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Is it possible to "stack" the outputs of 2 or more buck converters?

The buck converter circuit in question I am considering is the 48v to 5v typical example circuit from page 20 of the datasheet PDF for the "LT1339ISW#PBF" part found on mouser here.

LT1339ISW datasheet

I am posting from mobile, but if I can figure out how to directly post the datasheet circuit I will edit it in here later.

I wondered if it was possible to get the outputs of 2 of these buck converters to "stack" their output voltages by connecting the 48v inputs in series to a 96v voltage source and floating the grounds on the second buck converter.

I am thinking that if done correctly the 2 buck converters would share the 96v input source just like 2 equal valued resistors in series will equally share a voltage across them. And as long as you connect the load correctly, the outputs of the 2 buck converters would then act as if they were 5v voltage sources adding to an output of 10V at the up to 50A max rated output current.

For the sake of simplifying the question I am assuming it is possible to balance the inductance and capacitance of the the "stacked" buck converter circuits such that they function the same as the values in the single example circuit mentioned above.

I do realize that even if it is possible, this would not be the most practical circuit since real world components have varying tolerances and any significant imbalance or failure would likely cause cascading failures several other places in the combined circuit. At these voltages and currents it would make for a great risk of fire and injury to nearby people and objects.

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    \$\begingroup\$ Why not ? If the top one can handle the voltage to ground ? Regulation will be worse though \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 9:09
  • \$\begingroup\$ Nvm my comment. I was thinking of output stacking (see my answer), which is not what you asked. Input stacking with parallel outputs doesnt work as many answers demonstrated. \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 10:25
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    \$\begingroup\$ If you really need to go from 96V to 5V, then you're getting into flyback topology circuits. I'd start by looking at off-line converter chips. For a one-off, grab some USB wall-warts, and try them on 96VDC. If the internals are what I think they are, I think they'd Just Work. \$\endgroup\$
    – TimWescott
    Commented Apr 24, 2021 at 16:25
  • \$\begingroup\$ If you want to stack converters, you may be interested by this paper published in 2007. \$\endgroup\$ Commented Apr 24, 2021 at 17:51
  • \$\begingroup\$ In the project related to this question I am trying to figure out how to make a power supply that turns a 250V+ or 500V+ battery into 72v for the motor at currents of 100A to 700A (300A continuous) I figured if I could relatively simply add a few components and connect a bunch of off the shelf supplies together it would be a good way to implement in my project. I just simplified the problem such that it was much easier to understand conceptually. \$\endgroup\$ Commented Apr 25, 2021 at 22:03

4 Answers 4

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Usually: no.

Buck converters are usually not isolated, i.e. the output shares a potential with the input. Most often, that's ground. Therefore, "stacking" would just short the output of the lower one to ground. "Stacking" on the input simply makes no sense, these are not resistors that share the voltage drop. So that can't work out at all.

You could also build an isolated converter using the same or a different IC, but you'll need to come up with a rather complex method of feedback from the secondary side if you want competing, stacked regulation to still be good. Can't think of a case where that complexity would make sense more than you building a regulator meant for 96 V.

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  • \$\begingroup\$ LDO s can be stacked and are conceptually the same. What is the difference in Bucks that forbids it ? \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 9:12
  • \$\begingroup\$ the fact that switching regulators are, by nature, very dynamic loads, and your voltage division depends very much on identical load sharing – that sounds like a very bad idea. I wouldn't recommend stacking LDOs in the general case, either. \$\endgroup\$ Commented Apr 24, 2021 at 9:16
  • \$\begingroup\$ @tobalt, LDOs are linear. Buck converters are switchers. See my answer. \$\endgroup\$
    – Transistor
    Commented Apr 24, 2021 at 9:17
  • \$\begingroup\$ @transistor I know. But this doesn't explain any difference. In your example you connected them different that usual LDO stacking. \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 9:30
  • \$\begingroup\$ It does explain the difference. In an ldo, with constant output load, the input current draw is constant.. For switching power supplies, as the very name suggests, it's not! \$\endgroup\$
    – mmmm
    Commented Apr 24, 2021 at 16:07
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... just like 2 equal valued resistors in series ...

No. Buck converters have switching transistors.

enter image description here

Figure 1. Image source: Toshiba Semicon Storage.

Since your two buck converters will switch asynchronously you can be guaranteed that at various times one transistor will be on and the other off. You will not get even division of voltage across the two devices. The one that is off will have full line voltage and the one that is on will have zero current (because the other is off).

That's the first problem.

enter image description here

Figure 2. All the pink lines are connected together forming a nice short-circuit across the input terminals of the lower switcher. (Image by author using Toshiba artwork above.)

