If we want to create fully differential and balanced outputs, and if we don’t have a single ended to differential ended converter/driver, can we use the following scheme neglecting the voltage divider DC errors?:

enter image description here

The idea is to create same common mode noise at HI and LO nodes. VCM is the source model for the common mode interference capacitively coupling to the lines. Vs is the desired signal source.

Since the device inside the green box inherently does not have balanced outputs, can balancing be achieved by using a potentiometer R2 on the diagram?

Did anybody use this scheme before? Im wondering whether is a proper method.

In simulations the idea works but I don't know in practice would be issue: enter image description here

  • \$\begingroup\$ I don't think that will work. the impedance to the noise source is lower for Lo than for Hi (1 ohm vs. 1000 ohms) \$\endgroup\$ – Jasen Oct 21 '18 at 21:04

If you make Rx = Rout and R1 = R2 then yes, this is a fairly standard trick, but you must make the impedance match.

Note that higher input common mode impedance reduces the sensitivity to source impedance mismatch.

The search term you want is "Impedance balanced output" and you will find a ton of stuff from the audio industry (that likes this because it is cheap!), but it is just as applicable to any other sensor.

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