-You spent more components.

Yes, my circuit would need more parts than yours. But if you would design your circuit to really work reliably, then you would also need some more parts:

In your circuit current source is built arround 220k resistor and 3.1V zener diode. This looks nice and flowing current through these parts is very little, but nevertheless this circuit will not reliably work at low DC voltages arround 10V. Why?
Assume 10VDC is applied to your circuit, then maximum current through zener diode will be 10V / 220kOhm = 45µA. Again, a very little current, very nice, but unfortunately voltage drop of zener diode will be far less than 3.1V! Datasheet of standard 0.5W zenerdiode tells, that only about 1.5V is dropping across zener, which lets the current through opto led fall to less than 400µA!
Even increasing current through zener to 1mA, what I would call the absolute minimum, would result in an opto led current of no more than 700µA. A current, which by the way extremely changes with ambient temperature...

So, how will you be able to guarantee that enough current will flow through zener diode under all cirumstances, by the help of a simple high ohmic resistor???

Ok, one could make the voltage drop across the zener more stiffy, by the help of LM431 or similar. But again, some hundreds of µA need to flow to the enhanced zener. How will you guarantee this by the help of a simple high ohmic resistor? My calculation says, that at least 62kOhm resistor is needed to guarantee a current of 100µA to flow through enhanced zener, when applying about 10VDC to the circuit. But then, current through zener increases to 3.3mA, when applying 230VAC...

-You need extra code for activate - deactivate the circuit.

Ok, but that's not necessarily a disadvantage. By the way, you would also need extra code for reading the opto. Why?
It's not at all a good design practice to use optocouplers containing unshielded opto-Darlingtons to monitor mains voltage, because common mode rejection ratio is very poor. So, what does it mean, that a circuit does not provide a proper common mode rejection? It means, that due to common mode noise extra pulses can be generated, which eroneously trigger the opto! The only remedy in such a situation is to read out the opto a few times in a row, to be able to detect false triggers.

It's not a good idea to drive an opto with very little current, because then it becomes more and more difficult for the reciever to reject noise. A much more reliable way to use an opto in a mains voltage monitoring application is to use strong currents for turn-on, something in the range of 10...20mA. And, of course, an opto containing a shield between input and output side should be used!

-What hapend if you forget deactivate the circuit?

Nothing bad, if you add a simple temp fuse, like they are widely used with lightning protection devices to prevent trouble from damaged varistors.
Also, I would use a simple monoflop to guarantee, that opto is turned-on for only 5msec. Such a monoflop can easily be formed by a resistor, a capacitor and a Schmitt-trigger gate.

I'm not a fan of fet transistors

I'm a big fan of them, because they provide unique and unequalled properties!


Kai