We use a 1650 mAh capacity LiPo battery.

It has a 2.5volt undercurrent protection cutoff set. Our device uses a CONSTANT 5mA of draw UNTIL the battery drops down to 3.2V, at which time the system cuts its power draw in half.

Can someone help me write a FORMULA for calculating how many TOTAL hours of run time our device has?

Formula is what I'm looking for here because I want to create a spreadsheet from this formula so that we can use it to predict run times with different capacity batteries and different draw currents.


  • 1
    \$\begingroup\$ I'd suggest that you derive a curve-fit formula from your own test data (better) or the manufacturer's test data (often suspect in my experience). \$\endgroup\$
    – HikeOnPast
    Commented Apr 1, 2013 at 21:42
  • \$\begingroup\$ What does this "curve-fit" formula look like? What are its variables and multipliers? \$\endgroup\$
    – Tim Prince
    Commented Apr 1, 2013 at 21:46
  • 1
    \$\begingroup\$ @TimPrince, I think Hike means, you measure how long this particular battery lasts under different conditions (5mq, 10ma, etc), and plot that on a curve. Then fit a line to that plot, and use that to predict other conditions. At the very least, you ought to get a WYSIWYG ratio, for how many mAh you get vs how many the manufacturer says you can get. \$\endgroup\$ Commented Apr 1, 2013 at 22:13
  • \$\begingroup\$ Also, I hope your system also shuts down entirely when the voltage is below 2.5V Is that what you mean by an undercurrent protection cutoff? \$\endgroup\$ Commented Apr 1, 2013 at 22:18
  • \$\begingroup\$ The complexity of the curve fit equation will be a function of how complex your operating range is (inter-cell tolerance, temperature, discharge rate, cell age, etc) and what the resulting data looks like. If you test just one pack in your intended application and measure the discharge time, the equation is a single constant! \$\endgroup\$
    – HikeOnPast
    Commented Apr 1, 2013 at 23:43

2 Answers 2


According to the manufacturer's information, the 1650 maH capacity applies when the current drawn is 320 ma. This would last for 1650/320 or 5.16 hours at which time the voltage will have fallen to 2.75 volts (again from the data sheet). Since you plan to only draw 5 ma, the battery will last much longer even if your cutoff voltage is 3.2 volts instead of 2.75 volts. A rough number is to divide 1650 by 5 which gives 330 hours but there's no guarantee of this. Your best bet is either get more information from the manufacturer as to the battery behavior at low current drain, or make your own measurements. Apply your load and measure the battery voltage at intervals. This may take quite awhile (days) but you may be able to extrapolate the data if it appears to be following a simple curve.


The LiIon or LiPo battery discharge characteristics curves below are from the stack exchange question here - he gives no source or other information BUT these curves are liable to be generic enough to provide a guide.

Your battery's capacity is 1600 mAh nominal minimum at 0.2C.
You are discharging at 5 mA which is about 5/1600 = 0.03C so you are on the right and high side of the black curve by a goodly way.
According to the curves the battery has provided about 98% of capacity at 3.2V at 0.2C so you can expect that for practical purposes you will get 99%+ capacity from the battery at 0.03C by the time you reach 3.2V.

So you can expect that - Run time to 3.2V would be about Capacity/I drain, and that
Run time from 3.2V to cutoff would be ABOUT Capacity/Idrain_low x 0.01.
Note that I have not bothered taking off the 1% remaining from the first calculation

In this example:
Run time to 3.2V ~~~= 1600/5 = 320 hours.
Run time to 2.5V ~~~~~~~= 1600/2.5 * 0.01 = 6.4 hours

As others have noted, the latter figure is going to be horrendously ill conditioned.
If you ant some guaranteed reserve you would be advised to about 3.6V which, in this low load case, should give you about the same run time at 5 mA but much longer at 2.5 mA. eg by inspection C3.6V ~~~= 90%
T_5mA ~~~= 1600 x 0.9 / 5 =~ 285 hours ~+ 90 % of previous. T_2.5 mA ~~~~~~~~~~= 1600 x 0.1 / 2.5 = 64 hours ~+ 1000% of previous

Results will vary with temperature, cycle life of battery, battery quality, phase of moon, number of dead fish for sale in immediate locality and proximate bank holidays.
But this gives you some idea of where to start and an approximate idea of what to expect.

enter image description here


You are highly advised to stop before 2.5V. You gain very very little - as seen from the above curves - and you shorten the battery life by draining it to the dregs. In this case using eg 3V will make only a tiny difference and will do no harm but an unknown but finite amount of good.

Also of some use MAY be

My answer here

Battery university page re Li battery lifetimes

Solar charging:

Tim Prince said:

  • Our device is solar-powered.
    It charges in about 7 hours on a day of full sun.
    Once fully charged, the device will run for ~330 hours.
    With this design, there's almost no chance of it draining below the 2.5V lower limit.

If solar charging, be aware that while there may be 12 or more hours of daylight in summer, there will be fewer hours of equivalent full sun. So, if a battery needs 7 hours of full PV panel output to charge it will not achieve this almost anywhere on earth - the exceptions being locations near the poles in summer or a limited number of other locations favoured by combinations of lattitude, altitude and atmospheric clarity.

An excellent site for determining available "sunshine hours" by season is www.gaisma.com

Here is a table for Chicago, USA.
You get about 6 hours of full sun per day in June and July but only 1.5 hours on an average December day. If you need 7 sunshine hours to charge it will take about 5 December days to do so.

enter image description here

Here is the result for Kabul in Afghanistan
In Kabul in summer you will get 7 hours on an average day in 2 summer months and about 30% of this level in mid winter.
Do not expect to be able to solar-fast charge your product in NY NY in mid winter :-).

enter image description here

  • \$\begingroup\$ Just so you guys know. Our device is solar-powered. It charges in about 7 hours on a day of full sun. Once fully charged, the device will run for ~330 hours. With this design, there's almost no chance of it draining below the 2.5V lower limit. \$\endgroup\$
    – Tim Prince
    Commented Apr 3, 2013 at 4:39
  • \$\begingroup\$ @TimPrince - see additions re solar charging. You should have no problems :-) \$\endgroup\$
    – Russell McMahon
    Commented Apr 3, 2013 at 6:52

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