Design of LED Power Supply

1. LED current size We all know that if the LEDripple is too large, the life of the LED will be affected.

2. The power tube heats up. Regarding this issue, I have also seen someone post on the forum. The power consumption of the power tube is divided into two parts, switching loss and conduction loss. It should be noted that in most occasions, especially LED mains drive applications, the switching damage is much greater than the conduction loss.

3. Chip heating This is mainly for high-voltage driver chips with built-in power modulators. If the current consumed by the chip is 2mA, the voltage of 300V is applied to the chip, and the power consumption of the chip is 0.6W, which will of course cause the chip to heat up. The maximum current of the driver chip comes from the consumption of the driving power MOS tube. The simple calculation formula is I=cvf (considering the resistance benefit of charging, the actual I=2cvf, where c is the cgs capacitance of the power MOS tube, and v is the conduction of the power tube. Therefore, in order to reduce the power consumption of the chip, we must find a way to reduce c, v and f. If c, v and f cannot be changed, please find a way to divide the power consumption of the chip to the devices outside the chip, be careful not to Introduce extra power consumption. Simpler, just consider better heat dissipation.