This is a boost circuit.

 The MOS tube here is equivalent to a switch. As long as the speed is fast enough and combined with the output filter capacitor, a stable output voltage can be obtained.

How is this process achieved?

When the switch is turned on, the inductor L is grounded, the diode is cut off, the input voltage Vi charges the inductor L, and the voltage across the inductor is Vi.

 When the switch is not conducting, since the inductor L has been charged before, current flows through it and the current goes to the right, so the two ends of the inductor cannot change suddenly, that is, a voltage will be induced and the diode on the right will be turned on.

 The voltage at the right end of the inductor is equal to the output voltage Vo + the diode conduction voltage drop Vd, and the voltage at the left end of the inductor is the power input Vi.

Because this is a boost circuit, Vo+Vd will be greater than Vi, and the inductor will discharge at this time to supply power to the load and charge the output filter capacitor.

It should be noted here that when selecting the inductor, the peak current of the inductor cannot exceed the saturation current of the inductor.

Some people may ask, shouldn’t it be the effective value current?

Generally, we believe that the inductance of an inductor is constant, so when the current reaches a certain level, the inductance will decrease as the current increases.

As the inductor current continues to increase, the rate at which the inductance decreases will accelerate.

 The above is the simple working principle of the Boost circuit. We will continue to explore it in depth later!

Some of the above pictures and information are from "The Road to Success"