

 
Cheers

 

[davidsh]

 

Please, somebody put this thread out of its misery.

se

^^ I'm with you Steve. 

What's the problem?

 

[SilverEars]

Steve's point and the HE-6 impedance graph is pretty clear.  There should be no Back EMF since the impedance graph shows purely resistive load which means no induction. I don't know of a resistor that has Back EMF characteristics.    

 

[ab initio]

I really don't know what to believe. I don't quite see why current would be induced when the diaphragm/driver is moving because it is manipulated by the electrical signal of the amp. Wouldn't dismiss it though. I am kind of lost, honestly.

The fundamental principle is magnetic induction. You are correct that the amplifier provides an electrical signal that manipulates the diaphragm.
 
If you hold the diaphragm in place and apply a signal with the amplifier, the force you must exert to hold the diaphragm still is F = B*l*i  where (as we've been discussing above) B is the magnetic field strength, l, is the length of the voice coil wire through which the current is flowing, and i is the current supplied by the amplifier.
 
Because you're holding the diaphragm from moving, there is no motion of the diaphragm through the magnetic field and no back EMF (i.e., voltage) is induced.
 
However, when you release the diaphragm and allow it to move, there is a back EMF induced in the voice coil. In this case, the voltage is V = B*l*u, where B and l are the same as before and u is the velocity of the diaphragm. The current that flows as a result of the induced EMF depends on the total impedance of the circuit (including the driver impedance and the amplifier's impedance and anything else in between). If that total impedance is called "R", then the current that flows due to back EMF is V/R = B*l*u/R.  This current is flowing through the voice coil and adds (or rather, subtracts) with whatever the amplifier is providing, therefore resisting the motion of the of the diaphragm.
 
 
The total force applied to the diaphragm via electrical effects is the sum of the amplifier's current and the induced current due to diaphragm  motion:
 
F = B*l*i - B*l*u/R
 
Cheers

 

[castleofargh]

Please, somebody put this thread out of its misery.

se

well if we're only talking damping factor like the title suggest, then there is nothing wrong with this topic. if we think about the origin and the differences planar vs dynamic, then we kind of derailed ^_^.
 
 
so the idea I get from this:
Quote:

Consider the classing forced mass-spring-damper system (as described in previous posts) where F(t) = M*x'' + C*x' + K*x. Rewriting in terms of the undamped natural frequency w0 = sqrt(K/M) and the damping factor z = C/ ( 2*sqrt(K*M)), gives the an expression for the one-degree-of-freedom system:
 
x'' + 2*w0*z * x' + w0^2 * x = F/M [ * ]
 
Notice that when both the acceleration and velocity terms (the first and second terms in eqn [ * ]) are very small (namely, for frequencies much smaller than the natural frequency) then the force acts as a displacement controller:
 
w0^2 * x = F/M 
rearrange ->
x = F/(M*w0^2 ) [ ** ]

don't underestimate the power of the force! (at least for higher frequencies, but I don't know the formula to turn the quantity of midichlorian into force).
 
seriously I remember several people over time who mentioned on the interwerrb that for high freqs there was some effect that actually helped the driver maintain a fast enough movement (something to do with the air vibrating at the frequency kind of helping the membrane keep it's speed?) if someone knows what I'm talking about I would appreciate some light on this(don't know if it has a place in here, but if it impacts how long a movement is kept alive I guess it's part of it).
 
also again it might not seem directly related, but amps usually behave better when the damping ration is good. I don't remember reading much about an headphone amp that had better measurements with a 8ohm load vs the same amp with 150ohm load.

 

[ab initio]

Steve's point and the HE-6 graph is pretty clear.  There should be no Back EMF since the impedance graph shows purely resistive load which means no induction. So no Back EMF.  I don't know of a resistor that has Back EMF characteristics.    

 
I've been asking for typical parameters that describe a headphone driver (specifically orthodynamic drivers). That would be the most rigorous and satisfying way (at least to me) to figure out what's going on here.
 
With regard to your HE-6, the headphone driver's impedance has a contribution from the electrical components and a contribution from the mechanical components. Based on the graph you've given me I can conclude only one thing with absolute certainty and that is that you're impedance vs frequency curve is incomplete. Specifically, there *must* be a point where the headphone impedance begins to increase with further increase in frequency because of the inductance of the headphone cable+voice coil traces. The fact that that phenomenon is outside the audible frequency range is great because that won't affect frequency response of the headphone.
 
As far as the mechanical contribution to a (potentially) complex impedance in the audible range, there are different ways that this effect is not showing up in the graph. Until somebody gives me the driver parameters I've been asking for, I have no way to know a priori what that reason is.  The driver's mechanical resonance could be somewhere outside the range of frequencies shown on your plot, the coupling (B*l) could be sufficiently small that the mechanical contributions to the total impedance are negligible  compared to the electrical impedance of the driver, etc...
 
Does that clarify what is still unclear here?
 
Cheers