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Author Topic: planars and output impedance/damping factor?  (Read 1156 times)

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Armaegis

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planars and output impedance/damping factor?
« on: September 10, 2015, 05:51:53 AM »

The HD800 also has a natural bass hump. Our measurements here are done with my O2 amp. Tyll uses a SS amp as well I'm pretty sure. Both have negligible output impedance. About 92% of the tube amps (including transformer coupled) I recommend have an output impedance around 2-7 ohms.

Since we're on the subject, a subject which should probably be split off, any idea why planars like the LCDs and some Hifiman sound bassier on some amps versus others even though they have flat impedance curves?

I too am curious about this so I'll start the thread. I have some funny ideas in my head that lean more towards the material science aspect (sorta) rather than electrical, but I'm not quite sure how to put that into words yet. And it's all brain floof anyways.

I think its because of the fact that a lower impedance ortho receives only a fraction of power from a High output impedance amp compared to what it would get from a low output impedance amp. One has to remember that the lower frequency reproduction requires the maximum excursion from the driver and consequently would need much more power to be driven optimally in that region.
For instance, my 6AS7 OTL has an output impedance of 150 ohm and the LCD-2 sounds noticeably bass light on this amp. If I change he output tube to 6528 which lowers the output impedance to about 50ohms, the LCD2 gain that solid bottom end that they are famous for (almost comparable to Beta22 now). OTOH, my 150 ohm orthos sound the same FR wise regardless of the output tubes.
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Anaxilus

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Re: planars and output impedance/damping factor?
« Reply #1 on: September 10, 2015, 06:24:42 AM »

Thanks for the split. I'd love to hear your thoughts. Higher OI has sometimes resulted in flabbier woolier bass with some planars rather than actually less bass. I can buy less driver control rather than less bass, but I'd be curious to hear your thoughts on the materials science aspect.
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Re: planars and output impedance/damping factor?
« Reply #2 on: September 10, 2015, 06:57:37 AM »

At the lab we measured my LCD-2 at 0Rout and 120Rout. AFR wise there were no differences. We did register some change at bass frequencies with 32Ohm dynamic headphones, but it surely wasn't too big.
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Thujone

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Re: planars and output impedance/damping factor?
« Reply #3 on: September 10, 2015, 12:04:37 PM »

Additionally, I think it could be useful to discuss why dynamics would be expected to sound different based on their non-flat impedance curves. Maybe this is a simple explanation, but I haven't come across it myself.

I'd also like to hear your material thoughts, Armaegis.
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Re: planars and output impedance/damping factor?
« Reply #4 on: September 10, 2015, 12:51:54 PM »

Additionally, I think it could be useful to discuss why dynamics would be expected to sound different based on their non-flat impedance curves. Maybe this is a simple explanation


There is... it's called voltage division.
http://www.mediafire.com/view/?82kf0r5kdckdcer
How much 'effect' there is depends on:
1: The ratio between the top and bottom of the impedance differences of the headphone.
2: The ratio between the load impedance and output resistance.
Culmination of the effect is seen in current drive amplifiers (bakoon etc)
The frequency response is altered and follows the impedance plot exactly in these amps.

Only headphones with a varying impedance will measure/sound tonally different on other output resistances.



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OJneg

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Re: planars and output impedance/damping factor?
« Reply #5 on: September 10, 2015, 03:47:24 PM »

In what measurable aspects would a low damping factor appear for planar magnetic headphones? Obviously the load is a pure resistor electrically, but is there some sort of more complex acoustic impedance model that would be affected by higher Zout?
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Marvey

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Re: planars and output impedance/damping factor?
« Reply #6 on: September 10, 2015, 04:32:03 PM »

Thanks for the split. I'd love to hear your thoughts. Higher OI has sometimes resulted in flabbier woolier bass with some planars rather than actually less bass. I can buy less driver control rather than less bass, but I'd be curious to hear your thoughts on the materials science aspect.

Distortion / power. Amps will have a harder time delivering current when the damping factors / impedances go down. Softness, decreased volume, compression, wobbly bass.

Shouldn't be a problem if the amp can keep delivering current and probably why vintage amps like the Sansui despite super high output Z at the headout can still grip orthos well. Also explains why most lower power and/or higher output Z tube amps won't do HE1000 or LCD3 justice.

More than just effect on FR on non-linear headphone impedances or no effect on FR with flat headphone impedances.
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Armaegis

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Re: planars and output impedance/damping factor?
« Reply #7 on: September 10, 2015, 06:57:33 PM »

I have some funny ideas in my head that lean more towards the material science aspect (sorta) rather than electrical, but I'm not quite sure how to put that into words yet. And it's all brain floof anyways.

I've got all these scribbled notes that are hard to follow because I'm trying to draw analogies that are flawed due to the nature of the thing. Keep in mind this is also all just mental diarrhea. I could be wrong, so very very wrong.

