1
\$\begingroup\$

I'm trying to make simple lamp switching circuit to work, but it just doesn't want to work as intended - question is, why?

I'm using MOC3021 random phase optotriac with BT136/600E sensitive gate triac in such configuration:

schematic

simulate this circuit – Schematic created using CircuitLab

First thing - voltage V3 does nothing to this circuit - doesn't matter whether it's high nor low nor variable (zero-cross depending) - hot side never reatcs to it.

Second thing - depending on R3 (I've read somewhere that it's necessary if triac is sensitive) - if I plug this resistor, triac is always on, but if resistor isn't there - triac is always off. And it never reacts to optotriac U2.

Last thing - and very weird one - in configuration without R3 resistor, if I try to measure voltage on triac's gate, bulb sometimes flashes for half a second or so and the voltage drops from 230V to 30V and goes up again - same happens when I'm plugging/unplugging mains plug - looks like there's some connectivity problem, but there's none, checked it.

I haven't used a snubber for now, because lamp is resistive load so it shouldn't make any difference.. right?

And yes, I've checked the triac and optotriac, they seem to be good, even exchanged both with new ones - still the same.

What can be wrong with this circuit?

Regards

\$\endgroup\$
1
\$\begingroup\$

Usually the problem is that you have swapped MT1 and MT2 on the triac.

... depending on R3 (I've read somewhere that it's necessary if triac is sensitive) - if I plug this resistor, triac is always on, but if resistor isn't there - triac is always off.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. (a) What you intended. (b) What you've done.

The gate needs to be connected to MT2 to turn the triac on. Your R3 test confirms that this is the error. Your opto-triac is connecting the gate to MT1 and so it does nothing.

\$\endgroup\$
  • \$\begingroup\$ Yes, now it works like a charm, thank you very much! \$\endgroup\$ – Endlesik Sep 2 '18 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.