I have a Lithium battery pack circuit that converts 4.2V (two 4.2V cells in parallel) to 5V. Another option is to use a step-down converter from 8.4V (two 4.2V cells in series) to 5V. Considering both cited circuits are well implemented, which choice would be more efficient in terms of power dissipation?

I am looking for some general rule as "step-down is always preferable than step-up" or "the absolute voltage difference matters" etc.

  • \$\begingroup\$ Related, but not about efficiency: Step voltage up or step voltage down when using batteries? \$\endgroup\$
    – dawid
    Sep 30, 2013 at 2:56
  • 1
    \$\begingroup\$ Both are theoretically similar when powered from ideal voltage sources but you are using batteries and this makes a difference. You might find that the series batteries at 8.4v are better than parallel at 4.2v. \$\endgroup\$
    – Andy aka
    Sep 30, 2013 at 8:26

3 Answers 3


Boost converters are typically less efficient than Buck converters, but not by much. The fundamental reason has to do with the inductor current flowing directly to ground during the on-time, instead of through the load, like it does on Buck converters. There is an EE-Times article that mentions this: Unscrambling the power losses in switching boost converters

Generally, however, since inductor current flows to ground during the on time, only a fraction (off time to period ratio) flows to the output, as illustrated by the pulsing currents in Figure 2 (this is the reason why boost converters are generally less power efficient than buck converters)

Figure 2.  Current composition and distribution across a boost converter

Since the efficiency differences are not significant, you are probably better served deciding on additional criteria instead of regulator efficiency alone, including:

  • Charger complexity (single cell is simpler).
  • Cost
  • Size
  • Cells in series will be limited by their weakest cell.
  • Cells in parallel tend to charge one another, and there is an efficiency hit due to the chemical process.
  • Losses due to higher input current with lower voltages (parallel cells).

Batteries are often not terribly fond of being wired directly in parallel, since any mismatch in the open-circuit voltage will result in the higher-voltage battery feeding current into the weaker one. If one is using rechargeable batteries and the charge states are sufficiently close, this may simply result in the batteries trying to equalize each other. When using primary-cell batteries, however, or if the charge states are not particularly close, this current flow may be detrimental to both batteries.

Wiring batteries in series is often safer, provided that current shuts off before any battery's voltage drops below its minimum safe level. For primary-cell batteries, the minimum safe level is roughly zero volts (the concern not being "damage" to a dead and useless battery, but rather the possibility that a back-driving primary cell battery may dump harmful chemicals on nearby circuitry). For rechargeable-cell batteries, the minimum safe voltage is much higher (draining batteries below that point may greatly accelerate wear).

Any differences in the efficiencies of step-up versus step-down conversion are apt to be minor compared with issues resulting from series or parallel battery connections.

  • 6
    \$\begingroup\$ Old comment but the premise seems a little wonky. If you attach the batteries in parallel you will always do it so they're matched at time of connection. Then, they will remain matched because they are in parallel. Large sustained loads can mismatch them if they're very different IR, but it's unlikely to be significant. Meanwhile batteries in series need to be very similar to stay matched without balance circuitry. This makes series connections in many ways the harder ones. NiMH is a different beast, for reasons I don't know. Maybe you were thinking of them, but they're the exception. \$\endgroup\$
    – Tomek
    May 2, 2016 at 21:22

Step up and step down idealized converters are exactly the same thing.

The circuit is as in the image and instead of putting diodes an ideal 2 positions switch is used that is equivalent in scope. That circuit is equivalent to both a buck and a boost converter.

buck and/or boost converter not buck-boost

That circuit works in both ways from left to right as a buck converter and from right to left as a boost converter.

I suppose thus that there should not be any difference in efficiency between them, but more expert audience may have a different opinion based on other considerations.


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