Thanks for the replies. I don't work much with Zeners, so hence why I'm asking. I have been using optocouplers to trigger this part of the circuit, but they are expensive compared to a couple resistors, zener and cap.

My circuit is simple; main supply goes through a voltage regulator, so the supply to the PIC is always 5 volts. This other input is just to tell the PIC that this wire has been activated. I applied the Zener to the application since any kind of over-voltage situation will increase the voltage going to the PIC. I wanted to cap it off at 5 volts.

I chose the higher values so the PIC would read them as ones and zeros. If it was lower, I would have to do an ADC, right? I've modified my circuit design from the opto to the VD. If I did an ADC, I would have to reroute the traces to the ADC pins, and that's a cumbersome redo of my board.

My comments about the load was stating that I just couldn't have the supply come in and go straight to the PIC and have the Zener regulate it to 5 volts. It has to pass through a resistor first, or else the Zener would blow and then the PIC. Does that sound right?

Also, I am aware of the typical VD circuit where you have two resistors in series, supply is on one side of R1 and ground is on the other side of R2 and you tap in between R1 and R2 to get your divided circuit. I have found two such circuits with the same layout, but instead, Vin and Vout are reversed. The supply is applied between R1 and R2 and while R2 goes to ground, the other side of R1 is your output.

Trying to use the same formula but solving for the other unknown variable using algebra does not work. Is there a different formula for this method? In this instance, R2 is 10k and R1 is 100K.

All I want to do is trigger a pin on the PIC to let it know an input has power applied to it. There's optos, VD and transistors that can be used. Would a transistor be simpler to use in this instance?

Thanks,
Tony