Sampling an AC signal

A doorbell's resistance is 50-100 ohms per https://homeguides.sfgate.com/troubleshooting-doorbell-doesnt-work-36512.html

You usually have to have a LED+DIODE+RESISTOR for the lighted bell to work, so (16VAC half wave rectified) at bell or switch when off.
the bell should see 16VAC full wave rectified when switch pushed. The switch would see zero across it.

Now mess that up with a chime doorbell. You would probably have to add a resistor.

RING suggests putting a resistor (I think 50) across their terminals if you want to also use a regular bell. It power steals to charge the battery.

Here is one such device: https://www.elkproducts.com/product/doorbell-telephone-ring-detector-930/

Here https://www.legrand.us/networking/on-q-network-components/network-modules/mute-interface-mi1/p/nvmi1 is another device. Just don't ask legrand anything about it because it doesn't exist.

Visitor said:

When the button is pressed, the coil will see full-wave AC. When the button isn't pressed, there will be a current-limited (by the LED resistor) sine wave. But what will the voltage across the coil at that point?

Click to expand...

It would likely be in the region of 100mV or so, not enough to trigger an optoisolator - especially if it's feeding that via a bridge rec that drops an additional 1.2V.