is pull-push impossible for non-balanced headphone?

[1096bimu]

I understand what balanced is, and I always thought it should have no effect because the voltages that drive the headphone are way higher than any interference you could pick up, so the "interference resistance" part of it should be useless.

However, it has just occurred to me that pull-push signals should not be the same as a pull/push only signal? I tried looking for some answers but most results are still just people talking about what balanced is, so I thought it would just be easier to ask.

The way I understand this, a speaker would have a neutral position it sits in when unpowered, if you apply voltage in one direction it goes one way, another direction the other way.

For a normal unbalanced connection, you have one ground at 0v, and then the two wires for both drivers would have to have vary between 0 and the same peak voltage? Because for example if you have +5 for left, 0 for ground and -5 for right, then the current would just flow across both drivers rather than into the ground? But if both sides are pull-only then you can use diodes to prevent current crossing over both drivers.

This would mean the audio signal is pull only (or push only), so if I reach a maximum voltage and then the wave is supposed to come down, the voltage can drop but there's no force actually pulling the driver back to its rest position other than mechanical spring force from its construction? Wouldn't this introduce distortions in the audio? This would mean only the peak positions of the signal would be somewhat consistently reproduced, where as the dips would rely entirely on the driver's mechanical properties?

Wouldn't this also mean when playing a silent signal, there is still power passing through the drivers? Because if peak voltage is say 4v then silent should be a constant 2v not 0v?

However if you have a push/pull signal, once a wave maxes out and has to go the other direction, you would reach the opposite max voltage, and thus an electromagnetic force will try to pull the driver back into the correct position, and this should theoretically make the diaphragm better follow the desired wave motion?

And since the only way to have push/pull signals in a headphone is to have balanced wires, shouldn't that therefore make a difference? even a significant difference?

I also understand this question would be instantly answered if I had a oscilloscope and a balanced amp to check, but I don't have those. Essentially the question is whether the signal voltage from a unbalanced amp ever crosses 0v, if so how would that work over a common ground. And, does the signal from a balanced amp cross over 0v and if so wouldn't this make the headphone sound somewhat different?

 

[staticV3]

However, it has just occurred to me that pull-push signals should not be the same as a pull/push only signal?

A speaker driver has just two electrical connections and moves according to the voltage differential between them.

It doesn't matter if it's +2V and -0V or +1V and -1V.

All the driver sees is a 2V differential, hence "push-pull" serves no purpose.

The idea that one side pushes and the other pulls and therefore single-ended drive only has half the "force" is a misconception.

A positive differential pushes while a negative differential pulls.

Whether that differential is made up of signal and ground or hot and cold is irrelevant.

 

[unpluggged]

I understand what balanced is

If so, you would understand that the term "balanced" when applied to passive headphones is a misnomer. They don't have a balanced circuit at their inputs.

so the "interference resistance" part of it should be useless

You refer to common mode rejection, and it is achieved by differential amp subtracting the inverted signal from the non-inverted signal at the balanced input. Nothing of this sort exists at the input terminals of passive headphones, as I said in the previous paragraph.

"Balanced" when used in context of headphones refers to the amp's output configuration, specifically to differential or bridged amplification, where instead of a passive ground the amp generates a signal that is identical to the input, but inverted. Hence we get a doubling of the amplitude, though it is achieved not by actively subtracting one phase from another, but by the voltage between the + and − speaker terminals itself. The passive speaker does not care how the voltage is generated, by a single-ended amp or by a differential/bridged one.

 

[staticV3]

You refer to common mode rejection, and it is achieved by differential amp subtracting the inverted signal from the non-inverted signal at the balanced input. Nothing of this sort exists at the input terminals of passive headphones

Loudspeaker drivers inherently reject common-mode signals, no special circuit required.

Impedance-balance at the Amplifier output ensures that EM/RF noise is induced equally into each signal wire.

 

[MaxwellsEq]

I understand what balanced is, and I always thought it should have no effect because the voltages that drive the headphone are way higher than any interference you could pick up, so the "interference resistance" part of it should be useless.

However, it has just occurred to me that pull-push signals should not be the same as a pull/push only signal? I tried looking for some answers but most results are still just people talking about what balanced is, so I thought it would just be easier to ask.

The way I understand this, a speaker would have a neutral position it sits in when unpowered, if you apply voltage in one direction it goes one way, another direction the other way.

For a normal unbalanced connection, you have one ground at 0v, and then the two wires for both drivers would have to have vary between 0 and the same peak voltage? Because for example if you have +5 for left, 0 for ground and -5 for right, then the current would just flow across both drivers rather than into the ground? But if both sides are pull-only then you can use diodes to prevent current crossing over both drivers.

