Ideally, the switching waveform of the MOS tube should be consistent with the control signal voltage waveform. However, in practical applications, the switching waveform of the MOS tube is often abnormal. What is going on?

As shown in the figure below, when the input signal is a 1MHz, 5V square wave signal connected to the gate of the NMOS through a 200Ω resistor R1, the waveform shows that the minimum value of the Ud voltage is still around 5V and then starts to rise again, that is, the switch of the MOS tube is not fully turned on.

Why is this?

This phenomenon can be explained by the Miller platform. There is an equivalent capacitor between the MOS tube GS and GD, and the voltage across the capacitor cannot change suddenly. Therefore, when a switching signal is added to the gate, it is essentially charging and discharging these capacitors to make the Vgs voltage reach the threshold of conduction and shutdown.

Therefore, if a large resistor is connected in series with the gate, the charge and discharge current is small, and the capacitor charges slowly, the next shutdown signal will arrive before the MOS tube is fully turned on.

To solve the problem of incomplete conduction, the series resistor R1 is reduced to 20Ω. At this time, the Vd voltage signal is basically a square wave, and the MOS tube can be fully turned on. However, a new problem arises. There is a long "climbing slope" when Ud is turned off. This shows that even if the gate drive current is increased, the switching frequency of the MOS tube itself still has an upper limit.

 On the basis of the above, we further reduce the frequency of the gate drive signal to 100kHz. At this time, we observe that Ud and Ugs are basically a relatively standard square wave signal (the "Miller platform" can be observed through the Ugs signal). Generally, we believe that such a drive is qualified.

 What problems will arise when there is incomplete conduction and slow edge climbing?

From the power consumption calculation formula P=U*I, when the MOS tube is fully turned on, we can approximately assume that the loss of the MOS tube is 0, that is, the loss of MOS only exists in the switching process.

When the MOS tube is not fully turned on or the edge is slow, the loss of the MOS tube itself will increase significantly, which will increase the heating of the MOS tube. In severe cases, it may even cause the MOS tube to burn out.