DCD-1240 multifunctional rechargeable lighting

This rechargeable lamp can not only be used as a household lighting during power outages, but also as a fire emergency lighting lamp (automatically turns on the light during power outages). The rechargeable lamp has two 6V/4Ah maintenance-free batteries connected in series to provide a 12V DC power supply. Two 12v/7W electric beads are used for lighting (they can also provide a 12W/14W DC power supply to other electrical appliances). The circuit schematic diagram drawn for the convenience of maintenance is shown in Figure 1.

Working principle of charging lamp: 220v mains is stepped down by two non-polar capacitors C1 and C2 in parallel. Then it is bridge rectified by diodes D1~D4 to obtain pulsating DC voltage (R1 is the discharge resistor of the residual voltage on capacitors C1 and C2 after the 220v input voltage is cut off). The higher pulsating DC voltage after step-down rectification is added to the T1 and T2 poles of the bidirectional thyristor through resistor R2, the coil of relay J (JZC-7FA, 6v/DC) and resistor R5; at the same time, the voltage regulator DW breaks down. The G pole of the thyristor obtains the trigger voltage. The thyristor SCR (97A6) is turned on, the relay J is attracted, and the pulsating DC voltage charges the battery through the diode D5 and the contacts (Jl, J-2) of the relay.

We know that capacitor voltage reduction has a different characteristic from transformer voltage reduction. In the absence of voltage stabilization measures, the voltage at its output end is closely related to the size of the load current: when the load current is zero, its output end voltage is similar to the mains 220V (no filter capacitor added after rectification). As the load current increases, its output end voltage will decrease accordingly. The normal charging voltage of a 12V battery is approximately: 12V×1.15=13.8V. When the relay is closed and starts to charge the battery, the charging current required by the battery is large, so the voltage drop is also large. As the charging time increases, the battery charging current continues to decrease, and the voltage at point E of the circuit continues to increase. When point E exceeds approximately 14.5V, the voltage regulator DW breaks down, and the G pole of the thyristor obtains the trigger voltage and turns on. When the power supply is too weak, the thyristor turns off. In this way, no matter how large the battery load current is, the voltage at point E is always limited to about 14.5V to prevent excessive voltage from damaging the battery. Therefore. Resistor R5 and thyristor are both the path to ensure relay J is attracted when the mains power is turned on. They are also a load channel to prevent the voltage at point E from being too high when the battery charging current drops or even opens. The function of capacitor C3 is to effectively avoid a higher voltage being applied to the relay at the moment of connection, which may damage the relay coil (the working voltage of the relay coil is 6v/DC), because the voltage across the capacitor cannot change suddenly when the mains power is turned on and the thyristor is turned on.

Fault phenomenon: Plug in the mains power supply: both the red and green indicator lights are off.

Cause analysis: After disassembling the case for inspection, it was found that one of the two current limiting resistors on the circuit board was completely burnt. The other one was blackened. The PCB board was also severely burnt. This shows that the two current limiting resistors are very hot when working. After testing, the blackened resistor is R2 (51Ω), and the burnt resistor is R5 (open circuit). Judging from the appearance, the power of the two resistors is 0.5w. I replaced R5 with 51Ω, 100Ω, and 150Ω respectively for experimental testing: When the battery connection line is removed (no load): When using 51Ω, the voltage at point E is about 12v, which is too low; when using 150Ω, the voltage at point E is about 18v, which is too high: when using 100Ω, the voltage at point E is about 15v, which is more appropriate. At this time, in the experiments of the three situations, the green indicator lights are all on and the red indicator lights are all off, indicating that the thyristor is fully turned on. After testing, the maximum current through resistor R2 is about 150mA. Its power calculation: 0.15A×0.15A×51Ω=1.1475W: The power calculation of resistor R5: 15v×15V÷100Ω=2.25W. Obviously, the power of the resistors used originally was too small. And they were soldered on the board. The overheated resistors caused the resistors themselves and the PCB board to be "burned". Considering that disconnecting the battery connection is an extreme state, the power of resistors R2 and R5 are both 2w. The long pins are not shortened to facilitate heat dissipation. After welding, restore the original case. Plug in the power plug. The red and green indicator lights are on. After 15 hours of charging, the lights are turned on and the battery DC output is normal.