TimWilborne Posted February 15, 2006 Report Posted February 15, 2006 I need to convert 0 to 15 VAC to either a 0 to 10 VDC or 4-20 mA signal. Anyone got any ideas? Quote
robh Posted February 15, 2006 Report Posted February 15, 2006 (edited) I have never used one of these, but this was about all I could dig up. Is it as simple as a voltage transducer? Hope this helps. http://www.a-a-c.com/N3%20WS%205%20ACCD%20ACVT/108X-D.pdf http://www.a-a-c.com/6acvoltagexducr/6%20A...age%20Xducr.htm Edited February 15, 2006 by robh Quote
TimWilborne Posted February 15, 2006 Author Report Posted February 15, 2006 Thanks, also found one by Phoenix Contact. Part Number MCR-VAC-UI-0-DC/2811103. Anyone ever used one of these or any type these. Know some can be fragile. Guess I should mention this is going on a resistance welder Quote
Beuwolf_1 Posted February 15, 2006 Report Posted February 15, 2006 I would think a cheap and dirty way to do it would be to use a signal isolator. Lots of drive makers use them to condition signals of various voltages coming in, and prep it for a 0-10VDC or 4-20mA out put, usually user configurable. Check with Extron or Minarik for a very user friendly condtioner. Hope this helps, Beuwolf Quote
TERdON Posted February 16, 2006 Report Posted February 16, 2006 The cheapest possible solution I can find is this one (that doesn't necessarily mean it's the best): NB: This will only work if you can use -10 V to 10 V instead of 0-10 V. 1) Use a voltage divider to scale down the voltage to reasonable values (slightly below 1:2 seems reasonable - 15*sqrt(2) is just about 20 V, just use neighbor values in a resistance series or something). 2) Connect the output from the voltage divider to your analog input. 3) Do fancy signal analysis in the PLC software to calculate the mean absolute difference to 0 V over a couple of whole AC periods (100 ms, 200 ms, 300 ms etc = smart choice, multiple of 1/60 as well as of 1/50, so it works for us in Europe as well). Quote
benbrad Posted February 16, 2006 Report Posted February 16, 2006 (edited) Or use 4 diodes in a bridge rectifier formation, a capacitor of about 1000micro farad 40v+ rated and 2 500ohm resistors The bridge rectifier would make the voltage positive, the capacitor will remove the ripple and the 2 resistors are the potential divider to take the volts from 20 to 10. Total cost of parts = peanuts. Output = 0-10v. Can be found at any hobby shop or electronics parts supplier. No fancy software required, sorry TERdON Probably all thats in the units that robh suggested. Edited February 16, 2006 by benbrad Quote
TERdON Posted February 16, 2006 Report Posted February 16, 2006 Yeah, I said my solution probably wasn't the smartest. I considered your solution as well, but decided on mine instead. 1) I like programming. :) 2) We didn't learn what amount of current we can draw from the AC circuit - if low (ok, that's not very probable), the usual solution just won't work, you'll have to add an opamp as well (ok, that's peanuts as well, but connecting the parts should be counted as well IMHO). So yeah, unless the possible current draw is very low, your solution would probably be smarter as the cost is minimal compared to the PLC anyway. And if the current is limited, just add the above mentioned op amp (a 24V/-24V one would be suitable) before your circuit and we can disregard that as well. My solution might just be more suited in a microcontroller implementation though, where the cost of the controller easily goes below SEK 10 / €1 / $1 / £1... :) Quote
TimWilborne Posted February 17, 2006 Author Report Posted February 17, 2006 I do something similiar to Terdon's setup on Controllogix. But this one will have a Micrologix and I am worried that I will not be able to capture the sin wave and calculate the AC voltage. Response time must be fairly low. benbrad - You solution will work but I am looking for something more like robh suggested so maintenance can easily replace it if it goes bad. Thanks for all the suggestions Quote
panic mode Posted February 17, 2006 Report Posted February 17, 2006 chap and dirty DIY solutions will not produce good results because diodes used for rectifying will distort signal and input signal lower than some +/- 0.7V will be ignored. besides 1000uF filter capacitor with 500R resistors(s) would be way to slow for changing signals. if one still wants to do it himself, than rectifying should be done with better circuit. here is one using OpAmp. after rectification, one still has to filter and scale signal but this is very simple. Quote
hakko808 Posted February 18, 2006 Report Posted February 18, 2006 What are you listing requirements? Unless you are willing to send any homemade circuits in for testing and pay the appropriate fees, you may be stuck finding something off the shelf. Quote
TimWilborne Posted February 18, 2006 Author Report Posted February 18, 2006 The main reason I want to find something off of the shelf is so that it can easily be replaced by maintenance if it goes bad. Quote
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