Series Resistance in a mA Trip Relay Test Circuit

The purpose of a trip relay is to have two different states either side of a trip mA point in the current loop.

<...the series resistance to prevent overdriving the relay...>

What does that mean?

Your current sources are designed to increase output voltage to whatever value is needed to achieve the desired current output. One is capable of putting a higher voltage than the other in its' attempt to deliver whatever current you asked it to output. So naturally you require differing values of series resistors to limit the current.

Can you show us the circuit?

Guest,

I think you are misunderstanding the internal workings of your mA sources, and how the circuit should function. As was said by an earlier answer, these are current sources. They are not voltage sources. However, they will change the voltage they put out until the current flow is proportional to the process variable being monitored. This works just fine unless the total loop circuit resistance is too high for the source's voltage supply. Use Ohm's Law to calculate the maximum allowable loop resistance for a given source voltage supply. At the other end of your loop it sounds like you are using a current-sensing relay. You need to choose one that is rated for the range of currents your pressure transmitter can output. Its internal resistance must be less than the maximum allowable loop resistance you already calculated (remember to account for resistance of the wires in the loop, and give a modest safety margin). If you need to add an external resistor, then I suspect you have a relay that is not matched to the application.

--JMM

Here's a spec for a commercially available DC current alarm trip module:

Note that a properly selected model would be a current input, typically 4-20mA, but that there other models. Sometimes the input type is field selectable, sometimes not. It would depend on what brand & model it is. On this particular model the input range and type is not field selectable.

Note the spec for "Input Voltage Burden" for current input models.

If the device drops a maximum of 1.25V it would be when the current is at its maximum, 20mA.

Ohms Law tells us that the input resistance of this device is R = 1.25V/0.020A = 62.5 ohms. That's the value of this device's dropping resistor in the current loop. There should be no other reason to add a resistor in the loop.

Yes, I think more information needs to be provided by original Guest. As pointed out by Bud's post inputs are Factory Configured. If you can give specs. for exactly what you have, it would make figuring out what you're seeing easier.

I get the feeling that he is trying to alter the trip point of the relay by inserting a series resistor, the wrong thing to do. It demonstrates a fundamental lack of understanding of 4-20mA current loops.

Do you suppose that the OP is attempting to trip a "yellow" Safety Relay, a discrete in/discrete out device designed to be used in a safety circuit according to a standard like IEC 61508, with a 4-20mA loop powered signal?

Trying to substitute the current loop for the "safe PLC" output in the diagram below?

Bud, that may be it, I really don't know.