I am studying buck converters and have nearly grasped their ideal workings in the Continuous Conduction Mode however there is one last bit that I can't get a confident grasp with.
When the transistor switch is OFF, what makes the inductor voltage go negative? Is it that at that instant the diode is not in forward bias mode, it has therefore undergone a large negative
\$ di \over dt \$ according to
$$
V = L { di \over dt }
$$
and therefore a negative voltage is applied across it from say 5V to say -0.6V which is a typical diode forward voltage.
What restricts the maximum value of this voltage?
Best Answer
It's all down to Kirchhoff's voltage law. Let's look at a circuit:
Consider the instant the switch is opened. Your input is disconnected, which means:
$$V_\mathrm{out} + V_\mathrm{inductor} + V_\mathrm{diode} = 0$$
This is where the negative voltage is coming from. The output voltage cannot change instantaneously because of the capacitor. This means that the inductor voltage needs to become negative in order to satisfy Kirchhoff's voltage law.