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Phase shift detection circuit

Guest georgedb

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Guest georgedb

We use 230V. I was looking for a schematic that could be used as a simple oscilator: the phase shift (is that what it is called) in the power lines. Every 1/50th of a second the plus and minus are reverted (AC), so I'd say that's a 100 times per second. That would be a clean and easy counter, especially as the monent the voltage difference is zero, it's the best moment to trigger a Triac in order to switch 230V equipment.

But, this schematic looks a bit weird to me:

Input = 230V
The 4 rectifying 1N4007's create a DC voltage of SQR2 x 230V (230V x 1,41) = roughly 325V
200K resistance (is there a reason why there are 2 resistors, one resistor of 200K would have the same effect, wouldn't it?)
I = 1.6mA but ususally that would not be enought to power the LED. I checked the datasheet of the optocoupler (6N138) and it seem to me that it behaves like a normal LED.

Why would this schematic ever work, I'd say the LED is never lit?


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Texas Instruments say the 6N138 is obsolete.
Its datasheet shows that its minimum transfer ratio is 300% (it is typically 1300%) so with an LED current of 1.6mA its minimum output current can be 4.8mA. But with a 5V supply and a 4.7k output resistor its output current is only 1.1mA so it will be fine.

When an LED conducts ANY amount of current then it produces some amount of light. True, an LED at only 1.6mA will appear dim. But here the LED is extremely close to the very sensitive photo detector.

If you want to switch when the voltage is zero then you need a more modern "zero-crossing" optocoupler.

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OK, I had not expected that it would light at all with so few current.

I just lit a white LED and a red LED with a current of only 24uA. They were not very bright but were very obvious. Try it. You can see in sunlight and you can also see in moonlight that has MUCH less light.

What's this minimum transfer ratio? The stronger the light, the higher the output current?

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