1, chip fever
This is primarily for high voltage driver chips with built-in power modulators. If the current consumed by the chip is 2 mA, the voltage of 300 V is applied to the chip, and the power consumption of the chip is 0.6 W, which of course causes the chip to generate heat. 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 capacitance of the power MOS tube, and v is the power tube guide. The gate voltage is always 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, then find a way to divide the power consumption of the chip into off-chip devices, taking care not to introduce additional power. A little simpler is to consider better heat dissipation.
2, power tube fever
Regarding this issue, I have also seen people posting 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 for LED mains 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 capability and operating frequency of the chip. Therefore, to solve the heat generation of the power tube, it can be solved from the following aspects:
a: The MOS power tube cannot be selected one-sided according to the on-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, and the cgs of 5N60 is about 1200pF. The difference is too great. When the power tube is selected, it is enough;
b: The rest is the frequency and chip drive capability. Here we only talk about the effect of frequency. The frequency is also proportional to the conduction loss. Therefore, when the power tube is hot, the first thing to think about is whether the frequency selection is a bit high. Find ways 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 be increased or the inductance is also increased, which may cause the inductor to enter the saturation region. If the inductor saturation current is large enough, consider changing the CCM (continuous current mode) to DCM (discontinuous current mode), which requires adding a load resistor.
3, working frequency down frequency
This is also a common phenomenon in the debugging process of the user. The frequency reduction is mainly caused by two aspects: the ratio of the input voltage to the 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: the smaller point of the minimum current setting;
b: The wiring is clean, especially the critical path of sense;
We design and manufacture Linear Actuators For Racking System.
U10 is a heavy load Linear Actuator designed for industrial application, especially be used for solar racking system. U10 is a robust and powerful actuator, designed for heavy-duty applications in harsh environments like solar tracker. It features high load capability, long lifetime and low power consumption.
Electric Linear Actuator For Racking System, Linear Actuator For Racking System,Actuator For Racking System
TOMUU (DONGGUAN) ACTUATOR TECHNOLOGY CO., LTD. , http://www.tomuu.com