Isolated 12 VDC input to PLC

[cdenk]

I would like to measure say 0-15 vdc from a marine battery with a PLC (TRI-PLC T00MD-888+) that has 0 - 5 VDC ADC input. Concerned with developing a ground loop since there is already a common ground that the PLC is using to control other items that might have chassis ground. Thinking there should be an off the shelf transformer or something to do the isolation.

[Chris Smith]

Transformers only work with AC signals, not DC.

You can make a simple voltage divider out of a couple of resistors, and center tap them for the proper voltage and signal.

A 2000 ohm and a 1000 ohm resistor will center tap at 5 volts using 15 volts. Make sure the 2000 ohm resistor is at the top, and the 1000 ohm is to ground for a five volts center tap.

You can do the math on the current draw for 3000 ohms and 15 volts.

If you want less current draw over all, make them 20K and 10K resistors.

[philba]

A transformer only works with AC. To get true isolation, what you want is an analog opto-isolator like the PS8601

Basically, you will get a voltage out based on the current through the opto's LED. It works in a somewhat narrow range based on the output load resistor so you have to select the LED dropping resistor for that. See Output voltage vs forward current chart on page 8 of the datasheet and remember Ohm's Law.

Is the battery powering all the equipment? If so, wouldn't they all share the same ground? It doesn't hurt to use isolation but I wonder if its really needed here.

<small>[ October 26, 2005, 09:25 AM: Message edited by: philba ]</small>

[rshayes]

A differential amplifier should work for this. Since the source impedance of your signal is low, you should be able to use a single op amp.

Basically, you have two equal resistors on the input side of the op amp and two equal resistors on the output side. The input resistors should be about three time the output resistors to scale the signal down by a factor of three. Somewhere around 30K would be a good range for the input resistors. The output resistors would then be 10K.

The positive terminal of the battery would connect to a 30K resistor. The other end of the resistor would connect to a 10K resistor to groung and to the non-inverting op amp input.

The negative terminal of the battery would connect to a second 30K resistor. The other end of this resistor would connect to a second 10K resistor and to the inverting op amp input. The other end of the second 10K resistor would connect to the output of the op amp.

The output signal is between the op amp output and ground.

This will read the battery voltage even if the negative terminal of the battery is not at the same potential as your local ground. It is entively possible, depending on the wiring and the load currents, that these points are at different potentials, possible a few hundred millivolts to a volt or so.

The 30K resistors can be split into two 15K resistors in series. Capacitors and diodes can be tied to these points for filtering and transient protection if desired.

<small>[ October 26, 2005, 10:33 AM: Message edited by: stephen ]</small>

[Gorgon]

Hi Stephen,
This is a good idea. One thing to remember is to use a symetrical power of about +/- 8-10V to supply the op amp. Just to give it headroom to regulate properly. You might reduce the negative a bit to -5V to make it simpler.

TOK

[upsmaster]

A dc/dc converter with a 9 to 18 volt dc input with a regulated 5 volt out would work. These devices come in isolated fully ot not so isolated, price is mostly based on those features and wattage.

joe