1、 LED current size

The size of LED current directly affects the service life. It is recommended to reduce the rating for use, so try to control it as small as possible, especially if the LED heat dissipation effect is not good, the LED must leave enough margin.

2、 Chip heating

This is mainly for the high-voltage drive chip with built-in power modulator. If the current consumed by the chip is 2mA and a voltage of 300V is added to the chip, the power consumption of the chip is 0.6W, which will certainly cause the chip to heat up. The * current of the driving chip comes from the consumption of the driving power MOS transistor, and 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 turned 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, please find ways to divide the power consumption of the chip to devices outside the chip. Be careful not to introduce additional power consumption.

3、 Power tube heating

The power consumption of power transistor is divided into two parts, switching loss and conduction loss. It should be noted that in most occasions, especially in LED mains drive applications, the switching damage is much greater than the conduction loss. The switching loss is related to the CGD and CGS of the power tube and the driving capacity and working frequency of the chip. Therefore, the heating of the power tube can be solved from the following aspects:

1. MOS power transistors cannot be selected unilaterally according to the on resistance, because the smaller the internal resistance, the greater the CGS and CGD capacitance. For example, the CGS of 1N60 is about 250pf, that of 2n60 is about 350pf, and that of 5n60 is about 1200pf. The difference is too great. When selecting the power tube, it is enough.

2. The rest is the frequency and chip driving ability. Here we only talk about the influence of frequency. The frequency is also directly proportional to the conduction loss, so when the power tube is heating, first think about whether the frequency selection is a little high. However, it should be noted that when the frequency decreases, in order to obtain the same load capacity, the peak current must become larger or the inductance will become larger, which may lead to the inductance entering the saturation region. If the saturation current of the inductor is large enough, you can consider changing CCM (continuous current mode) to DCM (discontinuous current mode), so you need to add a load capacitance.

4、 Operating frequency reduction

This is also a common phenomenon during debugging. Frequency reduction is mainly caused by two aspects. The ratio of input voltage to load voltage is small and the system interference is large. For the former, be careful not to set the load voltage too high. Although the load voltage is high, the efficiency will be high. For the latter, you can try the following aspects:

1. Set the minimum current to a lower value;

2. Clean wiring points, especially the critical path of sense;

3. Select the small point of inductance or the inductance with closed magnetic circuit;

4. Add RC low-pass filter. This effect is a little bad. The consistency of C is not good and the deviation is a little large, but it should be enough for lighting. In any case, frequency reduction has no advantages but disadvantages, so it must be solved.

5、 Selection of inductance or transformer

Many users reflect that with the same drive circuit, the inductance produced with a has no problem, and the inductance current produced with B becomes smaller. In this case, look at the inductance current waveform. Some engineers do not notice this phenomenon and directly adjust the sense resistance or the working frequency to reach the required current, which may seriously affect the service life of the LED.

Therefore, before design, reasonable calculation is necessary. If the theoretical calculation parameters are far from the commissioning parameters, consider whether to reduce the frequency and whether the transformer is saturated. When the transformer is saturated, l will become smaller, resulting in a sharp increase in the peak current increment caused by transmission delay, and the peak current of LED will also increase. On the premise that the average current remains unchanged, we can only watch the light decay.