From the data sheet of the LM339 I cannot quite figure out what the maximum AC voltage is that can be applied when used as a single supply crossover detector. Can someone help?

I am looking to use the circuit to detect the zero crossover of a 220Vac circuit and drive a act as triger input to a 5V digital circuit?

http://pdf1.alldatasheet.com/datasheet-pdf/view/3064/MOTOROLA/LM339.html

 

The input common-mode voltage range is from 0V to 1.5V less than its positive supply voltage. If the input voltage becomes lower or higher then the IC will not work. It might be destroyed.

 

Is there something different between an 1x gain op-amp and a comparator?

What else, other than a transformer, can be used to divide the ac voltage to an acceptable level for the LM339?

 

Using the principal of an analog multimeter, 'n large shunt resistor, and resistor divider circuit, will it be safe and possible to implement?

 

The voltage divider will work, since you are sampling rather than trying to drive a load. But the transformer will provide isolation and therefore be much more safe.

 

Is a transformer the only way that I will be able to have the necessary isolation?

 

An opto-isolator is used to isolate lamp dimmer circuits and can be used instead of a transformer.

 

I know of an opto-isolator, but as I understand it can only 'transfer' on/off signals and not sine waves. Also, in this case my input is the high voltage, not as I believe an opto-isolator is 'normally' used. Am I wrong saying this?

 

If you modulate the current of the LED in the opto-isolator with a sine-wave then the photo-transistor will output a sine-wave.