# How do I calculate battery life allowing for duty cycle?

I have a battery rated at 2610mAh. The device it's powering transmits to the network for 35ms during this time it uses 1.95mA of power. the rest of the time the device draws 0mA. I want to workout how long the battery would last in seconds.

Here's what I've done to calculate this. I'm not too sure if I've done it right. First I converted the 2610mAh (battery rating) into mA seconds:

$$2610 \cdot 60 \cdot 60 = 9396000\ mA \cdot s$$

Then I calculated how many mA the device uses each second:

$$\frac{1.95}{35} = 0.055714 mA/ms \cdot 1000 = 55.714 mA/s$$

Now I use the formula $t=\frac{C}{I}$:

$$\frac{9396000}{55.714} = 168647s$$

Is this the correct way to do this?

• You haven't said how often the transmissions occur. – Dave Tweed May 6 '13 at 17:43
• I'm not sure how often. All I know is the active duty cycle is 4%. Does that help? – mintyfreshpenguin May 6 '13 at 17:50
• I wonder if it actually draws 0 mA when it's off. If you're using some microcontroller there is actually a tangible ambient (quiescent) current draw that needs to be accounted for. Sometimes this quiescent current takes more energy than your actual transmission. I wouldn't assume that there is 0 mA current draw just because you measure 0 mA with a volt meter. Often times your quiescent current will be in the low uA range but very real. – Joel B May 7 '13 at 15:25

If it's a duty cycle of 4%, then the average load is 4% of 1.95mA, or 78 uA. The battery should last 2610 / 0.078 hours, or 33461 hours. Over 3 years.

Battery capacity changes with temperature, time, and discharge rate (and the 2610 mAh rating isn't too precise either, in my experience.) These calculations are probably within ±20% of the real answer.

• that helped me out a lot! – greg121 Mar 24 '15 at 10:42

Your device needs 1.95 mA × 35 ms = 68.25 µA-s per transmission.

Your battery is good for 2610 mA-h / 68.25 µA-s = 137.65 million transmissions.

To turn this answer into a time value, you would need to know how many transmissions occur per some unit of time.

• Thanks for the quick reply Dave. I've got a transmit duty cycle of 4%. – mintyfreshpenguin May 6 '13 at 17:56

Useful battery life is a function of two main parameters:

• How quickly the chemicals in the battery will be depleted

• The level of depletion at which the device becomes unusable

The capacity figures for batteries indicate how much current they can supply before they are depleted to the point that the manufacturer does not expect them to be useful for typical applications. A battery's ability to supply power, however, will diminish throughout its useful lifetime, and even batteries that with half of what the manufacturer would regard as "useful" energy remaining may be unable to supply enough power to operate certain devices.

If one wishes to estimate how long a device will operate properly on a set of batteries, and the expected lifetime is sufficiently long that one doesn't want to wait it out, one might start with a fresh battery, drain it about 40% using a resistor or current sink (probably sized to drain over a period of an hour or two), let the battery rest for a little while, and test whether the device still works. Then drain the battery another 5%, test whether the device still works, etc. If one finds that the device starts malfunctioning after one has drawn e.g. 80% of the battery's nominal capacity, then assuming the current draw isn't affected by voltage, one should estimate battery lifetime in days as being 80% of its capacity, divided by the amount of charge the device consumes per day.

Note that depending upon the purpose of the device, it may be helpful to have it drop into a "lower-power" schedule when the battery gets low. For example, a temperature-monitoring device might transmit readings once every minute under normal battery conditions, but once every five minutes once the battery reaches a certain threshold. Such behavior could substantially extend the period of time at which the device could offer some degree of usefulness, at the expense of reducing the time during which it offered full utility.

## protected by KortukMay 10 '13 at 13:30

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