Why does MOS tube heat easily?
When designing the power supply or driving circuit, it is inevitable to use FET, also known as MOS transistor. There are many types and functions of MOS tubes. Use the power supply or drive, of course, using its switching function. Next, let's take a look at the five key technologies of MOS tube heating.
1. chip heatingThis content mainly focuses on the high-voltage driver chip with built-in power modulator. If the current consumed by the chip is 2mA and the voltage of 300V is applied, the power consumption of the chip is 0.6W, which will definitely 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 turned on. Therefore, in order to reduce the power consumption of the chip, we must find a way to reduce C, V and F. If C, V and F cannot be changed, please find a way to distribute the power consumption of the chip to devices outside the chip. Simply put, it is better to consider heat dissipation.
2. Power tube heatingThe power consumption of the tube is divided into two parts: switching loss and conduction loss. It should be noted that in most cases, especially in commercial LED power supply driving applications, the switch damage is far greater than the conduction loss. The switching loss is related to the cgd and cgs of the power tube, the driving capability of the chip and the working frequency. Therefore, to solve the heating problem of power tube, we can solve it from the following aspects: A, MOS power tube can't be selected unilaterally according to on-resistance because the smaller the internal resistance, the larger the capacitance of cgs and cgd. For example, the cgs of 1N60 is about 250pF, that of 2N60 is about 350pF, and that of 5N60 is about 1200pF. It's a big difference. When choosing a power tube, it is enough. B, the rest is the frequency and chip driving ability. Here, we only talk about the influence of frequency. It is also proportional to the frequency conduction loss, so when the power tube heats up, it is first necessary to consider whether the frequency selection is a bit high. Try 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 will inevitably increase or the inductance will also increase, which may cause the inductance to enter the saturation region. If the inductor saturation current is large enough, it can be considered to change CCM (continuous current mode) to DCM (discontinuous current mode), which requires an additional load capacitor.
3. The working frequency is reduced.This is also a common phenomenon when users debug, and the 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. Although the load voltage is high, the efficiency will be higher. For the latter, we can try the following aspects: A, set the minimum current to a smaller point; B, clean the wiring, especially the sense of critical path; C, selecting the small point of inductance or closing the magnetic circuit; D inductance, plus RC low-pass filter, this effect is a little bad, the consistency of C is not good, and the deviation is a little big, but it should be enough for lighting. No matter how to reduce the frequency, it is bad, only bad, so it must be solved.
4. Selection of inductor or transformerMany users responded that, with the same driving circuit, the inductor produced by A has no problem, and the inductor current produced by B has become smaller. In this case, look at the inductor current waveform. Have engineers noticed this phenomenon? Directly adjusting the inductive resistance or working frequency to reach the required current may seriously affect the service life of LED. Therefore, reasonable calculation is necessary before design. If the theoretical calculation parameters are far from the debugging parameters, it is necessary to consider whether the frequency is reduced and whether the transformer is saturated. When the transformer is saturated, L will become smaller, resulting in a sharp increase of peak current increment caused by transmission delay, and then the peak current of LED will also increase. On the premise of constant average current, we can only watch the light gradually fade.
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