Li-Ion Battery Array: V.out ≥ 3.5 [V], I.out ≤ 1 [A] : No voltage regulation? Boost? BuckBoost? SeriesxParallel — 1x4? 2x2?

Question

With a lithium ion cell array as follows:

• n.cell = 4.0 [-]
• V.cell.max = 4.2 [V]
• V.cell.nom = 3.6 [V]
• Q.cell = 5.0 [Ah]
  .
• V.load ≥ 3.5 [V]
• I.load ≤ 1.0 [A]

The load requires a 3.5 [V] or higher bus for powering LEDs and a 3.3 [V] bus for powering ICs (microcontroller, LED drivers, capacitive touch interface drivers).

Question

Is it preferable to setup the cell array (Series x Parallel) and conversions as:

1. 1Serx4Par, , shutoff at V.batt = 3.5 [V]
   .
2. 1Serx4Par, 3.5 [V] Boost, shutoff at V.batt = 2.7 [V]
3. 1Serx4Par, 3.5 [V] buckBoost, shutoff at V.batt = 2.7 [V]
   .
4. 2Serx2Par, 3.5 [V] buck , shutoff at V.batt = 5.4 [V], balancing IC

Considerations

1. seems dubious. Shutting off a V.nom = 3.6 [V] battery array at 3.5 [V] is a pretty serious loss of capacity, even with low droop due to low max current load and minimal conversion losses.

2. and 3. are straightforward responses to 1.. Either:

2. directs battery until below 3.5 [V] then boost to 3.5 [V]
3. buckBoosts to 3.5 [V] always.

• Do buckboost converters work well over a dynamic input range?
• What do they do when the battery naturally droops to near-unity gains (3.5 [V])?

[Edit: Yes, in a 4-switch buck-boost topology, and they buckBoost near unity.]
[Source1, Source2.]

4. stacks half of the cells, increasing the battery output voltage such that boosting is never necessary, but cutting the capacity in half and doubling the drawn current per cell. Doubling the drawn current per cell loads the cell more causing more rapidly reducing voltage drop .. which I believe proportionally increases the net current drawn slightly more in the voltage converter. Adding further complexity is the need to balance the stacked cells.

This leaves me hypothesizing that 2. and 3. are the more preferable, with 3. being the more likely to have an integrated, off the shelf solution as opposed to designing a switching solution in 2. between applying direct battery voltage and boost conversion.

Answer 1

I might consider either 3 or 4.

The advantage of 3 is that you can use a simpler BMS: a CCCV charger and a low voltage-cutoff on the load side. But you do need a buck-boost converter.

The advantage of 4 is that you use a simpler DC-DC converter. But you do need an honest-to-goodness BMS. By the way, when you say "shutoff at V.batt = 5.4 V", that's incorrect: the shutoff must be based on the individual voltages of each cell in series, not on the total voltage of the battery.

Consider also an inverting DC-DC converter: -3.6 V in, +3.5 V out. An inverting converter is just as simple as a buck converter.

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However, having said that, in practice I would not do any of the above. Instead, I would buy a ready-made power bank with a 5 V output (they are cheap, readily available, meet regulations, and have a warranty) and be done. Easy peasy.