Can someone explain why you need less duty cycle % to achieve the same output voltage in the discontinuous mode of operation in a buck converter?

This indicates to me that it's more efficient. If the current in the inductor goes to 0, then the diode turns off and there are no more losses associated with that freewheeling diode.

Also, the upper switch seems to turn on with less losses.

It should be the more efficient mode of operation.

Is this true? If so, what are the disadvantages that make it seem like it should be avoided at most times?

It seems that it isn't possible to accurately control the output voltage just using open loop in DCM.

  • 6
    \$\begingroup\$ You'll get a quick overview here. You may find some agreement there and some added thoughts to consider. I didn't want to duplicate what's already out there, so I'm not going to bother with a separate answer. I tend to reach for discontinuous mode designs, myself. But I'm a hobbyist and I happen to like knowing that the Webers go back to zero each cycle, when designing my inductors. \$\endgroup\$
    – jonk
    Aug 31, 2022 at 5:53
  • 5
    \$\begingroup\$ I think this is a fairly large over-simplification. But unfortunately, it would be difficult to answer without writing an awful lot of stuff about switching regulators. Maybe it will suffice to say that light loads may tend to favor discontinuous operation. But heavy loads tend to favor continuous operation. Everything enters into the picture including Vin/Vout ratio, inductor value, capacitor value, load current, and switching frequency. \$\endgroup\$
    – user57037
    Aug 31, 2022 at 6:36
  • \$\begingroup\$ How are the comparisons made? At the same output power? Or anything else? \$\endgroup\$
    – Antonio51
    Aug 31, 2022 at 12:37

4 Answers 4


Can someone explain why that in discontinuous mode of operation in a buck converter, you need less duty cycle % to achieve the same output voltage?

Let's say you have a buck converter capable of operating in both modes of operation. You will find that if it's operating in DCM, the load current has to be low. As the load current rises you reach a point (it's called the boundary point) where the converter is operating on-the-cusp of CCM. Above this point it fully operates in CCM.

So, low current loads are serviced in DCM and higher current loads are serviced in CCM. And, importantly, in CCM the output voltage is fully defined by the duty cycle whereas in DCM, the output voltage is defined by both the duty cycle and the load current.

Also it seems that it isn't possible to accurately control the output voltage just using open loop in DCM

So, if the current taken by the load is low, you MUST have a very low duty cycle or your output voltage rises out of regulation. This is because in DCM, the converter is a power regulator; in CCM the converter is a voltage regulator. Low output powers therefore require low duty cycle and, the point you may be missing is that it is impossible to force CCM on very light loads. Hence, in DCM you cannot run in open-loop without massive changes in output voltage as the load current changes.

This indicates to me that it (DCM) is more efficient?

DCM services low power/low current loads and CCM cannot do that. CCM services high power/high current loads and DCM cannot do that. Hence DCM cannot be more efficient on high load currents and CCM cannot be more efficient on low load currents.

what are the disadvantages that make it (DCM) seem like it should be avoided at most times?

  • Higher ripple voltage on the output i.e. it produces a noisier output voltage
  • Higher ripple current on the input
  • \$\begingroup\$ To clarify the logic: a converter operating in CCM at nominal load, must also operate in DCM at light load*. A converter operating in DCM at nominal load, always operates in DCM. You can make a powerful DCM converter, it's just not efficient, for reasons given by @jp314. *Synchronous converters can operate in forced CCM, which is usually done to reduce emitted noise at the expense of poor light-load efficiency. There's also BCM or QR mode, which may be worth a mention. \$\endgroup\$ Aug 31, 2022 at 14:24
  • \$\begingroup\$ @TimWilliams I stated in my opening line that I was considering a converter that operated across both modes if that helps clear up why you left some of your comment? I avoided dragging synch converters into the fray to keep this as simple as possible. For that same reason I'm avoiding the other things you mention but you can leave them as an answer. \$\endgroup\$
    – Andy aka
    Aug 31, 2022 at 16:23
  • \$\begingroup\$ Agreed -- just highlighting it more explicitly. \$\endgroup\$ Aug 31, 2022 at 16:44

Another way of looking at it is that in DCM, there is a '3rd' state -- both switches are off. They is why the active duty cycle can be (must be) lower.

