Question: Given a simple circuit of a draw around 28mA (when supplied 3.3v). I'm trying to maximize run time on a battery, but working with what I've got on hand... I have the following batteries:

* 5000mAh 3S LiPO (12.6V fully charged)
* 1700mAh 2S LiFE (7.2V)
* 600mAh 1S LiPO (4.2V)

I have the following regulators:

* MCP1700 3.3V LDO
* LM2931 5.0V LDO

The MCP1700 Can't handle the larger batteries. So I see my options are:

* Use the tiny 1S and the MCP1700
* Use the 3S and the LM2931 output into the MCP1700
* As above but with the 2S

Which should last longer?

Obviously the 5000mAh battery is by far the largest, but that's a pretty inefficient setup. The 2S is somewhat more efficient and still about 3x the size of the 1S. The 1S should be by far the most efficient, but the battery is tiny.

I can, of course, just try each and measure. Thought I'd ask here for insights.

Answer: 
```
                       +/-15%   Ah  Wh    TPS82150 Buck Wh-out  hrs @ 92 mW
5000mAh 3S LiPO (12.6V) 10.8  5000  54    86%            46     505 h
1700mAh 2S LiFE (7.2V)  7     1700  11.9  87%            10     113 h
 600mAh 1S LiPO (4.2V)  3.6    600  2.16  92% LDO         2     22 h

 Meanwhile a 3.3V LDO with 7V in avg is 60% efficient
                      with 10.8V avg is 30% efficient TPS71533

But chances the 5000mAh is only 2100mAh are very high unless proven. (Fake Chinese specs)

```

> 
> * A Lithium-ion cell starts at 4.2 but drops to 3.6 avg and drop out at 3V typ. So 3.3V out of a MOSFET LDO with 3.6 average is 92% efficient. Or more... since if Vin-Vdrop (50mV type) goes below 3.3 V , the output is now tracking battery. Ie 99% efficient best case
> 
> 
>
The rather quick estimation is this: the power output needed for your intended purpose is 92.4 mW.

Let's then assume that each LDO is 70% efficient. That means that for one LDO the input power drained from the battery is 132 mW. For two, the overall efficiency is 50% and will drain 188.57 mW.

Your three batteries can respectively output 63 000 mW, 12 240 mW and 2 520 mW for one hour.

Autonomy for each battery is therefore 334h, 65h and 19h. 

I'm not familiar enough with the inner workings of LDO, but I know that standard linear regulator are more wasteful when the output and input voltage are far appart (I.E that 7 in 5 out is more efficient than 15 in 5 out). Therefore, the two daisy chained losses of the bigger batteries are probably underestimated. 

In the end, I would either take the largest battery for longevity or the smallest battery for efficiency. Why ignore the 2s? You get roughly the lack of efficiency of the 3s combined with almost 6 times less autonomy.