This presented refrigerator door open alarm circuit which alerts you after a period of time whenever your refrigerator door is left open.
This circuit becomes very handy, because in case the door is left due to carelessness may cause significant increase in the consumption and affect the life of the fridge.
Simulation and Working
This circuit uses a photosensor LDR for detecting whether the door is open, or not. Whenever the sensor is illuminated by the light coming out from inside of the refrigerator, the circuit begins emitting an intermittent sound to alert you and bring the situation to your attention.
And as soon as the door is closed, and the fridge light goes off, the circuit and the alarm shuts off and stops emitting the sound.
For executing the entire operation a couple of timers ICs 555 are connected as shown in Figure.
When the LDR is not introduced to light the voltage on pin 2 (trigger) of the first IC 555 stays higher and its output (pin 3) is rendered low. Due to this the second IC 555 is rendered inhibited (low voltage level on pin 4) and the alarm is not allowed to activate.
When the LDR experiences an illumination (door is opened), the voltage level on pin 2 of the first 555 gets low causing the output (pin 3) to oscillate (square wave).
During the oscillation when output of the first 555 is at high level enables the second 555 to get triggered which also begins oscillating in tune with the first but at a much higher frequency.
A buzzer which may be seen connected with the output of IC2 now begins buzzing and alarming.
The circuit makes use of a PP3 9 volt battery, and should be placed as close as possible to the inner light of the refrigerator.
The circuit should be housed inside a box that may be waterproof and sealed to prevent moisture from affecting its operation.
Circuit Diagram
Parts list of the refrigerator door open alarm circuit
IC1 - IC2: 2 Timer 555
C1: 1uf 25V
C2: 100nF
R1: 10K 1 / 4W
R2: LDR (photoresistor)
R3: 2.2M 1 / 4W
R4: 1M 1 / 4W
D1:1N4148
Buzzer: Piezo type DC
Ram says
Ok, thank you. Can you suggest anything to step up 6 v to 7 volt to use 2 led in series to save the energy and make circuit more efficient.
Swagatam says
for stepping up the voltage you may have to employ a complex boost circuit, the easier option is to use single LEDs across the battery, all in parallel….by dropping the 6V to 3.5V through series diodes, or by using a emitter follower circuit as shown below:
https://homemade-circuits.com/2012/08/simplest-dc-cell-phone-charger-circuit.html
the 9V zener could be replaced with a 3.3V zener for dropping the voltage to the required limits
Abu-Hafss says
Hi Swagatam
Sometimes, the inner light of the refrigerator goes bad and most of the consumers don't replace them. In that case, the above circuit won't work.
I suggest, you can include in this article, another version of this circuit based on LM334, a thermistor or 1n4007 as temperature sensor. The sensor may be placed on the body of fridge just beneath the door. When the door is left opened, the cold air which is denser would escape from the lower part and come into contact with the sensor which will activate the buzzer after preset time. In this situation, it won't be required to place the circuit inside the fridge.
Swagatam says
Hi Abu-Hafss,
It's hard to imaging how somebody could keep the fridge without a light, but even in that case the external light entering the fridge would be sufficient to trigger the LDR whenever the door is opened.
Another option is to use a hall effect sensor or a reed switch to sense the door's rubber seal magnet while it's closed or opened, and trigger the alarm.
Abu-Hafss says
Swagatam
I had thought about the external light but, at night it won't work.
However, hall-effect or reed switch would be better option.
Swagatam says
I don't think at pitch dark the user himself would be able to see the fridge materials…there has to be some light to see things.
Abu-Hafss says
Suppose, the user opens his fridge (with inner light out of order) at night. He will use the room light for taking out the food. He closes the door and switches off the room light. But the door is not fully closed…………………………????
Swagatam says
that's a 0.1% scenario, in such a terrible case this circuit won't help after all it's a straightforward, cheap and built on an extremely basic principle.
because if the door is open by 1 or 2 mm then even the internal lights would be off disabling the LDR….and in a case where the door is left sufficiently open and the user completely ignores the buzzer, walks of switching off the room light then it's his problem not the circuit's problem.
Swagatam says
actually the circuit can be even more simplified by using just one IC….this circuit was not designed by me.
Abu-Hafss says
Actually, I misunderstood the circuit to warn when the door is not properly closed so that is not the circuits problem, it was my mistake š
Swagatam says
Ram, don't make it because it's not quite efficient, lot of power is wasted in the transformer. I have tested it
It's better to connect the LEDs directly with the battery through a calculated resistor.