The inverter gates are the ones in series. That could have been a single one. But, I figured I make use of the spare gates and have the LED wiring me symetrical.
OK the two diodes D2 and D1 act like an OR gate for active low signals. Active low means they goto negative or rather ground. SO if either are negative the shifetr is in N or F.
But, if neither are negative (actually float) the gate will go HI due its behavior internally.
On some gates the out put will flutter make false output. Float is a term to say the pin is neither
hi nor lo, sort of in limbo. Not good per se.
If it were the case; Here I'd put a pull up resistor in it. I got lucky and the IC does it. You wouldn't want to put that resistor outside of the diode since it then would possibly affect the circuits in the CDI.
And to keep the LEDs wiring similar for polarity I used a second gate. See the little circle on the output: that says the single inverts. Hi is lo, lo becomes hi.
So, showing Neutral is easy. its the straight shot with D3 and U1A. This was all needed to buffer the LED FROM the Switch itself. And that way its trying to not interfere with the CDI's internal logic.
The Gates really should be called U1a U1b U1c. Since they are individual gates on a common IC that supports 6 inverters. I see now I was lazy when I drew the diagram in Orcad.
The power supply. Good question. On the IC it self it needs a ground and a +5VDC. So the power supply wires over to the IC. If I hadn't used U3A then it would also wire to the red LED and that LED would be flipped around and not going to ground. I like my outputs signals active HI (+ output)drain to ground.
Hows that. I hope it helps.
MassOdy aka OverEngineeredOdy