I recently saw some DIY battery builds and something has been bothering me. So for example in one of the builds they needed 14V nominal so they put together 4x3.7v batteries in series so they achieved around 14V. But there exists a lot of voltage booster ICs that could boost the voltage from for example 2 batteries in series(7.4V) to 14V and the other two battery could then be connected in parallel and thus having a higher capacity. I know these ICs are around 85% or so efficient but wouldn't the double capacity of the battery outweigh this inefficiency? So my question is why doesn't anybody use voltage booster ICs instead of using a lot of series batteries to boost voltages? Or is there a rule that says that when the voltage gets doubled using a circuit the capacity is halved?

  • $\begingroup$ "... and the other two battery could then be connected in series ..." Do you mean "in parallel"? "... and thus having a higher capacity." With any arrangement of four cells you have the same amount of energy storage. Adding a voltage booster will, as you know, incur 15% losses so what is the point of doing that? Remember that P = VI. $\endgroup$
    – Transistor
    Apr 22, 2023 at 9:47
  • $\begingroup$ Batteries have been connected in series and parallel for many years before the fancy electronics even existed. $\endgroup$
    – Solar Mike
    Apr 22, 2023 at 10:38
  • $\begingroup$ Yeah sorry i meant to write parallel. So i know p = V*I but my question still stands. When increasing the voltage of a battery with a booster IC, does the capacity of the battery decrease (e. g. if the voltage gets doubled the capacity gets halved) or not (and i'm not talking about the 85% efficiency)? Because if not then adding more parallel batteries with a voltage booster IC is more worth it than using series batteries but then why doesn't any battery i've seen use this(some where new so it's not like the tech wasn't available)? $\endgroup$ Apr 22, 2023 at 11:06
  • $\begingroup$ Since P = VI then it should be clear that $ P_{in} = P_{out} $ and so $ V_{in}I_{in} = V__{out}I_{out} $. Therefore for a given output current at twice the input voltage you will need to supply twice the input current. You can't win. You can't even break even! $\endgroup$
    – Transistor
    Apr 22, 2023 at 12:01

1 Answer 1


Voltage booster circuits are inefficient. They spend energy in the conversions, effectively resulting in a lower capacity of the energy storage system overall. Worse, you get even less energy than a system that would have stored a higher voltage (Storage will inherently result in some losses so it's odd to say less than you put in)

Reducing (buck) DC voltage can be significantly more efficient compared to increasing (boost) if you can tolerate the waves introduced by a switching regulator.

If you already have to pay the storage penalty, you can theoretically get a boost cheaply(less additional cost) by charging in parallel, discharging in series setup, but changing hookups comes with its own costs.

Similarly, alternating current already pays penalties due to the nature of its electromagnetic emissions, so converting via coils of a transformer doesn't incur much additional penalty. DC boosting tends to convert to something closer to AC, incurring some pretty severe losses in the process.

In order for any change to occur, entropy must increase. That is the rule (2nd law of thermodynamics) and it prevents anything from truly being 100% efficient (There are things such as heat pumps that can get a coefficient of performance greater than 100% by claiming they didn't care about some of that entropy-penalty-related energy expenditure so it was free).

  • $\begingroup$ Thanks for the comprehensive answer! $\endgroup$ Apr 24, 2023 at 11:45

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