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*Disclaimer:*Some of you might easily recognize that the components and circuit I am talking about are related to one of my projects, on which I had posted some months ago. Actually, the circuit is the same as the one in my project, but the one I am posting in this thread actually uses high voltage AC mains, so I am

*never*going to implement it. This is just to satisfy my curiosity.

The schematic is given below:

As you can see, I am controlling the BT139 triac using the MOC3052 optoisolator. Both of them can handle high voltages. The circuit above is similar to the sample circuit shown in the datasheet of the opto. My question is, how do I calculate the value of R (boxed in red) in high voltages?

Say, I want to connect this Triac to my mains, which are at 220 V RMS, or 311 V peak. As per the datasheet, the opto can handle a maximum of 300 mW power dissipation at its detector. For safety, let's take a value of 250 mW. So, the peak current through the gate of the triac would be ##250~\text{mW} / 311~\text{V} = 0.8 \text{mA}##, and the value of ##R## would turn out to be ##311~\text{V} / 0.8 \text{mA} = 388.75~\mathrm{k\Omega}##. Using standard resistors, I can get ##330~\mathrm{k\Omega} + 47~\mathrm{k\Omega} + 10~\mathrm{k\Omega} + 1~\mathrm{k\Omega}##. Does this look sane?

Also, I had another question: what is the minimum gate trigger current (I

_{G}) required to switch on the triac? In the triac datasheet, the maximum I

_{G}is given as 2A, but nothing has been said about the minimum I

_{G}. If my above calculation is mathematically correct, will 0.8 mA be enough to turn on the triac?