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I'm still really new at working with triacs and understanding them, but I want to make a circuit that allows a microcontroller to turn on/off an AC signal (very low power, not mains, don't worry) to blink and sequence EL wire.

I sketched up this circuit: circuit

and tried to build a simple version of it. BAT +/- is a 3-4V (depending on charge level) DC battery, AC1 & AC2 are the output of an inverter fed by the same battery. the loads are connected across OUT1A/OUT1B, OUT2A/OUT2B, etc.

my issue is that when I connect the gate pin of one of these triacs to VCC, the el wire turns on (great!), but it also turns on when I connect it to ground (bad!). if I leave it floating, the el wire is off.

did I do something stupid here? I'm still learning the nuances of these things and I'm not sure where I messed up.

thanks so much!

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    \$\begingroup\$ "did I do something stupid here?" - you have the TRIACs wired up wrong? Please show us a photo of your setup. \$\endgroup\$
    – Bruce Abbott
    Aug 23, 2018 at 17:06
  • \$\begingroup\$ don't have the setup on me physically until later today, unfortunately. at worst I suppose I could have mixed up MT1 and MT2 on the triac -- would that produce this behavior? FWIW, I was trying to be really careful about this and triple-checked the datasheet to make sure the pins were correct \$\endgroup\$
    – user358829
    Aug 23, 2018 at 18:00
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    \$\begingroup\$ Define what you mean by " AC signal (very low power, not mains, don't worry)". What is this AC signal and is it DC biased relative to ground??? (Or does it swing fully +/- from ground equally in each direction?) This project may better be done with transistors depending on that. \$\endgroup\$
    – Kyle B
    Apr 7, 2021 at 14:50
  • \$\begingroup\$ Can you show link to Triac datasheet? \$\endgroup\$
    – SolarTec
    Apr 8, 2021 at 8:18

1 Answer 1

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If you connect the gate pin directly to Vcc you'll likely destroy the triac. After that, who knows..

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  • \$\begingroup\$ sorry, connected to VCC through a resistor (R1, R2, R3 on the circuit diagram) \$\endgroup\$
    – user358829
    Aug 23, 2018 at 17:48

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