I'm just curious if it's possible to power 2 devices at different voltages with a set of batteries while making sure that the 2 devices makes use of all the batteries. I have here an illustration of the connection that I would like to get verified. Each battery is rated ***n*** volts

Each battery is rated n volts

I used dashed lines in 2n volts connection for ease of seeing. I want to power each device with 4n volts and 2n voltsAn Alternative Illustration

An Alternative Illustration^

The solution that I propose is to connect 4 batteries in series first to power 1 device. Then since there can be 2 sets of batteries in series already I can connect these sets in parallel for the Voltage to not add up. However, just by looking at the circuit I think some portions are actually shorted.

Any help would be appreciated. Please don't be harsh if this is a dumb question.

  • \$\begingroup\$ Can you draw batteries in a straight chain, it's so difficult to imagine what you have drawn. \$\endgroup\$
    – Long Pham
    Jun 1, 2018 at 3:42
  • \$\begingroup\$ Yes, you can do that but batteries that power two devices will die faster. \$\endgroup\$
    – Long Pham
    Jun 1, 2018 at 3:44
  • \$\begingroup\$ @LongPham scontent.fmnl9-1.fna.fbcdn.net/v/t1.15752-9/… \$\endgroup\$
    – user152966
    Jun 1, 2018 at 3:54
  • \$\begingroup\$ I have done it. My bad \$\endgroup\$
    – user152966
    Jun 1, 2018 at 4:06
  • 1
    \$\begingroup\$ :( It still looks awful. \$\endgroup\$
    – Long Pham
    Jun 1, 2018 at 12:59

2 Answers 2


No. Your attempt to power a 4n load, and and 2n load from balanced batteries, shorts out each 2n battery.

You have two (or three) options.

a) Power the 4n load, and a 4n->2n converter to power the 2n load. This could be a linear regulator if the power level is low enough for you to not worry about the inefficiency and heat. Better though would be a buck regulator.

b) Configure the batteries as 2n, and power the 2n load, and a 2n->4n boost converter. Choose (a) or (b) depending on which load draws more power, or the convenience of having a buck or boost converter to hand.

c) Power the 2n load from a battery tap, and tolerate the consequences of the imbalance (replacing batteries earlier, or charging differently)

d) Power the 2n load from a switchable battery tap, so you spend 50% of the time powered from the 'top' batteries, and 50% of the time from the 'bottom' ones. Obviously you cannot common the grounds if using it like this.

e) Use a 2n set and a 4n set of batteries. You need more batteries, but get complete freedom.

Ok, so 5 options.


Wouldn't adding a pair of diodes solve the issue? If the current isn't something massive I think it would be fine no?Professional drawing

  • \$\begingroup\$ no <+ extra letters to make up the comment length> \$\endgroup\$
    – Neil_UK
    Jun 1, 2018 at 12:50
  • \$\begingroup\$ @Neil_UK how so? \$\endgroup\$
    – user152966
    Jun 2, 2018 at 7:56
  • \$\begingroup\$ Hello Red, and welcome. If you don't describe the issue, we cannot answer to your question. \$\endgroup\$ Jun 11, 2018 at 7:07

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