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I'm learning about step-down converter and I wanted to know what's the difference between continuous mode and discontinuous mode.

I know what they mean by deffinition (in the former the current through the inductor never reaches 0, while in the latter it does), but I would like to know what are the advantages and disadvantages that each mode present. What would lead me to design one or another?

EDIT:

After researching a little bit more, I read that, in discontinuous mode, higher current peaks could be needed. For the output average current to rise, the peaks in the inductor current would have to be higher in order to rise that average (given that there is a time where no current is flowing). So the transistor that acts as switch should be able to support that current that flows.

That is one point in favour for the continuous mode. I wasn't able to get completely other things I found out there, like something related to "pulse skipping" (which I don't know what it is by the way). Then, so far, I have one advantage of the continuous mode against the discontinuous one. Nevertheless, someone told me that discontinuous mode is more used than the another one, and that most flyback regulators operate this way. Is this true? If so, why is this? Why would someone choose discontinuous mode if transistors that support high currents (ergo, more expensive) are needed?

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  • \$\begingroup\$ I believe it's a great question. But where has your own thinking taken you, so far? Any thoughts about it, whatsoever? Or just a total blank? \$\endgroup\$ – jonk Nov 9 '16 at 23:55
  • \$\begingroup\$ @jonk Well actually I'm just starting to read about this topic so I'm not too familiar with it. Maybe something related with power dissipation? Or maybe DM would be better because it has less ripple? I don't really know. \$\endgroup\$ – Tendero Nov 9 '16 at 23:57
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    \$\begingroup\$ How would a continuous mode design cope if the load varied over a very wide dynamic range of currents? Can you see any interesting differences in sustained magnetic flux and/or its variations, one mode vs the other? What about the switch itself? How might it be differently characterized, one mode to the other (think 'peak current' here?) (Also, the control loop is in some important ways different.) \$\endgroup\$ – jonk Nov 10 '16 at 0:05
  • \$\begingroup\$ I'm actually kind of hoping one of the well-trained folks on this subject enters in here. I have a hobbist's perspective and that means a limited (not a comprehensive one) view. I know a few things but an expert can put things entirely in the right order and perspective. But if one fails to jump in after a time, I'll toss out some of my own limited views and hope it helps you. \$\endgroup\$ – jonk Nov 10 '16 at 0:07
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    \$\begingroup\$ This is not meant to be a "rude comment but a helpful one. People could discourse at length on this here BUT it is a complex enough subject AND well enough covered in MANY places on web that you really would be bettr served by reading more elsewhere first and then asking re unclear points here after that. \$\endgroup\$ – Russell McMahon Nov 10 '16 at 1:29
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As you say, in continuous mode, the inductor current never stops, while in discontinuous mode, there is a period of time each pulse when there is no inductor current.

To get into why each one is used, advantages and disadvantages of each, would take a whole book. Most likely, there are actually books written on this. If you really want to learn this, then you should go read one.

However, briefly, some of the issues of continuous versus discontinuous operation are:

  1. The ripple can be made lower in continuous mode. With high enough switching speed, the inductor current only goes up and down a little each pulse. This delivers more steady current to the output.

  2. The peak inductor current is less in continuous mode for the same load current. This means a smaller inductor can be used. The inductor is often the largest and most expensive component in a switching power supply.

  3. Continuous mode is difficult under low load. For any fixed switching frequency, inductor, and input and output voltages, there is a load current below which continuous mode can not be maintained. This means any continuous mode system either has to specify a minimum load current, or be able to deal with discontinuous mode for some load conditions. This is not always as simple as it sounds because the open loop transfer characteristics change between the two modes, and that may require the compensation scheme to change.

  4. Continuous mode is a bit trickier to make stable, at least when pushing the limits. This is grossly oversimplified, but consider reacting to a sudden decrease in load. Continuous mode always leaves energy in the inductor at the end of each pulse. It therefore can't stop dumping current to the output as quickly as discontinuous mode, which by definition stops dumping current at the end of each pulse. This implies a slower open loop response, which in general means its harder to keep stable, all else equal.

  5. In discontinuous mode, the diode is not conducting when the switch is first turned on. At low voltages, Schottky diodes can be used, which have very fast reverse recovery time. However, not having to deal with reverse recovery at high voltages can be useful. Consider that when the diode and switch are both on, they create a short across the input. That is not only bad for efficiency, but can abuse the switch and/or the diode.

  6. Really simple control schemes, like pulse on demand, aren't so simple anymore in continuous mode. Pulse on demand results in more ripple, but is inherently stable and very simple to implement. It's simpler to design something that stays nicely predictable and that you know won't saturate the inductor with discontinuous mode.

In practice, when I'm using a off the shelf chip to implement a switching power supply, I don't really care what mode it uses or when it switches between them. What you care about usually are things like ripple voltage, load and input regulation, and the like. How that's accomplished is (largely) irrelevant when plunking a chip on a board and following the datasheet recommendations for inductor and the like.

Sometimes I implement a switching power supply by having a microcontroller produce the pulses. That's often cheaper than a switcher chip, especially if the micro is already there for other reasons and it has a spare PWM output with maybe a shutdown input or comparator input, and ripple and the last possible bit of efficiency aren't that important. In such cases, a pulse on demand system or something that just shuts down the PWM when the output is above the regulation threshold is simple and good enough. If the system runs in discontinuous mode, it is easier to analyze and know the inductor won't be saturated.

You mentioned a flyback converter in your question. Flyback converters are generally discontinuous, else you're just wasting power by keeping a bias level of DC always running thru the primary.

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  • \$\begingroup\$ That was just what I was looking for. What an awesome answer! Thank you very much for it. I have just one doubt: in your item 5, you say the diode is not conducting when the switch is first turned on. Would this be an advantage or not? I didn't understand if this is a desirable effect or not, and why. \$\endgroup\$ – Tendero Nov 10 '16 at 18:21
  • \$\begingroup\$ @Tendero Agreed. This was exactly what I had hoped for with your question. A nicely laid out professional answer. I couldn't have done half as well. \$\endgroup\$ – jonk Nov 10 '16 at 19:07
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    \$\begingroup\$ @Tender: Diodes have a finite reverse recovery time. This is how long the diode still looks like a closed switch when it was conducting but then reverse voltage applied. In discontinuous mode, the diode has been off for a while before the switch turns on. In continuous mode, reverse recovery time is a real issue. Schottkys are almost instantaneous, but can't be used at high voltages. \$\endgroup\$ – Olin Lathrop Nov 10 '16 at 19:11
  • \$\begingroup\$ @Olin Lathrop:Why inductor is not saturated in DCM? \$\endgroup\$ – anhnha Sep 7 '17 at 7:28

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