I was reading the AN1403 "Determining Switching Losses of SEMIKRON IGBT modules" from SEMIKRON.
And I came across the following lines talking about the effect of the gate inductance on the Eon Switching losses:
Here is the circuit (Figure 7)
Here are their measures:
If I correctly understand, it is said that the higher the inductance is, the lower the Eon would be, where Eon are the energy loss for turning on the IGBT.
Well, from my mind, a higher inductance will limit the rise of the gate to source voltage, and it will slow down the commutation of the IGBT and so the losses would be higher.
I am not sur to understand what SEMIKRON is saying: "The effect of this inductance can be compared with a current source, which supplies additional gate charge at the beginning of the "Vge-plateau" period."
I think that as at the beginning of the turn on process the current is maximal and limited by the driver, the inductor is magnetically charged and when the miller plateau arises, the inductance is still a bit charged and so the inductance tends to reduce in time the Miller plateau. But clearly, I do not think that it is as simple as it.
Do you understand why a higher inductance on the gate reduces the Eon?
Thank you for your help.
---------------------------------E D I T ----------------------------------- I have done a simulation for trying to understand what SEMIKRON is saying, and the result of the simulation is not showing the SEMIKRON's measures.
Here is the simulation circuit :
Then I added to the circuit different inductances :
And here are the results showing the gate voltage and the power losses :
And here are the power losses :
I integrate it and the simulation with the higher inductance is the worse in term of Eon.