LED drive power design is not difficult, but in mind must first plan well.
As long as the calculation before debugging, measurement during debugging and aging after debugging, we believe that anyone can do well with the LED.
1. LED current size
We know that LEDripple is too big, and we haven't seen any experts say how much of an impact it has on LED life.
2. Chip heating
This is mainly aimed at the high voltage driver chip with built-in power modulator.
If the current consumed by the chip is 2mA, the voltage of 300V is added to the chip, and the power consumption of the chip is 0.6w, of course, it will cause the heat of the chip. The maximum current of the driver 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 capacitor of the power MOS tube and v is the gate voltage when the power tube is switched on. Therefore, in order to reduce the power consumption of the chip, ways must be found to reduce c, v and f.
If c, v, and f cannot be changed, then find a way to distribute the power of the chip to off-chip devices, taking care not to introduce additional power. Even simpler, consider better cooling.
3. Power tube heating
About this problem, also saw somebody to had posted in BBS.
The power consumption of power tube is divided into two parts: switch loss and on loss.
It should be noted that in most cases, especially LED mains drive applications, the switch damage is much greater than the conduction loss.
Switch loss is related to CGD and CGS of power tube, driving ability and working frequency of chip, so the heating of power tube can be solved from the following aspects: a. MOS power tube cannot be selected unilaterally according to the size of conducting 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, the CGS of 5N60 is about 1200pF, the difference is too big, when choosing power tube, enough can be used. B, the rest is frequency and chip driver capability, here only talk about the impact of frequency.
The frequency is also proportional to the conduction loss, so when the power tube heats up, the first thing to think about is whether the frequency is a little high.
Find a way to lower the frequency! Note, however, that when the frequency goes down, in order to get the same load capacity, the peak current has to go up or the inductance has to go up, which can cause the inductance to go into the saturation zone.
4. Frequency reduction of working frequency
This is also a common phenomenon in the debugging process. The frequency reduction is mainly caused by two aspects. The ratio 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 higher.
For the latter, you can try the following aspects:
A. Set the minimum current to a smaller point;
B. clean wiring, especially sense, the critical path;
C. Select small points of inductance or inductance with closed magnetic circuit;
D, add RC low-pass filtering, this effect is a little bad, C consistency is not good, the deviation is a little big, but for lighting should be enough.
In any case, there is no good, only harm, so must be resolved.
5. Selection of inductor or transformer
For the same drive circuit, the inductance produced by a is no problem, while the inductance current produced by b is reduced.
In this case, look at the inductive current waveform.
Some engineers did not notice this phenomenon and directly adjusted the sense resistance or working frequency to 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 away, to consider whether the frequency drop and whether the transformer is saturated.
When the transformer is saturated, L will become smaller, leading to a sharp increase in the increment of peak current caused by transmission delay, and then the peak current of LED also increases. With the average current constant, you can only watch the light decay.