1
\$\begingroup\$

I have a PLC (take the "P" with a pinch of salt), it has some analogue inputs (4-20 mA) that I need to use to detect if 24 V is present or not. I can program the logic to say above X mA 24 V must be present, below X mA it mustn't be present (the reason is I have some controllable relay outputs, but there is no feedback to say which state it is in and the only spare inputs are analogue). Is it simply a case of I = V/R, so on the analogue input common I stick 0 V DC and on the input I put a 1200 Ohm resistor and the 24 V DC into that. My logic could then be if >10 mA 24 V must be present, <10 mA it isn't?

\$\endgroup\$
4
\$\begingroup\$

Your idea is correct.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The equivalent circuit for your setup. Most 4 - 20 mA inputs use a resistor to convert the current to a voltage of either 5 V (250 Ω) or 10 V (500 Ω) at 20 mA.

Don't forget to add the series resistance of the 4 - 20 mA input for your calculation. With the configuration shown in Figure 1 you will get \$ I = \frac {24}{1450} = 16 \; mA \$ and a power dissipation in R1 of \$ P = I^2 R = (16m)^2 \cdot 1k2 = 330 \; mW \$.

If the device has a 500 Ω input the current and power will be reduced.

\$\endgroup\$
4
\$\begingroup\$

Connect the 4-20 mA input to the 24 V supply with around a 2 kΩ ½ W resistor in series. Even if the 4-20 mA input drops 10 V (unlikely), then it will still be driven with (14 V)/(2 kΩ) = 7 mA. That's well above the 4 mA minimum that you should be able to detect it with maybe a 5 mA threshold.

The ½ W rating is in case the 4-20 mA input drops no voltage at all, or you accidentally have a short. In that case, the power dissipation will be (24 V)2/(2 kΩ) = 290 mW.

\$\endgroup\$
  • \$\begingroup\$ Whoever downvoted this: Please explain what you think is wrong. \$\endgroup\$ – Olin Lathrop Jun 27 '17 at 21:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.