DC boost converter

Basically the mosfet shorts 1 side of the inductor to ground. The inductor charges through the resistor and eventually also becomes like a short with enough time, at which point it has the maximum amount of energy stored and the maximum current flowing through it. When the mosfet turns off, we have a sudden change in the current that was once flowing through the inductor into the mosfet to ground. Since V = - L dI / dt, that should mean that for this large negative dI /dt we see (the current sharply decreased , hence dI / dt < 0), the voltage across the inductor must increase positively, and by a lot. Since 1 side of the boost converter is tied to the supply, the voltage on the other side (the diode) must equal the supply + the new voltage across the inductor, which is going to be a lot higher than the supply voltage

Yeah. Basically 😑

what???