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Struggling with questions about balanced vs. single ended audio...

post #1 of 9
Thread Starter 
I was reading through steaxauce's thread "DIY balanced amp for Edition 9", where there was an interesting discussion about balanced vs. unbalanced audio, and it bugged me because I have been struggling with exactly what an audio signal is. I am a newb at this, and I have been reading a lot recently to get a better idea of how amplifiers work, starting with the Art of Electronics by Horowitz and also reading some of Self's books and articles, as well as several other books on audio and amplification. I am learning how to understand circuits and schematics, what each of the components do, etc., but I have yet to find a really clear definition on what a speaker "sees" from the amplifier. If I am wrong about this, please let me know; I learn best by asking questions and being wrong with my initial understanding of these kinds of complex problems.

I was under the impression that, effectively, a speaker "sees" two potentials at its inputs: the ground wire and the "live" wire. I thought that basically an audio signal was produced by holding the ground wire at a constant, reference potential and swinging the potential of the "live" wire with respect to the ground wire, creating a different voltage drop across the speaker. Current then flows through the speaker according to Ohm's law, which drives the speaker and produces the pressure changes in the air surrounding the speaker that we perceive as sound. The job of an amplifier then is to increase the amplitude of these voltage swings, and thus increase the current at any given time pushing through the speaker, and make the sound louder.

If I am close to being right about that then I can't wrap my head around how balanced headphones work. I thought in the typical balanced setup, you have three wires for a specific reason. I thought that effectively you have 2 "live" wires who push the audio signal with respect to the reference ground wire, only they push the signal 180 degrees out of phase to each other. Then, once the signals reach the speaker, their is some kind of active circuit that puts the two signals back in phase with each other. Theoretically, both wires would pick up the same noise as the signal is "sent" to the speaker. However when the two signals are returned to being in phase with one another, the noise is now 180 degrees out of phase with itself. When both signals are then played through the speaker the two audio signals undergo constructive interference, effectively making the overall audio signal much stronger, while the noise undergoes destructive interference, effectively removing any noise that was generated in the cable.

How then does this work for balanced headphones? There is no ground reference for both signals, only and L+ and L- for the left channel, and similarly only R+ and R- for the right channel. Also, there is no active circuit that recombines the two out of phase signals. It seems to me that if there *was* a ground wire, with no active recombination of the signal and both feeding into the headphone driver, the signals would completely cancel each other out and you would only get the noise picked up in the cable.

I know that I have something really wrong in my analysis, but I can't figure out what, exactly. I would really appreciate someone with the know how explaining exactly what is going on at a headphone driver in both the single-ended and balanced cases.


Edit: I realized part of my problem in thinking about this is that I am basing my conclusions about the noise canceling for the balanced configuration on sine waves instead of the actual transient nature of audio. nevertheless I am still very confused on exactly what a balanced signal is and how it works to reduce noise, or even if I am thinking about how a single-ended device works properly.
post #2 of 9
Your explanation is pretty spot on!

The only thing is, there is no need to convert signals back to single ended in order to drive the speaker - you can drive the speaker with a differential signal. However, unless I'm much mistaken, no noise cancellation happens this way. In fact, noise will interfere constructively in this fashion (ie it will be applied in the same way to both signals and they don't go through an opamp to subtract the noise or anything like that).

The reason differential speaker driving is done, in my experience, is to get more output power with the ever decreasing voltage supplies in portable devices. If you have a +/-5V power supply and drive an 8 ohm speaker with a single ended signal, the peak power you can get is V^2/R = 5^2/8 = 3.125W. If you're using a differential signal with the same power supply and same speaker, you can put 10V across it, which gives 10^2/8 = 12.5W.

With regard to your edit, you're quite right to note that audio isn't a sine wave, however pretty much everything including this works just the same for real music as it would for a sine wave.
post #3 of 9
Note that on a device such as a speaker or a headphone, using balanced cables won't do anything for noise as any signal strong enough to induce enough voltage to be audible will pull the fillings out of your teeth.

