MOSFET turn on process

Thread Starter

mirror_pole

Joined Nov 14, 2020
10
Hallo guys,

I have a few questions regarding the turn on process of a mosfet in switching applications such as SMPS.

From my understanding one common approximation is that for an inductive load the current through the inductor stays constant due to larger time constant associated with the inductor then the switching frequency of the mosfet.

But i still dont understand a few Things. How is it possible that after the miller plateau the drain current stays constant, while the drain source voltage is approximatly zero. I mean if i look at the I/V characteristics of the FET it is in deep triode region where the current should be zero. Propably i miss something, but i just dont get it if i look at the typical Level 1 mos equations.

I get the point that the current stays constant because of the approximation with the inductor, but how do i explain it from the point of few with mosfet characteristics?
 

Danko

Joined Nov 22, 2017
1,835
Yashraj Mehta
B.Sc in Physics & Electronics (Physics), St. Xavier's College, Ahmedabad (Graduated 2019)4y
Related
"Why does the drain current remain constant after the voltage across the drain and the source crosses the pinch off voltage in JFET?
This is due to limited number of electrons available for conduction of drain current through the channel.
The channel is made up of semiconductor material. The electrons in semiconductor material can be valence band electrons (bound to the atom; can not participate in conduction) or conduction band electrons (they are energetic; so either they jump from atom to atom as opposite of 'holes' or they travel freely within the volume bound by the surface of the material). At pinch-off voltage, all the conduction band electrons within the channel have already been used up to constitute the required electric current. So, after pinch-off voltage even if the drain-source voltage across the channel increases, there are simply no electrons available to increase the drain current! Hence, the drain current saturates. The magnitude of this current depends upon the doping of semiconductor used, the working temperature, the gate-source voltage etc.
If you are wondering “ what will happen to the extra electrical energy that went into the channel due to increase in drain-source voltage after pinch-off voltage?”. Well if this energy is above a certain threshold value it will cause breakdown of semiconductor material and results in burning of JEFT but for all practical purposes increase in drain-source voltage beyond pinch-off voltage has no significant effect.
This is not all but I hope you find this useful. ;)"
https://www.quora.com/In-a-JFET-or-...urrent-become-constant-at-pinch-off-condition
 

Papabravo

Joined Feb 24, 2006
21,228
As you can see from this example of an open loop buck converter. The current in the inductor is NEVER constant. It is ALWAYS either ramping up or ramping down. This is happening at the switching speed of the MOSFET. The component values have been chosen to guarantee Continuous Conduction Mode (CCM) operation.

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Thread Starter

mirror_pole

Joined Nov 14, 2020
10
Yes, but the ripple is so small that it can be approximated by the average constant current which flows through it. At least thats what textbooks say. Its also known as small ripple approximation.
 
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