But the doorbell is already running at 3.6V when idle, and I strongly suspect that pulling that line up to 12V could be bad for whatever the doorbell is connected to. It sounds to me like it's the equivalent of a microcontroller input with a pull-up resistor holding it high until the button press holds it low. If it's anything like that, 12V would fry it, right?

My bad, I wasn't paying attention. Let me try again and I'll be back. In the meantime, the original circuit can be simplified this way:



R1 is not necesssary for the circuit, but I placed it there so that you can measure the current being drawn to the 3.6V doorbell for both when the circuit is idle and when the button is pushed.

I changed the values of R2 and R3 to minimize that effect. I suggest you play with this circuit a little, and see how adjusting those values has an effect on the current flowing through R1

 

Thanks for that, but am I right in thinking that this keeps the relay on all the time and only off when the push button is pressed?

 

Thanks for that, but am I right in thinking that this keeps the relay on all the time and only off when the push button is pressed?

Aaarrrrgghhh!!! ...you're right! ... the signal is inverted ... back to the drawing board...

 

I tried the Opto-coupler but it activates my door bell when connecting the anode and cathode to the resistor and door bell 3.6 v and ground.

what is your input ? active low or active high ?