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This is a typical circuit for opto couplers:

MOC3020

What should be the appropriate wattage for 180 ohms resistor?

I think power dissipation should be minimal because in off state, there won't be any current and in on state, the voltage drop across resistor will be very small. As such is it advisable to use 0603 package resistor?

Power rating for 0603 was given as 1/16 watts - 1/10 watts in various articles.

Right now I am using 1206 package and it's working perfectly.

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Then that is fine. The resistor only conducts for a short period of time at start of each AC half cycle when the phototriac is on. And only until there is enough current flowing through it to turn on the main triac. The number determining the dissipation is the turn on current at the triac gate and as it is working fine this is low enough not to cause dissipation problems for the resistor.

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This answer has been edited to reflect my latest feeling about the turn-on time that all triacs exhibit. This turn on time can cause the peak voltage across the resistor in series with the gate drive circuitry to reach quite high values for a short period of time. I was going to delete this answer following discussions in the comment section but I do believe it is a real problem that cannot be dismissed so, the answer remains. Basically it boils down to: -

The peak voltage that the gate resistor can see might be circa 310 volts. 

When the main triac is not-yet-conducting this voltage can be up to \$\sqrt2\$ x 220 volts. It will only be short-lived but, that is all that might be needed to cause a breakdown. I'm thinking that if the opto-triac comes on at the peak of a cycle there will be a short-lived voltage rating problem.

A BT136 takes 2 us to turn on for instance and in this time there could be the full peak voltage across the resistor. I'm yet to be convinced that this is avoidable other than by using a photo-triac that is of a zero-crossing type. There are not many 0603 resistors that have a voltage rating this high: -

enter image description here

It could be argued that the 1206 resistor you are currently using may not have sufficient voltage rating. In which case, it may be more than a good idea to use two in series.

What should be the appropriate wattage for 180 ohms resistor?

Regards dissipation, it will be quite low for the reasons you state but you could use a sim to give a better estimate.

So, despite the down votes, I'm sticking by my belief that there can be a voltage across the resistor (no matter how short lived) that could very easily exceed the specifications for that resistor.

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  • \$\begingroup\$ Where is the current coming from to give this voltage when the phototriac is off? If there is any significant current through the resistor the main triac will turn on and there will be no voltage across the resistor. \$\endgroup\$
    – RoyC
    Commented Dec 23, 2016 at 14:21
  • \$\begingroup\$ @RoyC yes I have not stated that very well - I'll rephrase to mention the point I was trying to make. Thanks. \$\endgroup\$
    – Andy aka
    Commented Dec 23, 2016 at 14:36
  • \$\begingroup\$ Nope still not right. The voltage will never get to that as the AC voltage starts to rise the current through the resistor rises until triac turns on this will be in the 1-2V region if that high. You may remember me as OceanP in an earlier alias. \$\endgroup\$
    – RoyC
    Commented Dec 23, 2016 at 14:45
  • \$\begingroup\$ OK you've convinced me of my error. Deleting! \$\endgroup\$
    – Andy aka
    Commented Dec 23, 2016 at 14:48
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    \$\begingroup\$ I cannot delete until it is unaccepted! I suppose there is one circumstance when it will see a potentially big voltage and that is when the opto triac initially turns on at the peak of an AC cycle - this could be avoided by using a zero cross detection though. \$\endgroup\$
    – Andy aka
    Commented Dec 23, 2016 at 14:50

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