This presented refrigerator door open alarm circuit which alerts you after a period of time whenever your refrigerator door is left open.
该电路变得非常方便，因为如果由于疏忽而门被留下，可能会导致消耗的显着增加并影响冰箱的寿命。
电路操作
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.
一旦门关闭，冰箱灯熄灭，电路和警报关闭并停止发出声音。
对于执行整个操作，几个定时器555连接如图所示。
当没有引入LDR时，将第一IC 555的销2（触发）上的电压保持越来越高，并且其输出（引脚3）呈现低。由于该否则，第二IC 555抑制（引脚4上的低电压电平），并且不允许警报激活。
当LDR经历照明（打开门）时，第一555的引脚2上的电压电平降低，导致输出（引脚3）振荡（方波）。
在振荡期间，当第一555的输出处于高电平时，使第二555能够被触发，该触发也开始与第一个但以更高的频率振荡振荡。
A buzzer which may be seen connected with the output of IC2 now begins buzzing and alarming.
该电路利用PP3 9伏电池，应尽可能靠近冰箱的内部光。
该电路应容纳在一个箱内，该盒子可以防水且密封，以防止水分影响其操作。
电路原理图
冰箱门打开报警电路的零件清单
- IC1 - IC2：2计时器555
- C1：1UF 25V
- C2：100nf.
- R1：10K 1 / 4W
- R2：LDR（Photientistor）
- R3：2.2M 1 / 4W
- R4: 1M 1 / 4W
- D1：1N4148
- Buzzer: Piezo type DC
RAM，不要使它成为因为它没有非常有效，在变压器中浪费了大量的电力。我已经测试过它
最好通过计算的电阻将LED直接与电池连接。
嗨洼地
有时，冰箱的内部光线变坏，大部分消费者都不会取代它们。在这种情况下，上述电路将无法正常工作。
我建议，您可以在本文中包含，基于LM334，热敏电阻或1N4007作为温度传感器的另一个版本。传感器可以放置在门下方的冰箱上。当门被打开时，更密集的冷空气将从下部逸出，并与传感器接触，传感器将在预设时间后激活蜂鸣器。在这种情况下，将不需要将电路放置在冰箱内。
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.
Swagatam
我曾想过对外部光明，但晚上它不会工作。
However, hall-effect or reed switch would be better option.
我不认为天气黑暗的用户自己可以看到冰箱材料......必须有一些光看东西。
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…………………………????
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.
actually the circuit can be even more simplified by using just one IC….this circuit was not designed by me.
实际上，当门没有正确关闭时，我误解了电路警告，这不是电路问题，这是我的错误
好的谢谢。您是否可以建议将任何介于6伏到7伏逐步使用2 LED串联，以节省能源，并使电路更高效。
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://www.homemade-circulay.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
如果门是ajar，这将足以通过冰箱外的最微小的光触发的敏感。
是的，可以通过适当增加R1的值来完成。
I am a mom with 4 kids who don’t shut the fridge properly, which freezes up the fridge. Your directions look like greek to me. Do you make these for people to purchase? Instructions for circuit boards are the only thing I have found on line.
#NotAnEngineerJust(a)MOM
Thank you kris Morris! I am sorry I don’t sell readymade kits, however I can provide a much easier alternative with just a couple of parts, which you can buy and assemble yourself without going through any technical hassles, I’ll try to update the design soon in the above article…