Trouble with night light circuit

Discussion in 'General Electronics Chat' started by asdasd12345, Jul 24, 2005.

  1. asdasd12345

    Thread Starter New Member

    Jul 24, 2005
    8
    0
    I am just learning electronics and this is the first thing Im trying to build . I have a circuit that is turning on when the room is dark and off when bright. I have a photoresistor that is attached to a 1.5 volt terminal it turns on, but when I put it with a 6V terminal and use a resistor to lower the voltage the circuit doesnt work anymore. Anyone know how to get the voltage from 6 to 1.5.
     
  2. mozikluv

    AAC Fanatic!

    Jan 22, 2004
    1,437
    1
    hi

    a voltage divider will do it for you. the values that i will recommend is based on the assupmtion that your circuit will not draw more than 76ma. here's the circuit.
     
  3. asdasd12345

    Thread Starter New Member

    Jul 24, 2005
    8
    0
    Thank you for your help. How did you figure that out - is there a certain formula you use?
     
  4. ouabache

    Member

    Jul 12, 2005
    11
    0
    Voltage dividers are a common configuration to step down a DC voltage.
    The basic equation is Vout = Vin[R1/(R1+R2)]
    In this case Vout= 1.5V, Vin = 6V. and moziluv chose R1=22 ohms and R2=56 ohms.

    Vout = 6 [22/(22+56)] = 1.69V (open circuit voltage)

    This gave you 1.69v and you required 1.5v. As I mentioned, this is the
    open circuit voltage. Once you connect it to the rest of your circuit, it is under
    a load. Loading often pulls down the supply voltage. The 1.69v from the divider, gives you enough head room (≈ 13%) so that it can supply ≈ 1.5 volts to your circuit.
     
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