Small Tips for LED Drive Design: Five Key Points

1. 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 maximum 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 transistor and V is the gate voltage when the power transistor is turned on, 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 cannot be changed, please find a way to divide the power consumption of the chip into devices outside the chip, and be careful not to introduce additional power consumption. A simpler point is to consider Better heat dissipation.

2. Power tube heating

On this issue, I've also seen some people post in the Power Grid 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 in LED commercial power 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, so we should solve the problem of the power tube Heating can be solved from the following aspects: A. MOS power tubes 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, the CGS of 2n60 is about 350pf, and the CGS of 5n60 is about 1200pf. The difference is too great. When selecting power tubes, it is enough. B. 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. Find a way 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 become larger or the inductance will become larger, which may lead to the inductance entering the saturation region. For example If the saturation current of the inductor is large enough, it can be considered to change CCM (continuous current mode) to DCM (discontinuous current mode), so it is necessary to add a load capacitance.

3. Operating frequency reduction

This is also a common phenomenon during debugging. The frequency reduction is mainly caused by two aspects. The proportion of input voltage and 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: A. set the minimum current lower; B Clean the wiring, especially the key path of sense; C. select the small point of inductance or the inductance of closed magnetic circuit; D. add RC low-pass filter. The impact 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 is not good, but only bad, so it must be solved.

4. Selection of inductance or transformer

Finally, let's get to the point. I haven't started yet. I can only talk nonsense about the impact of saturation. Many users respond that the inductance produced with a is OK for the same drive circuit, and the inductance current produced with B becomes smaller. In this case, we should look at the inductance current waveform. Some engineers don't notice this phenomenon and directly adjust the sense resistance or working frequency to meet the needs This may seriously affect the service life of the LED. Therefore, reasonable calculation is necessary before design. If the theoretically calculated parameters and commissioning parameters are far from each other, 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 the LED will also increase With the increase. Under the premise that the average current remains unchanged, we can only watch the light decay.

5. LED current size

We all know that if the ledripple is too large, the life of the LED will be affected. We haven't seen any experts say how much the impact is. We've asked the LED factory about this data before, and they said it's acceptable within 30%, but it hasn't been verified later. It's recommended to control it as small as possible. If the heat dissipation is not solved well, the LED must be reduced. I hope some experts can give us a specific explanation Indicators, otherwise it will affect the promotion of LED.

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