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I am working with this circuit:

triac ckt

I am using a snubberless triac (BTA216X) and hence I am getting rid of snubber circuit (39 ohm res and 0.01 uF cap circled in red). Am I right in doing so?

Besides this, can I get rid of 330 ohm resistor circled in blue? As per my understanding, it works like a pull down resistor but I have seen some comparable circuits not using this resistor.

How critical is this component and what problems might occur if I don't use it?

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    \$\begingroup\$ What is the load? Using a 'snubberless' TRIAC doesn't always mean a snubber circuit isn't needed. Comments on the role and design of the gate resistor are here electronics.stackexchange.com/questions/248743/… \$\endgroup\$ – replete Jun 27 '18 at 8:36
  • \$\begingroup\$ Thanks for the link. It helped a lot. For my use case, the loads can be either capacitive, resistive or inductive. For ex - incandescent bulbs (Resistive), LEDs with their ac to DC converters (capacitive) and ceiling fans (inductive). \$\endgroup\$ – Whiskeyjack Jun 27 '18 at 9:01
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You have to analyze whether the triac in the optocoupler can be triggered by the dv/dt, not just the power triac, before you can draw conclusions about the snubber.

The 330 ohm resistor prevents leakage in the optocoupler from triggering the triac. Snubberless triacs tend to be fairly insensitive so you may not need that, but you can analyze the minimum trigger current vs. maximum leakage (probably at maximum temperature). You should try to keep the opto cool for lifetime reasons, but usually it ends up being mounted near the triac. In this case your triac is guaranteed not to trigger with 2mA gate current at 25°C Tj, and from Fig 7 we can assume that <1mA will not trigger it even when hot. The optotriac leakage appears to be <100uA at 100°C so I think you're fine without the resistor.

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