LDO confusion

Hi,

I'm confused designing a circuit where part is fed by an LDO to 2.8 V and part by a switching DC/DC to 3.8 V.

I have 8 batteries of 1.5 V and I'm considering 2 options:
2P4S: V = 6 V (DC/DC will be stepdown)
4P2S: V = 3 V (DC/DC will be stepup)

Is the following correct then?

- The lifetime for the part behind the DC/DC will be the same in both cases (assuming both ICs have identical efficiency in the operating zone)

- If the part behind the LDO draws 1 mA, every battery will be providing 0.5 mA in the first case and 0.25 mA in the second case, so for the consumption behind the LDO the second option is twice as efficient.

Conclusion: better 4P2S.

Correct?

Thanks!

Your "1.5V" battery cells will drop to 1.0V or less each during their discharge. Then the 8-cells circuit will have an input of only 8.0V or less (nowhere near 12V).
The 4 cells circuit will have only 4.0V or less and the 2 cells circuit will have an input of only 2.0V or less.

I know that but ignore it, it's the principle that counts. Both circuits are 8-cells, I think you may have misread.

The other way to look at it is in case 1 .0032 watts are lost in the LDO, in the second only .0002 watts.