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    What is a 4-20mA Current Loop?

    It can be very difficult to choose the right output from a sensor as there are many different options, and it is important to understand how they work in order to get a good working system.

    One of the amplified sensors commonly used is the two-wire 4-20mA current loop or 10-50mA system. This is easy electrical wiring, with only two wires, inherent noise immunity as it transmits current, a wide power supply range, 4mA current at zero and no dead wire. However, if you need more than a couple of metres of cable, it is advisable to choose an amplified output with built-in power supply regulation. This helps in overcoming electrical noise problems and ensures that the sensor has plenty of supply voltage.

    What’s in the circuit?

    Diagram showing a 4-20mA current loop circuit

    The sensor takes its power from the two wires in the form of a voltage and then regulates the current output (in the loop) in proportion to the sensor input: ie 4mA at zero input pressure and 20mA at 100% input. In order for the sensor to regulate, there is a minimum voltage needed, but also a maximum as the onboard voltage regulator can only dissipate a limited power.

    The second element of the circuit is a “drop resistor”, which is normally built into the indicator, controller, input card or whatever. This resistor converts the current to a voltage (normal ohm’s law) which then processed by the A/D convertor, microprocessor etc. The most commonly used value for a drop resistor is 250ohms and there can be several of these units on one loop.

    The third item in the loop is of course the power supply. This will be a standard DC supply: 24V DC is the most commonly used, but it is important to choose the right voltage.

    To calculate the power supply value, a minimum and maximum calculation is required:

    Vmin=sensor Vmin + (20mA*drop resistor)
    Vmax=sensor Vmax + (4mA*drop resistor)

    So, for example, if the sensor requires 12 to 35V and a 250ohm drop resistor is used:

    Vmin=12+(20mA*250)=17V
    Vmax=35+(4mA*250)=36V

    Therefore a 24V DC supply can be used.

    Read more: Index to all of our Technical Notes on Pressure

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