Buck Converter - Diode Forward Bias Theory Question

[ngmedaboard]

Assume a buck converter, switch opened state and in continuous mode.

I know when the switch opens, the diode forward biases completing the circuit and drawing energy out of the inductor.

But in order for that diode to forward bias, the input side of the inductor needs to swing about 0.5V below ground (assuming a Schottky diode).

Is that 0.5V below ground the cause or effect from the diode? In other words A, B or other?

A. The input of the inductor swings low enough causing the diode to forward bias.
or
B. The inductor is just agressively pumping current, it finds the path through the diode, which forward biases the diode thus causing the 0.5V swing below ground.

 

[BradtheRad]

The inductor must generate current, and it develops as great a voltage as necessary, to overcome the diode's forward voltage.

If the diode had a fwd V of 20V, then the swing would be -20V.

If it were 500V, the swing would be -500V.

The greater the amplitude of the voltage spike, the shorter time it lasts.

 

[ngmedaboard]

Sounds like B is the answer then. Current causes the diode to forward bias and voltage drop not the other way around. Thanks!