I am working on a design that needs to operate outside on two AA batteries for at least a year.

I have optimized the design to reduce power and using a spreadsheet analysis I now have an average current consumption. I should be able to verify that is right by monitoring a prototype for a while.

I have the following pieces of information:

How can I take this to create a realistic estimate?

I think it is perhaps reasonable to use the average environmental temperature \$Tave\$.

Note - I chose Duracell simply as an exemplar of a quality battery. I wouldn't need/want to consider poorer-quality batteries.

Many thanks!

Update - average power consumption is 1.11mW.

  • \$\begingroup\$ what temperature range? <0'C? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jan 2 '18 at 17:26
  • \$\begingroup\$ mAH is not a measure of power, it is a measure of electrical charge/battery capacity. For Pave I think you mean mW or equivalent power draw? Same goes for Psupply — it should be in mAH but is not a measure of power, but battery charge/capacity. \$\endgroup\$ – Murey Tasroc Jan 2 '18 at 17:31
  • 2
    \$\begingroup\$ 1.11mW sounds high. If your goal is roughly 10k hours, that would be 11 Wh of energy, or almost 4 Ah @ 3V -- a bit much for AA cells. You need to be about an order of magnitude lower than that. Typical alkaline AA capacity of 2500 mAh / 10k hours = 250 uA average consumption. \$\endgroup\$ – Dave Tweed Jan 2 '18 at 17:51
  • 1
    \$\begingroup\$ @Andy, don't bother worrying about the nonlinearity. Temperature effects, etc, mean you'll only get an order-of-magnitude estimate anyway. Just do a rough calculation and then give yourself a good engineering margin. \$\endgroup\$ – The Photon Jan 2 '18 at 17:56
  • 1
    \$\begingroup\$ I have alarm bells at mentioning of a DC-DC converter for something which needs run a looong time from batteries. What does your application (without converter!) need as voltage and what does it use as current? \$\endgroup\$ – Oldfart Jan 2 '18 at 18:01

First, you need to make a realistic ballpark estimate before going to intergraion of discharge curves and buck boost converter inefficiencies.

The estimate would be: at 3 V input supply and 1.11 mW, the device will take about 0.45mA average (assuming 80% efficiency of converter, if any). The Duracell discharge curve says about 650 service hours at 5 mA discharge rate. At 0.45 mA it will take about 11 times longer, or about 7200 hours. This is about 300 days of operation. Which is about 20% short of one year.

Conclusion: you can't guarantee this device to work for a year from 2xAA batteries. No amount of more accurate mathematical massaging of discharge curves would change this conclusion.

  • \$\begingroup\$ OK, so putting in my actual numbers. 3.3V, 1.11mA, 0.335mA, 94% converter efficiency, gives me 9114h = 379d \$\endgroup\$ – Andy Jan 2 '18 at 18:20
  • \$\begingroup\$ @Andy That's assuming something like a constant 20 degree C environment. Lower temperatures will reduce the energy that can be drawn from the cells. \$\endgroup\$ – Andrew Morton Jan 2 '18 at 18:22
  • \$\begingroup\$ @Andrew - yes, that is what I was originally worried about. \$\endgroup\$ – Andy Jan 2 '18 at 18:23
  • 1
    \$\begingroup\$ @Andy, 94% upconverter efficiency at 300 uA load is a bit optimistic, see TI Webench, 87-90% maybe. As of current state, this design has no margins on regular 2AA source. You might want to consider specifying lithium AA batteries instead, like data.energizer.com/pdfs/l91.pdf \$\endgroup\$ – Ale..chenski Jan 2 '18 at 20:02
  • \$\begingroup\$ @Ali Yes, but now I can play around with the firmware configuration to get the margin I need. :) \$\endgroup\$ – Andy Jan 2 '18 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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