Where is the current return path in balanced audio?

Thread Starter

CurlsOnKeys

Joined Jan 3, 2015
47
Confused confused confused...

The more I read about it, the more confused I get.

Unbalanced audio:
- tip carries signal/live
- sleeve has, as I see it, double function:
--> ties the grounds together, so it makes the grounds of outgoing and incoming device equal (which is necessary if one of both devices is "floating", like a 2-prong powered synth)
--> provides a current return path: since current has to flow in a closed loop, one has to provide a current return path for the audio signal.

Balanced audio (with opamps instead of transformers):
- tip carries signal
- ring carries signal inverted
- ground connects both chassis (thereby also making ground potentials of both devices equal, correct?) and should - as I've read - under normal circumstances not carry any audio current (asside from unwanted noise currents).

So if in balanced audio you have 2 wires carrying the signal (+ and inverted) and 1 (non-current-carrying) wire connecting both chassis and current has to flow in a closed loop, where's the audio current return path??

Got most info from here: http://sound.westhost.com/articles/balanced-2.htm and here: https://www.presonus.com/news/articles/balanced-unbalanced
 

crutschow

Joined Mar 14, 2008
34,420
................
So if in balanced audio you have 2 wires carrying the signal (+ and inverted) and 1 (non-current-carrying) wire connecting both chassis and current has to flow in a closed loop, where's the audio current return path??
................
The statement of 2 wires carrying the "+ and inverted" is a little confusing.
The plus and inverted really refers to the voltage with respect to the circuit ground.
The balanced audio is a differential signal with the signal appearing across the two wires and the receiver detecting the voltage between those two wires.
You can look at either wire as being the "return" for the other.

If the receiver inputs are reversed, then the phase of the received signal would be changed by 180° (inverted).

The receiver is designed to amplify only the differential signal and ignore any voltage common to both lines (such as noise caused by a ground loop) which is why a balanced line is used instead of a single-ended unbalanced line, especially for long cable runs.

Make sense?
 

Thread Starter

CurlsOnKeys

Joined Jan 3, 2015
47
Ah, yes, I think I get it!

By inverting one of the signal voltages, you basically allow that inverted path as a way for the signal current to return?
If wire 2 is at +1V and carries 1 mA of current in one direction, then wire 3 will be at -1V, carrying that same 1 mA in the other direction, so same amount of current goes in and out, right? At the receiving end both voltages are subtracted and you get +2V (which actually makes me wonder now if that means a balanced signal is twice as strong on the receiving end (1V - (-1V) going out = 2V coming in)?...

On the other hand, now I think of it, the inverting opamp on the receiving end actually won't draw any current, so is it correct to assume we're purely transferring voltages here, not currents?

See, more knowledge, more confusion :) But I feel I'm nearly there!
 

AnalogKid

Joined Aug 1, 2013
11,038
In a properly done balanced audio interconnect, there is no signal current in the ground or shield.

All circuits have an input impedance. In the case of balanced audio, that usually is set to 600 ohms by a physical resistor, which is bridged by a much higher impedance differential voltage amplifier. The important thing is that current does flow no matter how high the input impedance is at the receiver. The lower the input impedance, the stronger any external fields have to be to induce enough noise to be a perceivable percentage of the intended signal, but the bigger/stronger the intended signal source has to be to supply the signal current.

For dynamic microphones the input impedance frequently is set to 200 ohms. The input impedance is a load on the microphone, which is a moving coil generator just like in a power plant. The lower the impedance the greater the load current, which thanks to Faraday's law of induction makes it harder for the coil to move in the magnetic field, which dampens the diaphram's movement and affects the tonal quality of the mic. Low impedance = softer sound tending toward mushy or muffled. High impedance = bright sound tenting toward tinny or shrill. Microphone datasheets state the intended terminating impedance and the frequency response at that impedance.

ak
 

crutschow

Joined Mar 14, 2008
34,420
...............
By inverting one of the signal voltages, you basically allow that inverted path as a way for the signal current to return?
If wire 2 is at +1V and carries 1 mA of current in one direction, then wire 3 will be at -1V, carrying that same 1 mA in the other direction, so same amount of current goes in and out, right? At the receiving end both voltages are subtracted and you get +2V (which actually makes me wonder now if that means a balanced signal is twice as strong on the receiving end (1V - (-1V) going out = 2V coming in)?...

