Electronic – What’s wrong with the basic light/dark detection circuit

You need the LED to be lit when the photoresistor is high resistance. So replace the photoresistor with a fixed resistor R3, to supply the base current to turn the transistor on.

Then you need the LED to turn off when the light shines, and the photoresistor is low resistance. So connect the photoresistor from base to ground.

Now, when its resistance is low enough, it will drain the current from R2 to ground, and hold the base voltage below 0.6V turning the transistor off.

Say, at 3kilohms we aim to get the base voltage down to 0.3V. Then 0.3V/3k = I = 0.1ma. Then R3 must drop the remaining voltage 4.7V at 0.1ma, so R3 should be 47k.

Now the transistor will start to turn on when the photocell resistance exceeds 6 kilohms. If that's still too bright, increase R2.

Answer in a schematic

1. I expect that the supply voltage is not very critical - 6 volts should be fine.

2. You can determine using Ohm's Law, that the maximum current through R2 will be about .05 mA.

3. An LED advertised as "5 volt" will be a bare LED with a suitable resistor or current limiting circuit so that it can be operated from 5 volts with no additional current limiting components. (I believe the highest voltage bare LEDs are blue (or white, which is just blue with a white phosphor) which have a forward voltage of 3.0 - 3.5 volts)