Reading free potential contact

Previously you wrote
Jelf said:
If I understood correctly, that shouldn't be problem?
That more or less prompts me to verify that you circuit is OK. The point is that in order to figure out if yor circuit is OK or not, I have to know the details. It shouldnt be necessary to d.r.a.g them out of you.

To sum it up:
According to your last diagram, with the common being the "bottom" of the voltage splitter, you can hook up your 0-10V datalogger input to the "middle" and your 24V PLC input to the "top" of the voltage splitter. Power supply, datalogger input channel and PLC input channel must share the same common.

In an industrial application, this would have been done with a signal conditioning isolation amplifier, not a voltage splitter. But in principle there should be nothing wrong with doing it this way.
 
Jelf said:
I don't get your point. How does it affect to PLC what is the input voltage of datalogger?

But answer to your question. If I use for example 1000 ohm (first resistor) & 100 ohm (second resistor) resistor, Vin 24 V , so voltage to datalogger would be 2,18 volts when switch is closed.

Datalogger's range is 0-10 V.

Or did I misunderstood your question?

The issue is you are asking a question without providing details, noone can give you a correct answer without knowing the details. Can you tell me what flavor my birthday cake was?

Others may answer this better than I can but I will state what I think. The datalogger has an analog input (0-10v) for RPM but you plan to use a relay contact from a device that states RPM is above a certain value. On the input of the datalogger you plan to split the voltage, from the relay, to offer an analog value that matches the actual RMP i.e. 2.18v equals 218 RPM as an example, actual values may vary.

Yes this can work. The current aspect you would need to check the data sheet for the devices involved but 218ma should not be too high for most....BUT check the datasheet.
 

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