T8LED tube should be added to the withstand v […]
T8LED tube should be added to the withstand voltage test before it is manufactured and ready for aging? Some manufacturers skip this link and directly aging and shipping. Why? The answer is that the withstand voltage test will kill the lamp beads. To expand the concept, a considerable number of people in the entire LED industry believe that the withstand voltage test will kill the lamp beads. Therefore, a considerable part of the lamps on the market have not passed the withstand voltage test, and there are certain hidden dangers in the use of safety. Most of these lamps fail to pass the export commodity inspection or CE test, and the product quality is next to a new level. There is another problem: when the lamps assembled with a 3.75KV withstand voltage drive power supply, even 3KV cannot pass. What is going on here?
The mechanism of the dead lamp bead in the T8 withstand voltage test is analyzed below.
LED damage has two reasons. One is the voltage exceeding, and the other is the current exceeding. The leakage current of the withstand voltage test is set at about 10mA, which generally does not exceed the allowable current value of the LED. The most likely cause of damage to the LED is the overvoltage.
How did the voltage exceed?
The lamp is composed of three parts: driving power LED and heat sink. The withstand voltage test is generally to test the withstand voltage between the input terminal of the drive power supply and the lamp housing that the human body can touch. Withstand voltage is achieved by insulation. The insulation between the input terminal and the shell consists of two parts, one is the insulation between the primary and secondary power supplies (except for non-isolated power supplies), and the other is the insulation between the lamp beads and the heat sink (usually integrated with the shell). AC powered lamps are tested with AC high voltage.
Assuming that the applied AC high voltage is VC, the voltage divided by the power Y capacitor is VCY, and the distributed capacitor on the aluminum substrate is divided to the voltage VCY1/2, then VC=VCY+VCY1/2. We know that the capacitive reactance Xc=1 /(2 πf C), the larger the capacitance, the smaller the capacitive reactance, and the smaller the voltage divided in the series circuit, the opposite is also true. In this example, the Y capacitance of the power supply is a fixed value, between 1000p and 2200P. Assuming that the distributed capacitance value of the aluminum substrate is equal to the Y capacitance value of the power supply, the voltage divided by the aluminum substrate is equal to the power supply divided by the Y capacitor of the power supply, and both are 1/2 VC.
If CY1/2 is less than CY, VCY1/2 will be greater than VCY, that is to say, the voltage applied to the aluminum substrate is higher than half of the withstand voltage test high voltage value. The smaller the distributed capacitance, the higher the voltage. This is one of them. Second, the copper foil of the aluminum substrate has positive and negative poles, and the two copper foils are not connected together. The series-parallel light-emitting diode is connected between the two poles, and the distributed capacitance of the two copper foils is separate. The difference in the shape of the copper foil makes the distributed capacitance different, and the high voltage is also different.
If the voltage of the negative pole is higher than that of the positive pole, the secondary transformer winding and the rectifier of the power supply will be turned on, and the voltage will be forcibly flattened to the positive pole voltage. This situation will not cause the LED to overpressure and die, and if the positive pole voltage is higher than the negative pole voltage, At this time, there is current (this current is different from the leakage current) flowing through the LED, which is why the LED will flash during the withstand voltage test. The higher the voltage, the brighter the LED. When it exceeds the withstand voltage of the LED to a certain extent, the PN junction of the LED will be damaged.
Another situation where the positive electrode voltage is higher than the negative electrode voltage is that the negative electrode has broken down and is at zero potential. At this time, the positive electrode can be divided to a very low voltage to cause damage to the LED. Therefore, the main reason for the damage of the LED is that the pressure difference between the positive and negative electrodes is too large. This pressure difference is determined by the shape and position of the positive and negative copper foils on the aluminum substrate.