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triac/opto driver question

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Full Member level 2
Jan 26, 2006
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opto driver

I am designing a project that will use a triac to control a resistive load (fuser from a laser printer), for isolation, I plan to use a moc30xx (probably moc3032) to drive the triac.

Here is an example of the kind of circuit I will use:

My question is on sizing the 180 ohm resistor (from "hot" to pin 6). I haven't figured out how much current will flow. If I use ohms law blindly, it comes out huge yet I've seen projects that use 1/2W resistors for this application so I must be missing something. How do I calculate the actual current through the resistor in order to size it properly?

note, I have looked at a number of datasheets for this class of parts and looked at lots of tutorials but have found nothing on this point.

thanks for any help.



You don't need to put a too low resistor value, those 180 ohm is too low. Only need a resistor that provides enoght current to clamp the triac, with only a few uAmps its enoght. So, I don't know what is the value of "v hot", but if suppose VHot=313 V (PEAK), and you will put 500 uA to gate, the R could be: 313V/500uA= 616 Kohm. Even, you can choice, for example: 220K or 470K. Then power of resistor: V^2/R = 313 * 313 / 220k= 0.44 Wat. And 313*313/470k = 0.20 Wat.

Remember that 0.25 W resistor max tension is 150V, so you can put 4 resistors of 100K 0.25W in series.

Hope it helps

When the triac in the MOC3032 fires it turns on the external triac.
When the external triac conducts the voltage over it will drop to about 1,5V
(depending on the type) - the "rest" of the voltage will be applied to the load.


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makes sense. I was going down that path - noticing that when the moc3032 is not on, it's leakage is in the uA range. I assume there is some delay in the turn on when the resistor sees the full 120VAC so that needs to be factored in but even then the wattage will be very small (1.5*1.5/180 plus the latency wattage).


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