Correct way to measure a 4-20mA sensor?
That's exactly the way it is done on most PLCs including big names such as Allen-Bradley, etc.

Most that I have seen seem to use a 250 &Omega; resistor which will generate 5 V at 20 mA. I suspect that this is a compromise that while it reduces the resolution to 0.5 of full scale (1-bit) it also reduces the voltage drop at the device input leaving an extra 5 V for loop-powered devices.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. A typical 4 - 20 mA PLC input.

With this arrangement there is 19 V available for the loop to cover voltage drop on the cables and whatever voltage the transmitter requires.

If your transmitter is self-powered then 500 &Omega; would be fine and give you maximum resolution. See TE Connectivity UPW50 Series Wire Lead Wirewound Resistor 500Ω ±0.1% 0.5W ±3ppm/°C, for example, which seems to be specifically for this purpose.







simulate this circuit

Figure 2. A self-powered or externally-powered transmitter may be better able to provide the 10 V output required to drive 20 mA into 500 &Omega;.


  It is ok if I am not supplying the sensor from my device (as shown above with the 24 V)? As long as that sensors ground is connected into my input (as it is through my bnc connector)?


Yes, as long as the device can drive its output up to 10 V. This wouldn't be possible if, for example, the device was powered from a 9 V supply. You need to check the likely transmitters.