ac light dimmer light using triac and triac

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tht means I should put led in series with 1k8 resister
 

Yes, I have connected LED in series with 1k8,when I adjust the variable resistor (potentiometer) it goes dimmer and brighter.

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meaning I should connect it in parallel with the MOC internal diode?

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when I connect LED in series with the 1k8 resistor I can able to play with it.it shows brightness and when I switch to the other side it becomes dimmer
 

If you connect the red LED in paralled with the IR LED of the opto, it won't lite since the forward voltage of an IR LED (about 1.2V) is lower than of a red one (about 1.8V). So almost the whole current will pass thru the IR LED.

But you can connect the red LED in series with the IR LED of the opto (pin 1 and 2). It also is better, in this case, reducing 1K8 to 1K5.

Added:
It seems you did it already (perhaps I understood you wrong, you meant you put the LED in series without the pins 1 and 2), But as I said, it is better replacing the 1K8 with 1K5 when you add an extra red LED in series.

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With an Ohmmeter (better if set to check the forward diode voltage), check that the MOC output (pin 4 and 6) are not open.
 
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so, I shouldn't connect my REd LED in pin 1 and pin 2?
 

that meaning I should not use the 1k5 resistor and more?

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I mean where is this 1k8 coming from? is it the one on the Vcc? or should I substitute in 1k8 resistor?
 

that meaning I should not use the 1k5 resistor and more?
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Sorry I just said, replace the 1K8 with 1K5 resistor.

In this case, the current that will pass in the IR LED (pin 1 and 2) is:
I = ( Vcc - V_ir - V_red - Vsat ) / Rs
where:
Rs = 1K5, the series resistor
Vsat = 0.2, estimated for the saturation voltage of the opamp output when low.
V_red = 1.8V, typical forward voltage of a red LED
V_ir = 1.2V, typical forward voltage of an IR LED or can be found from datasheet
Vcc = 12V

I = 5.87 mA

Added:
I hope you have time to check (by using an ohmmeter) if pin 4 and 6 are conductive when the red LED is on (no matter its brightness is).
 
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You may like verifying first if your diode tester is functioning well.

(1) Try to measure the forward voltage of a diode you have.
(2) Try to measure the forward voltage of your red LED.
(3) You may like to also measure the forward voltage of the opto IR LED (pin 1 and 2 after removing the MOC IC from the circuit).

In these 3 tests, the polarity of the multimeter probes is important.
So if you get out-of-range voltage, exchange the connections of the two probes.
If in both cases you get the out-of-range display, then the device under test is faulty.

If your tester is ok, measure the voltage between pin 3 and 6 (the two terminals of the output internal triac) after returning the opto IC on your board.
The polarity of the multimeter probes is not important here.

If the red LED is on as expected when running your circuit and you get the out-of-range display on the multimeter, one of two reasons could be the cause:
(1) The MOC IC is faulty (its output triac is burned).
(2) The triggering current of 5 mA is not enough. In this case, we can test the MOC IC seperately as we will see.
 
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Hi, my fuse meter is damage I will have to go buy another one, I wasn't aware, that is faulty until I have tried to measure diode.
 

So you give me a break... Thank you
Try to buy more than one fuse, as I do
When do you expect you can be ready to go on with the tests?
 
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    dumi

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hi I have measured the diode and I got 6.4V and Red LED 1.92V,the voltage of opto 1.1V.Measurement on pin 6 and 4 is zero.
 

I guess you meant 0.64 V for the diode

Zero volt (which means zero Ohm) between pin 6 and 4, is not realistic! For instance, did you make this measurment before or after returning the MOC IC on the board?

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If the internal triac is burned and is transformed to a conductive piece (like a very short wire of copper) then the 25W bulb (the load of the second test) should always be on when it is connected in series between pin 6 and the mains voltage (pin3 to the second mains line).
 
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not before,after I returning the moc ic on the board.
 

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