# Triac power dissipation

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?

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.

• @G36 At max 8 K/W! – winny Jul 10 '19 at 12:50
• Look at fig 10 on the datasheet. "On-state current as a function of on-state voltage" and P = I_load x VT = 10A * 1.5V = 15W So you need a heatsink Rth < (150 - 30)/15 = 8 K/W – G36 Jul 10 '19 at 14:06 