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
    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

    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
    Thank you for your help. How did you figure that out - is there a certain formula you use?
  4. ouabache


    Jul 12, 2005
    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.