2
\$\begingroup\$

I'm wanting to nicely balance some unbalanced inputs to a power amp for good CMRR, and because I'm feeling too lazy to start trying to closely match pairs of resistors, I'm looking at using a hi-fi balanced line driver IC like this.

I know this can do the job for me, but is it appropriate or overkill? It's designed to drive balanced outputs of a device, including long cable runs, and this is obviously not needed because I'm using it to buffer / balance the input of a device.

Is there a more appropriate monolithic IC for me? (again, ignoring the obvious op amp solution that requires matched resistors)

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Not sure that this will give you any better CMRR because the unbalanced cable will already have common mode noise on it. Converting the signal to balanced doesn't eliminate what has already been induced in the cable. \$\endgroup\$
    – Martin
    Apr 8, 2011 at 16:39
  • \$\begingroup\$ Is this in the output stage of a product you're designing, or something you're going to add onto an existing product? \$\endgroup\$
    – endolith
    Apr 8, 2011 at 17:01
  • \$\begingroup\$ @Martin - Yeah, obviously, I won't be able to lose any noise that was already on the unbalanced line, but, as I understand it, I can prevent some noise received by the lines on the inside of the device from being amplified at the power amp stage \$\endgroup\$
    – Joe Mac
    Apr 9, 2011 at 2:04
  • \$\begingroup\$ @endolith - Neither, this would be at the input of a device I'm trying to design, to feed into a power amp IC: [TDA8920B](www.nxp.com/documents/data_sheet/TDA8920B.pdf). This is why I thought it might be overkill, because the IC is designed to drive an output stage. Plus David Kessner below believes that I may not even need to balance it at this stage. \$\endgroup\$
    – Joe Mac
    Apr 9, 2011 at 2:07

2 Answers 2

5
\$\begingroup\$

The simplest way to balance an unbalanced signal is with an isolation transformer. Unfortunately, a transformer that works well across the entire audio frequency range is going to be expensive ($>100 from Jensen).

Another way to do it is to use a chip similar to the one you found from THAT Corp. Is it overkill, probably. But it'll be cheaper and perform better than a transformer (assuming that you don't really need the isolation that a transformer will give you).

The "typical" way would be to use two op-amps. One is set as a mostly-unity gain non-inverting buffer and drives the + input of the amp. The other opamp is set as a unity-gain inverting buffer and takes it's input from the other op-amp and drives the - input on the amp. Vary the gain on the first op-amp if you need to.

Of course, the real question for this is: do you need to use a balanced signal or can you get away with going unbalanced straight into the amp? Of course, a balanced input will give you better audio quality but the op-amps or THAT chip will degrade quality. If you don't do a good job, the net effect could be a decrease in quality. And then you have to wonder if the change in quality (for good or bad) is even noticeable to your ears. It's very likely that simply going unbalanced into the amp is going to be the better choice.

\$\endgroup\$
2
  • \$\begingroup\$ The power amp stage will be dominated by [TDA8920B](www.nxp.com/documents/data_sheet/TDA8920B.pdf). Though not explicitly stated that I can see, the layout seems to suggest that the inputs are meant to be differential, balanced. If you still believe that they need not be, though, and no one else speaks up to contest you, I can consider this question answered for my purposes. \$\endgroup\$
    – Joe Mac
    Apr 9, 2011 at 2:10
  • \$\begingroup\$ @Joe Mac Right, the inputs don't have to be balanced to the NXP part. Just tie the + input to the audio signal, and the - input to GND. There is little point in doing a differential signal for the maybe 1 inch of PCB trace, when you have 3 to 20 feet of unbalanced cable feeding it. Any improvement would be lost in the noise of the NXP part anyway. \$\endgroup\$
    – user3624
    Apr 9, 2011 at 3:36
0
\$\begingroup\$

Oh, ok. The TDA8920 has a differential input. This is good for two things:

  1. Cancelling out interference in the cable
  2. Cancelling out common-mode ground loop voltages

For #1, you'd need a balanced line feeding it. This means the source impedance for the positive and negative signal pins needs to be equal, so that any interference creates identical voltages on each, which can then be cancelled out by the diff amp. (The signals do not need to be equal and opposite, despite popular opinion.) But since your signal is already unbalanced, you won't get this benefit.

The #2 case only matters if the source is grounded through the third prong on the power cord. If the amp and source are grounded through different outlets, there can be significant differences in their reference voltages. Then you can get 60 Hz power line hum in your amp, because ground loop currents flowing through the cable's shield will cause the amp to measure relative to a fluctuating ground. If so, you want to keep the power amp's ground disconnected from the source's ground, except the unavoidable connection through the wall. The ground from your source should connect to the - input of the amp, and nothing else.

On the other hand, if the source is floating, this will cause problems because the common mode voltage will be too great to cancel. To handle both cases (floating source and grounded source), you can use a ground lift switch on the input, or compromise with a smallish resistor (100 Ω) between the amp's ground and the source ground:

enter image description here

And follow the NXP schematic and PCB layout as closely as possible, including all the filter components shown above, or you'll get distortion, noise, EMI, etc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.