Increasing sensitivity of a light activated relay

Other Hardware Motors, Mechanics, Power and CNC
1

Hello,
I am attempting to make a fridge door alarm that will play random .wav files every few seconds when the fridge door is open.
I have already finished the programming part with the arduino, so that when the arduino has power, it will play random .wavs from an sd card.
I now intend to create a light activated relay circuit so that the arduino is only turned on when the door is open.
The main reason for setting it up this way was to avoid the large battery drain from the arduino when the door is closed.

I created the second circuit on this page Light / Dark Activated Relay (but with the photoresistor switched to make it light activated), but the 5v relay will only turn on under really bright torch light, not the normal fridge light. I have turned the pot but it will still not work as required.
How can I make the relay engage under less light?

Thanks in advance!

2

Hello;

Each LDR value differs from eachother even same series.
On your circuit LDR and Pot are simply a voltage divider. You just need to figure it out LDR's value with your light conditions.
You can follow this link to how to do that.
http://www.societyofrobots.com/schematics_photoresistor.shtml

3

Thankyou Omersiar.
I have looked at that site and calculated the required resistance for that resistor to be 150K.
However, I still have the same issue. It will not turn the relay on.

Here is my schematic:

4

You can test if the circuit will turn on the relay by removing R1. If the relay turns on and stays on for both light and dark conditions, then you could reduce R1 from infinite to 2M, 1M, 470K, etc until the desired operation is achieved.

5

R3 does nothing useful in an emitter-follower, replace it with a direct connection, save one
component!

6

Another thing to try is a logic level MOSFET to replace Q2 and eliminate R3, Q1 and R2.