I hope the title is acceptable under this website rules but it is exactly what happened. I wired up this circuit that I thought I tested in the past multiple times, but I might have wired up slightly different. However the following schematic is an accurate description on what I had.

The optocoupler triac driver is for the sake of this experiment always kept "on". When this works it will be controlled by a microcontroller, but for now it's tied to Vcc. NOTE: No, I do not want to use a zero-crossing sensing driver. This circuit will be used in a way that will make that inherent, but that part is not the subject of my tests and I want to understand why this circuit blew up.

The load is an oven element that works with the typical 240VAC split phase common in North America. The power triac is a beefy BTA40 mounted on a big heatsink. Datasheet links below.

When I was about to close the breaker to power this circuit, resistor R4 immediately became white incandescent. However, I have tested a variation of this circuit successfully in the past. What is really going on?

Circuit schematic

R4 at 330R should guarantee that even at peak voltage (240*SQRT(2)) there is never more than one amp through the MOC3053 (Itsm max = 1A according to the datasheet). In reality, as also the datasheet says:

The power dissipation of this current limiting resistor and the triac driver is very small because the power triac carries the load current as soon as the current through driver and current limiting resistor reaches the trigger current of the power triac. The switching transition times for the driver is only one micro second and for power triacs typical four micro seconds.

However it looks like that in this configuration the poor resistor had to dissipate so much more than that.

Datasheets: MOC3053, BTA40

EDIT: I was suggested to move the load, so I tried to put it "under" the triac; will this work?

enter image description here


2 Answers 2


Your wiring is wrong- the optotriac should go from MT2 to the gate.

It doesn’t matter whether the load is on one side of the triac or the other, so long as the optotriac + resistor goes from MT2 to gate.

Either of the below is okay:


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ I am sorry but I am confused on what MT2 is and ... well the triac driver has to go to the gate, is it not what happens there? \$\endgroup\$ Commented Jun 13, 2020 at 0:17
  • 1
    \$\begingroup\$ MT2 is the “top” triac power terminal. When the triac turns on, the voltage across the triac drops to a volt or so and the dissipation in the resistor drops to nothing. The way you’ve got it connected, the voltage across the resistor is the full mains voltage as long as the opto is getting drive, so the dissipation is enormous and it fries. \$\endgroup\$ Commented Jun 13, 2020 at 0:21
  • \$\begingroup\$ OK my triac (BTA40) datasheet calls it "A2". Should I "flip" the triac so that A2 is connected to "PHASE2" or should I move the load on the A1 side like I did in the edited circuit? \$\endgroup\$ Commented Jun 13, 2020 at 0:23
  • \$\begingroup\$ The load can be connected between MT2 (“top” terminal) and the mains, or between MT1 (“bottom” terminal) and the mains, but the opto must be connected between MT2 and gate only. \$\endgroup\$ Commented Jun 13, 2020 at 1:11

Connecting the load in the wrong place is what killed it. The way it is connected, the current is divided between the load and the 330 ohm resistor as if the two were connected in parallel.

As you revised the diagram in the question is correct.

  • \$\begingroup\$ so I guess in a triac the gate is "shorted" with the side of the main terminal when it is conducting. I really need to study these devices better. \$\endgroup\$ Commented Jun 13, 2020 at 0:13

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