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My circuit needs 3.3 volt, and I have 2 options with a Li-ion battery pack.

Option 1: I can put 2 cells in series and get approximately 4.4 to 6V (7.4 volt mean) and then using buck converter reduce it 3.3 V. In that case my battery capacity will be 2000 mAh.

Option 2: In another case, I can put the cells in parallel and get 4.2 to 3V (3.7 volt mean). In this case, I am going to get battery capacity of 2000+2000mAH i.e. 4000mAh.

My query here is, in Option 1, I will get more voltage so that I can use the battery pack from 7 volt to 3.7 volt. In Option 2, I have bigger capacity i.e. 4000mAh but cannot use the range from 3.3V to 3V without using a buck boost converter.

Which option will give me more battery life for my circuit?

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  • \$\begingroup\$ The problem you’ll have if you put them in parallel is that once they drop to the dropout voltage of your regulator, assuming you’re using either linear or buck, your 3.3 supply will start to drop in voltage. With a series configuration you’ll need some way (a BMS) to stop the cells getting out of balance, which would result in one of them getting over-discharged (this is bad). Personally I lean towards higher voltages because they can be more efficient in many scenarios- lower resistive losses and the odd diode drop being less significant. \$\endgroup\$
    – Frog
    Feb 12 at 5:35
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    \$\begingroup\$ When you say "Li-iron", is that a typo of li-ion, or are you referring to lithium iron phosphate (LiFePO₄) cells? \$\endgroup\$
    – Hearth
    Feb 12 at 5:52
  • \$\begingroup\$ Neither is inherently better. If you need really high currents, parallel will get you there more cheaply and easily. \$\endgroup\$ Feb 12 at 14:09
  • \$\begingroup\$ @Hearth 3.7V must be Li-ion, not LiFePO₄ which is 3.2V \$\endgroup\$ Feb 15 at 2:09
  • \$\begingroup\$ Exactly which cells will you be using, and what is the load's peak and average current draw? \$\endgroup\$ Feb 15 at 2:11

2 Answers 2

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The energy is the same in both cases. For a given load power, the time will be exactly the same in both cases. In both cases, the BMS stops discharge when the cell voltage reaches 2.9 V or so. Therefore, no difference.

If there is a DC-DC converter between the battery and the load, its inefficiency should be similar regardless of whether it's powered by 3.7 V or 7.2 V.

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The mAh capacity of a battery or group of batteries is really only a partial measure of capacity - you need to consider the voltage as well.

A 2000 mAh 7 volt battery contains the same energy as a 4000 mAh 3.5 volt battery - the two arrangements will run a given device for the same length of time (depending on the efficiency of DC-DC converters or other voltage conversion systems).

In your case, since you need 3.3 volts, I suspect (but could be wrong) that you may get slightly longer battery life with the parallel batteries particularly if you can manage without a DC-DC converter or voltage regulator, but the difference may be insignificant.

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  • \$\begingroup\$ Voltage converter is no 100% efficient, so that power loss should be considered too. \$\endgroup\$
    – user263983
    Feb 12 at 8:46

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