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Do headphones run on AC or DC? - Page 2

post #16 of 51

The last graph represents exactly the circuit diagram of 95% of the power amps. Note the common ground! It would have been useful to draw both ground connectors at the inside, both plus connectors at the outside, since most amps are designed that way.

 

My own power amp (Metaxas Solitaire) is an exception, since it uses bipolar transistors, thus true balanced drive.

.

post #17 of 51

JaZZ, I don't know anything about your amp, but the fact that it uses bipolar junction transistors (BJTs) is nothing special in itself indicating that it uses a balanced drive output.  BJTs are very versatile semiconductor devices used in all kinds of devices including many op amps.  Bipolar refers to the configuration of the three doped regions inside.

 

OP, as said by others, headphones and speakers do work the same way, and they're wired the same too.  (Well, to be more precise, powered speakers have their own amplifiers internally, etc., but the main difference is just the level of the power input and output.)

 

I think you should just consider the voltages in the system rather than the currents because the voltages are easier to understand.  In the end, the current through the headphone drivers is just the voltage across them divided by their impedance anyway.

 

What the headphone drivers do is convert the electrical voltage across them into vibrations matching that pattern.  The voltage across them is the audio signal (which is whatever the source is sending), and it changes rapidly over time.  The audio signal is mathematically just the sum of sinusoids at different frequencies with different amplitudes, changing over time.  Therefore, half of the signal will be a positive voltage, and half will be a negative voltage.  As others have said, the average of the signal, aka the DC component, should be zero or else the headphone drivers will be damaged.  Since this signal is changing positive to negative, you could say that it is alternating.

 

Here is what some music looks like, as taken from a clip I loaded in Audacity.  The top is the left channel output, and the bottom is the right channel output.  Note that the whole thing may be scaled up or down by multiplication by the amplifier, so this doesn't completely correspond to the voltage seen across your headphone drivers.  However, the shape of the voltage will be the same as what you see here:

Audio waveform

 

Again, this signal is the voltage seen across the headphone drivers.  In other words, the top line represents the difference between the L audio signal and ground, while the bottom line represents the difference between the R audio signal and ground.  Ground is at a 0 signal level, so the difference between the signal and ground is just the signal itself.  The ground can be shared and physically connected together because it's just 0.  It's the same reference point for both L and R audio signals.

 

The current from point A to B is the voltage difference between A and B divided by the impedance between the two points.  When the signal and thus the voltage is positive, the flow of electrons will be in one direction.  When the voltage is negative, the flow of electrons will be in the opposite direction.

 

Crosstalk is the phenomenon of what's on one channel being reproduced on the other channel.  So what one channel sees with crosstalk is the correct signal added to an attenuated copy of the other channel's signal.

 

Hopefully this explanation is more clear to you.


Edited by mikeaj - 7/23/10 at 5:35pm
post #18 of 51

Black cable is set at 0V, it's the voltage in the red cable that varies. It think, please correct me if I'm wrong.

post #19 of 51
Originally Posted by mikeaj View Post

JaZZ, I don't know anything about your amp, but the fact that it uses bipolar junction transistors (BJTs) is nothing special in itself indicating that it uses a balanced drive output.  BJTs are very versatile semiconductor devices used in all kinds of devices including many op amps.  Bipolar refers to the configuration of the three doped regions inside.


So I should have said bipolar transistors at the output stage... It effectively has no common ground, the ground is separated from the minus poles.

.
 

post #20 of 51

Why should there be severe crosstalk? Both speakers have a high impedance, the ground wire almost 0.

Which path would you choose if you were an electron? :D

 

The ground wire might "carry" double the current that is flowing through both speakers, or 0, depending on the L/R signal/voltage, so?

 

The only case you'd get into trouble is if the ground wire is very thin and very long.


Edited by xnor - 7/24/10 at 6:51am
post #21 of 51

can't you guys posting beyond your knowledge find a elementary "how things work" site to check your "facts" [edit: the post's got better even as I wrote so recent posters needn't get bent out of shape by this comment]

 

 

while "AC" literally stands for alternating current it is used more generally - alternating voltage is of course implied, and any signal that averages to zero may be described as "AC" - even when voltage and current aren't involved

 

sound is "AC" - sound waves propagate as alternating compression and rarefaction of the local air pressure - the average of the air pressure is not detected as sound - is not recorded and is not reproduced on playback - only the the "alternating" part of the air pressure (and velocity) is "audio signal" - usually only the AC components with frequency > 20 Hz (and less than ~ 20 KHz) are considered "audible"

 

dynamic transducers (most headphone drivers) create "alternating" sound pressure as their diaphragms move back and forth in response to the alternating audio frequency current in their voice coils - the average of the current and voltage should be as close to zero as possible - any small "DC" component to the drive signal forces the voice coil/diaphragm off center and may increase distortion; large DC signal may over heat the voice coil, melting plastic or glue and destroy the headphone

 

electronic amplifying devices amplify small audio input signals by modulating a DC, constant polarity power source in response the control signal - some circuits/devices have a DC component on their output that must be blocked/removed before the amplified alternating current/voltage signal reaches the headphone

 

the amplifier circuitry may get its power from a "DC" battery source or it may use the wall outlet "AC" mains line current and convert that into the DC needed by its internal signal amplifying components

but the signal reaching your headphone drivers is "AC"

 

 

a common gnd wire (and the standard TRS jack plug) does cause some L/R channel crosstalk - typically it is very low since the wire resistance is a small fraction of an Ohm and the headphone drivers many 10s of Ohms


Edited by jcx - 7/24/10 at 10:26am
post #22 of 51

I think of it like water.