- electricity is not this instantaneous thing, electrons have to move, there is a mass and momentum there
... particle-wave/energy duality, let's treat it like particles
- think of it like a fluid (voltage is pressure, flow is current, blah blah whatever analogy works, it won't be perfect)
- hydraulics: push on one end, the pressure isn't instantaneous, no equilibrium, there is a compression gradient that builds because pressure is particles pushing into one another, so it takes time for that compression to spread out across the rest

- but things like damping factor and output impedance... sorta like viscosity and the size of the reservoirs on either side of the conduit

- damping factor is stopping power derived from the input; low output impedance effectively shunts most of that energy back into stopping power (but how!?!), high output impedance means your energy gets wasted

- why is damping only considered purely from the output impedance vs load impedance? sure I can use a liquid with high viscosity and that works, but shear forces are way stronger with nearby solid walls. Push gushy liquid through a tube, it'll slow itself down. Push the equivalent amount of liquid through ten tubes that add up to the same cross section, you'll have a much harder time because there is a greater amount of surface area even if cross section remains the same
- also consider the energy coming out the end of the tube, it'll produce ripples at the surface of the tank, some energy can travel hit the other side and bounce back; but a big reservoir doesn't care, all that energy coming out is inconsequential compared to the bulk\

- sure electrons will travel through a wire, through a transducer, a coil, go through magnetic fields, etc, but an electron does not go in one end and magically come out the other; it pushes all the stuff in front of it; in AC that electron will never actually make it to the other end

- you have a tank of electrons on either side of the load, like a hydraulic

- how does an amp control the flow? strong force over a small area? low force over a wide area? The resultant pressure is the same but your implementations are different, and your distortions/implementations/complexity will be different, and the resultant energy dispersion on the other end will be different as well

- everything has momentum, electrons have momentum, but momentum is this weird mathematical concept because it's not actually a real thing (unlike energy) but the mathematical construct manages to remain constant
- so ok, tada you've got high damping factor, good stopping force, but force alone is not enough
- in physics, we need impulse which is a function of force and time
- fluid is gushing along a pipe, poof you flip a switch and reverse pressure and your pumps or whatever at the far ends start going the other way, but all that fluid in the middle is still going in its original direction even though it feels pressure going the other way (and that pressure takes a while to even get there since it starts at the ends and the compression wave takes a while to equalize over the entire body)

So the gist of it: stop treating electricity/voltage/blahblah as magical instantaneous things; there is a "mechanical" interaction to the electrons themselves
- even more complicated since all these reactions on the subatomic scale are also embodied within the macroscopic transducer which is a true mechanical result of the magical electricity that are operating on a completely different time scale

argh headache, gonna stop now
tl;dr: there's something we don't know how to measure/calculate yet


older more scattered version of my thoughts
(click to show/hide)
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Solderdude

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Re: planars and output impedance/damping factor?
« Reply #8 on: September 10, 2015, 10:56:25 PM »

- damping factor is stopping power derived from the input; low output impedance effectively shunts most of that energy back into stopping power (but how!?!), high output impedance means your energy gets wasted.
low output impedance... basically shunts the counterEMF entirely back onto the driver, so all that energy is used for control
like high viscosity? energy is converted to shear (stopping) force, rather than transmitting away.

Something to help you to ponder about the fluid analogy:

When you look at a driver that is in parallel to an amplifier (which is close to 0 Ohm when ideal) it is easy to think this is a parallel circuit.
For AC voltage it is so it would seem like it can 'damp' and 'grip' the driver.

However, the magnetic driver does not work on a voltage. The movement the voicecoil makes is determined by current  and not a parallel one as in voltage.
When we assume a resistive load the current is determined by the voltage applied/resistance.

Here is the thing though ... for currents (= movement) the amplifier and voice-coil resistance is a series circuit.

The 'stopping' power is thus determined by the current in the circuit and as the 'stopping force' that is determined by the back EMF has the same internal resistance as the  voicecoil the 'stopping force' (=current) can never be greater than the series resistance of the circuit.
It is NOT determined by the voltage across the amplifier.

So a headphone with a 300 Ohm voice coil will have about the same 'stopping' power when a 0.01Ohm series resistor is added or a 10 Ohm one.
310 versus 300.01 with the same voltage is about the same 'stopping CURRENT.
Damping factor difference 30,000 to 30 ... difference in stopping current 3% (0.3dB)

Of course, a 16 Ohm driver on a 120 Ohm source or 0 Ohm source does make a bigger difference in stopping currents.

- in physics, we need impulse which is a function of force and time

 Doesn't mass (a form of dynamic resistance) also play an important role ?



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briskly

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Re: planars and output impedance/damping factor?
« Reply #9 on: September 10, 2015, 11:46:18 PM »

Electrical-mechanical analogues are enough to make your head spin, aren't they? Force can be represented as either current or voltage, and the equations change for it. Would there be fluid/mechanical analogs for the Lorentz force or the rest of Maxwell's equations, I'm not so sure.

However, the magnetic driver does not work on a voltage. The movement the voicecoil makes is determined by current  and not a parallel one as in voltage.
When we assume a resistive load the current is determined by the voltage applied/resistance.

That follows from the Lorentz force, expressed as F = q(v x B) on a particle and F = I*L*B for current through a headphone voice coil. What's not so clear to me is how electrical resonances present themselves besides the obvious driver inductance. For example, the little bumps you can see on the PM1 impedance plot.
« Last Edit: September 11, 2015, 01:33:13 AM by briskly »
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