I'm designing a front-end for an ADC that will be measuring the voltage of a hall effect current transducer.

I have been reading up on differential vs single-ended measurement, and have become a bit confused as to how differential measurements work in a case like this.

I need to filter out noise that will be imposed onto the signal from the transducer because the wiring is passing through an electrical distribution box that has AC.

What I don't understand is this: if I want to reject this noise by using a differential amplifier, do I need to have an isolated power supply powering the current transducer? If not, then where do I make the ground connection in the differential measurement schematic, below?

Also, do I run all the wiring to the current transducer as a double twisted pair (i.e. all 4 wires?)


simulate this circuit – Schematic created using CircuitLab


The hall effect transducer that I'm thinking of using is CSLT6B100, which has a ratiometric voltage output of 2.5V ±1500mV over the range -100 to 100 A

It's block diagram is as follows:

enter image description here

and transfer function:

enter image description here

  • \$\begingroup\$ Can you provide a reference to the Hall Effect sensor you are using? Without this information it's impossible to tell what your sensor has in terms of regulators and amplifiers in it. A Hall device by definition IS a differential sensor, but it's unlikely you have a raw device. Read this to gain an understanding: allegromicro.com/~/media/Files/Technical-Documents/… \$\endgroup\$ Dec 9, 2016 at 5:36
  • \$\begingroup\$ @JackCreasey I added the reference to the transducer I'd like to use in the post \$\endgroup\$ Dec 9, 2016 at 13:05

1 Answer 1


In the schematic above, where I wrote "Does this connect to Vss", the answer is yes.

The differential measurement will be between Vout and Vss, and will travel in a twisted pair to the inputs of the differential amplifier.


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