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I’m having a problem working with TRIAC (BT-131) in my project. I’m trying to switch a low power LED with a 5V DC output from a sensor. I’ve used an Opto-coupler (MOC3021) in between the TRIAC and the sensor. When the sensor gives a 5V output (through a 470ohm resistor) to the Opto-coupler, it triggers the Gate of the TRIAC (through a resistor, R6). When R6 is 100k, the LEDs glows at the moment of switching, then turns off. If I use 10k or 470ohm resistor as R6, the LED glows continuously, but very soon R6 burns out and LED goes off.

schematic

simulate this circuit – Schematic created using CircuitLab

I don’t understand why is it happening and what is the remedy for this. I searched about the characteristics of the TRIAC in Datasheet, searched YouTube for similar projects, but didn’t understand the reason behind the phenomena.

Can anyone help me in this ?

My project setup

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    \$\begingroup\$ Please edit your question and draw a schematic with the tool \$\endgroup\$
    – Voltage Spike
    Jun 25, 2019 at 16:19
  • \$\begingroup\$ I hope this piece of content will help you to understand electronics-tutorials.ws/power/triac.html \$\endgroup\$
    – Rush PCB
    Jun 25, 2019 at 20:11
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    \$\begingroup\$ Add schematic please. \$\endgroup\$ Jun 26, 2019 at 4:31
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    \$\begingroup\$ Two capacitive droppers, mains on a messy breadboard, a rather pointless optotriac, and all that to switch on a couple of LEDs? Honestly, get a power supply and run your entire circuit on 12V. Far simpler, far safer. Probably more efficient, too. \$\endgroup\$
    – marcelm
    Dec 11, 2020 at 16:39

3 Answers 3

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Slightly tidied-up schematic to clarify the TRIAC switching arrangement.

When R6 is 100k, the LEDs glows at the moment of switching, then turns off. If I use 10k or 470ohm resistor as R6, the LED glows continuously, but very soon R2 burns out and LED goes off.

This doesn't make any sense. R2 is, effectively, on a separate circuit to the triac. The two circuits are connected at the mains (as though you had plugged them into separate wall sockets) and the PIR signal is opto-isolated from the triac so there is no reason for high currents in R2.

A photo of your setup may help us to diagnose further.

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  • \$\begingroup\$ Extremely sorry, my mistake.. I mean to say R6. The resistor R6 burnt out. R2 is alright. \$\endgroup\$
    – Roy
    Jun 29, 2019 at 5:30
  • \$\begingroup\$ How about the photo? \$\endgroup\$
    – Transistor
    Jun 29, 2019 at 7:45
  • \$\begingroup\$ My setup is not so arranged for the time being as I expected. I think it will not be clear from the photo. Still I'll try to upload a proper photo of the setup. \$\endgroup\$
    – Roy
    Jun 29, 2019 at 17:20
  • \$\begingroup\$ I have uploaded the photo of my setup. The colours of the connecting wires used here, are not according to the proper colour code. Just follow the connection paths. \$\endgroup\$
    – Roy
    Jun 30, 2019 at 16:22
  • \$\begingroup\$ I can't see anything obvious in the breadboard layout (other than the whole thing is very messy for a mains powered circuit). The breadboard is unlikely to have a mains voltage rating. \$\endgroup\$
    – Transistor
    Jun 30, 2019 at 17:42
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R6 and R7 don't have enough watts.

They are exposed to about the same voltage as the rest of the high power circuit. Because you list your frequency at 50Hz I assume you are in a 230V region.

At 100k you are on the cusp of not providing enough drive current to the gate of the Triac. The resistor doesn't burn because at that point it is sitting at about 0.5 watts. From your picture, it looks like you are using 0.25 or 0.5 watt resistors. At 10k the wattage goes up to about 5 watts and the resistor will burn.

Try putting 3 or 4 100k resistors in parallel. That should give you enough current and all the resistors would share the watt load. Also, step up to 1 or 2 watt resistors.

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Check voltage supply to PIR sensor. The capacitor C1 is only feeding half cycle, so no much current will flow through it. Another diode D4 is necessary to charge this capacitor in opposite polarity, or you can move the zener D2 in this place. Limiting resistor R2 should better placed in series with capacitor C1, so it will work in both polarities. I have modify the triac circuit T1 <-> T2, for better understanding instead of switching at hot side.

enter image description here

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