Five key factors to simplify LED driver design
This is primarily for high voltage driver chips with built-in power modulators. If the current consumed by the chip is 2 mA, the voltage of 300 V is applied to the chip, and the power consumption of the chip is 0.6 W, which of course causes the chip to generate heat. The maximum current of the driving 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 power tube conduction When the gate voltage, so 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 can not be changed, then please find a way to divide the power consumption of the chip into the off-chip device, be careful not to Introduce additional power consumption. A simpler one is to consider better heat dissipation.
Power tube heating
Regarding this issue, I have also seen people posting in 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 for 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 capability and operating frequency of the chip. Therefore, to solve the heat generation of the power tube, it can be solved from the following aspects: A, the MOS power tube cannot be selected according to 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 the power tube is selected, it is enough. B. The rest is the frequency and chip drive capability. Here we only talk about the frequency effect. The frequency is also proportional to the conduction loss. Therefore, when the power tube is hot, the first thing to think about is whether the frequency selection is a bit high. Find ways to reduce the frequency! However, it should be noted that when the frequency is reduced, in order to obtain the same load capacity, the peak current must be increased or the inductance is also increased, which may cause the inductor to enter the saturation region. If the inductor saturation current is large enough, consider changing the CCM (continuous current mode) to DCM (discontinuous current mode), which requires an additional load capacitor.
Operating frequency down frequency
This is also a common phenomenon in the debugging process of users. The frequency reduction is mainly caused by two aspects. The ratio of input voltage to load voltage is small and 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: a, the minimum point of the minimum current setting; b, clean wiring, especially the critical path of sense; c, the small point of the inductor selection or the inductance of the closed magnetic circuit ; d, add RC low-pass filter, this effect is a bit bad, C's consistency is not good, the deviation is a bit large, but for lighting should be enough. No matter how bad the frequency is, there is no advantage, only the bad, so we must solve it.
Inductor or transformer selection
I finally talked about the point. I haven't gotten started yet. I can only talk about the impact of saturation. Many users react, the same drive circuit, the inductance produced by a is no problem, the inductor current produced by b becomes smaller. In this case, look at the inductor current waveform. Some engineers did not notice this phenomenon, directly adjust the sense resistance or the operating frequency to reach the required current, which may seriously affect the service life of the LED. Therefore, before the design, reasonable calculation is necessary. If the theoretical calculation parameters and the debugging parameters are a little far away, consider whether the frequency reduction and the transformer are saturated. When the transformer is saturated, L will become smaller, causing the peak current increase caused by the transmission delay to rise sharply, and the peak current of the LED will also increase. Under the premise that the average current is constant, only the light fades.
LED current size
Everyone knows that if the LEDripple is too large, the LED life will be affected, how big the impact is, and no experts have ever seen it. I have asked the LED factory this data before, they said that it is acceptable within 30%, but it has not been verified. It is recommended to try to control the small points. If the heat solution is not good, the LED must be derated. I also hope that some experts can give a specific indicator, or else affect the promotion of LED.
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