Need help understanding 4-20 mA to 1-5vdc with 250 resistor

4-20mA to 1-5vdc question.

So you have a 4-20mA circuit . RTD gets power from transmitter. PLC can’t read read 4-20mA so you pop in a 250ohm resistor which turns it into 1-5vdc and now your in business.

Essentially the input is fully low side of circuit so you shouldn’t read anything with a voltmeter. I.e neg off the supply is the input + and neg off the RTD is input -

Now you put in a 250ohm resistor and that - to - is now …… 1-5vdc. I don’t get really
get that? I’m sure I’m overthinking it but wired one today and for some reason this is bugging me now.

Appreciate the help in understanding this

Below are pics this is for a chart recorder . Wiring is single mA input in schematic


Thank you in as advance

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Ohm’s Law. Current passing through resistance produces voltage. V = IR. 0.004A * 250Ω => 1.0V. 0.020A * 250Ω => 5.0V.

That’s it.

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This is perfect. I knew the formula and everyone I asked just keeps mentioning it with voltage drop. Which confused the heck out of me.

But it does in fact create voltage where there is theoretically none or very little. So it does amplify it.

No, no amplification. The voltage drop across the combination of the transmitter and the resistor at the input card is determined by the DC power supply. The transmitter will regulate its own apparent resistance to deliver the required current for the signal it is transmitting.

This is where I get lost. Right now if I take a meter (see below)
TB3 1-2 = 24vdc
TB3 1 -TB4 1 or 2 = 24vdc
But @pturmel how does the below turn into 1-5 if I clearly have 0 volts.
TB3 2 - TB4 1 or 2 = 0vdc

TB4 1-2 = 0vdc
so the input is getting 0 volts as it stands. We add resistance which doesn’t drop voltage but “adds” it? i.e. I can’t drop from 0 to get 1-5vdc…?

No, the transmitter changes its voltage drop to deliver the required current. If the downstream resistance is zero, its own voltage drop will consume the entire power supply voltage. After you add some downstream resistance, the drop will be split between the transmitter and the load.

If the resistor is missing, then there’s no path for the current to flow, so all of the drop will be across TB4-1 and -2.

I get it even with the resistor, there is no current so there is 0 volt. Once I apply current i can see the drop

@pturmel would this work on regular power supply with resistor? Or does it have to be a special transmitter for the voltage drop

No special power supply. 4-20mA transmitters are, by definition, current regulators. As long as there is enough DC voltage available for the load and for the transmitter’s own overhead, you’re good.

By adding the resister, what you are doing is creating a voltage divider. You have to go way back to you basic electronics classes to remember the example problems the textbook gave you! Here is a link to a Wikipedia article to refresh your memory:

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@bill.hyland this is not a voltage divider from my understanding because it has only one resistor. And a voltage divider has two.

The transmitter is the other resistor. A variable resistor, effectively.

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@pturmel thank you, looks like I have some googling to do.

The other half of the voltage divide is the field instrument.

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And of course the cable, with temperature variation, which is half the beauty of 4-20mA

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