I want to trigger a triac on or off by setting pulses instead of a constant gate voltage. The idea is that you provide the initial gate current for a short time and then somehow the triac keeps itself turned on. Is there a way to do so?

I've found similar circuits using relays (e.g. https://www.eleinmec.com/article.asp?24) and I am trying something similar with triacs (unsuccessfully so far).

I have also got to mention it is for AC use

Has anyone done it and willing to share?


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    \$\begingroup\$ TRIACs already latch. Did you underspecify your requirements? \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 28 '16 at 11:10
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    \$\begingroup\$ This should explain everything you need to know. And more. For free. \$\endgroup\$ – EM Fields Apr 28 '16 at 11:34
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    \$\begingroup\$ Is this for AC or DC power? \$\endgroup\$ – JIm Dearden Apr 28 '16 at 11:58
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    \$\begingroup\$ Triacs turn off (well, usually) every time the current drops below the 'holding current' typically some tens of mA. This happens twice per cycle on the AC line. If you want it to stay on for multiple AC cycles you need a flip-flop or something equivalent (and a power supply). If you are switching DC you should use an SCR not a triac but neither are easy to turn off for DC- there used to be devices called GTOs (not made by Pontiac- Gate Turn Off) but they are even more obsolete than Pontiacs. \$\endgroup\$ – Spehro Pefhany Apr 28 '16 at 12:20
  • \$\begingroup\$ @JImDearden This is for AC \$\endgroup\$ – Pet Apr 28 '16 at 14:12

You could use a photodiac to drive the triac with an SR latch (a couple of cross connected 2 input NAND gates) to drive the led.

This would respond to pulse inputs and isolate the driving circuit from the mains.

enter image description here

  • \$\begingroup\$ Thanks for this, it is close to what I am trying to achieve. However the problem is that the DC side is battery powered and it has to last. Ideally I would like the latching on the AC side \$\endgroup\$ – Pet Apr 28 '16 at 14:15
  • \$\begingroup\$ @Pet Using CMOS gates (4011) uses very little current - you could try increasing the value of R3 to reduce consumption \$\endgroup\$ – JIm Dearden Apr 28 '16 at 14:19
  • \$\begingroup\$ It is not the logic gates that wory me. If I am not mistaken, most common phototriacs (e.g. moc3040 and the likes) draw around 10-20mA. Out of a - say - 500mAh battery that gives 25 hours ON time... \$\endgroup\$ – Pet Apr 28 '16 at 14:23
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    \$\begingroup\$ @Pet The only current you need to draw from the battery is to supply the latch and drive the LED. The current through the diac is taken from the ac supply. This problem of battery and battery life wasn't mentioned in the original question and so isn't really a valid point to raise now. However, as you have access to the AC supply why not just add an (isolated) low voltage supply taken from it (small transformer, rectifiers, caps and regulator) and use this to drive the latch circuit. In that case your battery supplies no current to the circuit. \$\endgroup\$ – JIm Dearden Apr 28 '16 at 14:41
  • \$\begingroup\$ thanks for the help. I wasn't clear enough on what I want to achieve but I appreciate your effort to provide a solution. \$\endgroup\$ – Pet Apr 29 '16 at 13:57


simulate this circuit – Schematic created using CircuitLab

This is old school set/reset SCR circuit

  • \$\begingroup\$ I take it this is DC only?... What about AC? \$\endgroup\$ – Pet Apr 28 '16 at 14:17
  • \$\begingroup\$ @Pet Sorry, DC only. You didn't specified at the time of your questin that you want AC. \$\endgroup\$ – Marko Buršič Apr 28 '16 at 14:18
  • \$\begingroup\$ You are right, sorry about that. I have edited the original question. \$\endgroup\$ – Pet Apr 28 '16 at 14:25

You can trigger the gate of the triac using a latching relay that is rated for a couple A and for the mains voltage.

Just connect MT2 to the gate through the relay contact and a resistor of a few hundred ohms. The relay contact switches very little current on average so it will last a long time.

Make sure the relay isolation is appropriately rated for safety reasons.

Edit: Alternately, consider a dual transistor-output optoisolator, a supply derived from the mains to supply the circuit, and either direct drive or a triac driving optoisolator. Control a S-R latch (74HC74 using R and S inputs or a cross-connected gate) with the dual optoisolator.

Pulse the LED of one half the dual and it turns on, pulse the other LED and it turns off.

If the mains power gets interrupted it will lose knowledge of the state. If you want to avoid that, it might be easiest to replace the gate with a microcontroller that has EEPROM, but everything else could remain the same. It's pretty straightforward either way, but a fair number of parts.

  • \$\begingroup\$ This would be my last resort. I would rather not use relays... \$\endgroup\$ – Pet Apr 28 '16 at 14:17

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