To my understanding the reactive components in a converter should not store energy over a switching period, otherwise they will blow up. An in inductor for example abides by volt-second balance, which translates to zero average voltage over a switching period- no flux linkage/ no average energy stored. That being said, the inductor current waveform will be switching at rate Fs ( switching frequency) and will introduce a ripple on the current waveform. Now if we consider the transient time of buck converter. There will be no energy stored in the capacitor by the inductor at t=0. My question is how do we calculate the transient time of the buck converter? And what is the circuit analysis reasoning were the inductor current increases and settles at a DC offset? (Why doesn’t the waveform ripple go back to zero at period [DTs - Ts] - how does the capacitor hold the charge in period [DTs-Ts] and not discharge through the load connected on the output terminals
This is an inductor current graph, not an output capacitor voltage graph, so it's difficult to tell from it whether the output capacitor discharges through the load (it does).
The current between \$DT_s\$ and \$T_s\$ is not constant, but the derivative of the current is proportional to the voltage difference between the inductor terminals, which during "off" time is the voltage of the output capacitor at this point (not much) plus the forward voltage of the diode (also not much). Compared to the voltage during "on" time, the change in current isn't as pronounced, so it appears flat in this graph (but the \$i_L(T_s)\$ marker is applies only to the time \$T_s\$, not to the interval).