MEASURING 4-20MA

[DGR 0]

I'm using a 1762-OF4 analog output that I have configured for 4-20ma. Is there a way to measure the output without any load?

[Guest Guest]

Digital Meter...

[Wordman 0]

I always have a 'Fluke 175' V/mA Calibrator handy. you can use it for Input or Output, very helpful.

[Snerkel 0]

Use a digital meter set to mA but make sure you put a 1k resistor in series with the test lead. I have seen others just use a digital meter set to mA with no series resistor, however this is bad practice, and could potentially damage the analogue output. Check the documentation for the recommended maximum and minmum resistance for the loop.

[panic mode 246]

Snerkel is right... you could use analog meter as well but they seam to be harder to find these days. some digital era people don't quite know how to use them.

[JesperMP 2]

Why exactly is that ?An analog PLC input uses a resistor to convert the 4-20 mA to a voltage for the DAC. A digital meter uses a resistor to convert the 4-20 mA to a voltage for the DAC. What is the difference ? A 4-20mA output from a sensor or a PLC output should be designed to allow a shorted signal. The 20mA maximum current in itself more or less implies that 4-20 mA signals are short circuit proof. I have never come across this requirement before. But I could be wrong of course.

[TimWilborne 108]

I don't know if you have to put the resistor in there to protect the plc but it is always recommended. Also many Plcs quit driving the 4-20ma signal if they sense a short circuit and many voltmeters do not have enough resistance and the Plc thinks this is a short circuit so it quits driving the 4-20ma. This is why I have always heard you should put a resistor in line with the voltmeter. Otherwise you may not get an accurate reading

[Guest Guest]

Cheap Meter...

[Snerkel 0]

Ohms law gives you the answers. The multimeter has a very low internal resistance when set to Amps. The 4-20mA loop is driven by a voltage source, the current is controlled by varying the voltage. With a 1k load you need 20V to produce 20mA, if the load is only 0.05 ohm then you only need 1V to produce 20mA The PLC will struggle to give the true current loop value into such a low impedence, and a minor spike of the output could cause its destruction. A cheap digital meter is probably the best to use if connecting without a resistor as the leads and internal impedence are likely to be higher. It is far safer to just add a resistor in series with the meter to protect the meter and the PLC.

[panic mode 246]

depending on card design (cheap?) it may not be able to safely handle overload condition such as short circuit (well ampmeter is as good as short circuit) so when in doubt use of current limiting resistor will not hurt. i know i wouldn't try it without consulting manual. there are few things more embarassing for controls guy than explaining how some previously working component got burned. current limiting is so easily achived it's a shame that all DC outputs are not protected. then again, someone might want to measure 4-20MA (mega-amp...).

[DGR 0]

If I use a 1k resistor in series the max I get is 13.24ma. If I ever-so-reluctantly measure directly I get the full 20ma. My manual specifications does list an output protection of +/- 32ma and also has a continuous open and short-circuit protection so I guess I'm safe doing a direct measure. Thanks for all the input.

[Snerkel 0]

Try 500 ohm then, the documentation should give max resistance that the loop can have, but 500 ohm should work given that 1k gives 13mA. 1k is pretty standard as 24v will give you approx 24mA but maybe not for this unit. Even with a short circuit proof output I wouldn't risk it, especially on a regular basis.

[gravitar 3]

Maybe it just can't drive 20ma through a 1K? load! I wonder what you would see with a 10? or 100? series resistor..

[panic mode 246]

every analog output card should be able to drive 250 Ohm load in current mode (0-20mA or 4-20mA)

[Ghettofreeryder 0]

you dont need any resistors, just take your digital meter, set it to read current, and your done.

[gravitar 3]

What is the series resistance of a DMM in current mode? Isn't it 1 ohm? Can you count on an analog output to be short-circuit protected?

[Snerkel 0]

I just measured the resistance of my cheap meter set to mA with my expensive meter set to ohms and it gives 0.3ohm

[gravitar 3]

Ok, even worse.. so you'd be essentially presenting a dead short to the output terminals. Not something I would wanna do on a regular basis!