# Convert 24 V to 20 mA (none is variable)

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?

Your idea is correct.

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.

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.

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