Lifter fault alarm circuit principle

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As shown in the figure, the hoist fault alarm circuit. The alarm is composed of a sensor, a 555 monostable trigger, an alarm circuit, and the like. The sensor is a magnetron switch composed of a permanent magnet and a reed pipe, and the former is fastened to the driven wheel, and the latter is fixed at a specific position. The monostable flip-flop consists of IC (555), R, C2, and so on. The alarm circuit is composed of a relay, a light emitting diode, and the like.
When the rotation speed is normal, C1 is charged every time one revolution, because the magnetic control switch is closed. Adjusting the potentiometer W1 makes the charging voltage on C1 slightly higher than 1/3 VDD, and the output terminal (3 feet) of 555 is kept low, and the corresponding LED 1 is illuminated. When the driven wheel speed is lower than the normal speed due to the fault, the corresponding discharge time of the capacitor C1 is lengthened, so that the voltage on it is lower than 1/3 VDD, 555 is inverted, and the output is high. At this time, the relay J connected to the output terminal (3 feet) is sucked, thereby controlling the relevant process equipment of the intermediate relay. Simultaneously. The LED 2 that displays the fault is lit.
This circuit is based on the fact that when the hoist is overloaded or the drive pulley is loose, the speed of the driven wheel will be slower, and the relationship between the trigger level of 555 and the RC charge and discharge time constant is designed, thus having the function of fault alarm.

Hoist fault alarm circuit



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Parameter:

Input voltage: 100-277vac
output voltage: 25-143vdc
current: 100mA-8000mA.
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