This would mean the audio signal is pull only (or push only), so if I reach a maximum voltage and then the wave is supposed to come down, the voltage can drop but there's no force actually pulling the driver back to its rest position other than mechanical spring force from its construction? Wouldn't this introduce distortions in the audio? This would mean only the peak positions of the signal would be somewhat consistently reproduced, where as the dips would rely entirely on the driver's mechanical properties?

Wouldn't this also mean when playing a silent signal, there is still power passing through the drivers? Because if peak voltage is say 4v then silent should be a constant 2v not 0v?

However if you have a push/pull signal, once a wave maxes out and has to go the other direction, you would reach the opposite max voltage, and thus an electromagnetic force will try to pull the driver back into the correct position, and this should theoretically make the diaphragm better follow the desired wave motion?

And since the only way to have push/pull signals in a headphone is to have balanced wires, shouldn't that therefore make a difference? even a significant difference?

I also understand this question would be instantly answered if I had a oscilloscope and a balanced amp to check, but I don't have those. Essentially the question is whether the signal voltage from a unbalanced amp ever crosses 0v, if so how would that work over a common ground. And, does the signal from a balanced amp cross over 0v and if so wouldn't this make the headphone sound somewhat different?

Welcome to ASR.

Non-active headphone and loudspeaker drivers have no knowledge of earth or ground. They "float". This means you could operate your headphone amplifier output socket at +500V DC on both send and return and modulate this with up to a couple of volts of AC signal. Everything would be fine, unless you touched the wires.

 

[1096bimu]

Thanks for the answers, I picked up a cheap but good DAC/amp with balanced output and it sounded exactly the same as normal 3.5mm, probably unsurprisingly, after all the years of trying to rationally enter the hifi world.

I still don't fully understand it though, I'm thinking a driver diaphragm surely relies on spring force to center itself? if you made a driver that's just a linear motor with a free floating diaphragm it wouldn't work with a signal that only goes one way? it would just push the diaphragm out further and further away?
But if you have a "balanced" push-pull signal, and assuming everything is linear, and the diaphragm weighs nothing etc, it should work?

A speaker driver has just two electrical connections and moves according to the voltage differential between them.

It doesn't matter if it's +2V and -0V or +1V and -1V.

All the driver sees is a 2V differential, hence "push-pull" serves no purpose.

The idea that one side pushes and the other pulls and therefore single-ended drive only has half the "force" is a misconception.

A positive differential pushes while a negative differential pulls.

Whether that differential is made up of signal and ground or hot and cold is irrelevant.

ok but if I take a driver and simply connect a 2v DC to it, the diaphragm would not remain in the same position as no battery right?
so in that sense the center position is not the same when your signal is 2-0v versus +1-1v?

I'm thinking maybe I'm not understanding drivers correctly, maybe it's one of those AC only devices, so if you put +2v to it the diaphragm only momentarily moves and then immediately springs back to 0? But even if that was the case, if you only have + voltage then it can only go one way, it will never be forced to move the other way from the neutral position? Surely that should make a difference?

 

[staticV3]

ok but if I take a driver and simply connect a 2v DC to it, the diaphragm would not remain in the same position as no battery right?
so in that sense the center position is not the same when your signal is 2-0v versus +1-1v?

+2V to 0V and +1V to -1V will result in the same driver position.

if you made a driver that's just a linear motor with a free floating diaphragm it wouldn't work with a signal that only goes one way?

A Single-ended signal goes both ways, positive(=push) and negative (=pull).

 

[unpluggged]

a signal that only goes one way?

What is "a signal that only goes one way"? "Goes" which "way" exactly? Sound is oscillations. Oscillations are represented with the same waveform of electric signal (hence it's called "analog"). No oscillations = no sound.

 

[audio_tony]

I still don't fully understand it though, I'm thinking a driver diaphragm surely relies on spring force to center itself?

A speaker (or headphone) diaphragm is held in a 'center' position mechanically, by the spider (and also to a limited extent the surround.

See illustration below. The diaphragm is suspended in such a way that it can move to and fro, but always return to the mechanical center (in the absence of any DC voltage).

With regard to your 'push pull' question. Below is an illustration of a sine wave. The horizontal line through the center represents zero volts.

When the signal travels in a positive (+) direction, this causes the speaker diaphragm to move outwards, and when the signal moves in the negative (-) direction the speaker cone is drawn inwards (this is the 'push, pull' action you are referring to).

As others have explained, a balanced signal works in the same way. There is always a zero (0v) point where the signal 'swings' from positive to negative, whether it's balanced or not.

At the centre (0V or zero volts) - the diaphragm will return to it's mechanical center. I hope this helps.

 

[MaxwellsEq]

Thanks for the answers, I picked up a cheap but good DAC/amp with balanced output and it sounded exactly the same as normal 3.5mm, probably unsurprisingly, after all the years of trying to rationally enter the hifi world.