On the other hand, now I think of it, the inverting opamp on the receiving end actually won't draw any current, so is it correct to assume we're purely transferring voltages here, not currents?
...................
You need to remember that all voltages are relative.
Thus the differential voltage of interest is the relative value between the two differential wires, such as you measure if you connect the two leads of a voltmeter to the two wires.
The voltage from either wire to some arbitrary ground point is ideally of no interest.

The amount of current flowing in the differential line is determined by the differential signal voltage and the differential load impedance at the receiver end.

I don't understand what you mean by "which actually makes me wonder now if that means a balanced signal is twice as strong on the receiving end (1V - (-1V) going out = 2V coming in."?
Certainly -1V and +1V is the same as 2V.
 
Last edited:

Thread Starter

CurlsOnKeys

Joined Jan 3, 2015
47
You need to remember that all voltages are relative.
Thus the differential voltage of interest is the relative value between the two differential wires, such as you measure if you connect the two leads of a voltmeter to the two wires.
The voltage from either wire to some arbitrary ground point is ideally of no interest.

The amount of current flowing in the differential line is determined by the differential signal voltage and the differential load impedance at the receiver end.

I don't understand what you mean by "which actually makes me wonder now if that means a balanced signal is twice as strong on the receiving end (1V - (-1V) going out = 2V coming in."?
Certainly -1V and +1V is the same as 2V.
Thanks for your helpful insights so far! I understand that - in the end - we're interested in the differential voltage. With "twice as strong" I meant twice as strong as an unbalanced signal, or +6 dB louder, since you have two signal carrying wires which are inverted and then added together. But meanwhile I've found out that both wires only carry half of the signal strength, one positive, one inverted, and in the end you get the same value as an unbalanced signal with only one wire, carrying the full signal.

However, I'm still confused about the current path. I tried some current measurements just this morning, but I'm assuming the current is too small for my multimeter (20 mA minimum) to measure. I can imagine current in "closed" circuits, like that of an unbalanced connection: current goes out the hot lead of device A into device B's load, providing it with the desired voltage, "travels" through device B and flows back through the shield towards device A. Shield is there coupled to signal ground, loop is closed. However in a balanced connection: current goes "out" on 2 wires into the load of device B. The differential of the voltages in those 2 wires is amplified and the resulting current travels further in device B. However, how does it find its way back to device A. Through the shield wire? That's not supposed to carry current, no?
 

#12

Joined Nov 30, 2010
18,224
in a balanced connection: current goes "out" on 2 wires
No, it doesn't. The signal goes out on one wire and comes back on the other wire. The current reverses for every time the audio signal crosses the zero voltage point.
Consider a power transformer. The secondary has two wires. In your mind, the current goes out on both wires and never returns. It simply isn't true. The current takes turns going out one wire and coming in the other wire. The only difference between a power transformer and an audio transformer is the size.
 

HitEmTrue

Joined Jan 25, 2016
32
No, it doesn't. The signal goes out on one wire and comes back on the other wire. The current reverses for every time the audio signal crosses the zero voltage point.
Consider a power transformer. The secondary has two wires. In your mind, the current goes out on both wires and never returns. It simply isn't true. The current takes turns going out one wire and coming in the other wire. The only difference between a power transformer and an audio transformer is the size.
This all started in this thread where he could get audio signal could work with only one wire...I'm hoping to learn something from all this, too.
 

#12

Joined Nov 30, 2010
18,224
This all started in this thread where he could get audio signal could work with only one wire...I'm hoping to learn something from all this, too.
Yeah, this can get crazy real fast. The theoretical seems to be conflicting with the observed. The quirks obey the laws, but they are more detailed in their explanation. An electron can not lie. Take that to the bank. An electron can not lie. When you see the quirks, the goal is to find where the path is. Sometimes it's energy radiated through the air, as in power line hum. Sometimes it's capacitive coupling, as with a one wire audio signal. The path is there, but real world amplifiers have so many paths and some very sensitive amplifiers. People who don't have the basics down can get lost in the exceptions. They don't have the finesse yet to see the difference between picofarads and solid wires.
 

HitEmTrue

Joined Jan 25, 2016
32
Yeah, this can get crazy real fast. The theoretical seems to be conflicting with the observed. The quirks obey the laws, but they are more detailed in their explanation. An electron can not lie. Take that to the bank. An electron can not lie. When you see the quirks, the goal is to find where the path is. Sometimes it's energy radiated through the air, as in power line hum. Sometimes it's capacitive coupling, as with a one wire audio signal. The path is there, but real world amplifiers have so many paths and some very sensitive amplifiers. People who don't have the basics down can get lost in the exceptions. They don't have the finesse yet to see the difference between picofarads and solid wires.
That's me! :)

Good stuff, thanks.
 

