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.

  • \$\begingroup\$ You need to consider the behavior of the load at 5v vs 3.3v. Does it still draw 28mA at 5v or does it draw less? \$\endgroup\$
    – lucky bot
    Commented Jun 26, 2018 at 2:57
  • \$\begingroup\$ Please wait a day before accepting an answer, so that 1) every time zone can have a chance, and 2) you can see the community's answer and not just a single data point. \$\endgroup\$
    – AaronD
    Commented Jun 26, 2018 at 3:35
  • \$\begingroup\$ Better yet put in all acceptance criteria in question \$\endgroup\$ Commented Jun 26, 2018 at 3:59
  • \$\begingroup\$ @TonyStewartolderthandirt I did. My criteria just wasn't as in-depth as you'd like, I guess. "Which should last longer" was the extent of my criteria. \$\endgroup\$
    – bcsteeve
    Commented Jun 26, 2018 at 21:18
  • \$\begingroup\$ @Misunderstood from the question: "working with what I've got on hand" \$\endgroup\$
    – bcsteeve
    Commented Jun 27, 2018 at 19:06

3 Answers 3


LDO regulators don't "transform" energy, they are "linear" regulators. Whichever goes into the load, nearly the same current will be drawn from the battery.

So, within conditions of your task, the largest battery has 5 A-h capacity, the medium one has 1.7 A-h, and the smallest one has 0.6 A-h.

Since the same current (0.028 A) will be consumed from a battery in all cases, the run time will be 178 hours (5/0.028), 60.7 hours, and 21.4 hours correspondingly. I assume that regulators have negligible control losses (quiescent current) in their feedback/control circuitry. If not, this "control current" needs to be added to the 0.028 A load.

ADDITION: MCP1700 uses MOSFET as regulating element, with control circuitry consuming a negligible 1.6 uA (max 4 uA) additional current from input. For LM2931 the "ON" current is 2.5 mA typical, up to 30 mA, so LM2931 might last half of MCP1700 time. True, the MCP1700 can't handle input voltages above 6V, but there must be other low-quiescent-current regulators. And, as Tony Stewart hints, DC-DC switching converters would do a much better job.

                       +/-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
  • \$\begingroup\$ "But chances the 5000mAh is only 2100mAh are very high unless proven." - explain? \$\endgroup\$
    – bcsteeve
    Commented Jun 26, 2018 at 6:50
  • \$\begingroup\$ I bought these ones from Banggood. Fake specs \$\endgroup\$ Commented Jun 26, 2018 at 21:27

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.

  • 3
    \$\begingroup\$ You can't assume "LDO is 70% efficient". LDO efficiency is in simple proportion between input voltage and voltage delivered. For a 3S battery input power is 12.6 times 28mA, while the useful power is 3.3 V at same 28 mA, so 3.3/12.6=26% \$\endgroup\$ Commented Jun 26, 2018 at 1:41
  • \$\begingroup\$ thanks for the info @AliChen I was going for extremely rough estimates here. Using that info, then the clear cut winner is most likely the smaller battery. \$\endgroup\$ Commented Jun 26, 2018 at 1:56
  • \$\begingroup\$ Always measure LDO current across the whole voltage range including droput. Some draw substantial currents as they go through dropout. PNP regulators especially, but some cmos ones also have higher quiescent current at certain voltages. \$\endgroup\$
    – Henry Crun
    Commented Jun 26, 2018 at 2:05
  • 1
    \$\begingroup\$ I downvoted this for "assume the LDO is 70% efficient". For switch-mode supplies we often use efficiency estimations, but for linear supplies you should be using the fact that the input current equals the output current. Or in other words the mAh is the same before and after the regulator, not the mWh. \$\endgroup\$ Commented Jun 26, 2018 at 2:40
  • \$\begingroup\$ Also downvoted because of the efficiency assumption. LDO's are linear, just like non-LDO linear regulators. Only if it's explicitly called "switching" or "switchmode" is it even remotely useful to assume some kind of efficiency figure up front. \$\endgroup\$
    – AaronD
    Commented Jun 26, 2018 at 3:31

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