I want to add an LDR (Light Dependent Resistor) to my Motion activated stair lights, then On/Off switch can be removed as well as the single Led (On Indicator Only) from my project (I want the LDR to control the On/Off). This is basically the first project I've done, so if someone could explain/show me what I need and how to set it up that would be great. Also, does anyone know of a great place for a beginner to learn about electronics? I would love to be able to build my own projects with out asking for help, thanks.

This is what my PIR LED stair lights look likePIR LED Diagram

This is the link to video of the system working My Video


2 Answers 2


Update based on the use of the CDC (photocell) input already on the board (as discovered by the OP)

The schematic of the PIR board you refer to is similar to the following

enter image description here

It shows that the CDS pads (Cds2 in the schematic) connects to pin 9 of BISS0001

According to the datsheet that pin is:

Trigger disable input (VC)
VC >0.2Vdd=enable
VC<0.2Vdd =disabled)

Vdd supplied to the chip is 3.3v (from the 3.3v onboard regulator) so if you apply voltage >0.66v to that pin the controller turns on, if you apply <0.66v then it turns off.

One of the CDS pins is already connected to ground,the other one is connected in series with an onboard resistor of 1M Ohm to the 3.3v supply.
The CDC and the resistor form a voltage divider, if the CDS sensor has a resistance of 250K or higher the controller will turn on, if it has <250K then the controller will turn off.

You just need to find an appropriate CDS or add an external resistor to adjust the switching point.

The LDR (Light Dependent Resistor) has a resistance which increases in darkens and decreases with light.

Using that property you can connect it in a voltage divider to drive the base of a NPN transistor and turn on/off a load based on the room brightness.

Such a circuit that turns on at darkness looks like

enter image description here

If you are looking for the opposite effect you can use

enter image description here

Note that Vin is the positive supply line

You can replace the 500 Ohm resistor and led of the above circuits with a relay that replaces your switch.

enter image description here

Images from http://www.reuk.co.uk/Light-Dependent-Resistor.htm

  • \$\begingroup\$ Thanks a lot, Is there anyway this could be done without the relay as I want to keep this as small as possible \$\endgroup\$
    – Ridach
    Commented Jan 10, 2014 at 18:46
  • \$\begingroup\$ @Ridach I suppose an active device could be used. What is the max current through the switch? \$\endgroup\$
    – alexan_e
    Commented Jan 10, 2014 at 18:48
  • \$\begingroup\$ @Ridach I have added some additional info based on your solution, just is case you haven't managed to use the CDC input. \$\endgroup\$
    – alexan_e
    Commented Jan 20, 2014 at 21:01

I looked around and noticed a lot of PIR sensors, say "Photosensitive setting: Use CDS (Defaut: not welded)" so that would make me assume you can just add a LDR/CDS to the PIR board and I was right, There is also a spot for a Thermistor. I Edited this image to show simple placement Where you add the LDR


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