The average current is the total current consumption divided by the measured duration.

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As per the above image, there are two average currents.

Case 1: 866.82 uA over 2 mins of windows (complete cycle).

Case 2: 682.63 uA over 1 min of selected window (only sleep).

I need help confirming the calculation, as I have checked some examples and got confused. To calculate the battery life of two coin cells with a total capacity of 400 mAh.

Total current consumption is for Case 1:

enter image description here

Total current consumption is for Case 2:

enter image description here

When I check the units this seemed wrong, as units of the total current are mA-seconds, and for battery life calculation we need the current consumption reading in mA.

Because battery life is 400 mAh/(current consumption in mA, not (mAh/mAs)

Battery life would be in hours.

Please help me understand how the current consumption is calculated if not the average current multiplied by the time period. What current consumption is used in battery life calculation?

Is it simply:

Case 1: Battery life = (400 mAh/866.82 μA)

Case 2: Battery life = (400 mAh/682.63 μA)

  • \$\begingroup\$ "Average Current is the total current consumption divided by the measured duration" okay so if you have the total current consumption and you divide by the measured duration then you have the average. Do you have the total current consumption? Or do you have the average? If you already have the average why are you multiplying it by the duration? \$\endgroup\$ Commented Jan 3, 2023 at 7:13
  • \$\begingroup\$ @user253751 Yes I have average current consumption, but to calculate battery life I thought I need total current consumption. So that's why I'm multiplying with duration. \$\endgroup\$ Commented Jan 3, 2023 at 7:16
  • \$\begingroup\$ (I said total current consumption in my comment because that's what the question said, even though charge is the right word) \$\endgroup\$ Commented Jan 3, 2023 at 8:22

2 Answers 2


The average current consumption over a period t0 to t1 is the definite integral of the instantaneous current with respect to time over that period divided by the period.

\$\frac {\LARGE \int_{t0}^{t1} i(t)dt}{t1-t0}\$

If the current is constant during the entire period, the average current is just that constant current.

A more useful example might be if the current was 2mA for 1 second and then 150uA for 59 seconds during sleep. Let's say the whole cycle repeats every 60 seconds, so the average will be the same as long as we measure over an integral number of cycles.

So the average current is (2000uA * 1s/60s) + (150uA * 59s/60s) = 180.8uA.

Then with a 400mAH battery you could predict a life of 400mAh/0.1808mA = 2200h.

Of course battery capacity varies somewhat with current drawn and other factors, but this is a starting point.

  • \$\begingroup\$ You forgot to divide by the integrated time period in your description and formula. \$\endgroup\$
    – Mike
    Commented Jan 3, 2023 at 9:15
  • \$\begingroup\$ @Mike Good point, fixed. \$\endgroup\$ Commented Jan 3, 2023 at 9:27
  • \$\begingroup\$ @Spehro Pefhany What if that hi-lo proportion varies randomly? \$\endgroup\$ Commented Mar 14 at 15:15
  • \$\begingroup\$ @GrahamRounce The integral equation applies regardless of how i(t) varies with time. If you know the statistics of the variation you can calculate an approximate average over a period if the measurement period is long enough in relation to the cycle time. With something like a closed-loop controller where the demand and setpoints are changing that might not be straightforward. \$\endgroup\$ Commented Mar 14 at 15:27

Average Current is the total current consumption divided by the measured duration.

No, that should read, "Average current is the total charge consumption divided by the measured duration." Current is charge per unit time so total charge will be current × time with units of As (ampere-seconds).

When you divide your total charge by time (mAs/s) you will get the average current (mA).

  • \$\begingroup\$ Okay, so what average current is shown above is just a mean of samples that come under the windows I selected. I got the point. But how is useful to calculate battery life? \$\endgroup\$ Commented Jan 3, 2023 at 7:25
  • \$\begingroup\$ @JustdoinGodswork mAh divided by mA calculates h \$\endgroup\$ Commented Jan 3, 2023 at 7:29
  • \$\begingroup\$ @user253751 Thats what I understood, so the mA that's used here is the average current or the total current over the time period. That's what I'm confused with. I'll convert uA to mA but after that shall I use the average current i.e. 866.82uA directly. Or need to calculate total current consumption from the average? \$\endgroup\$ Commented Jan 3, 2023 at 7:39
  • \$\begingroup\$ @JustdoinGodswork there is no such thing as total current consumption. It's total charge consumption. Like, if my car has a fuel rate of 1 gallon per hour, do I just divide the 20 gallon tank by 1 gallon per hour and it says 20 hours? or do I need to calculate the total fuel rate? \$\endgroup\$ Commented Jan 3, 2023 at 7:41
  • 1
    \$\begingroup\$ @JustdoinGodswork Yes. And maybe sometimes you use 3.1 coulombs in one second, but them you only use 0.1 coulombs per second for the next 9 seconds, so it's 0.4 on average. It doesn't have to be exactly 0.4 coulombs in every single second, only on average. \$\endgroup\$ Commented Jan 3, 2023 at 8:22

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