Over time, the inductor will allow the current to slowly increase by decreasing its own resistance. In an ideal circuit the voltage drop across the inductor would remain constant. When the inherent resistance of wires and the switch is taken into account then the voltage drop across the inductor will also decrease as the current increases. Also during this time, the inductor will store energy in the form of a magnetic field. From t = 0 {\displaystyle t=0} to t = D T {\displaystyle t=DT} , the converter is in On-State, so the switch S is closed. The rate of change in the inductor current ( I L) is therefore given by

while in the On-state, the input voltage source is directly connected to the inductor (L). This results in accumulating energy in L. In this stage, the capacitor supplies energy to the output load.I L On = ∫ 0 D T d ⁡ I L = ∫ 0 D T V i L d ⁡ t = V i D T L {\displaystyle \Delta I_{{\text{L}}_{\text{On}}}=\int _{0} while in the Off-state, the inductor is connected to the output load and capacitor, so energy is transferred from L to C and R. d ⁡ I L d ⁡ t = V i L {\displaystyle {\frac {\operatorname {d} I_{L}}{\operatorname {d} t}}={\frac {V_{i}}{L}}}

If the current through the inductor L never falls to zero during a commutation cycle, the converter is said to operate in continuous mode. The current and voltage waveforms in an ideal converter can be seen in Figure 3.