Simulating PLC load w/Loop Powered Device

[robbm]

I have a 4-20mA transmitter that I need to power up just for using the display on the unit, there's no PLC in the application. I thought 250ohms was the minimum resistance i'd have to wire between the 24VDC supply and the transmitter to simulate the load of the PLC, but wanted to confirm if my thinking is right.
With a 24VDC supply, and a transmitter that requires at least 11VDC, would anything between 250 and 650ohms be usable? (13VDC max Vdrop/20mA=650ohms)and (5VDC min Vdrop/20mA=250ohms)

Thanks in advance

 

[IRstuff]

There IS a PLC forum on this site

TTFN
faq731-376

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[robbm]

This is for wiring a transmitter without a PLC, so that's why I posted it here...

 

[Compositepro]

You do not need any load resistor in the current loop unless you want to convert the signal from 4-20ma to a voltage like 1-10 volts. The instrument will use 4 ma to power itself and then control the loop current to be proportional to the measurement it is transmitting. There is a max loop resistance allowed which varies with the supply voltage. A higher voltage supply (up to a limit that the instrument manufacturer will specify) can drive 20 ma through more resistance (longer or smaller wire). This is a good topic for a Google search. Or read the manual.

 

[xnuke]

I agree with Compositepro - no resistor is necessary. Just wire your transmitter to the power supply.

xnuke
"Live and act within the limit of your knowledge and keep expanding it to the limit of your life." Ayn Rand, Atlas Shrugged.
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.

 

[danw2]

If you only need the display on the transmitter, then wire it to power supply, as others advise.

If you need to talk HART to configure the transmitter, then you'll need a minimum of 230 ohms in the loop, but 250 ohms is frequently used because that value is common in instrument shops, 270 or 330 ohms when you have to go to Radio Shack.