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The load is 227mA at 25ms (this is pulled 20times in a minute )and 0.001A for the rest of the time from 4 AA batteries. If this test is running for 20 days, I need to calculate the voltage left on the battery after 18 days.

http://www.toshiba-batteries-eu.com/sites/default/files/imagesfiles/Standard_Alkaline.pdf

My calculations for battery capacity are as follows

227mA *25mS + 1mA *2975mS = 8.65mA Typical AA Battery life is 2000mAH Battery hours 2000/8.65= 231hours 20 days is 480 hours 480/231% of battery consumed in 20days? is this right or am I looking at this incorrectly.

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  • \$\begingroup\$ You will stop being able to draw 227mA much sooner than 231h. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 7 '16 at 23:55
  • \$\begingroup\$ A typical AA has much higher capacity than that. 3000 mAh on average, unless you have a constant high drain. \$\endgroup\$ – Passerby Jan 8 '16 at 0:04
  • \$\begingroup\$ Great, Does 4AA mean times (4*2000mAH) of the mAH? \$\endgroup\$ – Nathaniel Jan 8 '16 at 0:06
  • \$\begingroup\$ Depends on whether they are connected in series or parallel. \$\endgroup\$ – Turbo J Jan 8 '16 at 0:07
  • \$\begingroup\$ They are in series \$\endgroup\$ – Nathaniel Jan 8 '16 at 0:08
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Your 8.75mA is high by a factor of three. You've effectively summed the charge transferred in a 3 second period and not divided by 3s in order to get coulombs-per-second, or amps.

So the average current is about 2.88mA and theoretically speaking, you could perhaps get (2000mAh/2.88mA)=693hr=28 days from the batteries. However, the problem is that batteries lose their capacity more-rapidly at higher drain currents than at lower drain currents. If you're using a very high-quality, high-current battery (e.g. an Eneloop NiMH) then you probably will get the rated capacity from it.

A cheap battery under 200mA+ peak load will discharge faster than indicated by its rated capacity, which means that the actual capacity available to you will be reduced. However according to that Toshiba datasheet, the discharge curve is pretty linear even down around 3-5 ohms load (where you are with your peak currents), so the performance calculation above is probably close to valid if you use the Toshiba cells. For a cheap cell or even worse, Zinc battery, it will be much much worse.

The second issue is that the rated capacity is for discharge to a fairly low voltage (0.9V), which may be a lower voltage than can support your circuitry. If your circuit fails at 4.5V (1.1V/cell) but the battery rating was computed for a discharge down to 0.9V, then you will clearly get less usable capacity than the rated capacity.

After 18 days, you will have used 18/28=65% of the capacity. If you take the 2 ohm 140-minute voltage discharge graph at bottom left as being representative of the cell's discharge-time/voltage curve (it may or may not be, because it's a very heavy discharge and you might have a constant-current instead of constant-resistance load) and look at it at 0.65*140=90 minutes, that gives you a voltage of about 1.05V/cell. So you might expect to see about 4.2V on your 4-cell pack after 18 days of use. There are lots of uncertainties there though.

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The "capacity" of an AA battery depends on the magnitude of the current drawn.

Here's a typical example from Duracell for their "Plus" AA batteries:

Graph Graph

If you take 1V as a lower cutoff and do the sums you'll see capacity varies from 1000 mAh at 1000 mA to 2700 mAh at 10 mA.

So to accurately calculate the voltage after 18 days with the duty cycle you mention probably requires a much more sophisticated mathematical approach than any I've seen so far.

In particular, using averages is unhelpful, short bursts of high current deplete the battery much faster than a low constant current of the same average value.

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Average current is basically 1mA plus 227mA x duty cycle. The duty cycle is 25ms every 3 seconds or 0.8333% therefore the average current taken is: -

1mA + (227mA x 0.8333%) = 1mA + 1.89mA = 2.89mA

Given that usually only 50% of the "Ah" of a battery is "usable" (generalism alert) you have available 1 Ah and this means 1000 mAh therefore you have 1000 hours divided by 2.89 = 346 hours = 14.4 days.

If you can truly use the full 2000 mAh then you'll get 28.8 days.

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  • \$\begingroup\$ How did you get the 0.8333%, thank you for your time. \$\endgroup\$ – Nathaniel Jan 8 '16 at 0:54
  • \$\begingroup\$ 0.025 \$\div\$ 3 = 0.8333% duty cycle. \$\endgroup\$ – Andy aka Jan 8 '16 at 9:16
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We can take some formulas to figure out the average current, then use that to figure out the battery life.

The first formula is used for both the standby and active currents.

Average I = I * (Time in Mode / Total Time Period)

In this case, the total time period is 3 seconds, based on your numbers. For Standby:

Standby Average I = 1 mA * (2975 ms / 3000 ms) = 1 mA * 0.9916 = 0.9916 mA

Same for Active:

Active Average I = 227 mA * (25 ms / 3000 ms) = 227 mA * 0.0084 = 1.9068 mA

The Total current is simply added up.

Active Average I + Standby Average I = Total Average I
1.9068 mA + 0.9916 mA = 2.8984 mA

Assuming average of 2800 mAh for any quality Alkaline AA Battery, we can calculate the total time:

Time = Capacity / Current

Time = 2800 mAh / 2.8984 mA = 966.05 Hours

Since you need days, divide hours by 24.

966.05 / 24 = 40.25 days

While there are some environmental factors that may limit the battery, the 2800 mAh rating for an AA battery typically already considers this.

You can read an in-depth look at the calculations in this Texas Instruments white paper: Benchmarking MCU power consumption for ultra-low-power applications.

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