How do I calculate how long a battery operated product will run?

Here's what I've got:

  1. 2 AA, 1.5V, 2700mAH batteries
  2. Voltage Regulator with a Iq of 25 uA
  3. Voltage Regulator Eff = 80%
  4. Active Current = 50mA
  5. Sleep Current = 1uA
  6. Duty Cycle = 99.9% (only active 0.1% of the time)
  7. Active Voltage is 3.3V

I've gone the current route and got an answer. I went the power route and got a TOTALLY different answer (days vs years different).

How do you do this?


2 Answers 2


My calculation, probably missing something, but here's what I did:

$$ 1 \mathrm{\ \mu A} + (50 \mathrm{\ mA} \times 0.1\%) + 25 \mathrm{\ \mu A} = 76 \mathrm{\ \mu A} $$

$$ \frac{76 \mathrm{\ \mu A}}{ 80 \, \% \mbox{ efficiency}} = 88 \mathrm{\ \mu A} $$

Round up to \$100 \mathrm{\ \mu A} = 0.1 \mathrm{\ mA}\$

$$ \frac{2700 \mathrm{\ mAh}}{ 0.1 \mathrm{\ mA}} \approx 3 \mbox{ years} $$

If you're using rechargeable batteries, they'll discharge on their own long before that. Or if any of your other calculations are off (like maybe it's a 98% instead of 99.9% sleep), that will affect it a lot too.

  • \$\begingroup\$ close enough for estimation purposes. The 51 uA is from the 3.3V which would get boosted up from 2.6-3V. The 25uA Iq is from the 2.6-3V battery. You're missing the increase in current (maybe 15-20%) due to step-up effects. But this brings it really close to 100uA, so you're pretty much spot on. \$\endgroup\$
    – Jason S
    Commented Jan 16, 2010 at 22:22
  • \$\begingroup\$ My concern is that your calculations don't include the number of batteries. There has to be a reason to put another battery in the product. \$\endgroup\$ Commented Jan 18, 2010 at 13:24
  • 5
    \$\begingroup\$ With batteries in series, they'll both drain at the same rate and provide the same amount of current. You've got a 3V 2700mAh battery instead of two 1.5V 2700mAh cells. \$\endgroup\$
    – edebill
    Commented Jan 18, 2010 at 17:52
  • 1
    \$\begingroup\$ +1 for adding the point regarding the self-discharge of the batteries. \$\endgroup\$
    – semaj
    Commented Apr 13, 2011 at 17:49
  • \$\begingroup\$ Excellent answer. As an improvement i'd say to multiply the battery capacity \$2700\$mAh with \$90\$% or \$95\$% due to the fact that the battery is considered dead after it has discharged about \$95\$% of it's power. \$\endgroup\$
    – KeyC0de
    Commented Sep 16, 2015 at 12:57

The output current required from the 3.3V regulator is

\$1 \mu A \times 0.999 + 50mA \times 0.001 + 25uA = 75.999uA \$

The output power is

\$ 3.3V \times 75.999 \mu A = 250.8\mu W\$

The input power to the regulator is

\$ \frac{250.8\mu W}{0.8} = 313.5\mu W \$

When the batteries are fully charged the input current to the regulator is

\$ \frac{313.5 \mu W}{3V} = 104 \mu A \$

If the batteries have a flat discharge curve then you will get a life of

\$ \frac{2700mAh}{104 \mu A} \$ = 25837 hours = 2.95 years

Since your batteries are 2700mAh 1.5V AA I am guessing that the discharge curve is not flat. You will need to draw higher currents as the voltage drops.

Also your regulator efficiency probably drops at lower voltages. Again I am guessing since I have not seen the design.

Be careful when calculating using currents. You may be inadvertently assuming that the input and output voltage of the regulator is the same. With an input of 3V and an output of 3.3V it is not a big error. If you do a more accurate estimate of the battery discharge curve it will matter.

  • \$\begingroup\$ we discussed this in the past, and it seems it must be discussed again. Signatures are not allowed on SE sites. I have removed your signature a second time after you rolled back to add it. Your site is interesting, put it in your bio so that all can see it easily. \$\endgroup\$
    – Kortuk
    Commented Apr 30, 2011 at 10:21
  • 2
    \$\begingroup\$ When this message was originally posted signatures were allowed on SE sites so the "Grandfather" principle should be applied. \$\endgroup\$
    – jluciani
    Commented May 6, 2011 at 1:18
  • \$\begingroup\$ We discussed this months ago in email. Signatures are not allowed. We are not active going through and removing your signatures, but as a question becomes active again or someone flags it we will remove the signature. We are not trying to actively go through and remove your signatures, but we also do not grandfather anything into rules. We removed your signature, you cannot roll it back, we discussed this and you agreed. \$\endgroup\$
    – Kortuk
    Commented May 6, 2011 at 2:07

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