I've just started playing with electronics, and wanted to make a light/dark detection circuit without using a transistor (they are too advanced for me at this moment.)

I decided to use these components:

  1. LED
  2. 10k Resister
  3. 9V battery
  4. LDR

I want my LED to turn off when I shine a light on the LDR. I also want my LED to turn on when I cover the LDR. But this does not happen. I'm not sure what is wrong.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ If that circuit does dim the LED, it only does so by wasting the power of the battery. You need a switch, such as a transistor. you should also really be using a comparator or op amp for this. \$\endgroup\$
    – dandavis
    Commented Nov 13, 2020 at 21:13
  • \$\begingroup\$ I have removed your pic and added the corresponding schematic, if that is okay. \$\endgroup\$
    – Mitu Raj
    Commented Nov 13, 2020 at 21:48
  • 2
    \$\begingroup\$ I have no idea how you expected this to work, so I think you need to explain why you think this should work before anyone can help you clarify the problem. We can of course show a real working circuit, but will that help you? There must be hundreds online. \$\endgroup\$
    – pipe
    Commented Nov 13, 2020 at 22:15
  • \$\begingroup\$ I agree with pipe's comment. You need to try to explain how you think you circuit would work in order for someone to be able to help you get on the right track. \$\endgroup\$
    – brhans
    Commented Nov 13, 2020 at 22:58
  • \$\begingroup\$ @pipe Very good question and I apologize for the ambiguity. This is my understanding: The light shines on the LDR would decrease its resistance significantly. Current follows the least resistance path and all or most of the current would flow through the LDR, and not the LED (because that path has a big resistance) \$\endgroup\$
    – wajmo
    Commented Nov 15, 2020 at 4:00

1 Answer 1


The circuit with LED and circuit with LDR are in parallel.

Assuming a normal 9V battery, the LDR won't have much effect on the LED, it would not shunt enough current to turn the LED off.

  • \$\begingroup\$ Thanks. What would shunt enough current and turn the LED off ? \$\endgroup\$
    – wajmo
    Commented Nov 15, 2020 at 4:01
  • \$\begingroup\$ The circuit needs heavy modification so there is no simple answer. Besides the concept of turning off the LED by shunting current is almost equal to turning off a device by using a switch to short-circuit the battery. Usually a thing is turned off by disconnecting it, not bypassing. \$\endgroup\$
    – Justme
    Commented Nov 15, 2020 at 9:52
  • \$\begingroup\$ I see. But I thought since the LDR when shined upon will have very low resistance, the current will take the path of least resistance and no or very little current will flow via LED. \$\endgroup\$
    – wajmo
    Commented Nov 15, 2020 at 13:40
  • 1
    \$\begingroup\$ You said that already, and the circuit does not operate like you thought. There will be 9V provided to LED+resistor regardless of LDR resistance or if it is even there. Because they are in parallel. \$\endgroup\$
    – Justme
    Commented Nov 15, 2020 at 13:54

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