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How to use SAADC to measure 4-20 mA current loop

Karlos

Hi,

Sorry for the newbie question but how am I going to use the SAADC to measure 4-20mA current loops? I've used the SAADC to measure temperature, voltage, and resistance but am not sure what input am I expecting from a 4-20mA output source.

Best,

Carlos

Parents
  • +1

    You need to scope the signal with an oscilloscope to see its voltage range, and take into account the capacitive loading of your probe. 

  • 0 in reply to haakonsh

    Thank you for the prompt response, Haankosh. So I should be measuring a voltage range against a fixed resistor from where I could derive the 4-20mA values, that makes so much sense.

    Kind regards.

  • 0 in reply to Karlos

    You need some form of impedance to measure a voltage drop over, a resistor works great. 

    What exactly do you need to measure? 

  • 0 in reply to Karlos

    It is intended to measure the 4-20mA current loop output of the device below. From the pdf "4 Optional isolated 4-20mA current outputs for re-transmission of conductivity, Redox, pH and temperature."

    https://lth.co.uk/storage/84/MTD75_Datasheet.pdf

  • 0 in reply to Karlos

    That did not really say much about the circuit driving the signal. I guess you'll have to run the current through a resistor and probe the voltage drop over the resistor terminals with an oscilloscope to see what you're dealing with. 


  • 0 in reply to haakonsh

    No, it doesn't, unfortunately. But your suggestions are a good starting point for me to start the code. 

    Thanks and best regards.

  • +1 in reply to Karlos

    An excitation voltage source is required, the magnitude of which depends on the minimum voltage differential required across the sensor; in the old days that might have been as high as 4 volts drop but now you can probably drive the 4-20mA loop directly from Vdd of 3.3 volts through the sensor and back to Gnd through a (say) 100R resistor. 100R gives 2.0 volts at 20mA, 0.4 volts at 4mA and 0.0 volts in a fault (disconnected) condition. A lower value resistor allows more voltage for the external sensor, but reduces the signal level at the nRF52; 100R (100 Ohm) is often used, but 10R would work fine unless you need maximum signal:noise ratio; 10R at 20mA is 0.2 volts signal. A design goal would normlly allow up to 25mA input so excess voltage monitoring can also be checked, as well as broken-wire (open-circuit).

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  • +1 in reply to Karlos

    An excitation voltage source is required, the magnitude of which depends on the minimum voltage differential required across the sensor; in the old days that might have been as high as 4 volts drop but now you can probably drive the 4-20mA loop directly from Vdd of 3.3 volts through the sensor and back to Gnd through a (say) 100R resistor. 100R gives 2.0 volts at 20mA, 0.4 volts at 4mA and 0.0 volts in a fault (disconnected) condition. A lower value resistor allows more voltage for the external sensor, but reduces the signal level at the nRF52; 100R (100 Ohm) is often used, but 10R would work fine unless you need maximum signal:noise ratio; 10R at 20mA is 0.2 volts signal. A design goal would normlly allow up to 25mA input so excess voltage monitoring can also be checked, as well as broken-wire (open-circuit).

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  • 0 in reply to hmolesworth

    Hi hmolesworth, I appreciate the detailed explanation of requirements to complete a 4-20mA current loop. I will need to ask the vendor who will be the one to supply the excitation voltage. 

    Kind regards.

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