Simplify the Five Principles of LED Design
This is mainly for the high voltage driver chip with a built-in power modulator. If the chip consumes a current of 2mA and a voltage of 300V is applied to the chip, the power consumption of the chip is 0.6W, which of course will cause the chip to heat up. The maximum current of the driver chip comes from the consumption of the driving power MOS transistor. The simple calculation formula is I=cvf (taking into account the charging resistor efficiency, the actual I=2cvf, where c is the cgs capacitor of the power MOS transistor, and v is the power transistor conduction When the gate voltage, so in order to reduce the power of the chip, must find ways to reduce c, v and f. If c, v and f can not be changed, then think of ways to split the chip's power consumption to off-chip devices, pay attention not to Introduce extra power consumption.The simpler one is to consider better heat dissipation. For the Led auto light design, chip cooling is a key consideration.
Power tube heating
On this issue, I also saw someone posting on the forum. Power tube power consumption is divided into two parts, switching loss and conduction loss. It should be noted that in most applications, especially LED mains-driven applications, switching damage is much greater than conduction losses. The switching loss is related to the power supply's cgd and cgs, and the chip's driving ability and operating frequency, so to solve the power transistor heating can be resolved from the following aspects: A, can not be selected one-sided according to the size of the on-resistance MOS power transistor, because The smaller the internal resistance, the greater the capacitance of cgs and cgd. Such as 1N60's cgs is about 250pF, 2N60's cgs is about 350pF, 5N60's cgs is about 1200pF, the difference is too big, when choosing the power tube, enough is enough. B. The rest is the frequency and chip driver capability. Here only the frequency effect. The frequency and the conduction loss are also proportional to each other. Therefore, when the power transistor is heated, the first thing to consider is whether the frequency selection is a bit high. Try 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 larger or the inductance becomes larger, 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 a load capacitor.
Frequency reduction of operating frequency
This is also a common phenomenon in the debugging process. The down-clocking is mainly caused by two aspects. The ratio of input voltage and 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. Set the minimum current to a smaller point; b. Clean the wiring, especially the key path of the sense; c. Select the inductor or select the inductor of the closed magnetic circuit. d, plus RC low-pass filter it, this effect is a bit bad, C's consistency is not good, the deviation is a bit big, but for the lighting should be enough. In any case, there is no advantage in lowering the frequency, only the disadvantages, so it must be resolved.
Inductor or transformer selection
Finally I talked about the key points. I haven't started yet. I can only say the effect of saturation. Many users react, the same drive circuit, no problem with the inductance produced by a, and the inductor current produced with b becomes smaller. In this case, look at the inductor current waveform. Some engineers did not notice this phenomenon and directly adjusted the sense resistor or the working frequency to reach the required current. This may seriously affect the service life of the LED. So, before the design, a reasonable calculation is necessary. If the parameters of the theoretical calculation and the debugging parameters are a little different, we must consider whether or not the frequency reduction and the transformer saturation. When the transformer is saturated, L will become smaller, resulting in a sharp increase in the peak current increase caused by the transmission delay, and then the LED peak current will increase. With the average current constant, you can only see the light decay.
LED current size
Everyone knows that if the LEDripple is too large, the LED lifetime will be affected, how much it will affect, and no one has ever said. Previously asked the LED factory this data, they said that within 30% can be accepted, but later not verified. It is recommended to try to control small points. If the thermal solution is not good, the LED must be derated. It is also hoped that experts can give specific indicators, or else they will influence the promotion of LED.
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