#12

Joined Nov 30, 2010
18,224
Well, the good side of this is my employment. I was with a band in L.A. for several years and I finally asked if they appreciated the fact that I could make electrical problems go away. The answer was, "Electrical problems don't develop when you're here."
I had never thought of it in that way. I fixed several problems over the years and had no idea that it would have gotten worse if I wasn't there. Makes me wonder how bad it gets when a band has no electrician.:eek:
The bottom line is, when you know electricity backwards and forwards, you can take one look and know what to do. When you're a beginner, it's all some horrible, magical prank that the machines are doing to you.:mad:

The tweeters are burned out.
You have an ultrasonic oscillation.
WTF? How can you know that?
It's the only possible explanation when you know everything about how 2-way speakers are built.

The microphone shocked me.
You need to synchronize the 3 position ground switches on your amplifiers.
Why would anybody install those switches instead of building all the amplifiers so they can't shock you?
That stuff was designed when there were only 2 holes in a wall outlet.

My FM receiver isn't working for any of the channels.
You have water in the antenna socket.
But it hasn't rained here for 6 months.
But this is the first morning gig we ever worked. It's dew.

Just another horrible prank the machines play on beginners.:D
 

Thread Starter

CurlsOnKeys

Joined Jan 3, 2015
47
No, it doesn't. The signal goes out on one wire and comes back on the other wire. The current reverses for every time the audio signal crosses the zero voltage point.
Consider a power transformer. The secondary has two wires. In your mind, the current goes out on both wires and never returns. It simply isn't true. The current takes turns going out one wire and coming in the other wire. The only difference between a power transformer and an audio transformer is the size.
Actually in my mind the transformer is quiet easy to visualise and understand, but it's a good analogy because it makes me see the balanced line a bit better. In a transformer where the secondary is for example 5V peak, I find it easy to see how current flows, since 5V means that wire A has 5V "more" potential than wire B, so current flows out of wire A and back into wire B. At zero crossing, no current flows and when wire A is at -5V, it has 5V "less" potential wire B, so current out of wire B and back into wire A. At every time there's a closed loop and one direction of current.

Understanding the balanced line was harder, since I kept thinking of +ve and -ve voltage, but actually it's just as what crutschow said that makes sense now after your transformer analogy: all voltages are relative. So if in a balanced line wire a is at 5V peak (compared to ground reference in device A), then wire B is at -5V peak (inverted), so wire A a has 10V "more" potential than wire B and current flows out of wire A and back into wire B. And vice versa. I've seen the light :) Thank you both!

This all started in this thread where he could get audio signal could work with only one wire...I'm hoping to learn something from all this, too.
Thanks for bringing it all together :)

Actually, to expand on this, I tested this a bit more extensively:

- 2prong powered synth, unbalanced, connected to battery powered amp: connecting only +ve output lead gave nothing but a very loud 50 (Europe) Hz noise hum. My guess was because of both grounds were at entirely different potential, so audio current has no intention to flow. Induced mains noise in the (ungrounded) synth finds it way to the amp through the lead = hum.

- 2prong powered keyboard, unbalanced, connected to battery powered amp: connecting only +ve gave mains noise + (low) audio sound. See above: both grounds at different potential, not as "different" as with the synth, so small audio current flows. Induced mains noise also finds its way through the lead. Audio return current has nowhere to go, so my guess it it flows "back" through the lead, opposing itself. If I also connect ground, mains noise disappears and audio is louder.

- 3prong powered audio interface, balanced, connected to 3prong powered speaker: connecting only +ve = perfect sound. No mains hum. No audio loss. Also connecting -ve or connecting shield made no difference. My guess is both devices are grounded from same mains socket, so they're at same ground potential (= no mains hum) and audio current finds a return path to the ground wire of the speaker.

Difficult thing is that almost none of this can really be measured or viewed with a scope (I don't have a differential probe). Can't measure tiny currents with my voltmeter, can't "measure" DC voltage difference between 2 grounds, since power caps also discharge towards ground, so I don't get a steady reading etc... So a lot of it comes down to guessing.
 
Top