I'm driving a high power load with the BTA440Z (datasheet):
line voltage is Vac = 230 AC, load power is P = 1.33 kW, thus load current is ~ I = 5.8 A.
If i swap the load with another that draws P = 2.33 kW, then I becomes ~ 10 A; in this condition the triac overheats and it doesn't un-latch due to thermal runaway.
To properly size the heatsink, I need to know how much power the triac is dissipating, and I realized that in the datasheet there is no such thing as an equivalent to what in a MOSFET would be "Rds on", and that I don't actually know how to find such a parameter.
The circuit is as follows, taken from this datasheet, without the snubber network:
Initially R1 was 680 ohm, because i thought that it could work anyway and less current would pass through the opto; but i subsequently dropped it to 360 ohm because i figured that more current through the gate would make for a wider channel between the main terminals: is this assumption correct? Does this current also contribute to the power dissipated on the triac?
Based on what i read, what i know, (electronics-tutorials.ws, link 1 and link 2, and the Wikipedia page about Triacs), and the following image, my guess is this:
for the most part, the power dissipated by the triac is given by the current to the load I, times the triac on-state voltage VT.
Is this correct?
Please note that without a zero-cross detector, my microcontroller won't turn off the opto after the triac has been fired into conduction, so assume the opto is always on.