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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: enter image description here

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 suitable?

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    \$\begingroup\$ Did you mean 120VAC? \$\endgroup\$ – Phil Frost Jan 9 '13 at 22:25
  • \$\begingroup\$ @PhilFrost - good question, it's quite possible he did given the 200W heater mentioned. \$\endgroup\$ – Oli Glaser Jan 10 '13 at 3:51
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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:

Triac Circuit

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.

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    \$\begingroup\$ I found my answer at the bottom left corner of page 2 of the cited by you document. IMO the answer will be improved, should that information be edited into it, with simple explanations. \$\endgroup\$ – Vorac Jan 22 '13 at 15:29
  • \$\begingroup\$ @Vorac - Good stuff, will edit the info into the answer. \$\endgroup\$ – Oli Glaser Jan 23 '13 at 5:00
  • \$\begingroup\$ I'm a bit confused, and unclear about original question. Is R1 wattage in this case VinT*IGT=20V*100mA=2W or R1*IGT*IGT=150E*100mA*100mA=1.5W or maybe something completely else, since 2W seems quite a lot. \$\endgroup\$ – domen Mar 5 '15 at 22:06
  • \$\begingroup\$ The first formula gives you a lower bound: R1_min = Vin(pk) / I_FSM. The second formula an upper bound: R1_max = (Vin - V_GT - V_TM) / I_GT. Finally you can put various commercial resistor Watt ratings into R1 = Vin² / W and see which ones give you a R1 within the other bounds. \$\endgroup\$ – Tobia May 13 '15 at 21:46

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