As told at the beginning of this section, the converter operates in discontinuous mode when low current is drawn by the load, and in continuous mode at higher load current levels. Evolution of the voltages and currents with time in an ideal buck converter operating in discontinuous mode. Therefore, a fraction of the power managed by the converter is dissipated by these parasitic resistances. Since, when the engine is running, these dynamos (powered by the on-board electronics of Euro 5 and 6 engines) do not always emit the correct charging current, a DC-DC converter is essential for charging the service battery properly. The only difference in the principle described above is that the inductor is completely discharged at the end of the commutation cycle (see waveforms in figure 4).

When we do this, we see the AC current waveform flowing into and out of the output capacitor (sawtooth waveform). 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.That means that the current I L {\displaystyle I_{\text{L}}} is the same at t = 0 {\displaystyle t=0} and at t = T {\displaystyle t=T} (figure 4). the current at the limit between continuous and discontinuous mode is I o lim = V i T 2 L D ( 1 − D ) = I o lim 2 | I o | D ( 1 − D ) {\displaystyle \scriptstyle I_{o_{\text{lim}}}={\frac {V_{i}\,T}{2L}}D\left(1-D\right)={\frac {I_{o_{\text{lim}}}}{2\left|I_{o}\right|}}D\left(1-D\right)} . This assumption is acceptable because an inductor is made of one long wound piece of wire, so it is likely to exhibit a non-negligible parasitic resistance ( R L). From this, it can be deduced that in continuous mode, the output voltage does only depend on the duty cycle, whereas it is far more complex in the discontinuous mode. Browse our comprehensive portfolio of DC/DC buck converters and leverage corresponding design tools to address power-supply design challenges for any application.

As can be seen in figure 4, t on = D T {\displaystyle t_{\text{on}}=DT} and t off = ( 1 − D ) T {\displaystyle t_{\text{off}}=(1-D)T} .From the initial state in which nothing is charged and the switch is open, the current through the inductor is zero.