Under normal circumstances, when a MOSFET is in a switching state, the switching waveform should be the same as the control signal voltage waveform.

However, in reality, when a large resistor is connected in series with the gate, the charging and discharging current on the gate is very small, causing the charging speed of the capacitor to slow down. This results in the MOSFET not fully conducting, even though the next signal to turn off the MOSFET has arrived.

We are familiar with the Miller plateau, which tells us that there is an equivalent capacitance between the GS and GD of the MOSFET, and the voltage across the capacitor cannot change abruptly. Therefore, fundamentally, it is about charging and discharging these capacitors to make the Vgs voltage reach the threshold for conduction and turn-off.

We use this circuit to explain: here, when the input signal is 1 MHz, a 5V amplitude square wave signal is connected to the NMOS gate through a 200 ohm resistor R1. However, the Ud voltage, which should reach its minimum value, starts to rise again when it reaches 5V, indicating that the switch is not fully conducting.

In this case, the series resistor R1 can be reduced to about 20 ohms, and the Vd voltage signal remains a square wave, which allows for full conduction.

However, when Ud is turned off, there is a long "ramp-up" period, meaning that even if the gate drive current is increased, there is a limit to the switching frequency of the MOSFET itself.

So, we reduce the gate drive frequency to 100 kHz, and then the Ud and Ugs become a relatively standard square wave signal.

When a MOSFET is not fully conducting and the edge slope is slow, what happens?

In fact, when a MOSFET is fully conducting, the loss of the MOSFET can be considered as 0, meaning that the loss of the MOSFET only occurs during the switching process.

On the contrary, the loss of the MOSFET will increase, leading to more heating and ultimately causing the MOSFET to burn out.