Where does the negative inductor current go during CCM at no load in synchronous converters?

Question

As you all know, there are two primary mode of operation for dc-dc converters : CCM and DCM. At no load, the inductor current stops at zero at DCM. However, if CCM is enforced (such as Forced PWM), then inductor current can go to negative at no load or light load condition if the inductor ripple is large.

I had a simulation of a synchronous buck-boost controller at work and I saw that the current goes back to the input during negative inductor flow phase.

My questions:

1. Can I confirm that this is true and that if I use forced CCM, then my power supply would need to be one that can absorb this power during negative inductor current flow? If not, then where would this negative inductor current go?

2. Is this type of power supply common? Because there are quite a bit of synchronous converters that runs in CCM all the time out there

Thanks

Answer 1

There are several reasons why ensuring the buck converter remains in continuous conduction mode (CCM) in light- or no-load conditions:

1. 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).

2. 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.

3. 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.

4. 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 : )

Answer 2

Energy is transferred back and forth between input and output caps via the inductor. But, as long as no power source is connected to the output, the net current flow will be from input side to output side. Or, I guess with ideal components, the net flow will be zero.

It is not super common, but some synchronous bucks actually are bidirectional converters, capable of sinking or sourcing current at the "output." So if you connect a power source to the "output," then net current actually can flow backwards to the input side.