Hey All,

I'm struggling to calculate what resistor power rating would be suitable for the gate input of the BTA24. The Igt for the TRIAC is 0.025A and trying to figure out if I need to find R by using R= Vs/Igt , R=V(out)/Igt, or P=I^2(R)[adjusting R to meet P]. I just want to make sure the heat dissipation at the gate isn't exceeding the resistors power rating and fail. I'm trying to use SMD resistors on the designed PCB. Any help or direction is much appreciated!

 

Also, if the max current (Igt) is 0.025A = 25mA, I imagine I'd need to size the LPF resistors to limit the potential current to below 25mA @ 208V similar to an LED? When I do the math I'm getting about 8300 ohms, clearly I'm missing a concept somewhere.

 

No. 25mA is the minimum current that is guaranteed to trigger the triac. If you keep it below 25mA it won't trigger.
Use 220Ω, which will give a gate current of about 1.5A.
Don't forget that the triac triggers within about 1us, shorts out the supply of gate current, so that the 1.5A pulse only lasts for the time it takes to trigger.
You need to get a triac triggered as quickly as you can, which means using as much gate current as you can.
Don't bother with C6, it only slows down the gate pulse, unless you are in a truly dreadful EMC environment.
Also, look at the voltage rating of the resistors in the gate circuit. Surface mount resistors and some through-hole types are only rated at 200V, so you need two in series.