Electronic – MOSFET turn on/off curves misunderstanding


The MOSFET's turn on/off curves have a part where the drain current and the voltage across Drain-Source are not zero. That's where most of the switching dissipation happens.

The problem I have is in T1-T2 section (see figure below). If the source voltage (Vin) is constant and the current is flowing through the load (ILOAD), how can we still have Vin across Vds?

We shouldn't have the same current flowing through the load when Vds=0 and Vds=Vin …

Can you please explain what am I missing?

Does this logic differ with a different source type, a different load type … What are the parameters I need to properly predict the MOSFET's behavior when turning on/off, other than the gate driving?

enter image description here


Image taken from a Vishay application note

Best Answer

The answer is that all these plot shows in application notes are made by assuming the inductive load plus a flyback diode is switching by a MOSFET. As shown in fig 6.

Additionally, the authors assume the MOSFET operates as a switch in the application when the inductive load is in "continuous current mode"(we replace the induction with a constant current source at the drain for this purpose). Thus, the current in the flyback diode and in the inductor is not at 0A at the beginning of a turn-on process.

And due to the flyback diode current, the MOSFET Vds will start to decrease only when diode current reach 0A (the MOSFET needs to take over all the inductor current from the diode) before Vds starts to decreases.

And the simulations confirm this:

enter image description here

Try read here


But of course, for a pure resistance load, the situation looks different.

enter image description here

As you can see for a purely resistive load everything looks as we expect to be looking.

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