Where does the negative inductor current go during CCM at no load in synchronous converters?
There are several reasons why ensuring the buck converter remains in continuous conduction mode (CCM) in light- or no-load conditions:

the duty ratio is the same regardless of the load: when the buck transitions to the discontinuous mode (DCM), the relationship linking \$V_{out}\$ to \$V_{in}\$ changes and involves frequency, load and the inductor. If the converter remains in CCM in no-load conditions, you still have \$V_{out}=DV_{in}\$ (simplified formula).

there is no change in the control-to-output transfer function: the buck in voltage-mode (VM) is a second-order system in CCM and befomes an over-damped second-order in DCM. With the synchronous rectification at work, the transfer function is almost the same in light-load conditions.

in forward converters, typically those used in the dc-dc bricks, going into light-load conditions usually means skip cycle and lost of the auxiliary voltage: the rectified pulses are extremely narrow in skip and, without precautions, the \$V_{cc}\$ collapses and you need to increase the capacitor at the \$V_{cc}\$ pin. If the converter remains in CCM, this does not happen and the auxiliary voltage is always there.

finally, and this was true in post-regulated converters, like with the former mag-amp killers, synchronous rectification was letting you nicely implement leading-edge modulation to regulate the secondary outputs like a 3.3-V output made of a main 12-V one.


A typical circuit in SIMPLIS would look like this:

It simulates quickly and delivers the below waveforms:

You can see the inductor average current sets to 0 A while it remains continuous with a regulated 5-V output. During the freewheel phase involving S1, the current circulates from the ground up as with a classical diode: the current in the inductor depletes with a slope equal to \$\frac{V_{out}}{L}\$. When it reaches 0 A, a classical diode would spontaneously block and this is DCM. But here, S1 is bidirectional (the MOSFER operates in quadrants I and III) and the current reverses to now flow from the upper side of S1 to ground and through the load and the capacitor. At the end of the period, the controller instructs S1 to turn off. When the upper-side switch S2 turns back on, the current it sees is negative. This lasts until the inductor current crosses 0 A again and goes back to its positive peak at which moment the S2 switch turns off and S1 turns back on. The below drawing valid for a no-load situation should hopefully explain this text with less French in it : )