There are 3 major losses in a DCDC:

  1. Quiescent and bias currents. These can be reduced in DCM if pulse skipping is used because certain blocks can be turned off until the next pulse is needed.
  2. Gate drive power -- this is to charge the switching FETs. In pulse skipping DCM, this is lower.
  3. Resistive losses (I²R) in the switches (and inductor). At higher currents, this dominates.

DCM has higher inductor ripple and higher peak currents for a given load than CCM would have. At low overall load current, the I²R losses are negligible and it is more beneficial to save on quiescent current and gate drive losses. At high currents, the I²R loss dominates and so it is more beneficial to run CCM.

In addition, in DCM, the inductor current has to peak at at least 2x the load current; in CCM there is less inductor ripple and so peak inductor current is lower. This means the inductor can be (physically) smaller as it doesn't need as high a saturation current rating.

  • 1
    \$\begingroup\$ Don't forget bypass caps (input and output) -- the higher ripple current makes their ESRs more important, too :) \$\endgroup\$ Aug 31, 2022 at 14:13

Just to round out the answers -- @jp314 and @Andyaka gave great explanations, and @Antonio51, waveforms -- there is some merit to operating on the boundary between them as well, hence: BCM (boundary conduction mode). Or CrCM ("critical"), or QR (quasi-resonant), or sometimes, "valley mode switching". (Always check the waveforms to get a clear definition of what some term is being used to mean.)

Anyway, consider DCM: as the inductor current falls to zero, the catch diode recovers softly (i.e., at dI/dt defined by the inductor, not the switch), and voltage rings down freely between VIN and GND. If we turn on the switch just as that ringdown reaches an upward peak, the voltage across the switch is minimized (or indeed zeroed, if approx. Vin < 2 Vout), reducing or eliminating switch turn-on losses as well. This is BCM operation.

I suppose, strictly by what the words mean, BCM means, switching on at any time after inductor current reaches zero, but before the ringdown has dissipated (which is just plain old DCM); quasi-resonant (QR) specifically is switching on during one of those peaks. Or perhaps BCM, QR and DCM all overlap, with the former two being limited to the edge of that case. Anyway, that's just semantics; what's important is the physics, not what we call it.

The advantage to this mode is, reduced switching loss versus DCM (due to the reduced switch voltage), and much less than CCM (which incurs full switch voltage at turn-on plus diode recovery).

The disadvantage to BCM and DCM, is the inductor (and input and output capacitors, to a lesser extent) needs to be very high quality. QR boost PFC is a very popular application, and typically requires a gapped ferrite-core inductor, wound with litz wire. Solid wire on a cheap powdered-iron core will not do.

This is kind of not even applicable to buck, in practical terms, as I can't remember having seen a QR mode buck controller; it is fairly popular for flyback and boost, though. To be clear, the scheme works out exactly the same (buck and boost are ultimately the same circuit, operated with forward or reverse current flow), it just happens to not be popular for buck mode.


Here is an example of simulation for devices quasi "ideal".
Simulation is made with voltage output the same.
Same schematic. Made with Duty = 0.3 in left picture, 0.7 at right picture.
Voltage input supplies are therefore adapted for the same voltage output.
No really "big" difference (efficiency) between the two cases, except certainly for other "variables" (example: peak inductor currents).

enter image description here

enter image description here

And with some losses in "switches", diodes, inductors, capacitors for example ...

enter image description here

  • \$\begingroup\$ If you care about efficiency, then you use a synchronous rectifier, not just diode. Your sim seems to have an 'ideal' switch -- it ignores RDSON losses as well as gate drive power. It also doesn't have ESR in the L or C's. \$\endgroup\$
    – jp314
    Aug 31, 2022 at 15:55
  • \$\begingroup\$ Right. As @Jonk said (link in comment), OVERVIEW of problem. I always begin with a "theoretical" background ... with quasi-all "ideal" components ... And then, one can add all "parasitics" he wants to make the simulation more "real". ... and found "conclusions" in that case. Will add later in the answer some "losses" ... \$\endgroup\$
    – Antonio51
    Aug 31, 2022 at 16:25

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