ac light dimmer light using triac and triac

i am referring to the circuit U1:A

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shouldn't I take the output of U1:A and connect it to Rs of MOC pin 1 and then take GND and connect it to pin 2 of Moc. What should the output Voltage of U1:A be in order to power the MOC?

Let us start from the beginning
U2:A pin 1 is high only when the voltage of pin 2 is close to zero (actually be lower than the voltage at pin 3). So we get a short high pulse at pin 1 at every zero crossing. Agree?

Yes. I agree.

Ooh,I jst saw the mistake dat I have done,yes I have swap the inputs of U1:A,thnx.Can we move to the next problem which is moc3023 and the rest.

This high pulse, thru R7 and Q1, should discharge C2 and the output of U2:B (pin 7) becomes high too (but a bit lower than 12V).
After the pulse, if V(in-) at pin 6 > V(in+) of pin 5, Vout (pin 7) decreases linearly. But in your circuit V(in-) < V(in+). Could you see it? In this case, Vout stays at its high level.
If you exchange the connections at V(in-) and V(in+), V(in-) becomes higher than V(in+) and we get a ramp (from high to low) at Vout (as shown on the attached image). Agree?

Yes,I saw it then I change it.

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what i am stack with,it is when I connect my pulse width modulation to MOC 3023,can u be able to test it for me?

Now, the attached image is for the full circuit.

Added:
Sorry, I forgot adding the power triac

Meaning that my circuit is fine. What is the gate current needed by the triac?

Do you know the type of the triac that you can use?

Added:
From the datasheet of MOC3023, the outut current peak is 1A (short duration).
The worst case is when Vac is at its peak. Assuming Vrms(max) is 250V:
Vpeak(max)= 250 * SQRT(2) = 354V
The series resistor (Rs), with the gate of the power triac, could be calculate as:
Rs = 354V / 1A = 354 Ohm
We choose Rs = 390R (in your circuit it is 12R).

Assuming the gate sensitivity of the power triac (as BTA06-400) is 25 mA typically, the triac turns on at:
Vt = Rs*Ig + Vtm + Vg
where:
Ig = 25 mA, assumed as the minimum triggering current
Vtm = 1.3 V, the output drop of MOC3023 (datasheet, "On-state Current vs. On-state Voltage")
Vg = 0.7, the gate-M1 drop of the power triac which is equivalent to the typical forward voltage of a silicon diode
Therefore, the triac turns on when its voltage on MT2 and MT1 (output) reaches the voltage:
Vt = 390 * 0.025 + 1.3 + 0.7 = 11.75V

Let us find out the power drop due to this delayed on state (neglecting the power triac on voltage):

P_load/Pmax = [ π - β + sin( 2*β ) / 2 ] / π
where:
π = PI = 3.14
β = arcsin( Vt / Vpeak ), in radian

For Vrms=220V, Vpeak=311V
β = arcsin( 11.75 / 311 ) = 0.0378 rad

P_load/Pmax = 99.9%

So, when there is triac,should I connect 12V to pin 1 of MOC and the output of U3 I should connect it to pin 2 of MOC?

So, when there is triac,should I connect 12V to pin 1 of MOC and the output of U3 I should connect it to pin 2 of MOC?

Click to expand...

It is not about triac, it is about LM358 (or LM324 and alike). This opamp can sink current better than sourcing it. Since most ICs including MCUs (mainly the older ones) have this feature, I personally used to arrange the design so that the load could be connected between Vcc and the output pin.
In the circuit we discuss, the load current is relatevily low, just about 5 mA. So you can connect the load (here the load is the LED of the opto IC) between the opamp output and ground (always with a series resistor to limit the current). But if you do it (as in your first circuit), my connections to the two inputs of the opamp should also be exchanged to reverse the high/low state of the output (otherwise the load will be turned on always). I hope you get it.

I'm at lost.

Is it a real project or just a study?

It is a real project. My Mentor said I should test it using a light bulb before I will connect it using an AC Motor,so that is why I was asking if,that the current coming out frm the MOC will it be enough to power Triac?and should I connect the 12V through to pin 1 of MOC and the output of U3 through to pin 2 of MOC?

Please note that controlling the speed (power) of an AC motor is not exactly the same as dimming the light of a bulb (I assume it is a simple bulb that has a resistive filament).
I hope you know the power of the motor that will replace the bulb of the test.
About the circuit, it is better that each of its elements has a name (label) and not only the opamps so that we are sure we are talking about the same resistors for example.

should I connect the 12V through to pin 1 of MOC and the output of U3 through to pin 2 of MOC?

Click to expand...

Please remember that there is no single way, in electronics, to design a circuit for a specific job. So if one likes changing a part of a working circuit, he has to change other parts or connections as well so that the result (the output) can be the same as it is of the original one. Please note the changes at the inputs and output of U3 on the attached circuit here in comparison with the one attached to post #27.
https://obrazki.elektroda.pl/4883173900_1381818518.png

About replacing your 12 Ohm (between the MOC and power triac) with 390 Ohm, it seems you couldn't get how I did the calculations. I wonder why

Added:
Note: if it is my circuit, I would add a diode in series with R4 (10K, on my attached image). Its direction is from the output (PLS) to the base of Q1. This prevents applying a relatively high reverse voltage (about 9V) on the base-emitter junction of Q1 when PLS returns to zero volt (trailing edge of the pulse).

Added:
Yesterday, I tried completing the circuit by adding the power triac. Unfortunately, the simulation speed becomes very very slow! Unless I find out a way to let its speed normal after adding the model of a power triac as BTA06-400 (it is good for 60W bulb but perhaps not enough for your motor), I can only attach the image of the circuit but without its current/voltage traces.

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At last, I managed to have a model of triac that restored the simulation speed to normal.

I will test it physically and I will let u knw of the results.

my bulb doesn't want to turn on,is it because I am using BTA16-600?

I don't think it is the cause.
When the delayed phase of the triggering current is close to 90 deg (5 ms), the voltage on the power transistor reaches about 311 V (Vrms*1.414). At this moment, the output triac of the opto is on and the current passing thru it to the gate is:
Ig = 311 / 390 = 0.797 A (about 800mA)
This current triggers even the most powerful triac

Even if the delayed phase is relatively small therefore the triggering current is small too leaving the power triac in the off state, soon after while the AC voltage raises there will be a moment when this current becomes big enough to trigger the power triac.

For instance, if the power triac is off, check if the 390R resistor is hot. It should become too hot if the MOC IC is not defective and its small triac is properly turned on.

Added:
You can test the output of the MOC as follows:
1) the power triac is removed and its load.
2) the 390R is replaced with a 25W filament bulb.
(see image of post #27)
The MOC should dim the light of the 25W bulb without being damaged.
The load current thru the opto triac will be:
I = P / V = 25 / 220 = 113 mW (not big for the small triac of the opto)

Added:
Try inserting a red LED in series with the 1K8 (at the input of the MOC). If it is off, chech the connections and the polarities.

No,is not hot.

It means the triac of the opto is also off

I suggest:
1) you remove from the test everything after the MOC... I mean you remove the mains supply, your load, the power triac and the 390R.
2) Then you replace the input of the MOC with a red LED (note the polarity).
3) when you play with the trimmer (potentiometer), the brighness of the LED should change.
4) you can also measure the DC voltage on the 1K8, it should varies between minimum and maximum limits.

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Check the connections of Q1 (mainly its emitter and collector). I guess you have a datasheet of your NPN transistor.

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Which tools you have to measure the voltages on your circuit?