AC Light switch using relay

 

There are alot of circuits on how to do this. Most of them require you to place the photosensor between the base (B) of the transistor and the Emitter (E). But i have found this method not to be working well so i decided to place the photosensor between the resistor and the power supply (DC).

So how is this method different. First the photosensor is already sending power even at low lighting. So if the source is 12v, that means that power though the photosensor under low light is approximately 8v. This means that 8v won't be able to course an effect on the relay since the relay is 12v based. When a bright light is put in a close distance with the photosensor, There will be low resistance on the photosensor hence the desired power which will follow to the relay

The point where I have labeled as " AC power out" is the off point at default on the relay. What I mean is that the point with AC power out is the one is not connected inside of  the relay at first.

I have used a "bc547" transistor, a photosensor, 400 ohms resistor , 100 capacitor this causes a delay when power is off or when the photosensor resistance increases, and 5 legs 12v relay.

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