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I'm trying to estimate the battery life of my device and just need a sanity check since numbers I'm getting are wildly different depending on the method I use to calculate battery life. For brevity I'm going to write first method that I've tried.

Device will spend 7 seconds in active mode, during which average current is 5mA. After that unit spends 600 seconds in sleep mode, where average current is 70uA.

So total period is 607 seconds, meaning ~0.012% of the time unit is drawing 5mA, while rest of the 99.988% it's drawing 70uA. So assuming this logic is sound (which might not be the case) my unit is drawing

(0.012*5 + 99.988*0.07)mA = ~7.06mA

On 1000mA battery with 0.7 factor (700mAh), that means my unit will run for ~99 hours?

Does this makes sense? Is there a better way to calculate/estimate battery life if your unit is not drawing constant current like in this case?

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  • \$\begingroup\$ You definitely made some kind of mistake. There is no combination of 5mA and 70uA that can give you an average of 7mA... \$\endgroup\$
    – user57037
    Commented Aug 24, 2018 at 4:59
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    \$\begingroup\$ It is probably easiest to just calculate \$\frac{7\:\text{s}\cdot 5\:\text{mA}+600\:\text{s}\cdot 70\:\mu\text{A}}{607 \:\text{s}}\approx 127\:\mu\text{A}\$. \$\endgroup\$
    – jonk
    Commented Aug 24, 2018 at 5:09
  • \$\begingroup\$ Very nice, Jonk! Much simpler way to see it and calculate it than what I usually do. And as long as the time units are consistent, you could use seconds, or us or ms or whatever (must be same on top and bottom). \$\endgroup\$
    – user57037
    Commented Aug 24, 2018 at 5:21
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    \$\begingroup\$ As far as what did you actually do wrong, it looks like the main problem is that you didn't deal with the percentages correctly. 7/607 = 0.0115. So that is OK. But 600/607 = 0.988. So the full equation is 0.0115 * 5 + 0.988 * 0.07 = 0.127 mA. So your method is OK. You just made some calculation errors. \$\endgroup\$
    – user57037
    Commented Aug 24, 2018 at 5:53
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    \$\begingroup\$ @mkeith Just total Coulombs divided by total time. Pretty much can't go wrong that way. \$\endgroup\$
    – jonk
    Commented Aug 24, 2018 at 7:33

2 Answers 2

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"... So total period is 607 seconds, meaning ~0.012% of the time unit is drawing 5mA ...*"

No, it's 0.012 of the time (with no '%' symbol). That's 0.012/1, not 0.012/100 of the time.

That means that the remaining time is 0.988, not 99.988.

$$ 0.012 \times 5 + 99.988 \times 0.07 = ~7.0 \ \text {mA} $$

should have been

$$ 0.012 \times 5 + 0.988 \times 0.07 = ~0.127 \ \text {mA} $$

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It's always good to do a 'back of envelope' approximation to any complicated calculation, to get your own sanity check. It's far too easy to drop a decimal place or three and get totally the wrong answer.

It's using 70uA for nearly the whole time. So its total current will be 70uA plus the other contribution.

It's using 5mA for 7 seconds per 600 seconds. 7 per 600 is about 1%, and 1% of 5mA is 50uA. But 1% underestimates 7/600, so it's going to be slightly more than that.

The total current is therefore a little more than 70uA + 50uA = 120uA.

This corresponds well with Jonk's accurate calculation of 127uA, well enough to reassure that we probably haven't made a mistake.

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