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My Understanding

As far as my understanding of MOC3021 optocoupler goes, it should latch pins 4 and 6 whenever there's a current flow between pins 1 and 3.

The Confusion

Image 1 - D1 not glowing and D2 glowing

So, in this image no current flows between pins 1 and 3, however pins 4 and 6 is latched. Image 1 - D1 not glowing and D2 glowing

Same happens on the breadboard except that even the LED between 4 and 6 isn't glowing either on the breadboard ( unlike the simulation ).

Breadboard circuit

Image 2 - D2 not glowing and D1 glowing

Image 2 - D2 not glowing and D1 glowing

Same happens on the breadboard

Image 2 - D2 not glowing and D1 glowing ( Breadboard )

Update : Yes increasing current works out fine

Actually it's part of an arduino circuit in which I would be giving an output from an arduino pin to this, and there won't be any LED ( not white one for sure as I came to know now that it requires more voltage ).

So, perhaps it would be fine with an arduino pin, this was just to check the components before hand. Also , since MOC3021 has 15 mA forward current, so 5V would do the work ( as high in arduino pin would give 5V ). I suppose even 3.3V would do the work.

Thanks all .

That's great piece of info Finbarr and jonk ( about white LED more forward current and choosing a specific opto isolator like why to choose MOC3021 or MOC 3022).

I am a newbie in this field , so it's great to get those tidbits of info.

working

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  • \$\begingroup\$ Try to increase the input current to 20mA \$\endgroup\$ – G36 Nov 22 '18 at 17:52
  • \$\begingroup\$ Your diagram shows a red LED but it looks white in the picture, White LEDs have a much higher forward voltage than red, it's almost a miracle it lights up at all. \$\endgroup\$ – Finbarr Nov 22 '18 at 17:59
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    \$\begingroup\$ I only buy the MOC3023 because it requires less current. My memory says that you need \$5\:\text{mA}\$ for the MOC3023, \$10\:\text{mA}\$ for the MOC3022, and \$15\:\text{mA}\$ for the MOC3021. And in all cases I plan on \$1.5\:\text{V}\$ for the emitter diode. \$\endgroup\$ – jonk Nov 22 '18 at 17:59
  • \$\begingroup\$ That's a great piece of info Finbarr and jonk, I am a newbie in this field , so it's great to get those tidbits of info. \$\endgroup\$ – Sizu Taylorventuresllc Nov 22 '18 at 18:15
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You quite clearly say you are using a 3 volt battery so I would expect the current into the gate to be quite small circa 1 mA and, given that the MOC3021 is typically rated to switch at 8 mA you just may not be providing enough current to the gate.

The gate current is too small because the device itself will need about 1 volt across its input terminals and the LED in series is probably dropping about 2 volts. Hence you have used up all 3 volts in bearly overcoming the blocking voltages in the circuit.

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