China Emergency Light Manufacturers introduces the trou […]
China Emergency Light Manufacturers introduces the troubleshooting methods of emergency lighting:
1. The main power supply is normal, but the indicator light is off. The fault is mainly caused by the damage of the light-emitting diode. It can be welded, and a multimeter can be used to judge whether it is in good condition. The larger the resistance of the current limiting resistor R1 or the open circuit, the greater the resistance, it will also cause no light.
2. When the power is cut off, the indicator light is still on. The failure is caused by deterioration of reverse characteristics or breakdown. As long as you replace an IN4004 rectifier diode, the fault can be eliminated.
3. After the power is off, the brightness of the lamp will decrease after a period of normal operation. This indicates that the battery may be faulty. Because the circuit has no protective measures to prevent the battery from being charged, if the power supply is cut off for a long time, the battery cannot be disconnected. When the battery terminal voltage is discharged 10v, the battery will discharge and its plates will be damaged prematurely. The better way to solve the battery over-discharge is to establish a protection circuit. The circuit is relatively simple. In the emergency lighting process, when the battery terminal voltage is lower than 10V, VD is cut off, the relay KA is released, and the load circuit of the battery is cut off to avoid over-discharge and shorten the service life of the battery.
4. The function of emergency lighting circuit failure switch SA2 emergency light circuit: when SA1 is closed, SA2 is placed in the "start" position, the winding voltage oscillating transformer L1T2 is applied to one end of the lamp filament through SA2 and the ground power supply (battery ), the voltage L3 on the winding is applied to the other end of the filament tube. After preheating for about 3 seconds, turn the switch SA2 to the "on" position. At this time, the high-frequency pulse voltage induced on the L2 winding acts on both ends of the lamp tube, and the lamp tube lights up.