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A twist on the impedance adapter

post #1 of 38
Thread Starter 
This is the original drawing by Mrvile for the impedance adapter and seems to be the traditional way of thinking about the impedance adapter.



So I modified the drawing to explain more clearly the question I had in mind and which nobody answered yet on the headphone forum.

Traditionally, the current is perceived this way. Current flowing towards the ground (this is a historical mistake that was never really fixed in electronics) so it makes sense in that context to place the resistance in the "path" of the current as seen in the original drawing.



The thing is that current doesn't exist in itself. Electrons actually flow in the opposite direction; from the ground to wherever (Ground tab to drivers to left and right tab).



So the question is: would a single resistor on the ground side be equivalent to 2 resistors on tip + ring side like in your original drawing. (I think the answer is yes).
post #2 of 38
Sorry I can't contribute (my electrical knowledge is Ohm's law on a good day) but I'm very interested to see if this is right. Would save me a bit of time, at least.
post #3 of 38
the answer is "sort of." what you draw will reduce current, and make PART of the heapdhone act like it has a resistor on it, but this is something to be avoided.

putting a single resistor on the shared ground line can cause weird crosstalk issues.

in the case of my balanced headphones and ones where i put the resistor in the cups (before the ground line is shared...), i prefer to use a resistor on the in and out of phase lines each, but thats a different story.

edit:
it should also be noted that since the voltages (and relatively currents) in the various "signal" wires will ALWAYS swing above and below ground, the current flows both ways...
post #4 of 38
Thread Starter 
Quote:
Originally Posted by nikongod View Post
the answer is "sort of." what you draw will reduce current, and make PART of the heapdhone act like it has a resistor on it, but this is something to be avoided.

putting a single resistor on the shared ground line can cause weird crosstalk issues.
Hmmm, why? I'd like to understand what you perceive the underlying problematic is.

Quote:
Originally Posted by nikongod View Post
in the case of my balanced headphones and ones where i put the resistor in the cups (before the ground line is shared...), i prefer to use a resistor on the in and out of phase lines each, but thats a different story.
Hmmm, I don't know for balanced headphones, i'm not certain what the difference is.

Quote:
Originally Posted by nikongod View Post
edit:
it should also be noted that since the voltages (and relatively currents) in the various "signal" wires will ALWAYS swing above and below ground, the current flows both ways...
Huh? Aren't we talking about DC here? I seriously doubt that what you say is true (but maybe I don't understand completely). Just from the fact that we have a gound tab and a positive tab on the drivers points out to the flow being DC and not AC. If we were dealing with AC it would be another story (but I've never seen or heard about any drivers being driven by AC current. Or maybe they all are and I never knew before ).

For dynamic headphones, we're only dealing with a coil making a magnet move a piece of cardboard (or whatever material the driver is using to emit sound waves).

edit:
Ahhhhhh, nevermind, it does indeed seem like the polarities change. Drivers are AC driven!!! Darnit, I always thought that drivers were DC driven. In any case, i still don't understand the crosstalk issue you're mentioning.
post #5 of 38
Quote:
Originally Posted by El Condor View Post
Hmmm, why? I'd like to understand what you perceive the underlying problematic is.
the problem comes in the specfic (and very real case) that follows. to make the example the worst case, i will use the best example

first off, for the example here, we add about the same value as each driver to the ground line as opposed to the "signal" line. this is very realistic for grados, where adding some impedance will get an amp with a capacitor coupled output to go deeper in bass.

so we have the problem that we are playing a stereo sound. for the best example, say all of the bass is in your left ear (my example, i pick the scenario that works best to prove it.) the right driver at "r" resistance is exactly paralleled with the ground line also at r, and the "return path" of the bass signal into the left driver becomes BOTH the ground AND the right driver. this plays the signal out of phase and at a lower level in your right ear. then we have the fun that ensues when you have signals SLIGHTLY out of phase with eachother. do they go through the ground path, or do they mix in the drivers?
Quote:
Hmmm, I don't know for balanced headphones, i'm not certain what the difference is.
in a balanced headphone there is NO shared connection from one headphone to the other.
post #6 of 38
Quote:
Originally Posted by El Condor View Post
Huh? Aren't we talking about DC here? I seriously doubt that what you say is true (but maybe I don't understand completely). Just from the fact that we have a gound tab and a positive tab on the drivers points out to the flow being DC and not AC.
Electrons flow one way (I beleve - to + in american labels) and protons (or holes if you start looking at transistors) flow the other (again should be + to - in american labels), so it does go both ways.

