Simple and practical elevator power outage alarm circuit diagram

As we all know, if the elevator power supply suddenly fails, if the personnel in the distant power distribution room duty room cannot detect it immediately, the passengers on the elevator may panic. Here we introduce an acoustic alarm circuit, as shown in the figure below. As long as the elevator power supply fails, an acoustic alarm signal will be emitted in the elevator's power distribution room. The circuit works as follows: the circuit is powered by a 6V battery. Once the elevator AC power supply fails, the circuit will sound an alarm and the alarm time can be preset.

After the alarm reaches a certain time, the audible alarm ends, and the personnel on duty can immediately find the fault. Once the fault is eliminated, the power supply resumes operation, the alarm circuit is cut off, and the AC power supply charges the battery at a constant voltage.

When charging the battery, the 230V AC power is stepped down by transformer

When the battery is charging, the transistor T1 is turned on through the diode D4, causing the T2 tube to be turned off, and the relay RL1 will not work because there is no current flowing through it. As a result, the battery is always in a charging state. The maximum charging voltage is 6V plus the tube voltage drop of D5.

Once the AC power supply fails (such as open circuit), the transistor T1 is cut off, T2 is turned on, the relay pulls 1 to work, its normally closed (N/C) contact is opened and the closed contact (N/0) is connected, and the battery is charged. The circuit is cut off. At the same time, timers IC2 and IC3 are connected to the battery through the closed contact (N/0) of the relay RLI. The output pulse signal of pin (3) 1KH2 of timer IC3 (NE555) excites the reset pin (4) of IC2 (NE555). IC2 works in the pulse state, and the output signal of pin (3) of IC2 is amplified by the audio amplifier IC5. Push the speaker to sound an alarm, indicating that the AC power supply is faulty.