# Driving MOSFET/IGBT - Why Miller Plateau is a plateau?

I'm currently trying to understand what the Miller plateau is: what the different parameters that influence the Miller plateau are, when it begins, and when it ends. I'm reading document "Fundamentals of MOSFET and IGBT gate Driver circuit" from Texas Instrument, but it misses some details for my understanding.

Miller plateau seems to begins when the diode (consider a buck converter) stops conducting. Then the drain voltage is now not clamped to Vout by the diode and the source to drain voltage can diminish. The Cgd capacitor previously charged between Vgate and the drain voltage which was equal to Vout must release some charge as the drain voltage diminish! So some current is flowing from the driver to the drain "through" Cgd capacitor. Here is the question: why is the current steady? If the driver output current is able to source more current than the Cgd capacitor is able to discharge itself, the current should not be a plateau and inversely, why this plateau is so long? Did I make a mistake?

The current through the capacitor is nearly steady since it's related to the (assumed-constant) current that the gate driver can supply. Assuming that other parasitics are insignificant, the gate driver current cannot avoid the plateau, only shorten it.

If the driver output current is able to source more current than the Cgd capacitor is able to discharge itself, the current should no be a plateau and inversely... why this plateau is so long ... Did I do a mistake ?

The $$\C_{gd}\$$ parasitic can be discharged with arbitrarily high current. It shunts gate current away from the gate, preventing $$\V_{gs}\$$ from rising appreciably, which creates the plateau.

Here's a detailed transient response (using a voltage source + Thévenin resistance), and a resistive load:

1. The input voltage begins to rise, at marker V1. $$\V_{gs}\$$ also starts rising at this time and so does $$\V_{ds}\$$ due to feedthrough via $$\C_{gd}\$$.

2. At marker V2, the MOSFET begins to conduct. This causes Vds to fall. As Vds falls, current is conducted via Cgd (which looks like a short circuit to fast transients), drawing current away from the gate. This arrests the fall of Vds. In my simulation which used a large Cgd for emphasis, 99% of the available current was shunted via that cap.

3. Throughout the plateau, we continue to see high current through Cgd. Vgs can only rise slowly--a small-signal slew rate of Vgs by $$\\frac{dV}{dt}\$$ requires a current of approximately $$\C_{gs}\frac{dV}{dt} + (1 + g_mR_{load}) C_{gd}\$$. In this region of operation, $$\g_m\$$ is quite large since the transistor is turning on and hence in its saturation region (where current is controlled primarily by Vgs).

4. At marker V3, the transistor is turned on fully. Since the drain voltage no longer needs to slew, the plateau is finished and Vgs can now rise quickly.

• Thank you very much !
– Jess
Dec 25 '19 at 16:40