The debugging technology of LED driver chip is mainly in the following aspects.
1. Chip heating
This is mainly aimed at the high voltage driver chip with built-in power modulator. If the current consumed by the chip is 2mA and a voltage of 300V is applied to the chip, the power consumed by the chip is 0. 6W, of course, will cause the chip heat. The maximum current of the drive chip comes from the consumption of the drive 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 GATE voltage when the power tube is on. Therefore, in order to reduce the power consumption of the chip, we must find ways to reduce C, V and F. If C, V, and F cannot be changed, then find a way to divide the power of the chip into off-chip devices, taking care not to introduce additional power. More simply, consider better cooling.

2. Power tube heating
The power consumption of the power tube is divided into two parts, the switching loss and the conduction loss. Note that in most applications, especially LED mains drive applications, switch damage is much greater than conduction loss. The switching loss is related to the CGD and CGS of the power tube as well as the driving capacity and working frequency of the chip. Therefore, the heating of the power tube can be solved from the following aspects: A: MOS power tube cannot be selected unilaterally according to the size of on-off resistance, because the smaller the internal resistance is, the larger the capacitance of CGS and CGD will be. For example, the CGS of 1N60 is about 250pF, the CGS of 2N60 is about 350pF, and the CGS of 5N60 is about 1200pF. The difference is too big. When choosing the power tube, it is enough. B: All that's left is frequency and chip power. I'm just talking about frequency here. The frequency is also proportional to the conduction loss, so when the power tube heating, the first thing to think about is whether the frequency choice is a little high. Find a way to reduce the frequency! Note, however, that as the frequency decreases, the peak current must increase or the inductance must also increase in order to achieve the same load capacity, which may cause the inductance to enter the saturated region. If the inductor saturation current is high enough, consider changing CCM(continuous current mode) to DCM(discontinuous current mode), which requires adding a load capacitance.

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