ac light dimmer light using triac and triac

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which pin 3 are u referring to,MOC/Triac?

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I measured it in DC
 

Oh sorry... I meant pin 4 of the MOC... You see I also do mistakes

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I measured it in DC
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So it is natural that the voltage is zero since it is the average of an AC voltage if the two half cycles (positive and negative) are symetrical.

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Could you also measure the DC voltage on 1K5 resistor when the red LED is at its maximum brightness?
The voltage will be less than 12V.
 
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DC volt at 1k5.

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how to measure the voltage between pin 6 and pin 4
 

If the red LED is at full brightness and the voltage on 1K5 is only 1.27 volt, there is something wrong in your circuit.

The average (hence DC) of this voltage should be close to:
V_1k5 = ( Vcc - V_ir - V_led - Vsat ) * dutyCycle
V_1k5 = ( 12 - 1.1 - 1.9 - 0.2 ) * dutyCycle = 8.8 * dutyCycle

So even at full brightness the duty cycle (of the U3 output pulse) is 0.7 only
V_1k5 = 8.8 * 0.7 = 6.16 V

For a reason I cannot guess, your voltage is very low. This also means that the triggering current is very low; surely less than 5 mA.
You have to find out why it is low

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I suggest you remove all the connections on the side of the high voltage (220Vac) when you examine and test your board to get a reasonable voltage on 1K5 as we calculated it above.
 
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ya,is very low.0,85mA.I think maybe it is the diode we've added in series with the transistor
 

ya,is very low.0,85mA.I think maybe it is the diode we've added in series with the transistor
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I don't think it has any effect on the U3 output. Since you are not convinced why it is added (to avoid the voltage breakdown of the Q1 base-emitter junction due to the high reversed voltage at the end of the pulse of U1), you can remove it (by shorting its two terminals for example) anytime and see how the circuit works.

Are you sure Vcc that supplies the MOC and LED is 12V. Which opamp are you using?

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You can measure at the same time the DC voltages:
(1) between the pin 1 and 2
(2) of the red LED
(3) on 1K5
(4) between the opamp output and ground
(5) Vcc that supplies them all.
 
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u know whats happening,during the night the voltage is changing in my country,now is about 270Vac,and when I measure the voltage output from my regulator is now sitting at 15 V

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pin 1 and pin 2=1.27V
red LED=2.21V
1k51.32V
Vcc=14.58V

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opamp output and Ground=9.75V
 

now is about 270Vac
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Wow... here it goes sometimes to 250Vac (and as low as 140V), Many appliances may be dammaged if they are powered by this high voltage.

my regulator is now sitting at 15 V
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So your power supply is not regulated, right?

Anyway, could you measure (1) to (5) as I suggested on my previous post and at the position of the potentiometer (preferably making the LED bright)?
 

1 to 5 where? position of potentiometer, you mean all resistor connected to it?

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oh,i get it.

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(1)=1,28V
(2)=2.23V
(3)=1.33V
(4)=0.89V
(5)=14.59V
 
Please repeat these measurements but after disconnecting the 1K5 from the opamp output and connecting it to ground instead (this makes the duty cycle 100%, no pulses).
Of course, there is no need for (4) since the opamp output is open (not connected to anything, floating, but its voltage varies by changing the position of variable resitor behind it).
 
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1)=1,41v
(2)=2.42v
(3)=10.93v
(4)=0.006v
(5)=14.7v

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(1) between the pin 1 and 2=1,41v
(2) of the red LED=2.42v
(3) on 1K5=10.93v
(4) between the opamp output and ground=0.006v
(5) Vcc that supplies them all.=14.7v
 

so I_1k5 = 10.93 / 1.5 = 7.28 mA, this is a reasonable value. And this is the instantanuous current that passes the IR LED when U3 output becomes low (and it is zero current when high).

Unforunately you cannot see the waveform of the pulse at U3 output as I do in my new tests by using an oscilloscope.

For instance, in this case (if we keep 1K5 to ground), if you connect the MOC, the 390 Ohm, the power triac and the 60W bulb (as you did before), the bulb SHOULD be "on" always when the mains voltage is applied.

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Did you tell me the type of your opamp?
 

its what I get when I connect it like that. The load is turn on
 

its what I get when I connect it like that. The load is turn on
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This means there is something wrong in the driving pulse that lets the load be off always!!!

If you like we can continue tomorrow, here it is 1:34 AM
 
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ya, I also suspect that,but it doesn't make sense.

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Yes,please. here is 00:38 am

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:smile:good night.
 

Hi,have u tried 2 simulate this circuit of mine and test it?
 

Of course I did, it didn't work because of polarities. The opamp (U2:B) cannot discharge the capacitor C2 after it is reset by Q1 because its IN+ > IN-. The output 7 stays at high level after the high pulse of pin 1 (output of U2:A) shifts it (via Q1) with IN- to a high voltage (a bit lower than 12V).
Pin 4 (IN+) is close to 12V (thru R6) while Pin 6 (IN-) is below it by the voltage divider R4 and R5.

Sorry to ask you again and again... which opamp are you using? I don't think you use LM324, right?
 
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