# LDR working differently than usual

I observed something different in LDR (Light Dependent Resistor) today. I made a "magic" LED circuit such that when lights are off, the LED glows, else when the lights are on, LDR creates resistance and the LED stops glowing.
Later when I connected just the LDR and LED (in series) to the battery, LED glowed very dim when there was light in the room, and stops glowing when no light reaches the LDR. Why does this happen?

I know just the basics of electronics. I request you to provide an answer I can understand.

• It actually is by magic and it takes magic to read circuit digrams that are invisible. – Andy aka Mar 24 '17 at 16:40

Typical LDRs (light-dependent resistors) increase their resistance when dark. Therefore connecting one in series with a LED should give you exactly what you observed. When dark, there is too much resistance to light the LED noticeably.

It takes a active circuit to use the resistance of a LDR to turn something on when it gets dark.

I go into more detail, include a complete circuit, at https://electronics.stackexchange.com/a/53681/4512.

LDR creates resistance and the LED stops glowing.

You didn't share your circuit or your LDR part number and datasheet, but you are probably wrong about this. LDR's generally decrease in resistance when light is present.

Using some kind of transistor circuit, this can easily made to decrease the current flowing to an LED, giving your "magic LED" circuit.

Later when I connected just the LDR and LED (in series) to the battery, LED glowed very dim when there was light in the room, and stops glowing when no light reaches the LDR. Why does this happen?

Because the LDR resistance increases when light is removed.

As Mr. Lathrop stated, most LDRs have a quite high (a few hundreds of kOhms) dark resistance. When it comes to light, this value can drop to a few kOhms depending on the light intensity (e.g. you can find a value in datasheet as $R_{100}$ meaning resistance under 100lux). Even this "low" value can be high to drive a LED directly from a 9V battery.

Suppose you have a 5mm red LED and an LDR having $R_{100}=5k$. Under 100 lux of light intensity, LED current will be (9V-2V)/5k=1.4mA. This amount of current will cause the LED to glow dimmed.