I still don't fully understand it though, I'm thinking a driver diaphragm surely relies on spring force to center itself? if you made a driver that's just a linear motor with a free floating diaphragm it wouldn't work with a signal that only goes one way? it would just push the diaphragm out further and further away?
But if you have a "balanced" push-pull signal, and assuming everything is linear, and the diaphragm weighs nothing etc, it should work?



ok but if I take a driver and simply connect a 2v DC to it, the diaphragm would not remain in the same position as no battery right?
so in that sense the center position is not the same when your signal is 2-0v versus +1-1v?

I'm thinking maybe I'm not understanding drivers correctly, maybe it's one of those AC only devices, so if you put +2v to it the diaphragm only momentarily moves and then immediately springs back to 0? But even if that was the case, if you only have + voltage then it can only go one way, it will never be forced to move the other way from the neutral position? Surely that should make a difference?

You are mixing up lots of different things. You would gain some benefit from reading some basic books or articles on signals.

You hear things when your eardrum moves in and out. The faster it moves in and out the higher the frequency. If it just moves in and doesn't move out (e.g pressure change in an aircraft), you don't hear a sound, but you may feel a sensation.

A loudspeaker and headphone driver are the same technology as eachother. If they move in and out, they make a sound you can hear because they move your eardrum in and out. If they just move out and stay there, they make no sound (and you would not be able to hear it anyway).

Loudspeakers and headphone drivers are AC (alternating current) devices. An alternating current goes up and down. A DC (direct current) is fixed and doesn't go up and down. A battery is a DC source. If you want to visualise this, a DC circuit has all the electrons hopping along in one direction continuously. An AC circuit has the electrons hopping in one direction and then reversing and hopping in the opposite direction. How quickly this reversing happens depends on the frequency. So the electrons hop in to your driver and pull it inwards, then they hop back out and push it outwards.

A single-ended output on a headphone amplifier is FULLY AC. A "balanced" output on a headphone amplifier is FULLY AC. Because the driver is also fully AC it has no knowledge of what type of amplifier it's plugged into which is why it behaves identically and sounds identical.

 

[MaxwellsEq]

You are mixing up lots of different things. You would gain some benefit from reading some basic books or articles on signals.

You hear things when your eardrum moves in and out. The faster it moves in and out the higher the frequency. If it just moves in and doesn't move out (e.g pressure change in an aircraft), you don't hear a sound, but you may feel a sensation.

A loudspeaker and headphone driver are the same technology as eachother. If they move in and out, they make a sound you can hear because they move your eardrum in and out. If they just move out and stay there, they make no sound (and you would not be able to hear it anyway).

Loudspeakers and headphone drivers are AC (alternating current) devices. An alternating current goes up and down. A DC (direct current) is fixed and doesn't go up and down. A battery is a DC source. If you want to visualise this, a DC circuit has all the electrons hopping along in one direction continuously. An AC circuit has the electrons hopping in one direction and then reversing and hopping in the opposite direction. How quickly this reversing happens depends on the frequency. So the electrons hop in to your driver and pull it inwards, then they hop back out and push it outwards.

A single-ended output on a headphone amplifier is FULLY AC. A "balanced" output on a headphone amplifier is FULLY AC. Because the driver is also fully AC it has no knowledge of what type of amplifier it's plugged into which is why it behaves identically and sounds identical.

The next thing to know @1096bimu about is Single-Rail amplifiers and Push-Pull amplifiers. NB this is NOT the same as single-ended connections (3.5mm, RCA) and balanced connections (4.4mm, XLR).

Single rail amplifiers have the gain element (e.g. a transistor) suspended (biased) between 0 and a voltage rail (let's say +15 V DC). The AC signal swings on its output from +15V to 0V with a DC offset of 7.5V. This DC voltage would cook the loudspeaker so it's removed by a capacitor.

In a push-pull amplifier, a transistor is hung between 0 and 15V DC and another is hung between 0V and -15V DC. When the AC signal is positive it uses the the positive biased transistor and when negative it uses the negative biased transistor. There's no DC component on the output, so no output capacitor is needed.

BUT because a loudspeaker is an AC device, it doesn't know any of this, so it behaves identically and sounds identical with single-ended and push-pull amplifiers.

 

[1096bimu]

At the centre (0V or zero volts) - the diaphragm will return to it's mechanical center. I hope this helps.


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Thanks, I think the only confusion was, for a 3.5mm connector where is that center point, is it still 0v or is it some raised voltage such that only the bottom part of that wave is at 0v.

Now I know it's at 0v, so theoretically there should be no difference between balanced and unbalanced, but practically if you have a USB powered amp it can do up to 10v amplitude with balanced but only 5v on 3.5mm assuming not using inefficient boost converters.