A 4 to 20 mA loop current is a popular terminology with Instrumentation/Electronics engineers in process industries. Field transmitters like pressure,temperature,flow, etc., give out 4 to 20 mA current signals corresponding to the respective process parameters.

Industrial equipment, such as plant control rooms (situated at a distance from the field), will house a distributed control system (DCS) or programmable logic controller (PLC) to monitor, record, and control these process parameters. This equipment will supply 24 VDC to a typical transmitter through one wire and receive current proportional to the process parameter through another wire.

Typically, two wires are needed to connect the supply voltage and ground, and two more wires are needed to connect the current signal. Thus, a two-wire system cuts cable cost by 50%. Hence, all field devices must conform to this two-wire system in process industries. DCS/PLC should receive a current in the range 4 to 20 mA. A current of zero indicates the cable has been cut.

Still, there is equipment, like gas analyzers, which give out a conventional 0 to 20 mA current output. These signals are to be converted into the 4 to 20 mA loop current format to feed the DCS/PLC in the control room.

Figure 1’s circuit does exactly this.

Figure 1 A 0 to 20 mA current source to a 4 to 20 mA loop current converter module circuit. The SPAN & ZERO potentiometers can be multiturn PCB mountable types for precision adjustment. Q1 should have a heatsink.

How it works

Connect the 24-V power supply, digital ammeter, and a load resistor to J2 as shown in Figure 1.

Then, connect a current generator to the J1 connector. This current flows through R3 and is converted to a voltage.

The output of U1B is this voltage multiplied by (1+(R10/R11)), which is nearly one. Let us call this Vspan. The output of U3 is Vreg.

There are three currents at pin3 of U1A. Let us analyze the basic equation of this circuit:

The third current through R4 is:

The total current at pin3 of U1A is:

In this circuit, R4/R6 is chosen to be 99; therefore:

Both U1A and Q1 adjust the current flow through R6, satisfying the above equation in closed-loop control. U3 generates 5 VDC from the 24 VDC input for circuit operation.

R12 loads the regulator to draw a small current. Q2 and R1 limit the output current to around 26 mA.

How to calibrate this circuit

Connect a 24 VDC power supply to J2, a load resistor of 200 Ω, and a digital ammeter

to J2 as shown in Figure 1. Connect a current generator to J1 as shown.

Keep the current as zero. Adjust Rzero until Ioutput reaches 4 mA.

Now, set the current generator to 20 mA. Adjust Rspan until Ioutput shows 20 mA.

Repeat this a few times to get the correct values. Now this current converter is calibrated.

How to improve accuracy

This circuit gives an accuracy of < 1%. To improve accuracy, select components with close tolerances.

You may introduce a 2.5-V reference IC after U3. Connect R2 and Rzero to this reference. In this case, R2 will be 50 KΩ and Rzero will be 20 KΩ.

Figure 2 illustrates how this current converter module is connected between the field transmitter and the control room’s DCS/PLC. Make sure to introduce a suitable surge suppressor in the line going to the field.

This module does not need a separate power supply. This can be kept in the field near the equipment giving out 0 to 20 mA.

Figure 2 A block diagram that shows the connection of the current converter in process industries.

Jayapal Ramalingam has over three decades of experience in designing electronics systems for power & process industries and is presently a freelance automation consultant.

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