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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.

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  • \$\begingroup\$ what temperature range? <0'C? \$\endgroup\$
    – D.A.S.
    Commented Jan 2, 2018 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
    Commented Jan 2, 2018 at 17:31
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    \$\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
    Commented Jan 2, 2018 at 17:51
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    \$\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
    Commented Jan 2, 2018 at 17:56
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    \$\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
    Commented Jan 2, 2018 at 18:01

1 Answer 1

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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.

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  • \$\begingroup\$ OK, so putting in my actual numbers. 3.3V, 1.11mA, 0.335mA, 94% converter efficiency, gives me 9114h = 379d \$\endgroup\$
    – Andy
    Commented Jan 2, 2018 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
    Commented Jan 2, 2018 at 18:22
  • \$\begingroup\$ @Andrew - yes, that is what I was originally worried about. \$\endgroup\$
    – Andy
    Commented Jan 2, 2018 at 18:23
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    \$\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
    Commented Jan 2, 2018 at 20:02
  • \$\begingroup\$ @Ali Yes, but now I can play around with the firmware configuration to get the margin I need. :) \$\endgroup\$
    – Andy
    Commented Jan 2, 2018 at 22:25

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