Electronic – Optocoupler Question

If the two circuits need to be isolated, then an optocoupler, relay, or optical relay would be preferable. This should prevent ground loops or other dangerous interactions.

A relay may require more current than the gpio can supply, but then again I haven't used them as often as opto's.

The LED has a voltage drop of 2v and a runs well on about 10-15ma, but how do I handle the capacitor in parallel?

The capacitor won't have any effect on the LED current after the circuit has been "on" for a few microseconds, so you don't need to worry about it when calculating the resistor value.

do I need a resistor between the optocoupler and B as the video would suggest?

You don't need a resistor.

You don't need to limit current through the optocoupler --- it won't draw any more current than what's stimulated as photocurrent by its LED (driven by the 15 V AC)

You aren't using the signal at the optocoupler's 'C' pin as a logic signal, so you don't need it to pull to a low voltage when the opto is active.

A current path into the 'B' pin could speed up the operation of the phototransistor, but for this kind of application, you don't need nanosecond response times.

It looks like there's a pathway, when the optocoupler is closed and the transistor is allowing current from E to C, that directly connects the 3.3v source to the GPIO pin without a resistor on the route (via the transistor) - is that a problem? Wouldn't it cause too much current to the GPIO pin?

Assuming you have the GPIO configured as an input (and this should be the power-up default), it will have a high input impedance, and won't allow more than a few microamps to flow into it.

A few k-ohm resistor would likely not change the behavior any and would protect you from overcurrent in the event you accidentally program the GPIO as an output and the driven value conflicts with the value generated by this doorbell circuit.