How does AC in a stereo headphone work?

Discussion in 'Homework Help' started by LX2010, Jun 15, 2010.

  1. LX2010

    Thread Starter New Member

    Jun 15, 2010
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    Hi there, sorry if this is a silly question but i was stumped on it for quite some time hoping you guys can help me out over here.

    As you guys know, there are 3 wires in a headphone, Left , Right and Ground.

    As far as i know (might be wrong,) AC is used to produce sound from the vibration of the voice coil in the headphone caused by the repulsion/attraction from the magnet in the headphone.

    My question is this.. since Left and Rright have different currents to produce different sounds, and that L and R share the same ground wire (saw that when i cut up my earphones), is it true that coupling of the currents occurs?

    However, i don't think that's the case since if coupling occurs then both L and R would play the same sound. So how exactly does AC work such that both L and R play different sounds and yet share the same ground cable?

    As of now, i'm guessing that there's some kind of short circuit going on which prevents the audio from L and R to mix. However, since AC is non directional (correct me if i'm wrong), does coupling still occur at the point where the ground wire combines?

    i drew a pic for easier visualization..
    [​IMG]
    L is the red wire, R is the green wire and yellow is the ground.

    Would really appreciate it if you guys could help me out. :) (i might have quite a few misconceptions here so if you guys spot any, feel free to point them out.)

    Thanks again!
     
  2. Georacer

    Moderator

    Nov 25, 2009
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    The two channels are actually independent! (that is the required condition for operation at least). Think about two lamps, one wich runs at 5 volts, and one at 220 volts. At first imagine them works as two separate circuits, each with its own voltage source. You can define one wire to be the ground for each circuit, that's arbitrary. Now connect these two ground wires in a single point. The lamps now are still powered by their respective voltage source, but this time, they have a common ground, wich doesn't affect their operation.
    Check my attachment. Notice that the different volatage applied in each lamp represent the different audio signal each speaker can carry.
     
  3. Markd77

    Senior Member

    Sep 7, 2009
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    The important thing is that the voltage of ground is always zero.
    The voltage of L and R fluctuate either side of zero and the current in the headphone is proportional to that voltage (more or less - it's an inductive load).
    If both channels were identical sine waves and in phase then the current in the ground wire would be double that of each channel. If they were perfectly out of phase then there would be no current in the ground and it would flow from L to R and vice versa.
     
  4. LX2010

    Thread Starter New Member

    Jun 15, 2010
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    So essentially, is it true to say that all this works because the ground wire has 0 voltage?

    I'm still slightly confused on why the AC from both channels do not mix. Is it because of the ground wire, which has 0V, which is present?

    I read somewhere that...

    The term "ground" can mean..

    1. A 'common' connection, but not connected to Earth
    2. A connection to the power supply (usually to the negative terminal
    3. A connection to the inside of a shielded metal box
    4.A connection to a metal stake driven into the earth (or a connection to a metal water pipe which extends out of the house into dirt.)

    However, it is only number 4 which is connected to the ground.


    In the case of the headphone, which definitions of ground apply? Also, is it possible to make the ground wire constantly have 0 voltage without connecting it to the ground?

    Once again, sorry if the questions here are silly. My class just touched on AC not too long ago and i'm reading on because i'm quite interested in this. Thanks for your time!
     
    Last edited: Jun 16, 2010
  5. sceadwian

    New Member

    Jun 1, 2009
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    Coupling of the L and R signals together is often called crosstalk, or common mode interference. It is almost always inaudiable in headphones, but not always so in stereo leaders from signal level equipment.
     
  6. Ghar

    Active Member

    Mar 8, 2010
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    There definitely is coupling.

    The wire has a finite impedance and this couples the channels together.
    However, the impedance of the wire will be less than the speakers letting the sound be fairly independent.

    stereo.png

    The 'ground' voltage (between voltage sources) may be zero but the 'ground' at the speakers is not 0, it is (IL + IR)*Z

    The thing is that the speaker impedance is high (at least a few ohms to a few hundred ohms) while the wire impedance will be less than an ohm. The wire's resistance will be much less than ohm (see 28 AWG for example, 65 milliohms per foot, http://www.powerstream.com/Wire_Size.htm) and the inductance will be a few hundred nanohenries, with 500 nanohenries being a mere 63 milliohms at 20 kHz (the extend of audio).
    That ~0.5 ohms is small compared to an 8 ohm speaker, making the coupling small but it still exists. Headphones are generally much higher, like 150 ohms making the coupling less significant.

    If you want an equation, this would be it for my schematic there:

    Voltage across left speaker:
    V_{Left speaker} = \frac{Z_{spk}}{Z + Z_{spk} + Z||(Z + Z_{spk})} V_{LSrc}+ \frac{Z||(Z + Z_{spk})}{Z||(Z + Z_{spk}) + Z_{spk} + Z}(\frac{Z_{spk}}{Z_{spk} + Z})V_{RSrc}

    Plug in the numbers I made up:
    Z_spk = 150
    Z = 0.5

    Obviously Z << Z_spk
    Simplify:
    V_{Left speaker} = \frac{Z_{spk}}{2Z + Z_{spk}} V_{LSrc}+ \frac{Z}{2Z + Z_{spk}}V_{RSrc} \\<br />
V_{Left speaker} = 0.993 V_{LSrc} + 0.003 V_{RSrc}

    The coupling is pretty small in this case... it's actually -50 dB.

