I have a circuit that consumes 1.6μΑ during sleep mode and 40mA during awake mode. The awake period lasts for 3 seconds and the circuit is on awake mode every 30 minutes. The circuit is powered by this battery (the link leads to the datasheet).
I used this battery life calculator because i do not have the knowledge to do the maths on my own.
I know that as the circuit working the capacity of battery is decreasing as well as the voltage. But as the battery voltage is decreasing there are a few components that cant work under 3.0 volts.
I am trying to say that there is point where the battery will be capable of feeding power the circuit but some components are not going to work because the voltage will be lower than their voltage threshold.
For example, if i have a battery at 8500mAh 3.6V and at 2.9V are still left 1000mAh i dont care because my circuit doesn't work at 2.9V.
Is the rest capacity "wasted"?
Is that thought wrong?
Does the attached calculator take care about that too?
Because i didn't see in the equations nothing about voltage.
Many thanks
Best Answer
This isn't that hard that you need an on line calculator.
First determine the average current consumption:
In sleep mode, the current (1.6 uA) is so much smaller than the wake mode current that we can assume that this 1.6 uA is used 100% of the time.
In wake mode we have 40 mA for 3 seconds every 30 minutes = 30 * 60 = 1800 seconds. That is a duty cycle of: 3 / 1800 = 1/600 = 0.00167 So those 40 mA peaks average out to: 0.00167 * 40 mA = 66.7 uA
Total average current consumption: 1.6 uA + 66.7 uA = 68.3 uA
Now we look in the datasheet to see what battery capacity that gives us given a discharge to 3.0 V.
The nominal capacity of this battery is 8.5 Ah. Let's look in the graphs if that is the value we can use. Battery capacity decreases with increasing load current but the load current for that 8.5 Ah is 4 mA, a lot less than the 68.3 uA we need. So yes, we can use the 4 mA value, our 68.3 uA is so small that the battery's capacity is not deteriorated by it.
From graph 1 we can determine to which voltage the battery is discharged in the capacity test. All curves are quite flat above 3.0 V so when the voltage is below 3.0 V then the battery is quite empty already. So that 8.5 Ah is valid for discharging to 3.0 V.
So let's continue with that 8.5 Ah. 8.5 Ah means the product of current and hours is 8.5 Ah. So: 8.5 Ah / 68.3 uA (from above) = 124451 hours = 5185 days = 14 years !
In practice batteries for long life applications are often only guaranteed for a 10 years lifetime. The 14 years exceeds that. So when used in a product you should instruct the user to replace the battery every 10 years for optimal performance.