That said, for most anything the standard hobbiest works on, this point will never bother them as in Metal Wire the protons don't flow and while the detail is of interest when dealing with semiconductors it isn't greatly needed to understand how to design around them in a basic sense
post #7 of 38
post #8 of 38
Quote:
Originally Posted by El Condor View Post
Hmmm, why? I'd like to understand what you perceive the underlying problematic is.
The shared ground return is exactly that -- shared between the two channels. If they both go directly to the amp's ground reference (which presumably is quite "stiff" and does not fluctuate, there is no problem). However, when you insert a resistor there, the return current from one channel will cause voltage drop across that resistor (which dynamically vary with the signal), and that in turn will modulate the reference for the other channel (and vice versa). This will grossly reduce stereo channel separation.
post #9 of 38
Thread Starter 
Quote:
Originally Posted by Blind Tree Frog View Post
Electrons flow one way (I beleve - to + in american labels) and protons (or holes if you start looking at transistors) flow the other (again should be + to - in american labels), so it does go both ways.

That said, for most anything the standard hobbiest works on, this point will never bother them as in Metal Wire the protons don't flow and while the detail is of interest when dealing with semiconductors it isn't greatly needed to understand how to design around them in a basic sense
On a side note, protons don't move in a wire, only loosely coupled valence electrons do .
post #10 of 38
Thread Starter 
Quote:
Originally Posted by amb View Post
The shared ground return is exactly that -- shared between the two channels. If they both go directly to the amp's ground reference (which presumably is quite "stiff" and does not fluctuate, there is no problem). However, when you insert a resistor there, the return current from one channel will cause voltage drop across that resistor (which dynamically vary with the signal), and that in turn will modulate the reference for the other channel (and vice versa). This will grossly reduce stereo channel separation.
AHHHHHHHH!!! Very clean explanation. So this raises the question in my mind. Doesn't the same effect happen at a lesser scale even without the resistor since both channels share the same ground? If the left and right channel are varying in a sensibly different way, wouldn't there be some form of crosstalk anyways (this would explain the interrest in balanced connections)?
post #11 of 38
Quote:
Originally Posted by El Condor View Post
On a side note, protons don't move in a wire, only loosely coupled valence electrons do .
I said that. Note where I said "This is a moot point because protons don't move in a metal wire"
post #12 of 38
Quote:
Originally Posted by El Condor View Post
AHHHHHHHH!!! Very clean explanation. So this raises the question in my mind. Doesn't the same effect happen at a lesser scale even without the resistor since both channels share the same ground? If the left and right channel are varying in a sensibly different way, wouldn't there be some form of crosstalk anyways (this would explain the interrest in balanced connections)?
As I said, assuming that the amp's ground reference is stiff, then the only crosstalk induced in this manner is from the contact resistance of the headphone plug/jack and from the headphone wiring itself. That is usually in the milliohms scale so it's very, very small.

The assumption that the ground is stiff is true for most amps (Except resistor rail-splitter cmoys for obvious reason), but only to a certain degree. In conventional 2-channel, dual-rail amps, the "ground" is the center point between the positive and negative PSU rail capacitors. Thus, the capacitors' ESR plays into this. Depending on the caps used, there could be many tens of milliohms. Still low, but if the headphones are low impedance types then the increased current flow would cause enough ground "wiggle" to become quite measurable.

In a 3-channel active ground amp (e.g., M³, PPA and the new β22 amp configured as such), the active ground channel's output impedance is easily an order of magnitude lower than a passive ground, and the return current gets diverted to the regulated supply rails rather than signal ground, so this problem is eliminated. Similarly, a fully balanced amp has the same advantage in that the two channels' grounds are in fact separated, and also actively driven. This is why one of the most immediate and measurable improvement of active ground or balanced amps is in the stereo crosstalk performance.

Edit: Btw, all this talk about electrons vs. holes in current flow is totally irrelevant with respect to where you put the resistors in the original post. Whether you think in terms of electrons or holes is just that -- a way of thinking about it, and no more. Most engineers tend to prefer thinking in terms of current flow from + to - (i.e., holes) because of convention. Nevertheless, either way you think, replacing the two resistors in the L and R wires with one in the ground line is not equivalent.
post #13 of 38
Thread Starter 
Quote:
Originally Posted by amb View Post
As I said, assuming that the amp's ground reference is stiff, then the only crosstalk induced in this manner is from the contact resistance of the headphone plug/jack and from the headphone wiring itself. That is usually in the milliohms scale so it's very, very small.
This is quite interresting. Hmmm, how can the ground reference be stiff (I'm not completely certain I understand what you mean by stiff) when the flow of current, as I understand, has to go in both direction as far as the drivers are concerned (both the ground and the signal alternating between + and -)?