The second is that usually there is no isolation between input and output. Connecting the outputs in parallel will short-circuit the power inputs on the lower of the two units.


Series connected inputs + series connected outputs

enter image description here

Figure 3. An attempt at series connecting the inputs and series connecting the outputs won't work either. The lower switcher is completely bypassed.

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  • \$\begingroup\$ What if you connect the negative Vin of the top regulator to the positive Vout if the bottom regulator. The load will be between the positive Vout of the top regulator and ground \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 9:29
  • \$\begingroup\$ Post a schematic in a new answer. I can't picture what you are describing. \$\endgroup\$
    – Transistor
    Commented Apr 24, 2021 at 9:50
  • \$\begingroup\$ Yes sorry. I will I think I misunderstood the question anyway. But will post it later. \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 9:52
  • \$\begingroup\$ This would be connecting both circuits to ground. Yes it would have this effect with how you showed them connected, but rather than connecting the top converter to ground i spoke about leaveing it floating so that when both converters inputs were connected in series to the 96v source the top converter would see 48v potential instead of the 0v of gound. I know I did not explain what I was thinking that well, so the misunderstanding was my fault. \$\endgroup\$ Commented Apr 24, 2021 at 16:08
  • \$\begingroup\$ @John, see Figure 3. \$\endgroup\$
    – Transistor
    Commented Apr 24, 2021 at 16:18
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See the topology image of the buck converter:

enter image description here

You are planning something like this:

enter image description here

It's still one converter except the control electronics will not get any more power from the input and the diode path of the upper converter is blocked, both are blocked by the lower converter. Or the diode path is not blocked, if the switching transistor has a drain-source reverse diode. But without proper simulation the function of the control circuit is a puzzle which cannot be reasoned beforehand.

When the system is started the output capacitor of the lower converter is empty, the upper one gets all of the input voltage. It's maybe not designed for +100% overvoltage and probably puffs the smoke out.

Does not work, forget it!

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  • \$\begingroup\$ Remove the connection from top in- to bottom in+ and it will work. Bottom in+ is bat+ instead. \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 9:40
  • \$\begingroup\$ If one regulator stands the full voltage one could simply change the voltage control feedback and use the other regulator for something else. \$\endgroup\$
    – user136077
    Commented Apr 24, 2021 at 10:38
  • \$\begingroup\$ 1) well one might have fixed voltage regulators on hand and would want to combine them for another voltage. 2) I misunderstood the question as stated in my answer. So my comment above is irrelevant and your answer is perfectly sound. sorry. \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 10:50
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    \$\begingroup\$ No harm caused. This is not a nuclear power plant. \$\endgroup\$
    – user136077
    Commented Apr 24, 2021 at 10:53
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I believe now that I misunderstood the question, but as I was very insisting in several comments, here is what I had in mind, namely stacking the output voltages of several regulators (which was not asked).

Barring internal details, both an LDO and a Buck look like this:

enter image description here

The internal details really do not matter for this fundamental analysis. However, the switching nature of the Buck will demand more care in a real implementation.

Output voltage stacking of such regulators is possible:

enter image description here

But of course, the output transistors of the top regulator have to tolerate the voltage to Ground. Also the regulation will become worse as more regulators are stacked, as each of their misregulations add up. Also, problems with oscillations can be harder to fix. So I agree, it is not a straightforward or usual thing to do, but is not fundamentally impossible.

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  • \$\begingroup\$ What if the current should be downwards in the wire between the regulators? The control circuit of the upper regulator at least needs it. \$\endgroup\$
    – user136077
    Commented Apr 24, 2021 at 10:31
  • \$\begingroup\$ Yeah this is a usual "detail" that such regulators cant sink current. In practise one would add slight resistive load to the bottom regulator which allows for the quiscient current of the top regulator to be sent to Ground \$\endgroup\$
    – tobalt
    Commented Apr 24, 2021 at 10:34
  • \$\begingroup\$ Yes, this. Rather than misunderstanding the question I was thinking about the problem a bit backwards when explaining things in the question. I was thinking of your circuit diagram, but with an added connection between the Vout of the bottom regulator to the node above the load. Although I had it backwards as after a good night's sleep i realized this would stack the current rather than the voltage. \$\endgroup\$ Commented Apr 24, 2021 at 15:58
  • \$\begingroup\$ #Correction: in my example ground of the top regulator would see 48v, and the output of the bottom regulator would be connected to the node above the load in your diagram. \$\endgroup\$ Commented Apr 24, 2021 at 16:22

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