Inductive load: why can D1 stay on?
Use the load-type button to compare noninductive and inductive loads, keeping the other settings fixed. With inductance selected, increase L. Predict what happens when the source voltage becomes negative.
Reveal the explanation
For the ideal circuit without inductance, io = vo/R. With L > 0, current cannot change abruptly. Without DF, D1 remains on until its current reaches zero; if current persists after the source voltage becomes negative, the output voltage is also negative.
With E or VF: without inductance, io = (vo − E)/R. Turn-on from zero current requires Vm > E + VF, and the opposing voltage and diode drop can shorten or remove the current tail. With no current, vo = E, not necessarily zero.
Further explanation · Extended
Without DF, D1 turns off at the first subsequent current zero β. While it conducts after 180°, the inductor releases energy to R, to the DC load (Eio), to any diode loss, and back to the AC source. For E = VF = 0 and L > 0, β > 180°. A positive E + VF can move β before 180°; there is then no energy-return interval.