# How long would this battery device last?

How long could this device be expected to last before needing a new battery?

If that is too subjective, is there some kind of calculation that can be done from the given information (draw, battery capacity)? Approximation? Some real world device worth comparing to?

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No shift key on your keyboard? First word of a sentence starts with a capital. I fixed it for you. You're welcome –  stevenvh Apr 9 '12 at 7:46
@stevevh, your name needs a Cap too. ;) –  kenny Apr 9 '12 at 12:37
@stevenvh the SS called, I heard they are recruiting grammar nazis. but thanks for the edit I suppose –  CQM Apr 9 '12 at 20:46

It's absolutely impossible for anybody to tell you how long the battery will last given the information you gave. We have no idea how much this device will be transmitting, receiving, or in sleep mode.

Further, you can't simply take the mAH rating of the battery, divide it by the power consumption, and arrive at how long the battery will last you. For the simple case this does somewhat work, but this isn't the simple case. You need to take into account the internal resistance of the battery itself and these coin cells tend to have a relatively high internal resistance. What this means is that under high current loads the battery will waste more energy than under low current loads. We don't know what battery you'll be using or, as I stated before, what the ratio of high vs. low current loads is.

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I agree with David: impossible to say. You'll have to make a usage profile. How many hours of usage during a day? And when it's in use, does it transmit all the time?
Say you want to use it for fitness (one of the suggested usages), for instance to transmit heart rate. How long is that? 3 hours a day? Your update rate doesn't have to be high, once a second is enough. The device wakes up from low-power mode in a "few hundred microseconds" (that's a dangerous spec) and let's say you need 10 ms to transmit a new measurement. Then your duty cycle is

$d.c. = \dfrac{3}{24} \cdot \dfrac{10ms}{1s} = 0.125 \%$

Then average current is

$I = 47mA \cdot 0.125\% = 58.75\mu A$

That's 27mA for the transmitter, 20mA for the reciever (sic Josh, I didn't see that figure anywhere). We'll ignore standby current. (This is an estimation, you can't make exact calculations.)

Let's say an AAA NiMH battery has a capacity of 1000mAh, then the batteries should last two years. In practice that will be a lot less because of the battery's self-discharge, but you have an idea.

Bottom line: it depends on usage. Josh used it continuously, and his batteries were dead after a day. Make a usage profile.

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