RTD transmitter to CpLX

Hello all. I am having a time trying to get an RTD online. Instead of an RTD module one has opt'd to use an RTD to 4-20mA transmitter. I am pretty sure I have it wired correctly however, it can't be, because I am getting nothing in the plc. The PLC module is 1769-IF4, configured for 4-20mA. Let's see if I can import this diagram.......

View attachment goofy transmitter.bmp

ceilingwalker said:

Hello all. I am having a time trying to get an RTD online. Instead of an RTD module one has opt'd to use an RTD to 4-20mA transmitter. I am pretty sure I have it wired correctly however, it can't be, because I am getting nothing in the plc. The PLC module is 1769-IF4, configured for 4-20mA. Let's see if I can import this diagram.......

View attachment 55935

Click to expand...

Do you have the unit wired as page 2-20 of the manual below ? Manual Link

rajy2r said:

Do you have the unit wired as page 2-20 of the manual below ? Manual Link

Click to expand...

For current, yes. I am pretty sure my error lies with how I have wired this transmitter.

I have wired these modules plenty of times however, I have always used an RTD module for an RTD, not a transmitter like this.

ceilingwalker said:

For current, yes. I am pretty sure my error lies with how I have wired this transmitter.

Click to expand...

Do you have 24VDC being supplied to power the transmitter on the 2 wires ?

Looks like a puck that you can wire it either for a 2 wire or 3 wire RTD.


You may need a 250ohm resistor. It does have the thermocouple attached correct.


What is the scaling set for. The transmitter should tell you the measuremnt range like 0-300,50-250 Etc. Then you will need to set the scaling or enginerring units in the plc to match what the transmitter is set for.


Are those pots on it for zero and span to calibrate it?


May need to use a 4-20mA generator and multimeter with a mA input to see if you are getting a reading from the transmitter.

The original wiring diagram (in post #1) is incorrect. Polarity on the analog input is wrong. Polarity only reverses for 4-20ma below the equator (I'm joking).

Revised diagram (below) shows the correct analog input polarity.

You would use the 250 ohm resistor if you were trying to connect a 4 - 20 mA signal to an analog voltage input. If your connection is to an analog channel configured as 4 - 20 mA, omit the resistor.

If you have the 250 ohm resistor in place, disconnect the wires to the analog module and use your voltmeter to measure the voltage drop across the resistor. If the resistor is not in place, connect your meter's mA leads in series with the rest of the circuit to measure the current.

Steve Bailey said:

You would use the 250 ohm resistor if you were trying to connect a 4 - 20 mA signal to an analog voltage input.

Click to expand...

Thank you, that's what I thought.

danw said:

The original wiring diagram (in post #1) is incorrect. Polarity on the analog input is wrong. Polarity only reverses for 4-20ma below the equator (I'm joking).

Revised diagram (below) shows the correct analog input polarity.

Click to expand...

Thank you

danw said:

The original wiring diagram (in post #1) is incorrect. Polarity on the analog input is wrong. Polarity only reverses for 4-20ma below the equator (I'm joking).

Click to expand...

This is particularly important for equatorial installations. Those are likely to have your transmitter located one side of the equator, but wired back to an input card on the other.

We use plenty of these types of probes, the original looks correct to me, +ve is from 24vdc power supply, not the AIN module, -ve goes to analogue input, input card is to 0vdc to complete the circuit. No need for the resistor.

I do not understand why you need the resistor, I have wired many of these and just put supply to + of the transmitter - of transmitter to + of analog input and 0v to - of analog input, I would only use a 250 ohm resistor if converting from 4-20 to 1 to 5v

Not sure if you fixed it or not, but have you tried to wire in the third leg on the RTD side? Some of these transmitters have an option in the programming that requires them to see that signal there for them to output anything.

There's also the change that it's a dud...

Why a resistor? problem with the diagram

Why a resistor?

All electrical current measurements (except for analog moving coil galvanometers) are actually voltage measurements - current flows through a resistor creating a voltage drop which is then measured by the amplifier-A/D combo.

Sometimes the resistor is built into the analog input, sometimes it is not and is an external resistor.

Most PLC's seem to use 4-20mA AI modules with internal resistors, so an external would not be used with an AI that already has an internal resistor.

But not all 4-20mA AI modules use internal resistors.

The original drawing derives from the 'external' resistor genre.


I suspect that use of an external resistor comes from the DCS world because the DCS world drove the development of the two-wire loop-powered 4-20mA signal. DCS's deal with hundreds, if not thousands of 4-20mA loop signals coming into marshalling panels. DCS systems originally used large 24Vdc power supplies to power dozens of loops from one power supply.

If there is short circuit in the pair of any loop's "two-wire" field wiring, that short circuit puts 24V (+) on the one conductor directly on the conductor running to the Analog input's (+) terminal. The input's dropping resistor then has the full 24Vdc power supply potential across it. The input resistor burns up.

The dreaded Ohm's Law tells us that 24V across 250 Ohm resistor produces 96mA of current. 96mA of current at 24Vdc is 2.3 watts. The resistor is probably a 1/4W or maybe 1/3W resistor if it is external, but is more likely an 1/8W resistor if it is internal. Regardless, 2.3W will rapidly heat up and burn out even a 1/2W resistor.

If the burned out resistor is external, then the burned out resistor is the only damage and replacement is relatively easy because it is external, connected to the wiring terminals.

If the burned out resistor is internal, inside the AI module/card, then that channel is toast. Repair involves a costly DCS input board/module.

Since DCS's deal with hundreds and thousands of loops, a DCS might experience field shorts and view the external resistor as a means of protecting the AI cards/modules from field short circuits.


Problem with the diagram

The polarity of the connections at the analog input are backwards on the original drawing.

The power supply (+), by convention, connects to the (+) terminal on the field transmitter.

The field transmitter's (-) then connects to the analog input's (+) terminal.

Whether the resistor is internal or external, the field transmitter's (-) terminal connects to the AI's (+) terminal. That is not case in the original diagram where the field transmitter's (-) terminal connects to the (-) terminal on the AI (-).

Dan