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120 VAC voltage detection

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With your original circuit if you add four 45V zener diodes in series with the resistor - two forward facing, two backward facing, it should fix the problem. Maybe drop the resistor value to around 20k. Note that doing this will give you narrower pulses on the output but they will be at 120Hz.

Mike.
 
65VAC instead of 120VAC. I believe this is an induced voltage because of the length of the wires.

65 VAC to ground I can explain easy enough, but not across L1, L2. Induced, maybe.

65 V is easily explained when a particular problem exists. Picture an outlet strip with a computer ( L1, L2 and G) plugged into one outlet. Now picture that the outlet that the strip is plugged into has a bad ground.

Inside the computer is an RFI filter with generally 4 capacitors to ground. They are "identical" and each have the :same"leakage, approximately. You now have ground at approximately 1/2 120 VAC. Been there, done that at home and work and it's not a good sign, You can easily deliver a few mA.
 
This looks really promising. I believe the following will work.

Vth(+) = 5V (From Datasheet)
Ith(+) = 2.77mA (From Datasheet)
V+ = 98V (My desired turn on voltage)

Rx = (V+ - Vth(+)) / Ith(+)
Rx = 33,574 ohms

Rx/2 = 16,787 ohms

View attachment 107963

I'm having a hard time testing this configuration in the office. What is the best way to get a 65VAC input to the chip? I'm using a AC fan voltage control, but it cannot have a open circuit and work. If it is open, the terminals meature about 96VAC all the time. When I put an AC inductive load on it (AC Fan or AC Relay), as soon as the load turns on, the voltage changes.

Also, I'm using a 3.3VDC MCU instead of a 5VDC MCU. Will this affect my Vth(+)?
 
I'd say variac, variable autotransformer, as my first choice.

A 2400 ohm potentiometer or close to that. parallel a resistor to a 5K pot to get around 2400. Check currents and power. Use a combination of a fixed resistor and potentiometer. You need to be able to supply the current the OPTO nees without exceeding the power dissipation of the variable resistor.
 
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