Quote:
Originally Posted by amb View Post
The assumption that the ground is stiff is true for most amps (Except resistor rail-splitter cmoys for obvious reason), but only to a certain degree. In conventional 2-channel, dual-rail amps, the "ground" is the center point between the positive and negative PSU rail capacitors. Thus, the capacitors' ESR plays into this. Depending on the caps used, there could be many tens of milliohms. Still low, but if the headphones are low impedance types then the increased current flow would cause enough ground "wiggle" to become quite measurable.

In a 3-channel active ground amp (e.g., M³, PPA and the new β22 amp configured as such), the active ground channel's output impedance is easily an order of magnitude lower than a passive ground, and the return current gets diverted to the regulated supply rails rather than signal ground, so this problem is eliminated. Similarly, a fully balanced amp has the same advantage in that the two channels' grounds are in fact separated, and also actively driven. This is why one of the most immediate and measurable improvement of active ground or balanced amps is in the stereo crosstalk performance.
I somewhat figured out the advantage of having a balanced connection (stereo crosstalk), but I'm not certain I completely grasp what is involved when the ground is shared by both drivers. I have in mind that if both drivers keep alternating the direction of the flow of current, then some important form of crosstalk will occur, and from what I'm reading, it doesn't seem to be the case.

Quote:
Originally Posted by amb View Post
Edit: Btw, all this talk about electrons vs. holes in current flow is totally irrelevant with respect to where you put the resistors in the original post. Whether you think in terms of electrons or holes is just that -- a way of thinking about it, and no more. Most engineers tend to prefer thinking in terms of current flow from + to - (i.e., holes) because of convention. Nevertheless, either way you think, replacing the two resistors in the L and R wires with one in the ground line is not equivalent.
Understandably, but this was stemming from my misunderstanding of how things work. Although things are clearer, I'm still not fully understanding how the I can reconcile the idea of the ground being stiff and the flow of current alternating between the two wire ends on the coil.
post #14 of 38
By "stuff" I mean that the ground voltage doesn't vary with the amount of current flowing in/out. It should ideally be 0V at all times because all other voltages are referenced from it. Given that, it shouldn't matter that the grounds are shared, because the return current current from one transducer will simply be "sunk" into ground rather than being "seen" by the other. As soon as the ground becomes "wiggly" rather than "stiff", then bad things happen to stereo separation.

Introducing resistance into ground makes the reference wiggly, thanks to Ohm's Law: V = I * R. R is the resistance and I is the current. The higher the current or the resistance, the wigglier the ground becomes.
post #15 of 38
Thread Starter 
Quote:
Originally Posted by amb View Post
By "stuff" I mean that the ground voltage doesn't vary with the amount of current flowing in/out. It should ideally be 0V at all times because all other voltages are referenced from it. Given that, it shouldn't matter that the grounds are shared, because the return current current from one transducer will simply be "sunk" into ground rather than being "seen" by the other. As soon as the ground becomes "wiggly" rather than "stiff", then bad things happen to stereo separation.

Introducing resistance into ground makes the reference wiggly, thanks to Ohm's Law: V = I * R. R is the resistance and I is the current. The higher the current or the resistance, the wigglier the ground becomes.
Bah, I can understand digital circuits well, but I always had trouble wrapping my brain around analog circuits . We'll have to change that because this is pissing me off hehehe ..

I also understand Ohm's law but I absolutely don't understand how the ground can be a stiff reference at 0V and make possible AC current circulation. Wouldn't the ground have to switch to a positive potential to for the signal to reverse and the current to flow in the opposite direction so that we can have AC through the driver?

How does one alternate between these states gracefully?

+ -------Ldriver (coil) ----\
.....................................--------(negative)
+ -------Rdriver (coil) ----/
__________________________________________

(negative)----Ldriver (coil) ----\
............................................------ +
(negative)----Rdriver (coil) ----/

__________________________________________

+-------------Ldriver (coil) ----\
............................................------ -/+
(negative)----Rdriver (coil) ----/

__________________________________________

(negative)----Ldriver (coil) ----\
............................................------ +/-
+-------------Rdriver (coil) ----/

Or maybe they don't?

edit: I just had a flash. If we have a stiff 0V reference for the ground and then if the + tab of the driver was to alternate between a positive and negative voltage, this would effectively reverse the current without affecting the potential at the 0V reference?

But then I still don't understand this case:

+5V -------Ldriver (coil) ----\
.........................................-------- 0V reference
-5V -------Rdriver (coil) ----/

I see that current would want to circulate in both direction and that no current would circulate at all? To push this idea further, if the voltages were to stay in sync, but one positive and one negative, wouldn't we have a cancelling wave with nothing coming out of the drivers?
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