    If you lower Z_spk to 8 ohms and make the wires super long to increase Z to 1, you would get:

    V_{Left speaker} =  0.8 V_{LSrc} + 0.08 V_{RSrc}

    The coupling went up to -22 dB and you're losing some amplitude even (it's only 0.8 for the source you want)

    I simply wouldn't call it a ground, it's the signal return. The word ground gets confusing and abused. A ground is not meant to carry any current.

    Edit:
    Err these calculations ignore phase (purely resistive). It works fairly well when Z << Z_spk but it drifts further away otherwise. The non simplified equation is correct however.
     
    Last edited: Jun 15, 2010
    LX2010 likes this.
  7. Georacer

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    Nov 25, 2009
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    Why does ground have to carry no current? Do you refer to the specific application or talking generally?
    My conception is that since voltage level is not independently defined, but is relative, in any circuit we need a voltage level as referrence for all the other voltages. This is the "electronic" definition of ground.
    There is of course the electric power system's ground wich is driving a node's voltage to the earth in order to avoid this specific node obtaining voltage relatively to the earth (and ourselves consequently), but in the end of the day, this is a common referrence point too.
     
  8. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Ground is often the return. It's tough to get around kirchoff's current laws.
     
    Last edited: Jun 16, 2010
  9. Ghar

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    Mar 8, 2010
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    Ground, as in earth ground, is for safety where it makes sure there is no large voltage between the equipment and the environment. It is only supposed to carry current during a fault. Within a non-earthed system it is meant to prevent voltage differences across your different boxes etc. since you can still have a shock hazard.

    Ground, as in signal ground, provides a reference for your circuit. Having current flowing through the non-zero impedance of your ground makes it a poor reference. Putting in a 'ground' plane or grid significantly reduces the impedance letting it stay a decent reference.
    Differences in reference voltage, which are guaranteed if you run current through the reference conductors, cause common mode current.

    I'm talking generally about grounding... it has a different objective than returning current, though as was said it's not that obvious how to do it.
     
  10. JoeJester

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    In this stereo system, what they are calling "ground" is really the return, and it carries current ... unless you wish to revise Kirchoff's Current Law.
     
    Last edited: Jun 16, 2010
  11. Ghar

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    How is this at all related to Kirchoff's law?
    What I said is that the word ground is abused and this shouldn't be called a ground. Like you said "it's really the return"
     
  12. JoeJester

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    Whoever said there was no current in the return is sadly mistaken. Granted it will be zero potential if you use it for both the positive and negative probes, which is what you do when you connect the oscope neg lead to the chassis or signal ground.

    How is it related to Kirchoff?

    You have two outputs sharing a common return. The speaker(s), Left and right, are in series with the L or right source. Are you suggesting the current flowing in that return is zero?
     
  13. Ghar

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    No one said there is no current in the return.
    I said a ground is not meant to carry current and the word ground should stop being used to refer to a return because it confuses everybody.

    Am I suggesting the current in the return is zero? You're seriously asking me that? I'm the one who wrote the equations and drew a diagram showing the current in the return.
     
  14. JoeJester

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    Then my comments weren't directed at you ... but others.
     
  15. Georacer

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    This is a little weird: In my college, (and pretty much everyone I have met outside it), we use ground as a means to create a reference point for the voltage in digital and electrical circuits, and by all means, it is allowed to carry as much current as it needs for the circuit to work. That of course doesn't mean that we connect the circuit to the deeps of the earth, rather than call "ground", this common node for voltage reference. Is that true in anyone else's country or college specificaly?
     
  16. retched

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    Dec 5, 2009
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    I have heard ground used as many things.

    I have considered ground like ground floor.

    No matter how many stories you rise in a building, the ground floor is always at 0ft

    All the people that enter the building to get work done, always leave at the ground floor. (Unless something VERY BAD happens)

    When you measure how high you are in the building, it is being measured from the ground floor.

    If you are at sea level, and your on the 10th floor, you are x feet in the air above the ground. (not sea level)

    If you are in Denver, Colorado, the ground floor is 5280 ft up. But according to the building, you are still only x feet above the ground.

    the ground is your reference point you do your measuring from, and the point you leave. Same as in a circuit.

    So if you jump off a 100ft tall building in Denver, you will fall 100ft before you hit the ground.
    If you jump off a 100ft tall building in New Orleans, (Which is negatively biased, by the way) you still fall 100ft before you hit the ground.
     
  17. Georacer

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    I looove metafors, and officially from today, this is one of my favorites!
     
  18. Ghar

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    Don't get me wrong, it is definitely a common usage and pretty much everyone uses it.
    The problem is that it's confusing and it leads to people doing strange things. People think 'ground' has some special property and imagine that it can suck away noise or be this nice solid reference or a magical safety device just because they label it as a ground in the schematic.
    It's a conductor like any other and this needs to be kept in mind.
    For many purposes you don't need to worry but get into high frequency or high performance circuits and you will run into trouble because generally a return is not a good reference.
     
  19. Georacer

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    You are right about misuses of the term ground. But I think that if you draw your schematic and keep true to the Kirchoff's laws, no matter how you call your return, you can't go wrong.
     
  20. BillB3857

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    Feb 28, 2009
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    If you fed identical signals to both the left and right earphones, the common would carry twice the current as compared to either the left or the right feed wire. If you were to send exactly opposite signals (180 degrees out of phase, same frequency), the common wire would carry no current. The current through the common wire, while playing normal stereo, will represent the differences in phase and frequency between the two channels.
     
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