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Industry
Num：75
See：19542021-07
13
Five key points for successful design of LED driver power
LED current size
The size of the LED current directly affects the service life, it is recommended to reduce the use, so try to control the small point, especially if the LED heat dissipation effect is not good, the LED must leave enough room.
Chip fever
This is mainly for high voltage drive chips 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 heat of the chip.
The * 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 gate voltage during the power tube's switching on. Therefore, 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, then please find a way to distribute the power of the chip to the devices outside the chip, and be careful not to introduce additional power.
Power tube heating
The power consumption of power tube is divided into two parts, switching loss and conduction loss. It should be noted that in most cases, especially in LED mains drive applications, the switching damage is much greater than the conduction loss. Switching loss is related to the CGD and CGS of the power tube, as well as the driving ability and working frequency of the chip, so to solve the heating of the power tube can be solved from the following aspects:
1. MOS power tube can not be selected according to the on-resistance, because the smaller the internal resistance, the greater the capacitance of CGS and CGD. 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 the power tube, it is enough.
2, the rest is the frequency and chip drive ability, here only talk about the influence of frequency. Frequency is also proportional to the conduction loss, so the power tube heating, the first thing to think about is whether the frequency selection is a little high.
Note, however, that as the frequency decreases, either the peak current must increase or the inductance must increase in order to achieve the same load capacity, which may cause the inductor to enter the saturated region.
If the inductor saturation current is high enough, consider changing CCM(continuous current mode) to DCM(discontinuous current mode), which will require an additional load capacitor.
Operating frequency down
This is also a common phenomenon for users in the debugging process, and the frequency drop 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 load voltage is high, the efficiency will be higher.
For the latter, try the following:
1. Set the minimum current to a smaller point;
2. Clean wiring, especially the critical path of SENSE;
3. Select the small point of inductance or the inductance with closed magnetic circuit;
4, add RC low pass filter, this effect is a little bad, the consistency of C is not good, the deviation is a little big, but for lighting should be enough. In any case, down frequency has no benefits, only disadvantages, so we must solve.
Selection of inductor or transformer
Multiple users report that with the same driving circuit, there is no problem with the inductor produced by A, while the inductor current produced by B becomes smaller. In this case, look at the inductance current waveform. Some engineers do not pay attention to this phenomenon and directly adjust SENSE resistor or operating frequency to meet the required current, which may seriously affect the service life of LEDs.
Therefore, before the design, reasonable calculation is necessary. If the theoretical calculation parameters and debugging parameters are a little far from each other, it is necessary to consider whether the frequency is down and whether the transformer is saturated. When the transformer is saturated, L will become smaller, leading to a sharp increase in the peak current increment caused by transmission delay, and then the peak current of LED will also increase. With the average current constant, just watch the light fade.
The size of the LED current directly affects the service life, it is recommended to reduce the use, so try to control the small point, especially if the LED heat dissipation effect is not good, the LED must leave enough room.
Chip fever
This is mainly for high voltage drive chips 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 heat of the chip.
The * 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 gate voltage during the power tube's switching on. Therefore, 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, then please find a way to distribute the power of the chip to the devices outside the chip, and be careful not to introduce additional power.
Power tube heating
The power consumption of power tube is divided into two parts, switching loss and conduction loss. It should be noted that in most cases, especially in LED mains drive applications, the switching damage is much greater than the conduction loss. Switching loss is related to the CGD and CGS of the power tube, as well as the driving ability and working frequency of the chip, so to solve the heating of the power tube can be solved from the following aspects:
1. MOS power tube can not be selected according to the on-resistance, because the smaller the internal resistance, the greater the capacitance of CGS and CGD. 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 the power tube, it is enough.
2, the rest is the frequency and chip drive ability, here only talk about the influence of frequency. Frequency is also proportional to the conduction loss, so the power tube heating, the first thing to think about is whether the frequency selection is a little high.
Note, however, that as the frequency decreases, either the peak current must increase or the inductance must increase in order to achieve the same load capacity, which may cause the inductor to enter the saturated region.
If the inductor saturation current is high enough, consider changing CCM(continuous current mode) to DCM(discontinuous current mode), which will require an additional load capacitor.
Operating frequency down
This is also a common phenomenon for users in the debugging process, and the frequency drop 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 load voltage is high, the efficiency will be higher.
For the latter, try the following:
1. Set the minimum current to a smaller point;
2. Clean wiring, especially the critical path of SENSE;
3. Select the small point of inductance or the inductance with closed magnetic circuit;
4, add RC low pass filter, this effect is a little bad, the consistency of C is not good, the deviation is a little big, but for lighting should be enough. In any case, down frequency has no benefits, only disadvantages, so we must solve.
Selection of inductor or transformer
Multiple users report that with the same driving circuit, there is no problem with the inductor produced by A, while the inductor current produced by B becomes smaller. In this case, look at the inductance current waveform. Some engineers do not pay attention to this phenomenon and directly adjust SENSE resistor or operating frequency to meet the required current, which may seriously affect the service life of LEDs.
Therefore, before the design, reasonable calculation is necessary. If the theoretical calculation parameters and debugging parameters are a little far from each other, it is necessary to consider whether the frequency is down and whether the transformer is saturated. When the transformer is saturated, L will become smaller, leading to a sharp increase in the peak current increment caused by transmission delay, and then the peak current of LED will also increase. With the average current constant, just watch the light fade.
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