I was asked by the boss to purchase or build a PLC I/O test panel. Some people call them PLC trainers. Iíve seen many of them sitting in other PLC programmers offices. For the most part, most of them are home made and are just pieces of wood or a NEMA box with holes drilled out so that the programmer can mount four to eight pilot lights and switches to be used as discrete inputs and outputs. I donít know if you know what Iím talking about but these I/O testers are used to test software integrity for the proper operation of both software and hardwired Permissives and Interlocks. You would just run a circuit between the I/O power supply, the switch, to the pilot light and then into the PLC input module channel. By closing the switch on your tester you could then simulate a field generated discrete input such as a limit, pressure or flow switch. In the case of an output you would just connect the PLC output module channel to a light so that when the output was active the light would come on. By doing this you could monitor the PLC program and weed out the bugs that were in the original program design. So I built a 32 channel I/O tester with a variable 4-20mA loop source. In the tester case were everything is mounted,
I installed a 24VDC 1Amp power supply. I also installed a 4-20mA panel meter (Fox), a 5K resistor and a 1K resistor. Now I know to reduce line Amps you need to install the resistors in parallel and to reduce voltage you would install the resistors in series. For both voltage and Amp reduction you would use a combination circuit. Also the mA meter has a 8 to 26VDC 100mA power requirement.
To reduce the power supplyís 1Amp level to the required panel meters power requirement of 100mA:
Should I use a 250 Ohm resistor in between the + and Ė of the panel meters power connectors? Do they make a 240 Ohm? If so should I use this instead? Or should I use a 10K and a 250 Ohm in parallel?
Do I connect the two variable resistors (5K / 1K) in parallel? If so how is this done? There are three connectors on each, left, middle and the right side. I know that the right and left connectors come off the ends of the resistance coil and the center one is the slide. I think I connect the two slides together and then tie them to the Pos. (+) line between the 24VDC 1Amp power supply and the panel meters 4-20 mA signal input. Then I would connect the two right hand side connectors together and tie them into the Neg. (-) line between the 24VDC 1Amp power supply and the Neg. (-) terminal screw that will be used to connect the PLC A/I module channel. The Pos. (+) terminal that connects to the PLC A/I module channel will run to the Neg. (-) connector at the panel meters 4-20 mA signal input. Is this right?
Does it matter which variable resistor is down stream (closest to the power supply)? If question #2 is correct and I do I connect the two variable resistors (5K / 1K) in parallel? Then when both of the variable resistors are at full potential I will have 833.33 Ohms which will bring the 1 Amp down to about 28mA. If this is true then how do I adjust for the 16mA between 4 and 20 mA? Should I put a resistor in parallel up stream between the first variable and the power supply? If 1200 Ohms =20mA and 6000 Ohms = 4mA then should I use two variable resistors with a total parallel impedance of 6000 Ohm in order to get down to 4mA.
And one last Question?
I'm worried that if I should accidentally turn the resistance down to low that I could burn out the meter with 1Amp being sent into the 4-20mA signal connection. Do you have any idea of what I could do to prevent this from happening. I'm thinking a resistor in parallel between the variable resistors and the power supply. I know I will have to redo my math if this is the case. Finally what resistor wattage would you use.