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If I connect two DC-DC Modules (example module in the figure below) in parallel, can I increase its current capacity?

DC-DC module

Also, I have another doubt out of curiosity. If I connect a module in the following configuration, instead of supplying current, it needs to sink. Will it work? In this case, the module is having input of 7V and out of 5V, 2A output. I connected the output such it closes a loop with a 7V battery and 1 ohm resistance.

DC DC module along with a battery

Are these modules designed to deliver or can also sink the power to maintain the 5V at their terminals?

Links to the datasheet https://datasheet.lcsc.com/lcsc/1912111437_HI-LINK-HLK-10D2405A_C465413.pdf

I am unable to find an English datasheet. However, I found a datasheet for AC-DC module. https://datasheet.lcsc.com/szlcsc/1912111437_HI-LINK-HLK-5M05_C209907.pdf

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  • \$\begingroup\$ Link to datasheet, please, in your question rather than in the comments. \$\endgroup\$
    – Transistor
    Aug 3, 2021 at 14:48
  • \$\begingroup\$ @Transistor, I am unable to find a datasheet in English. I updated the question \$\endgroup\$
    – SKGadi
    Aug 3, 2021 at 15:13
  • \$\begingroup\$ Put in parallel, one will do all the work because it will have a slightly higher regulation voltage than the other. Worst case scenario is that both will burn. Don't try and use the output as a shunt for an external 7 volt battery either. Not recommended. \$\endgroup\$
    – Andy aka
    Aug 3, 2021 at 15:20
  • \$\begingroup\$ I got curious and asked another follow-up-question. Please look into it as well. \$\endgroup\$
    – SKGadi
    Aug 4, 2021 at 13:57
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    \$\begingroup\$ @SamGibson, Yeah, you are right. I am new to this community. Sorry for not paying attention. I asked it as an entirely new question. Here is the link: electronics.stackexchange.com/questions/580204/… \$\endgroup\$
    – SKGadi
    Aug 4, 2021 at 14:42

4 Answers 4

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About the paralleling of DC/DC converter: in general no, you can't, unless they are designed to work in that way (look for current sharing in the feature, or feedback sinchronization or something similar)

They can't usually do that essentially for two reasons: first of all usually due to tolerances one of the converter will try to deliver most of the current; and many compensation schemes aren't really happy when someone other supplies power from the output. Or more simply they can't source in a prebiased output.

For bigger converters these day usually the solution is a polyphase converter: multiple power stages synchronized from the same clock. I've seen brochures for controllers expandable to up to 24 phases so there's a lot of power to be handled.

Another huge benefit is that EMC is greatly simplified being all synchronized.

As for the second question, about converters used as `sinks'. I don't really think it would work, in general. Many converters can only source current. Synchronous topologies could in principle sink but the controllers doesn't necessarily handle regulation (some only use that to handle gross overvoltage situations).

Of course it is feasible to do a two quadrant converter (even four quadrant ones, they are used for 42V car supplies); simply it isn't a commonly designed use case.

In your case your best course of action would be to ask HiLink (or whoever makes the converter brick) if your configuration is supported or not.

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I don't see anything in that datasheet (using machine translation from Chinese) that indicates it's okay to parallel them. Unless explicitly stated, it's not permitted.

Same with sinking current.

Note it's listed as "Not recommended for new designs" by lcsc.

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@Lorenzo has answered both points. I'll add a comment:

The output of the converter's internal transformer will most likely have a diode and smoothing capacitor. There is no path for current to be absorbed by the device.

If there was a path and you pushed 2 A into the device while it's holding 5 V then it would have to dissipate 10 W as heat. The device itself is rated at 5 V, 1 A so it has an output rating of 5 W. Its efficiency is rated at ≥70% so it would normally dissipate \$ (\frac 1 {0.70} - 1)5 = 2 W\$. It should be clear that the package would not be rated for 10 W.

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In order to sink current, the design must be capable or an output greater and less than the input voltage. aka buck-boost regulator. For the same reason, LDO's cannot sink any current. This device is Buck convert ONLY.

Current sharing involves current sensing and balancing the output voltage in order to share current and possibly adding a series R to allow for compliance of sharing two controllers to a common load. This often leads to stability problems with oscillations of circulating voltage and current and requires special designs.

Your schematic shows you want to sink 2A from a 1 ohm load with a 2V drop. A better solution is to use a step-down "buck" converter from the 7V battery.

Some types only cost $2 to $0.75 , so it is best to choose the right part with your specs include surge current than try to boost with 2 wrong parts.

LM2596 DC-DC Buck Converter Step Down Module Power Supply DIP Output 1.25V-30V 3A

Find ones with adjustable Vout and current limit.

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