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Old October 10th, 2005, 02:54 PM   #1
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What does it mean when the audio is balanced???

I tried a search on this but could not find in simple terms what does it mean when you are talking about "BALANCED" audio? Thankyou.
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Old October 10th, 2005, 03:02 PM   #2
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Balanced is referring to how the audio signal is carried over the cables. A balanced signal requires three wires instead of two. Two of the three wires carry the signal 180 degrees out of phase with each other (the third is a common ground). Along the way, noise may be introduced into the signal. When the signal reaches its destination the two wires are flipped 180 degrees again putting them back in phase with each other BUT out of phase with any noise that may have been introduced. So balanced cables can run for very long lengths (100's of feet) without introducing any noise.

Unbalanced uses just two wires and any noise that gets introduced can be heard at the other end. These two wires can act like a radio antenna and pick up all sorts of interference. Unbalanced cables should only be run for short lengths. 10-15 feet.

~jr
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Old October 10th, 2005, 03:11 PM   #3
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Thanks John,
That was an easy to understand explanation. If I remember correctly, XLR mics have three prongs so this automatically makes all XLR mics balanced...is this correct. Also, is the "on-camera" mics built into a camcorder balanced since it is built in and no cables involved? Thanks again John.
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Old October 10th, 2005, 04:02 PM   #4
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The noise benefit, nicely descrbied above, is call "common mode rejection". That is, the noise is common to both of the balanced lines, and is rejected by the 180º cancelation characteristic.
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Old October 10th, 2005, 04:09 PM   #5
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Yes, Having an XLR adapter or cable is a good indicator of being a balanced audio device. On-camera mics don’t need balancing because the wires are so short.

~jr
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Old October 11th, 2005, 10:16 AM   #6
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Having a 3 pin XLR connector does not necessarily mean that the circuit is balanced. As an example of this I have cables (purchased - I didn't have to make them up) with RCA connectors on one end and three pin XLR's on the other that I use to connect the XL-2 audio output to a mixer input. Obviously only 2 of the pins on the XLR are used in this case and that is fine. It is quite possible to connect a single ended (unbalanced) circuit to a balanced one in this fashion. Note that while what I'm describing does work it is better to use a "balun" (balanced-to-unbalanced circuit: a simple transformer) in applications that require this.
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Old October 11th, 2005, 12:35 PM   #7
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Quote:
Originally Posted by A. J. deLange
Having a 3 pin XLR connector does not necessarily mean that the circuit is balanced.
Thanks for pointing that out. That’s why I said it was a “good indicator” and not that it “means” they are balanced. I guess my choice of words was too subtle and I should have been more specific. XLR is just an adapter, you can wire it any way you want. The only way to be sure that a device is balanced, is to check the specs.

~jr
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Old October 11th, 2005, 01:25 PM   #8
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Balanced means the two signal conductors have the same impeadance to ground. This eliminates hum and buzz. The shield is for RF protection. You can have a balanced output without driving both signal lines but both lines need to have the same impedance to ground.

Sam
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Old October 11th, 2005, 03:04 PM   #9
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Hope you all don't mind me butting in here for a moment, but to echo Steve's reply: John Rofrano, your post (#2) is the BEST explanation I have ever heard regarding balanced/unbalanced cables. Finally, someone made it understandable--and easy to remember--for "regular folks" like me! Thank you!!
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Old October 16th, 2005, 04:44 AM   #10
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Output impedance

John,
I agree. Your explanation of balanced cables is great and easy to understand. Could you do the same with output impedance? What does one need to know about that value when looking at mics? Any advice would be much appreciated.

Ramdas
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Old October 16th, 2005, 09:09 AM   #11
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Impedance, as the word itself suggests, is the degree to which a circuit resists the flow of current. If 1 volt is applied to a circuit the amount of current which will flow is 1/impedance (note that this is a simplification as impedance equals resistance + j*reactance where j is the square root of -1 but reactance should be small in audio circuits and if you are saying "Wait a minute, there is no such thing as the square root of a negative number" you'll understand why we don't want to talk about this aspect of impedance any further here). What is important with microphones and especially loudspeakers is that if two circuits are interconnected the maximum power is transferred between them when their impedances are the same. If they are connected by a transmission line (cable) that cable must have the same impedance as the source and load. Thus a microphone with 50 ohms output impedance will deliver the maximum possible amount of power to a mixer if the mixer has 50 ohm input impedance and the connection is made via a cable with 50 ohms characteristic impedance. It would also be acceptable to use a center tapped transformer to change the single ended 50 ohm impedance of the microphone to 600 ohms balanced, run through a 600 ohm transmission line, then use another transformer to go back to 50 ohms single ended and connect that to a 50 ohm input mixer. Thus the important thing is matching the impedances of the items being connected and the cables connecting them. In audio this is not so terribly important with short cable runs but with long ones (really long) mismatches may have an effect on the frequency response of the system. The problem is that the power which is not delivered to the load is reflected back down the cable. The inefficiency associated with power reflection in a microphone circuit is not significant because it is easily recovered by amplification. The power loss with a loudspeaker system is. In either case the waves reflected back down the cable add with the ones headed towards the load and it is this that causes the frequency response effects.

