Mike Posted March 25, 2013 Report Share Posted March 25, 2013 One day I had the need to measure the high voltage using the controller V350-35. And he came to such a strange decision. Accuracy, of course, is not high - a few volts. But in this case was sufficient. Link to comment Share on other sites More sharing options...
MVP 2023 Joe Tauser Posted March 26, 2013 MVP 2023 Report Share Posted March 26, 2013 Simple, cheap, and old-school. The only problem is if one of the 100 ohm pots gets disconnected you'll get 500V on the input of the PLC, which I don't think it will like. Putting components like resistors in control systems always makes me nervous. If I was doing it I would spend extra money and err on the side of caution- http://www.nktechnologies.com/voltage-transducers/vtd-dc-voltage-transducers.html At $180 they're not the cheapest things in the world but they won't come back to bite you. Literally. Joe T. Link to comment Share on other sites More sharing options...
Mike Posted March 26, 2013 Author Report Share Posted March 26, 2013 Thanks. Not so all is terrible ). Through resistor 5.1Meg current flows only 100 uA max. I also forgot to mention that all the analog inputs were engaged, so I had to use a thermocouple input. Link to comment Share on other sites More sharing options...
MVP 2014 Simon Posted March 26, 2013 MVP 2014 Report Share Posted March 26, 2013 I agree with Joe, using resistor dividers is a simple way to get a basic measurement but it exposes you to catastrophic damage if something goes wrong. While your circuit is 100% healthy, yes, only a small current flows through the resistors, and the voltage at the measuring point is within the range that the PLC is happy with. However remove one of the 100Ohm pots from the circuit and analyse the result. The input impedance of a thermocouple input is about 10MOhm. So the PLC now becomes a significant part of the voltage divider and you get about 333V on your thermocouple input. The 5.1MOhm resistor is not going to prevent destruction of that input, and perhaps the entire PLC. You also have a safety risk, as there is reasonable chance for high-voltage to enter the low-voltage circuit under fault conditions. It looks like your 500V is DC, which is far more dangerous than the the same voltage in AC. Use an isolated transducer or signal convertor as Joe suggested. I would look for solutions in the following order: look for a programmable transducer, that may give you 500V input and millivolts output Use a transducer with 500V input and 0..10V or 4...20mA output. Use appropriate resistors on the *output* side to drop the signal voltage down to mV for your PLC Use your resistor bridge to drop the voltage on the high-voltage side, then feed that through an opto-isolated signal convertor the PLC. Look for a programmable convertor with the option for mV outputs. For example the WAS4 PRO DC/DC (item number 8560740000) from Weidmuller (www.weidmueller.com) accepts up to 200V input and can be configured for mV output. I won't post a link, as it looks like a temporary link that may not work when you go to click it. Go to the Weidmuller page and search either the description or item number. In any case, make sure the transducer is opto-isolated, and check that the power supply to the module is also isolated from the high-voltage side. I hope this helps. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now