How to determine the ratings of radio parts on the diagram?

Question

Arduino

tell me the values ​​\u200b\u200bfrom walking from the task - as a maximum
or teach how to count - as a minimum

Answer 1

keldish , about R4 your estimate is similar to the truth. Further, everything is simple:
The current of the LED in the optocoupler is usually a few mA, for definiteness we take 5. Divide 220/5, we get ~ 43 kOm (this is total for R1 and R3, the power dissipation for them should be at least 1.5 W).
Diode D1 in your circuit is turned on incorrectly, with this inclusion, the control current will not reach the LED. Therefore, we turn it upside down with respect to the drawn one. On R2, at a current of 5 mA, it should fall, say, with a dozen volts - 10/5 = 2 kOm. It is necessary to check on the datasheet whether the LED of the optocoupler is able to withstand a reverse voltage of these 10 volts (after all, 220 is a change!), And if not, then recalculate R2 to such a voltage that it can withstand with a margin.
And the last one is C1. If you want each individual half-cycle of 220 volts to reach the MK, then C1 should be. no more than 0.01 uF. If it is necessary that the MC react only to the very fact of pressing the button, then the half-periods must be smoothed out, then C1 should be electrolytic, at least a few microfarads, plus up.

Answer 2

I ’ll start answering myself
R4 - 10kOm
I think R1, R2, R3 is a divider from 220V to what voltage it’s not yet known depends on the optocoupler
until everything stops at this point

Answer 3

The optocoupler LED glow voltage is greater than the voltage drop across the diode. With a "reverse" half-wave, mains voltage will be applied to the LED of the optocoupler. R1+R3 changes to one. R4 is superfluous, C too. In general, the scheme is not successful