Delayed light off device principle circuit diagram

As shown in the figure, the circuit working principle of the device is shown in the figure. When the button switch AN is not pressed, the transistor VT has no bias current and is cut off. At this time, the resistance between c--e of VT is greater than 50K, so the space between the ① and ② pins of LSE is equivalent to an open circuit, and the ④ pin of LSE outputs low level. , relay J is in the released state. When the switch AN is pressed, its current quickly charges the capacitor C3 through AN , and the charge is fully charged within 1 second. At this time, the transistor VT is turned on, and the resistance between its poles is reduced, so LSE is turned on, and its ④ pin outputs high level, relay J is energized and closed, and lamp H lights up. After releasing your hand, the charge in C3 is discharged to VT through resistor R, maintaining the conduction of VT. When the charge in C3 is discharged, VT is cut off, J is automatically released, and H goes out. The lighting time of the lamp H depends on the values ​​of R and C3. According to the values ​​shown in Figure *, the delay time after testing is approximately 1 minute and 30 seconds.

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