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1、 LED current size
The size of LED current directly affects its service life. It is recommended to reduce the rating for use, so try to control it as small as possible, especially if the heat dissipation effect of LED is not good, LED must leave sufficient margin.
2、 Chip heating
This is mainly aimed at high-voltage drive chips with built-in power modulators. If the current consumed by the chip is 2mA and a voltage of 300V is applied to the chip, the power consumption of the chip is 0.6W, which will naturally cause the chip to heat up. The maximum current of the driving chip comes from the consumption of the driving power MOS transistor, 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 on. Therefore, in order to reduce the power consumption of the chip, it is necessary to find ways to reduce c, v, and f. If c, v, and f cannot be changed, please find a way to allocate the power consumption of the chip to devices outside the chip, and be careful not to introduce additional power consumption.
3、 Power tube heating
The power consumption of a power transistor is divided into two parts, switching loss and conduction loss. It should be noted that in most situations, especially in LED mains drive applications, the switch damage is much greater than the conduction loss. The switch loss is related to the cgd and cgs of the power transistor, as well as the driving capacity and operating frequency of the chip. Therefore, to solve the heat generation of the power transistor, the following aspects can be taken into account:
1. We cannot choose MOS power transistors solely based on the size of the conduction resistance, as 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, and the cgs of 5N60 is about 1200pF. The difference is too large. When selecting a power transistor, it is sufficient.
2. The rest is frequency and chip driving ability, and we will only talk about the impact of frequency here. The frequency is also proportional to the conduction loss, so when heating a power tube, the first thing to consider is whether the frequency selection is a bit high. However, it should be noted that when the frequency decreases, in order to achieve the same load capacity, the peak current must increase or the inductance must also increase, which may lead to the inductance entering the saturation region. If the saturation current of the inductor is large enough, it can be considered to change the CCM (continuous current mode) to DCM (non continuous current mode), which requires an additional load capacitor.
4、 Operating frequency reduction
This is also a common phenomenon among users during debugging, and frequency reduction 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, as high load voltage may result in higher efficiency. For the latter, the following aspects can be attempted:
1. Set the minimum current to a smaller value;
2. Clean wiring points, especially the critical path of sense;
3. Select a small point for inductance or choose an inductance with a closed magnetic circuit;
4. Add RC low-pass filtering, this effect is a bit bad. The consistency of C is not good, and the deviation is a bit large. However, it should be sufficient for lighting. Regardless, frequency reduction has no benefits, only drawbacks, so it must be addressed.
5、 Selection of inductance or transformer
Multiple users have reported that for the same drive circuit, there is no problem with the inductor produced with a, while the current of the inductor produced with b decreases. When encountering this situation, it is necessary to check the waveform of the inductance current. Some engineers do not pay attention to this phenomenon and directly adjust the sense resistor or working frequency to reach the required current, which may seriously affect the service life of LEDs.
So, before design, reasonable calculations are necessary. If there is a slight difference between the theoretical calculation parameters and the debugging parameters, it is necessary to consider whether to reduce the frequency and whether the transformer is saturated. When the transformer is saturated, L will decrease, causing a sharp increase in the peak current increment caused by transmission delay, and the peak current of the LED will also increase. Under the premise of keeping the average current constant, we can only watch light decay.