Light Deactivated Relay Circuits?

Discussion in 'Homework Help' started by Mike Hunt, Feb 25, 2010.

  1. Mike Hunt

    Thread Starter New Member

    Jan 6, 2010
    3
    0
    Hello All,

    I'm working on a project.

    To make a LED turn on using a photo resistor and a relay. The LED is supposed to turn on when there is light. But how would I do the opposite? When the lights are turned off the LED will go on? HELP

    Basically the whole class has been trying to do this and no one knows what to do. Could you guys explain or draw out something I can use as a guide?

    Thanks

    [​IMG]
     
  2. JDT

    Well-Known Member

    Feb 12, 2009
    658
    85
    These circuits will basically work - but not very well!

    The resistance of the Light Dependant Resistor will increase as it gets darker. So the voltage on the base of the transistor will rise as it gets darker. The emitter voltage will follow the base voltage (less 0.65V).

    So at a certain level of darkness the voltage across the relay coil will be sufficient to energise it.

    As the light level increases, the resistance of the LDR decreases. The voltage on the base and emitter decreases. At a certain point, the relay will de-energise.

    To do the opposite just reverse the positions of R1 and the LDR.

    The problem with this simple circuit is that the switching thresholds depend on the relay. A better circuit uses positive feedback to get a well defined sudden switching action. A small increase in input voltage causes the output to suddenly change state. This is called a Schmit Trigger.

    I have attached a typical circuit. This is light de-activated. Depending on your LDR you may have to experiment with your variable resistor value.
     
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