Need some help... Interesting.

[Chemdawg 0]

Hello everyone. I'm working on something, and I'm trying to figure out if I can even do it. Here is what I have. SLC500 with 5/04 processor, 3 relay output cards, 1 DC sink input card, 1 triac output card, and 1 NI4 analog input card. I am also using a Panelview 600 display. This setup is controlling a home model railroad setup. So far everything works phenomenal, but I always find other things I would like to be able to control with it. So here is what I am trying to do. My railroad being the size that it is, is broken down into 8 zones. The track voltage if read from each side of the track will show as 14v AC. if you read each individual track to ground, it's 16v DC. I want to be able to see on the panel view what the voltage is at each zone of the layout. Now it doesn't matter to me if I'm reading the 14v AC or the DC. I would just like to know. I was thinking of the analog card, but that only goes 0 to 10 volt. The 14 is too high. Is there a card I can get that can read this and display it to my panel view? Thanks in advance. Pat

[alan_505 24]

You could use a 0-10v card and then use a resistor to reduce the voltage to the card. Alan

[MrAutomation 20]

Specifically, use a voltage divider. You would probably want to set up the ratio so that your card can read from 0 - 20 VDC (For simplicity, and also in case the voltage goes up for some reason). Connect 2 x 5K Ohm resistors, in series, between the track voltage and DC common (The input impedance of the analog card is 760K Ohms. If you want to get really accurate, factor that into your calculations and get high precision resistors. Typically, you can just ignore this high value). Connect the point between the two resistors to the 0-10V analog input card. You can then scale the input in the PLC so that 0% = 0 Vdc and 100% = 20 Vdc. Done. Easy as that. Bear in mind that this circuit will draw 1.6 mA with a 16V supply. This should also be non consequential, but you can increase the resistors to reduce the current draw if needed. Just don't go too big, or you will have to start factoring the 760K input impedance into the calculation. Edited 12 Nov 2014 by MrAutomation