I'm working on a project that requires low energy consumption. In standby mode, it will consume about 11uA. However, when it starts working (it will transmit data in RF), the current goes up to 121mA. This is going to happen 4 times a day, and the peak current duration is 2 seconds, at most.

Because of it, I'm searching for a battery (rechargeable or not) that gives me a high autonomy. I'm expecting 1 year of autonomy, at least. However, most batteries datasheets don't give the information about the duration of a peak current, and how much the battery gets discharged for it. I'm looking for batteries like CR2032, LIR2450 and AA types.

Let me give an example: LIR2450 has a nominal capacity of 100mAh. With dc current of 11uA, the autonomy goes for about 9100h. However, if in the middle of the day (all over the week) my circuit requires about 121mA for 2 seconds, how much does the autonomy drop? Is the battery going to be killed by that consumption?

  • 2
    \$\begingroup\$ If the manufacturer doesn't provide this information in a datasheet then you can only get anecdotal data here, and past history (as they say) does not guarantee future results. Different manufacturers or changes in the manufacturing process can invalidate historical observations. If you really need to know you should purchase a batch of batteries and characterize them yourself. \$\endgroup\$ Commented Jul 11, 2019 at 14:29
  • \$\begingroup\$ Also, temperature and charge state have large influence on this. You'll really want to look for batteries that come with extensive datasheets. \$\endgroup\$ Commented Jul 11, 2019 at 15:43
  • \$\begingroup\$ If it going to be used somewhere with light, you could use a little solar cell to keep a small rechargeable NiMH button cell topped up. Then the only test you need to do is to check if the cell can supply 121 mA for the required time. Suggest viewing: Big Clive with a solar light on YouTube. Suggested search: "NiMH button cell maximum current". \$\endgroup\$ Commented Jul 11, 2019 at 17:39
  • \$\begingroup\$ @AndrewMorton. Nice! I was thinking about using a solar cell either with rechargeable batteries. LIR2450 is one of those batteries. It has a cycle life fo more than 500. It would last more than a year. But the problem remains in the autonomy and discharge of them with a peak current of 120mA (in my case). I will probably go for testing these batteries, anyway. It looks like there is no official document talking about it. \$\endgroup\$ Commented Jul 11, 2019 at 18:00

2 Answers 2


Alkaline or NiMH AA batteries are certainly OK at 121 mA. The current draw in a speedlight/photoflash is higher than this based on the big cap charge time.

4 AA batteries can charge a 1000 uF, 300V cap in about 5 sec -> 0.5*C*V^2 = 45 Joules = 6V * 1.5A * 5 sec if my math is correct.

121 mA is more than I would draw from a coin cell. This Energizer "pulse" test only draws 7 mA. http://data.energizer.com/pdfs/cr2032.pdf Even if they aren't damaged at 121 mA, the internal resistance is likely to be an issue.

Edit: OK, here is some data on the autonomy. An Eneloop NiMH rated at 1900 mAh was tested to have a capacity of 1877 mAh at 200 mA. Therefore, there is minimal loss in capacity at 200 mA, slightly more loss at higher currents. https://lygte-info.dk/review/batteries2012/Eneloop%20AA%20HR-3UTGB%201900mAh%20(White)%20UK.html

Edit2: Discharge Time Calculation:

enter image description here

It is easiest to determine how many 6 hour cycles it will last. The calculation isn't hard, but you need to be careful with your units. Here everything is in hours and mA. If I made a mistake, the eagle-eyed people here should catch it.

Note that batteries will discharge even with no load. Modern batteries are quite good, they will go for several years with minimal loss. Still, I would shoot for 2X margin in your situation (calculation should show > 2 years).

If physical space is at a premium, consider AAA batteries.

  • \$\begingroup\$ So how much does the autonomy drop because of the heavy current loads? That's the real question and I don't think you addressed it. \$\endgroup\$ Commented Jul 11, 2019 at 17:06
  • \$\begingroup\$ @ElliotAlderson - added into on autonomy. \$\endgroup\$
    – Mattman944
    Commented Jul 11, 2019 at 17:18
  • \$\begingroup\$ @Mattman944. Thank you! I guess you answered one question. There's another one that has to do with the battery discharge with these current peaks. Does it discharge a lot, or the voltage drops not much? \$\endgroup\$ Commented Jul 11, 2019 at 17:39
  • \$\begingroup\$ The voltage will drop during discharge. Here are some more curves. powerstream.com/AA-tests.htm For Alkaline, the autonomy is good at 100 mA, not so good at 500 mA for most battery brands. \$\endgroup\$
    – Mattman944
    Commented Jul 11, 2019 at 18:02

As you saw, small batteries are rated in mAh. For example, a AA cell is about 1500mAh. So basically you need to find your overall current * time to determine when the mAh of battery will be exhausted.

You can get an approximate model of your consumption by breaking the usage into device states. You have two you’ve described: an 11uA idle, and a 120mA transmitting one. Multiply each current by the time spent in each current-state for a given interval and you will have an estimate for the mAh demand over that time.

You can refine your model further by taking into account the voltage degradation as your battery is used up. This will work in your favor as the currents will go down too.

Both alkaline and lithium cells handle peak currents well as they have a relatively low internal resistance so you could choose either one for your application. Lithium has a lower internal discharge rate than alkaline, but I’ve seen improvements in alkaline that gets them pretty close (like 10 years shelf life.) That said, 120mA is a bit much for a coin cell and will degrade it more quickly, so the AA size seems to be more appropriate.

  • \$\begingroup\$ The capacity rating is only valid at one specified discharge rate. Higher discharge rates cause lower capacity. You can't just multiply current times time and add up the values. \$\endgroup\$ Commented Jul 11, 2019 at 17:07
  • \$\begingroup\$ For a first-order approximation that’s exactly what you do. The rest is refinement. \$\endgroup\$ Commented Jul 11, 2019 at 17:12
  • \$\begingroup\$ What i'm doing now is this: i gotta wakeup the system 4 times during a 24h-day. So, there's a pulse of 121mA for 2 seconds each every 6 hours. Said that, i'm calculating the current dc value through the Mean Value Theorem. Basically, i'm calculating the area under the curve and dividing by the period (T = 6h). I'm not sure that i can do this for autonomy calculations. What do you think? \$\endgroup\$ Commented Jul 11, 2019 at 17:35
  • 1
    \$\begingroup\$ Yes, that’s the approach. A piece-wise integration of current * time, which you calculate as a sum of products. This gives a first-order model for your consumption based on mean value theorem. As for actual battery life, such a calculation with the stated mAh capacity will get you in the ballpark, but to model the life more precisely you probably need to characterize the battery using this profile in its ambient operating conditions, or have a good model for the exact battery you’re choosing. How much effort you put into this depends on how much it costs to service the battery in the field. \$\endgroup\$ Commented Jul 11, 2019 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.