# 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.