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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?

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2 Answers

6

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

  1. 1uA + (50mA*0.001) + 25 uA = 76 uA
  2. 76 uA / 80% efficiency = 88 uA
  3. Round up to 100 uA = 0.1mA
  4. 2700mAh / 0.1mA = ~3 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.

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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. – Jason S♦ Jan 16 at 22:22
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. – rdeml Jan 18 at 13:24
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 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. – edebill Jan 18 at 17:52
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The output current required from the 3.3V regulator is

1uA * 0.999 + 50mA * 0.001 + 25uA = 75.999uA

The output power is

3.3V * 75.999uA = 250.8uW

The input power to the regulator is

250.8uW / 0.8 = 313.5uW

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

313.5uW / 3V = 104uA

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

2700mAh / 104uA = 25837hours = 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.

(* jcl *)

http://www.wiblocks.com

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