Simple (for some) maths problem that has been denting my damaged brain all day :-
An electronic device uses 0.001 mA when asleep.
It sleeps for 58 seconds.
Then it wakes up and consumes 15 mA for a total of 2 seconds.
After that the cycle repeats, i.e. it returns to using 0.001 mA for 58 seconds etc
In 1 hour, how many mA has it consumed ?
The need is to calculate the life of the battery, which is given in mAH (between 30 and 250, depending on the battery).
The consumption figures above are arbitrary - the actual consumption has to be measured in each state.
It's the Maths that are baffling at the moment.
I think i have a solution but the results make no sense.elementcollector1 - 18-10-2016 at 11:58
Total current consumed per minute = 15 mA + 0.001 mA = 15.001 mA
Total current consumed per hour = (current/minute)*(60 min/hr) = 900.06 mA, or about 0.9 A wg48 - 18-10-2016 at 12:28
You are both need to swot up on charge/coulombs.
In this case you need the charge measured in mAH (ie mA x hours) thats the integral of current wrt time.
So in one minute the charge is (0.001 x 58)+(15 x 2) = 30.058mA.s
Then x 60 for the charge in an hour and divide by 3600 to convert to mA/h
Oops, I have corrected my error with the units hopfully its correct now
[Edited on 18-10-2016 by wg48]aga - 18-10-2016 at 12:52
Total current consumed per minute = 15 mA + 0.001 mA = 15.001 mA
Total current consumed per hour = (current/minute)*(60 min/hr) = 900.06 mA, or about 0.9 A
I think i got an answer the same early on today.
That can't be right as it's only using 0.001mA for most of the time.Cryolite. - 18-10-2016 at 13:15
I believe you are confusing units here. An ampere is a unit of current, but to measure the capacity of a battery, you need the corresponding unit of
charge, the amp-hour. The quantity "mA per hour" is not really a well-defined quantity in this context: amperes already measure the "flow rate" of
electrons past a given point, and mA/h to me indicates something along the lines of "change in current draw per hour", similar to the relation between
velocity and acceleration.
What I think you are asking for is the average current draw of such a gizmo. You can then multiply this value (in miliamps) by the number of hours you
want the battery to power the device to get the required charge (in mAh) that the battery must hold. (Of course, the voltage of the battery must match
the operating voltage of the gizmo.)
This is easy to do. The device operates on a 60-second cycle, so we can just compute the total amount of charge drawn over a minute and divide this by
60 seconds to get the average current draw over that one-minute window.
In our 60 seconds, we spend 58 seconds drawing 0.001mA (for 0.058 mAs of charge drawn), and 2 seconds drawing 15 mA (30 mAs of charge drawn). Thus
over the entire 60-second window the device draws 30.058 mAs of charge, for 0.501 mA average power draw. If you want the battery to last x hours, you
consequently need 0.501x mAh of charge capacity in the battery.Marvin - 18-10-2016 at 13:37
So in one minute the charge is (0.001 x 58)+(15 x 2) = 30.058mA.minute
You mean 30.058mC per minute.
There are a few ways to do this, probably the easiest to work with is average current.
0.001*58/60 + 15*2/60 = 0.5ma and change. Magpie - 18-10-2016 at 13:54
I think wg48 is correct now in edit and agrees with Cryolite.aga - 18-10-2016 at 14:27
Ah feck.
I've been through several 'answers' today and wg48's agrees with what i eventually thought might be the 'right' answer.
Experimentation will certainly work, but if it comes out as a few days with a 250 mAH then the project dies right there.
This is part of some actual IoT.
I took a break from cramming the encryption code + application code into 2048 bytes to take a look at device life-expectancy.
If wg48 is right, the battery lasts for a very short time if the device transmits every 30 seconds.
36 days is a bit short of the required 3 years on a CR2032 battery.Cryolite. - 18-10-2016 at 14:46
3 years is a very long time, but half a milliamp is a minuscule amount of current. If you can guarantee constant access to sunlight, maybe you could
use a small solar cell and a lithium polymer battery?
A typical CR2032 cell is around 225 mAh in capacity-- this hypothetical device would last 450 hours on the battery, under 20 days. aga - 19-10-2016 at 00:01
Thanks all for your help in this.
Seems that with the the actual timings/currents the poor thing will last just 18 days
