Please look at the buck -boost converter below. It is 2 outputs, one buck and one boost output. Would the circuit work as describe in the picture? In what condition the circuit will it work? If so how would you control those switches?

EDIT: I know about conventional buck, boost and buck-boost converter. However, here I am looking for a specific topology as in the figure. I want to know what conditions so that can work like at some load conditions.

enter image description here

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    \$\begingroup\$ Your circuit is neither a buck nor a boost nor a buck-boost converter. It is like you draw a resistor and say it's a capacitor and asking if it will work... \$\endgroup\$ – Huisman Jun 7 '19 at 6:00
  • \$\begingroup\$ You have a battery connected to an inductor. Then two switches with a capacitor and a current source on the other side? Are I1 and I2 supposed to be self contained buck and boost converters? I don't understand what you're trying to do? \$\endgroup\$ – hekete Jun 7 '19 at 6:44
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    \$\begingroup\$ @anhnha but that definition is wrong. A buck converter is using a switched inductor, plus some rectifying element (typically, a diode) that conduct during the off time of the switch. Your schematic is missing the diode, so, it's not a buck converter. Same for boost, just that your circuit is even less like a boost converter. So, to answer your question: \$\endgroup\$ – Marcus Müller Jun 7 '19 at 8:53
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    \$\begingroup\$ I didn't. Interesting technology. It is interleaving between the outputs. I think you drew that using S1 and S2? But there is way more involved than you drew in your schematic. \$\endgroup\$ – Huisman Jun 7 '19 at 11:22
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    \$\begingroup\$ Read more about how Buck-boost works \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 7 '19 at 21:23

Here is a research paper on a 5 output version: 5 Output SIMO. As you can see the control system is fairly complex.

I think I understand what you want to do. You want to take this standard single inductor, multiple output design and reduce it to just the switches before each capacitor like so:


simulate this circuit – Schematic created using CircuitLab

You want to have M4 buck in discontinuous mode, but then M5 will drain Lind to boost and it will be like M4 is actually in continuous mode?

I think this might be possible if you have very stable loads that would allow you to precisely calibrate your values?

I don't have enough experience to give you a definite answer, we will need someone more knowledgeable on the subject. I'm mostly posting this to give someone else more information to figure this out, since the comments suggested people didn't know what you were asking.

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  • \$\begingroup\$ well, this is a normal topology and easy to do. My idea above is just use two switch and ask in what condition will it work. \$\endgroup\$ – anhnha Jun 8 '19 at 7:47
  • \$\begingroup\$ So you want to get rid of all the switches except M4 and M5? \$\endgroup\$ – hekete Jun 8 '19 at 7:54
  • \$\begingroup\$ Yes, I am asking about that special topology. And ask can it work under some conditions. \$\endgroup\$ – anhnha Jun 8 '19 at 9:02
  • \$\begingroup\$ Yes, only M4 and M5. M4 charges the inductor and M5 discharges it. I think it can work in some load conditions. \$\endgroup\$ – anhnha Jun 8 '19 at 10:44
  • \$\begingroup\$ I think it might be able to work with very stable load conditions, but I will leave it open for someone that has better knowledge on the subject. \$\endgroup\$ – hekete Jun 8 '19 at 11:08

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