How to calculate battery life if mAh and current draw of device is known

[211]

I'm trying to find a suitable coin cell battery to run a small PIR motion detector module.
The specs on the module say it operates on 4-15V @ 45uA (microamps) when no signal is detected (standby), and 20mA when it senses motion and sends a logic output.
Since this operates at just above the standard 3 Volts that most coin cell batteries are I'm looking at this 6V battery from Digikey, capacity rating on this guy is 165mAh @ 30µA.

My question is, at 45uA how long will this battery last? I expect to see motion only a dozen or so times a day so the majority of it's life will be in standby mode.

 

[cobra1]

165mAH = 165,000uAH. if you have a drain of 45uAH, then its 165,000uAH / 45uAH = 3666.6 Hours

Or approximatly 5 Months.

There will most likely be other drains on the battery that have to be considered but for what you have given above this would be your answer.

 

[alec_t]

I'm guessing the capacity will be proportionally less if you draw 45uA continuously (as it is for sealed lead acid bats), so let's say 165*30/45 = ~110mAh. Not knowing how long the 20mA draw lasts makes precise calculation impossible, but if we guess 5secs at 10 times a day then the average sensed-motion current is 5*10*20/(60*60*24)mA, i.e. ~ 12uA. So the battery life should be 110*1000/(45+12) = ~1900hrs = ~80 days.

 

[carbonzit]

If you can estimate the amount of time the detector will actually detect something, you can get a more accurate number. Let's say it's activated 5 percent of the time: your consumption is then

i = n (20mA) + (100-n) (45µA)​

where n is the percentage of time it's on.

 

[211]

Impossible to say how long it will be in 'detection' mode for the life of the battery. I do know the detection dwell time and output signal is less than 1 second for each occurance (couple dozen occurances a day???).
I'm basically going off the standby time. From there I can deduct an estimated percentage to account for the time it will be in detection mode.

Based on the guesstimate above I was hoping to get more than 5 months. Hoping for a year or more.

 

[carbonzit]

5 months is the best-case time, which includes zero activation time. Each activation will shorten that time somewhat. (20mA is almost 450 times 45µA.)

 

[CraigHB]

You could use these batteries and get a year a half on standby. Just need a PCB mount socket that can hold two of them like this.

 

[ronv]

Coin cells have high internal resistance so may have trouble supplying the 20 ma without a large voltage drop. If you can spare the space I would look for something like a AAA type or a couple of lithium ion types.