Collection: LED Drive Debugging Summary

key word:

LED, transformer, current, chip

1. Chip heating

This is mainly aimed at the high-voltage drive chip with built-in power modulator. If the current consumed by the chip is 2mA and the voltage of 300V is added to the chip, the power consumption of the chip is 0.6W, of course, it will cause the chip to heat up. The maximum current of the drive chip comes from the consumption of the drive 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, V is the gate voltage when the power tube 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 to devices outside the chip, and be careful not to introduce additional power consumption. To be simpler, consider better heat dissipation.

2. Power tube heating

On this issue, I've also seen someone post in this area. 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, as well as the driving capacity and working frequency of the chip, Therefore, to solve the heat generation of power tube, we can solve it from the following aspects: A: MOS power tube 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 tube, Enough 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 decreases, in order to obtain the same load capacity, the peak current must increase or the inductance will also increase, which may lead to the inductance entering the saturation region. If the inductance saturation current is large enough, you can consider changing CCM (continuous current mode) to DCM (discontinuous current mode), so you need 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 small proportion of input voltage and load voltage and large system interference. For the former, pay attention 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: lower the minimum current; b: Clean wiring points, especially the critical path of sense; c: Select the small point of inductance or the inductance with closed magnetic circuit; d: 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.

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 react that there is no problem with the inductance produced by a in the same driving circuit, and the inductance current produced by 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 reach the required current, This may seriously affect the service life of LED. Therefore, reasonable calculation is necessary before design. If the theoretically calculated 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, Then the peak current of LED also increases. On the premise of constant average current, 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 data of the LED factory 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 specific indicators, Otherwise, it will affect the promotion of LED.

Having said so much, it seems that the design of LED driver is not difficult. We must know it well. As long as we calculate before debugging, measure during debugging and aging after debugging, I believe anyone can do led.

Collection: LED Drive Debugging Summary 1

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