Balanced cabling is essential for microphones because you have long cable runs with very low signals. Induced signals can overwhelm the audio, but by balancing you reduce this to nil. In line-level signals balancing is less important but it's still cheap insurance. Most professional audio equipment uses balanced inputs and outputs, because they're often in electrically noisy environments.
post #4 of 9
Thread Starter 
But how does it work? What I mean is, you have 2 different signals, L+ and L-, but no reference ground. To me it seems like you have two signals with zero returns, which simply doesn't work at all.

Anyway, I am fairly accurate on how the single ended system works then? It's bothered me that I couldn't really define exactly what is going on at a driver.
post #5 of 9
Since noone has answered yet, I will see if I can describe it so it makes sence. You were doing good with the balanced circuit with the understanding that the '+' and '-' signals are 180 degrees out of phase, which is good for microphone signals and line level signals, especially for long distances ( as LC pointed out ). However when you are talking powering speakers or headphones the last step you spoke of where the '-' signal is inverted and then combined with the '+' to cancel out common mode noise, does not happen. The '-' is the reference ground you are looking for. So if the '+' is +1V and the '-' is -1V, you would actually get a difference of 2V at the driver. Does that clear it up at all? Balanced audio can be tricky to understand.
post #6 of 9
Current flows when there is a different in voltage potential. Whether this is between a positive source and ground or between a positive and negative source does not matter.

A balanced setup also has twice voltage swing, twice the slew rate, and four times the total power.
post #7 of 9
Thread Starter 
*Smacks head* Thanks, Gross, a light just went off. Part of my problem is that I was thinking about taking a wave, like a sine wave, inverting it, then adding the inverted wave to itself. This gives you a straight line... ie., no signal. But that's NOT what a driver does, is it? Its more like taking the L- inverted signal and "forcing" it to be constant. Since a voltage drop is just the potential difference between 2 parts of the circuit this basically means that "forcing" (in the mathematical sense, not the physical one) the L- wire to be a reference constant potential effectively doubles the voltage drop from the L+ to L- wires compared to the single ended case where ground is actually held (relatively) constant. Makes SO much more sense now, just didn't really think correctly about how the driver "sees" two wires on either side of itself carrying a signal... in effect it sees what it always sees, a voltage drop of some kind from the + to the - side of the driver. I guess balanced configurations just allow a larger voltage swing (is that what slew rate is, btw?) than an analogous single ended system.

Edit: Thanks also Cauhtemoc. And I think I understand slew rate now... voltage swing is basically saying it can have twice the amplitude of single ended transmission, right? Slew rate is effectively how fast an amp can change the voltage swing. Yay. Learning is fun!
post #8 of 9
technically speaking, the only thing that's actually necessary for something to be balanced is that the impedance for the two connections is the same. This would allow any noise to be induced equally on the output and return lines, thus making removing the noise as seen by the speaker/headphone. So the signals don't actually have to be inverted and you don't always need two copies of an amplifier for a balanced configuration.

Practically, though, it usually ends up that the easiest way to get everything to work is to use inverted signals. A ground is also still used, but it's completely isolated from the signal and only used to shield out a bit more noise.
post #9 of 9
Originally Posted by Gross View Post
You were doing good with the balanced circuit with the understanding that the '+' and '-' signals are 180 degrees out of phase [...]


This is not entirely correct. 180 degrees out of phase means the signal is delayed in time, equal to half a period of the signal ("moving" the signal half a wavelengt back).

The correct term is inverted, where the signal is simply flipped upside down. This is a common mistake, as an inverted sine and a sine 180 degrees out of phase look identical.

But if you consider an actual audio signal, the difference is big. Inverted means positive is negative and the other way around, and you have two identical signals that always have the same value as each other, only opposite. Combining the two will always result in zero.

A 180 degree phase shift will mean that the signal arrives slightly later than the one with no phase shift. Combinig these two signals will not give zero for all values, but will be a distorted jumble with no use. Different signals will give different, but equally useless jumbles. What's more, the phase shifted signal will (theoretically) not be identical to the original, as the time delay (equal to half a wavelength) depends on the frequency at any specific point in the signal. Low frequencies will be delayed more than highs, thus distorting the sigal.


I know this is nitpicking, and I assume you were talking about the concept of an inverted signal, merely using the wrong terms.

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