Impedance balancing a "black box" output

Is there any direct measurement technique to determine the actual impedance of an unbalanced output to determine the best resistor value to use to impedance-balance the cold side--without opening and tracing the circuit of the unit under test, that is.

One idea I had was to make a little pigtail cable with a TS plug on a cord going into a little metal box with a TRS jack. A pot with a scale calibrated in ohms would take the place of the cold-to-ground resistor. Connect and adjust for least noise to determine the proper resistor value to hide inside the shell of the eventual permanent plug. How big would this pot need to be? (ie. what is the highest output impedance I'd likely encounter?)

Would that work or am I missing something else?

--Bob

[quote author="stickjam"]Is there any direct measurement technique to determine the actual impedance of an unbalanced output to determine the best resistor value to use to impedance-balance the cold side--without opening and tracing the circuit of the unit under test, that is.
[/quote]

I'm no expert on this, but I would guess that a simple way to do this would be to connect a sig. gen. sya at 1Khz sin to the input of the device and load the output with a specified resistor. You could then calculate the output impedance by measuring the voltage dropped on the resistor.

Someone who really knows will probably drop in.

I wouldn't bother. You'll double your output impedance and will probably obtain no benefit worth mentioning. You can run an unbalanced output into a balanced input, no problem.

Pretty much agree with NYD. Most of the time what works best: use two conductors under a shield and connect the shield and "-" to the unbal common, connect "+" to hot. If the two pieces of equipment are already at about the same potential then you may get better results leaving the shield connected at only one end (I think Whitlock has a white paper on this problem available on the Jens*n site). The advantages of quasi-balanced would only be if you were in a very high magnetic noise environment, where the induction into each conductor would be equal and therefore rejected by your diff input.

But the answer to the question: yes, as rodabod suggests, note the level change when you load the output with a resistor vs. very light loading. For example, when the loading is equal to the output impedance the level will drop in half.

But...if the output is coming from an op amp with a little series R, which is often the case, do the measurement at a low signal level or you will be seeing current limiting in the op amp instead of just the reduction due to loading of the series R. 100mV rms unloaded 1kHz might be a good start.

I do have the equipment connected like that (TR on mixer connected to TS on source with the shield tied only to S on the mixer) I'm still getting a profound 60Hz hum. Many are 80's-vintage synth modules.

My understanding is that putting a matching impedance on the cold side to ground results in any induced noise to be applied equally to hot and cold, thus cancelled by the differential inputs.

I'm gonna try the pot idea. I've got the stuff handy--what's to lose?

--bob

[quote author="stickjam"]
My understanding is that putting a matching impedance on the cold side to ground results in any induced noise to be applied equally to hot and cold, thus cancelled by the differential inputs.
[/quote]

Is the idea of impedance matching both hot and cold not so that you just get an equal amount of noise pulled through each source - eg. 1mA of current noise will be common on both sides and therefore cancel?

I'm not sure if it would help 60Hz hum, or at least I don't understand how it would.

The 60Hz hum is common on your hot signal, you aldo need it common on your cold signal if you want it to cancel - I can't see how impedance balancing the cold side would do this.

Have you tried a decent DI box - they obviously solve impedance and balancing issues.