 

All the pipes are filled with water.

Remember, the electrons are -always- there, even when the cable in unplugged and on the floor, they are just moving.. (or in this case, since it's alternating current they are moving back and forth.)

 

In the case that pressure is put to the water, it will go where there is least resistance.

It can't flow through the speaker unless there is equivalent positive charge on the other end.

 

 

I am a computer scientist.. so who even knows if this makes sense =p

post #23 of 51

The analogy with water, pipes with different diameters and pressure (water tanks) works pretty well. A nice introduction to electronics for laymen.

post #24 of 51
Thread Starter 

It's a series of tubes! 

 

I think I finally get it now. My problem was that I thought that both the red and black speaker wires would take turns carrying the active signal. In other words, during the + phase, the red wire would push electrons towards the speaker. And during the - phase, the black wire would push the electrons. And seeing how the black wires are crossed, I figured that the L and R electrons would get mixed together, creating crosstalk.

 

But now my understanding is that only the red wire carries the active signal. During the + phase, it pushes electrons, and during the - phase, it pulls electrons. The black wire just goes with the flow.

 

Is that right? Thanks for helping me with this everyone.

post #25 of 51

Really, do some searches to learn about  this stuff, it's a fascinating and enjoyable subject.

You coudl get drawn into the dark side..........DIY

post #26 of 51

Regarding your wiring diagram above - both are correct. The common terminal, is tied to the chassis ground of the circuit.  Any voltage within the amplifier circuitry - INCLUDING THE RED SPEAKER TERMINAL - is measured in terms of the difference above (or below) this circuit ground. For example, when you measure a voltage on a point in the amplifier, the black lead, or ground lead of your volt meter would be tied to ground. The red lead would be placed on the point in the circuit that you want to measure. If you are measuring a DC point in the circuit, say a positive 12 volt supply, your meter would read +12VDC. You can also have negative points in the circuit, which simply means that the voltage is at some voltage level BELOW the circuit ground, which is considered 0 volts.

 

To understand AC, think of the same scenario where the circuit ground is at 0 volts. In the image below, the ground reference is the straight line. The curvy line is the voltage signal, showing strength (amplitude) on the vertical axis, and time on the horizontal axis. If you were to hook up an oscilloscope (think visual voltmeter) to an AC outlet in your home, this is exactly what you would see.

 

 

images?q=tbn:ANd9GcS4U153mibcAwtWECDetArh70nZDw7GwjZvIMgSA2w0bmV4bgk&t=1&usg=__K-vXtbuxbmPbX13wP-v3ty-f5ik=

 

 

Now for your speaker question. Both the left and right channels require a common ground and the signal lead. The ground lead (black in your diagram) is common to both left and right channels. The red leads in your diagram carry the musical signal and are different between the two channels. The signal itself is an AC current, meaning that it alternates direction. If you were to look at an audio signal on an oscilloscope, you would see that it moves both above AND below the 0 volt ground reference constantly. At any given point in time, the voltage may be above zero volts, or below zero volts. This voltage swing dictates not only current flow, but direction of current flow as well. When you hear a term like 1000 hertz, that means that the voltage is moving above and below the zero reference level at a frequency of 1000 times per second.

 

To answer your question simply - speakers are powered by alternating current. If the signal lead has a voltage above ground, the speaker cone moves forward. If the signal lead has a voltage that is below ground, the speaker cone moves backward. If there is no voltage on the signal lead, the speaker would be at its resting position. So you see, by varying the frequency, we can control how FAST the cone moves (pitch), and by varying the amplitude, we can control how FAR the cone moves (volume).

 

Hope this helps.


Edited by FlatNine - 7/24/10 at 4:38pm
post #27 of 51

Sorry I'm digging out this thread, but I have a simple question:

 

So, if I click "play" to this signal, am I gonna cause damage to my speakers?

 

soundsignal.png

(made this on Soundforge)

 

This is a DC current, right? Is my soundcard really going to output a DC signal if I play this?

post #28 of 51
Quote:
Originally Posted by Vitor Machado View Post

Is my soundcard really going to output a DC signal if I play this?


If it does, then yes it could damage your speakers. Just try it out! Use a loopback cable to record the soundcard's output. If you're on windows ensure that the "DC offset cancellation" option is disabled for your recording device (line-in).

post #29 of 51

Depends on if you're in the time domain or frequency domain.  I can't tell because I can't see the X-axis of your graph.

post #30 of 51

Quote:

Originally Posted by tlniec View Post

Depends on if you're in the time domain or frequency domain.  I can't tell because I can't see the X-axis of your graph.

 

It's clearly a waveform with a DC signal at -4 dBFS.

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