Analysis of five key technical points related to heating and burning of MOS tubes in LED fluorescent lamp power supply design

　　1. Chip heating

　　This content is mainly for high-voltage driver chips with built-in power modulators. If the current consumed by the chip is 2mA, and a voltage of 300V is applied to the chip, 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 effect 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, you must find a way to reduce c, v and f. If c, v and f cannot be changed, then please find a way to distribute the power consumption of the chip to devices outside the chip, and be careful not to introduce additional power consumption. To put it simply, just consider better heat dissipation.

　　2. Power tube heating

　　Regarding this issue, I have also seen someone posting on the power network 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. The switching loss is related to the cgd and cgs of the power tube, as well as the driving ability and operating frequency of the chip. Therefore, to solve the heating of the power tube, you can solve it from the following aspects: A. You cannot choose the MOS power tube based on the size of the on-resistance, because the smaller the internal resistance, the larger the cgs and cgd capacitance. 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 a power tube, just use enough. B. The rest is the frequency and chip driving ability. Here we only talk about the influence of frequency. The frequency is also proportional to the conduction loss, so when the power tube is heated, the first thing to think about is whether the frequency is a bit high. Find a way to reduce the frequency! However, please note that when the frequency is reduced, in order to obtain the same load capacity, the peak current must increase or the inductance must also increase, which may cause the inductor to enter the saturation region. If the inductor saturation current is large enough, you can consider changing CCM (continuous current mode) to DCM (discontinuous current mode), which requires adding a load capacitor.

　　3. Reduce the operating frequency

　　This is also a common phenomenon for users during the debugging process. 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 efficiency will be higher if the load voltage is high. For the latter, you can try the following aspects: a. Set the minimum current to a smaller value; b. Make the wiring cleaner, especially the key path of sense; c. Select a smaller inductor or choose an inductor with a closed magnetic circuit; d. Add RC low-pass filtering. This has a bad effect. The consistency of C is not good and the deviation is a bit large, but it should be enough for lighting . In any case, frequency reduction has no advantages, only disadvantages, so it must be solved.

　　4. Selection of inductor or transformer

　　Finally, we are talking about the key points. I haven't gotten started yet, so I can only talk about the impact of saturation. Many users have reported that for the same drive circuit, there is no problem with the inductor produced by a, but the current of the inductor produced by b becomes smaller. In this case, you need to look at the inductor current waveform. Some engineers did not notice this phenomenon and directly adjusted the sense resistor or the operating frequency to reach the required current. This may seriously affect the service life of the LED. Therefore, before designing, reasonable calculations are necessary. If the theoretically calculated parameters are a bit different from the debugging 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 the transmission delay, and the peak current of the LED will also increase. Under the premise that the average current remains unchanged, you can only watch the light decay.

　　5. LED current

　　Everyone knows that if the LEDripple is too large, the LED life will be affected, but no expert has ever said how big the impact is. I asked the LED factory about this data before, and they said that 30% or less is acceptable, but it has not been verified later. It is recommended to control it as small as possible. If the heat dissipation is not well solved, the LED must be used at a reduced rating. I also hope that an expert can give a specific indicator, otherwise it will affect the promotion of LED.