Impedance matching is very important at higher frequencies (video, sync lines, SDI, FireWire...) where the interconnecting cables can be an appreciable fraction of a wavelength or indeed several wavelengths long. Impedance mismatches will result in appreciable signal distortion in such cases.
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Old October 16th, 2005, 10:22 AM   #12
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Impedance is the measure of signal resistance. Less resistance (lower impedance) is better because it means the signal can travel longer before the signal deteriorates (usually in the high frequencies first).

The only thing you really need to remember is that, in general, a low impedance microphone should be connected to an input with the same or higher impedance. That means you should check the impedance of equipment you’re going to plug the mic into as well. If a microphone is connected to an input with lower impedance, there will be a loss of signal strength because the mic has more resistance than the equipment it is being plugged into. (i.e., don’t plug a high impedance mic into low impedance equipment) In this case you should use a line-matching transformer.

Bottom line: lower is better, matched with the equipment is perfect.

~jr
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Old October 16th, 2005, 10:48 AM   #13
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Balanced Cable

John's post #2 in this thread is a convenient way to think of this subject and except for the idea of "flipping the wires 180 degees" is correct in that it is the way XLR cable is used to carry a balanced mono signal and cancel interference without the need for active circuits. I've used that explanation myself in an XLR FAQ at camcorderinfo.com.

But as far as the true definition of balanced conductors, Sam Gates's is the one that's correct. All that's required to carry a balanced signal is that both conductor have the same impedance to ground. Electronic circuits (differential amplifiers) can do the noise cancellation at the receiving end.
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Old October 16th, 2005, 11:13 AM   #14
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John,

You are confusing impedance with loss. A signal will travel further in a high impedance transmission line with less loss than a lower impedance line with more loss. Telephone lines, for example, are 600 Ohms. A signal travels with less attenuation in RG/8 than it does in RG/58 even though both are 50 Ohms. The former is less lossy (because it's bigger). Given that things are kept matched it doesn't matter what the system impedance is. If things are not matched and actual power transferred is not important (as would be the case with a mic) it is generally a good idea to have the source at a low impedance. The fraction of the source voltage appearing at the load is Rl/(Rl + Rs) with Rl being the load impedance (assumed to be resistive) and Rs the source impedance (also assumed to be resistive). This also assumes the interconnection is very short relative to a wavelength. If the source impedance could be set to 0 all the voltage from the source would appear at the load. An "ideal voltage source" is a source with 0 output impedance (there is no such thing in the real world but it can be approximated closely).

OTOH you can very well plug a high impedance mic into a low impedance device if you are not concerned with frequency response distortions (shouldn't be an issue with short cables) and if the low impedance device has enough gain to overcome the voltage division loss (and low enough noise relative to the dropped incoming voltage). In fact this might be a good way to deal with a "hot" mic.

It is always best to stay matched!
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Old October 16th, 2005, 11:29 AM   #15
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Balance

Balance actually means what it says - that if the voltage on one signal wire exactly balances (equals) the voltage on the other signal wire (with respect to ground in both cases) they will cancel or in other words that the circuit described as balanced subtracts the voltage on wire 1 from the voltage on wire 2. The subtration may be done with transformers or differential amplifiers - it doesn't matter and it doesn't matter what the impedances are though they are, naturally, usually equal in both sides. Thus if a cable picks up voltage Vi on both signal lines from an unwanted source the receiver produces Vi - Vi = 0 at its output. In fact some of Vi makes it through. The ratio of the amount that makes it through to Vi is called the "common mode rejection" and is a measure of how good the balance is.

The way to get a signal through this system is to put Vs on one wire and -Vs on the other. The receiver output is then Vs - (-Vs) = 2Vs. To get -Vs from an AC signal Vs one inverts the phase 180 degrees. That's what John was talking about in his first post on this subject. To subtract two (analog) signals from one another one flips the phase of the one and combines it with the other. That's what he meant when he talked about flipping 180 degrees again.
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