I need to power an Arduino and four 4.5v-9v 2A Li-ion charger boards with a single wall adapter. Not wanting a shared circuit between the chargers, I'm thinking maybe a 12v 600mA (or something..) plug with a buck converter for each charger and powering the Arduino directly from the plug. I have no clue how to calculate how much amperage the buck converter will output given the input/output voltage range possible. This one states it only ouputs 1A when stepped down to 5v from 12v

Need each charger circuit to have 2A available whenever needed but cant figure out which buck converter/voltages/input current I need?

Im not fussed about exact voltages once its within range of the Ardunio (5v-12v) and chargers (4.5v-9v) of course.

Edit: According to this and taking into consideration 80% efficiency, it seems my application would need a 12v plug that delivers at least 1.04A for a buck converter to step down to 5v 2A output. Does this seem correct? If it is then I guess the only question left is will any old buck converter that is rated for 2Amp or more be suitable?

sketch up

  • \$\begingroup\$ Are you sure your wall charger is able to deliver the required 40+ Watts for your worst case scenario? Does the buck need to be integrated or could you work with a chip and build the circuit yourself? \$\endgroup\$
    – po.pe
    Feb 19, 2018 at 7:09
  • \$\begingroup\$ I have no idea how much wattage they can handle. I can build some kind of a circuit on a breadboard but Id need detailed instructions for such a job. My talent pretty much ends at drop in breakout boards and such. If i was to (hypothetically) use a fast phone charger (1.67A @ 9v or 1.25A @12v) what most likely happen? \$\endgroup\$
    – sryan2580
    Feb 19, 2018 at 7:18
  • \$\begingroup\$ Your supply voltage will not be stable. And your assumption that you need only 1.04A @12V is only correct if you just operate one charger at a time. Voltage may be used for parallel devices, but your current splits up into the different paths and therefore you ned 4 times the current one buck converter needs to provide 5V@2A. You would need something like this ti.com/lit/ds/symlink/lm2576hv.pdf but still your wall charger needs to be something like 12V@4A \$\endgroup\$
    – po.pe
    Feb 19, 2018 at 7:36
  • \$\begingroup\$ okay then, how about a laptop charger @ 19.5v, 3.33A. Would i be right in saying that my system would have nearly 1A of headroom \$\endgroup\$
    – sryan2580
    Feb 19, 2018 at 7:48
  • \$\begingroup\$ Yes as 19.5V*3.3A = 64.35W and 4*5V*2A = 40W you have approximately 24W input power left (assuming the arduino's consuptions is negligible) \$\endgroup\$
    – po.pe
    Feb 19, 2018 at 7:54

1 Answer 1


it only makes 1A

No, It can output 3A with carefully heatsinking.

I have no clue how to calculate how much amperage the buck converter will make

"will make " or "will consume "?

Here is the graph showing efficiency vs input voltage :

enter image description here

As you can see, the efficiency is about 80 % if input voltage is 12V. The only problem is if your 12 V adapter can provide enough wattage for your load.

  • \$\begingroup\$ so the ebay listing is incorrect, it in fact will make 3A once thermals allow? Also when i said 'make' i meant be able to output, my bad. I assume the graph is from the LM2596 from what i linked to above? \$\endgroup\$
    – sryan2580
    Feb 19, 2018 at 7:50
  • \$\begingroup\$ what do you think of a laptop charger @ 19.5v, 3.33A. Handles 65W so maybe good for the job? \$\endgroup\$
    – sryan2580
    Feb 19, 2018 at 7:53
  • \$\begingroup\$ yes, of course. \$\endgroup\$
    – Long Pham
    Feb 19, 2018 at 8:02
  • \$\begingroup\$ "do you think " is enough, not "what do you think" :) \$\endgroup\$
    – Long Pham
    Feb 19, 2018 at 8:03

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