Should the T8 LED tube be added to the withstand voltage test before it is ready for aging? Some manufacturers skip this link and directly aging shipments. why? The answer is that the withstand voltage test will kill the lamp. Then expand the concept, a considerable number of people in the entire LED industry believe that the withstand voltage test will kill the lamp beads, so a considerable number of lamps on the market have not passed the pressure test, and there are certain hidden dangers in the use of safety. Most of these lamps cannot pass the export inspection or CE test, and the product quality has a grade. There is also a problem: the lamp assembled by the driving power supply with a voltage of 3.75KV can't pass even the 3KV when it is under pressure. What is going on here? This article attempts to talk about personal opinions through analysis.
The mechanism of the T8 withstand voltage test dead bulb is analyzed below.
There are two reasons for LED damage. One is that the voltage is exceeded, and the other is that the current is exceeded. The leakage current of the withstand voltage test is set at about 10 mA, and generally does not exceed the allowable current value of the LED. The biggest cause of LED damage is the voltage excess.
How is the voltage exceeded?
The luminaire is composed of three parts: a driving power LED and a heat sink. The withstand voltage test is generally to test the withstand voltage between the input end of the drive power source and the outer casing of the lamp that the human body can access. The withstand voltage is achieved by insulation. The insulation from the input to the outer casing consists of two parts, one is the insulation of the primary and secondary of the power supply (except for the non-isolated power supply), and the other is the insulation between the lamp bead and the heat sink (generally integrated with the outer casing). AC-powered lamps are tested for withstand voltage with AC high voltage. See below.
In the figure, CY is the Y capacitor of the driving power source, and CY1 and CY2 are the partial capacitances between the positive and negative copper foils of the aluminum substrate and the aluminum plate. The high voltage during the withstand voltage test is applied directly between CY and CY1/CY2. Equivalent to the figure below.
Assume that the applied AC high voltage is VC, the voltage divided by the power source Y capacitor is VCY, and the distributed capacitance of the aluminum substrate is divided into voltage VCY1/2, then VC=VCY+VCY1/2. We know that the capacitance tolerance Xc=1/(2Ï€fC), 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 fixed, between 1000p and 2200P. Assuming that the distributed capacitance value of the aluminum substrate is equal to the Y capacitance value of the power source, the voltage divided by the aluminum substrate is equal to the power source to which the power source Y capacitor is divided, and both are 1/2 VC. If CY1/2 is less than CY, VCY1/2 will be greater than VCY, that is, the voltage applied to the aluminum substrate is higher than half of the withstand voltage test high voltage value, and the smaller the distributed capacitance, the higher the voltage. At this time one. Second, the copper foil of the aluminum substrate has positive and negative poles. The two copper foils are not connected together, and the LEDs connected in series are connected between the two poles, and the distributed capacitance of the two copper foils is discrete. The difference in shape of the copper foil makes the distributed capacitance different in size and the high voltage to be divided is also different. If the voltage of the negative electrode is higher than the positive electrode, the secondary winding of the power supply transformer and the rectifier are turned on, and the voltage is forcibly leveled to the positive voltage. This situation does not cause the LED to overshoot, and if the positive voltage is higher than the negative voltage, At this time, there is a 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 is too large to exceed the LED withstand voltage, it will damage the PN junction of the LED. Another case where the positive voltage is higher than the negative voltage is that the negative electrode is broken down and is at zero potential. At this time, the positive electrode is only enough to be divided into a very low voltage to cause LED damage. Therefore, the main cause of LED damage is that the voltage 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 foil on the aluminum substrate.
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