opto isolator - Optotriac+triac: how do I calculate the gate resistor

I am building a small circuit to turn on/off a 200W heater. I am using a MOC3023 and a BT136. At the datasheet of the the optotriac, this circuit is recommended:

My question is how is the 180Ohm resistor value determined. I understand that it conducts for a very short time, but not much more. Is this a valid value for both 230VAC and 12VAC? Is 1/4W resistor always fuitable?

Answer

The 180Ω resistor will probably need to be altered for 12VAC, what value it needs to be depends on the minimum trigger/latching current needed for the main triac. Also, a 0.25W resistor is only suitable if the wattage doesn't exceed 0.25W, so you need to calculate max voltage across it times current through it.

This Fairchild app note is worth a read for deciding on the values of components to use.

For a resistive load the circuit shown below:

The relevant part of this note with the formulas is:

Triac Driving Requirements

Figure 2 shows a simple triac driving circuit using the MOC3011M. The maximum surge current rating of the MOC3011M sets the minimum value of R1 through the equation:

R1 (min) = Vin(pk)/1.2A

If we are operating on the 115 Vac nominal line voltage, Vin (pk) = 180 V, then:
R1(min) = Vin(pk)/1.2A = 150 ohms.
In practice, this would be a 150 or 180 ohm resistor.

If the triac has IGT = 100 mA and VGT = 2 V, then the voltage Vin necessary to trigger the triac will be given by:

VinT = R1 * IGT + VGT + VTM = 20 V.