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I'm measuring the 0-10V pressure output of a pressure sensor with a LOGO! smart relay (PLC). I would like to make sure that the pressure sensor is working even when I'm measuring nearly zero values with the device. My idea was to connect the other output (4-20mA) of the sensor to the LOGO!'s digital input to make sure that the sensor is working. I would do this beacause the LOGO! only has voltage analog inputs so I can't use the 4-20mA directly in the analog input. The input resistance of the digital input is 80k and the 4-20mA output of the sensor provides nearly 24V (which is the supply voltage) when such a high impedance is present between the output pin and the common ground. So my purpose would be to make sure if the pressure sensor has supply voltage therfore it is probably working. (Unfortunately when a nearly zero pressure is measured the 0-10V output gives 10mV which is a very small value to decide if the pressure sensor is working at all.)

I've tried and it was working. The current was only 0.33mA coming from the 4-20mA output and flowing into the digital input. The voltage dropped at the output by 0.2V. Eventually I measured a logical high value at the digital input so it was working, but I have concerns that in the long term this usage may demage the sensor's output. I dont't have the particular circuit diagram of the pressure sensor's current output but as I understood there aren't major differences between the implementations. Here is an exaple I found: https://www.analog.com/en/design-center/reference-designs/circuits-from-the-lab/CN0314.html#rd-overview

Does someone know if this soulution could work in the long run?

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3 Answers 3

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I used the LOGO! units successfully on a number of industrial projects and interfaced with 4 - 20 mA signal by using a current to voltage converter, commonly known as a resistor. For 20 mA to 10 V conversion you need a 500 Ω.

I was delighted to find that TE Connectivity make a range of 0.5 W, 500 Ω, 0.1% tolerance resistors available in single units from RS, starting at about €3.50 at the time of writing.

This gives a simple, cheap, reliable solution with the advantage of a "live zero" (2 V) signal to detect sensor loss or fault.

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  • \$\begingroup\$ Did you just put it for exemple in the I1 and M terminals of the LOGO? I suppose the wire coming from the sensor should be in the same ferrule as one of the resistors endpoints which should be placed in the I1 and the other endpoint of the resistor (also in a ferrule) connected in the M. \$\endgroup\$
    – Gurul_01
    Commented Jan 6 at 9:32
  • \$\begingroup\$ Or is it possible to place the resistor in a DIN rail terminal block reliably, maybe without soldering? Are there DIN rail terminal blocks for this purpose? Thanks for your answer! \$\endgroup\$
    – Gurul_01
    Commented Jan 6 at 9:35
  • \$\begingroup\$ The easiest is to use two DIN terminals side by side. Connect the resistor between the two top terminals and wires on the bottom. Watch out for small voltage drops along the ground wiring. Wire the resistor negative directly back to the LOGO! M terminal for the inputs to minimise this. (Don't forget to accept one of the answers when you get what you're looking for.) \$\endgroup\$
    – Transistor
    Commented Jan 6 at 9:58
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Your high impedance of the digital input could create a voltage near the sensor's supply voltage, which might exceed its output stage's voltage rating, leading to damage. Not to mention, the 4-20mA output isn't designed to drive high impedance loads, and this could cause potential instability in the output stage.

I would recommend looking at the sensor's datasheet to design your solution to match the recommended load impedances. I also agree with @Sphero's idea for a precision resistor. If you're also concerned about noise, you should isolate the sensor from the PLC to prevent any ground looping.

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We have no way of knowing how the pressure sensor is implemented. There are a huge number of ways it could be done.

If the manual for the sensor indicates that both outputs can be used simultaneously then that would be a sort of endorsement for at least a portion of what you are doing.

I may suspect it's okay, or you wouldn't be getting the results you are getting but I might also guess that there could be problems. 80kΩ is a rather high impedance and I would rather shunt that with a resistor of perhaps 3kΩ to avoid potential issues with noise or leakage.

You might also consider using a precision resistor of 500Ω (or 503Ω to compensate for the 80kΩ - if that's also true for the analog inputs!) directly with the 4-20mA output and leaving the 0-10V output open.

This is exactly how (and why) the 4-20mA standard is intended to be used.

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    \$\begingroup\$ Be sure your DIO can handle a failure mode of 24VDC (from the 4...20 mA output), or else you may unintentionally damage the DIO input. Slap on a reverse biased Zener across the input, or a forward biased diode from the input to Vcc to protect the DIO input from over-voltage. You may also want to investigate how your DIO input handles signals loitering the 'gray area' voltage range as you pressure signal is increased. If the signal ends up in a transition range, and there's suitable noise present, the dio signal may 'bounce'/oscillate a bunch if hysteresis is not present. \$\endgroup\$ Commented Jan 5 at 20:58
  • \$\begingroup\$ Aside from damage, if the DIO clamping circuit is not well (and a bit more expensively) designed it's possible for it to subtly affect other things such as analog inputs when it is out of range. This is kind of like a 'soft error' in that it does not cause damage, but it also will not be exactly working properly. Since your intention is to operate continuously in that mode, it could conceivably be an issue. There are many ways things could go pear-shaped. Using a 4-20mA analog signal as it was intended to be used (see penultimate and final paragraphs above), those things are much less likely. \$\endgroup\$ Commented Jan 5 at 21:05
  • \$\begingroup\$ Thank you! How do I connect a parallel resistor? I was also thinking about that. Should I use a DIN rail mountable resistor maybe? I would like it to meet the standards of an industrial application. \$\endgroup\$
    – Gurul_01
    Commented Jan 5 at 21:08
  • \$\begingroup\$ You could sleeve it and put it right on the terminal block. Add ferrules or lugs if you want to make it really nice. \$\endgroup\$ Commented Jan 5 at 21:11

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