The problem experienced is that when applied the AC voltaje to the SSR it produces an OFF current about 800microA which is enough to trigger the PLC
input so always the PLC is having a high signal.
What I need to know is what to agg to the SSR to avoid the 800 microA current so the input works properly.
I'm assuming you have 110 V inputs on your AC input PLC.
Place an 8.2K, 3 watt resistor between the input terminal and neutral. The triac will turn off (no leakage) as soon as it sees the 8.2K load.
I've had to do this hundreds of times, and it works great.
Best regards ...
You need to add a 'pull down' resistor parrallel to your input. You did not indicate the ACV level you are working with but using ohms law (R=E/I), you should be able to calculate the proper resistor for your voltage. You need several pieces of information to determine this:
1. Min on level for your input module.(E)
2. Actual leakage current of Solid state device into 0 ohms. (I)
Using these parameters will give you the turn off voltage, however you will want to drop to a voltage about half of the min off voltage for consistant operation, so half your resistance.
Make sure you consider the wattage of the resistor. When the Solid state device is on, you will have full voltage across the device. Again, use ohms law to determine current at full voltage I=E/R, then EI for power consummed. Size up from there.::
What I have used is a Pilot Light, Normally a transformer type. This will keep the leakage drained and the input working correctly. Some inputs have better suppression and this is not a problem.::
William R. Good firstname.lastname@example.org
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