I need to read the values from several analogue sensors mounted on a circuit breaker in an electrical substation. I'm not working with any high voltages, but the circuit must still work as the breaker switches normal operating current (not fault current). The cables to the sensors will be up to four metres long. Some of the sensors have two wire 20mA current loop interfaces. Others are potentiometers. I'm aiming to sample at around 2kHz with 8 bit precision or better.

I'm not familiar with the dealing with noise from other equipment. I'm assuming I'll need to consider interference from both electric and magnetic fields.

  1. Do current loops sensors tend to perform well in the vicinity of electric and magnetic fields? Should I use a twisted pair to reduce the net "loop area" that magnetic fields pass through? Should I have a grounded sheath to reduce capacitive coupling to the HV equipment?

  2. What might I need to do for the potentiometers? Might a single ended signal work? Should I use a voltage dependant current source to convert the output to a 20mA current loop? Or perhaps use an inverting/non-inverting op-amp pair to create a differential output and use twisted pair cable? Or even mount discrete ADCs next to each one and use RS485 to send the signal digitally?

  • \$\begingroup\$ What are "current loops sensors"? \$\endgroup\$
    – JimmyB
    Jun 21, 2019 at 12:11
  • \$\begingroup\$ "the breaker switches normal operating current"? A circuit breaker should not 'switch' in non-fault conditions. What do you mean by "circuit breaker"? \$\endgroup\$
    – JimmyB
    Jun 21, 2019 at 12:15
  • 3
    \$\begingroup\$ Speak with your fellow sub-station engineers and get the proper deal on information. If they say it's a bad environment then take all the steps you have listed unless there is some aggressive cost control imposed. \$\endgroup\$
    – Andy aka
    Jun 21, 2019 at 12:17
  • 1
    \$\begingroup\$ Agree with @Andy. An idea to keep in mind: It is relatively easy to get 60dB rejection from nearby circuits using good emc practice as described in the question. More than 80 dB is hard to do. 60dB means that a 20mA signal in the presence of 20A currents is do-able. 20mA signals in the presence of 200A currents is getting difficult. If you're in the 200A range, then I'd go with the RS485 option. \$\endgroup\$
    – scorpdaddy
    Jun 21, 2019 at 12:39
  • \$\begingroup\$ @JimmyB With a two wire current loop sensor, you supply a voltage across the terminals, and the sensor sinks a current in proportion to the value it reads. The advantage is that current is the same everywhere in a loop. instrumentationtoolbox.com/2012/03/… \$\endgroup\$
    – William
    Jun 21, 2019 at 15:59

1 Answer 1


Lets examine the magnetic interference.

The Vinduce into a rectangular loop, from a straight wire located in the plane of the loop, and ignoring natural-log coefficients that may obfuscate the causality, is

Vinduce = [ MU0 * MUr * Loop_Area / (2 * pi * Distance) ] * dI/dT

In air, this simplified to

Vinduce = 2e-7 * Area/Distance * dI/dT [volts]

Assuming some evil spikes on the current waveform (from motors, or poorly filtered switching regulators) at the 100amp/100 microsecond size, using non-twisted pair sensor cables of physical nature 1meter long by 2mm signal-return spacing, and this sensor cable located 1cm from the evil spiky breaker&cables, we compute this level of trash

Vinduce = 2e-7 * (1 meter * 2milliMeters) / 1cm Distance * 100 amp/100uS

Vinduce = 2e-7 * 0.002 / 0.01 * 1e+6 amp/sec

Vinduce = 2e-7 * 0.2 * 1e+6 = 0.4 * 1e-1 = 0.04 volts of trash.

In a 2.5 volt sensor signal, the (unfiltered) trash causes 2.5/0.04 = 62:1 ratio, thus about 6 bits may be possible. Heavy filtering may improve the accuracy.


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