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



2 Answers 2


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:

enter image description here

It simulates quickly and delivers the below waveforms:

enter image description here

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 : )

enter image description here

  • \$\begingroup\$ Hi yes, the average inductor current is 0. but half of the period the inductor current is negative though. Where does this negative inductor current go? Won't current flowing from the inductor back to the input a problem? \$\endgroup\$
    – helloguys
    Jul 11, 2020 at 16:30
  • \$\begingroup\$ especially in a boost type topology. The way I see it, there is nowhere for negative current to flow besides back to the input. \$\endgroup\$
    – helloguys
    Jul 11, 2020 at 17:25
  • 1
    \$\begingroup\$ The source sees the neg. current when the upper-side switch turns back on as the current in the inductor is negative at the end of the freewheel session. The input capacitor sources or sinks current without problem. The low-side switch S1 "sees" the current reversing direction when the inductor current hits 0 A and goes backward, pushed by the output capacitor. It keeps circulating in the same mesh involving S1 but as S1 is bidirectional, it allows the circulation in both ways. It's only when S2 turns on again that it "sees" the negative valley. \$\endgroup\$ Jul 11, 2020 at 19:36
  • \$\begingroup\$ I think i understand it now. I took what you said and looked back at my simulations (boost mode of a buck boost synchronous converter). I did see that during negative inductor current flow time, the negative current goes into both the input caps and the dc source. So both the input caps and dc source help sinks the current. Let me know if I am wrong. \$\endgroup\$
    – helloguys
    Jul 11, 2020 at 20:16
  • \$\begingroup\$ Man! I have been searching for this information for weeks, and here it is in cogent and plain understandable language. \$\endgroup\$
    – RodB
    Dec 31, 2021 at 22:55

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.


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