Fred, Bob and Linda have built an an electronic device. The gig runs in 3 minute cycles, it stays awake for 0.4s then sleeps for the remaining time in the cycle, that is 179.6s and then it starts again 420 times a day.
They measured that the gig draws 6mA in active mode and 4.3µA in sleep mode.
They are given a battery rated at 800mAh (that is considered to be "perfect" for this exercice)
They are being asked what will be the autonomy?
Fred's approach:
You have 24h * 60 / 3min = 480 measurement cycles per day.
Each measurement lasts 0.4s / 3600 = 0.00011 hours.
so in awake mode the gig draws 6mA * 0.00011h * 480 = 0.3mAH per day.
and in sleep mode (the rest of the time), it draws 0.0043mA * 23,947h = 0.1mAH per day.
Total = (0.3 + 0.1) = 0.4mAH.
so with an 800mAH battery: 800 / 0.4 = 2000h = ~66 days.
➜ Fred says 66 days.
Bob's approach:
Per cycle:
Consumption in awake mode = 6 mA × 0.4 s = 2.4 mA⋅s.
Consumption in sleep mode = 0.0043 mA × 179.6 s= 0.77 mA⋅s.
Total Consumption: 2.4 + 0.77 = 3.17 mA.s.
Average consumption per second = 3.17mA.s / 180s = 0.0176mA.
Average consumption per hour = 3600 x 0.0176mA = 63.4mA.h.
Battery autonomy = Battery capacity / Average consumption per hour.
➜ For an 800mA/h battery, we'll have 800 / 63.4= 12.61 hours.
Bob says ~12h30
Linda's approach:
800mAH theoretically means the battery can supply:
800mA for 1 hour,
400 mA for 2 hours,
200mA for 4 hours,
...
6mA for 133 hours, which is 478,800 seconds.
You take 480 measurements of 0.4s each day, meaning you are awake for 480 x 0.4s = 192s in a day.
Neglecting standby consumption, we would have 478,800 / 192 = 2493 days, so probably a bit less when not neglecting standby.
Linda says ~ at least 5 years (the maths says 6.8 years).
So the results are :
Fred says 66 days.
Bob says ~12h30
Linda says ~ at least 5 years.
That's quite a difference...
QUESTION:
➜ Is one of them right and if so where is the reasoning mistake for the other 2 (or are the 3 wrong and you have another solution ?)
The Fred's and Bob's approach both contains confusion with units that made their calculations incorrect.
Fred's error
Since all this figures are "mAH per day" (??),
so with an 800mAH battery: 800 / 0.4 = 2000 not hours, but days
Bob's error
mA * s divided by sec will be mA. It is not a consumption, but a current, average current.
The current 0.0176mA get us consumption 0.0176mA * s per second, 0.0176mAH per hour and 0.0176mA*day per day.
For an 800mAH battery, we'll have 800 mAH / 0.0176 mAH = 45 454 hours or about 1894 days
The Linda's approach doesn't contains gross mistakes, it's just not very accurate.
All three calculations (after correction the tipos) give us approximately the same time - about 2000 days or about 5 years.
Using unsigned longs with working units as microamps and microseconds there would be plenty of room to lose roundoff places... the numbers will be bigger and battery self-drain won't look so small.
just saying. Microamp-microseconds power!
When i think of the speed of light, per microsecond makes it easier to grasp for me. I know what 300m looks like and 299.792485 is pretty close!
I would like to ask for the Battery related missing data?
1. Is is Lead-Acid? 2. Ni-Cd? 3. Or any other type? 4. How many cells are there? 5. What is the terminal voltage? 6. What is float volatge if type 3? 7. Link to manufactuer's specifications
From my experience of working with industrial battery plant:
Lead-Acid Battery: Float Volt = 2.2V/cell, Final volt = 2.0V/cell.
Ni-Cd Battery: Float Volt = 1.2V, Final Volt: 1.1V
Final volt is the value at which point, the battery must be stopped dischrging and recharging should start immediately to avoid boost charging to retain the normal useable life of the batter pack.
If I know the Terminal voltage and the type of Battery, the MCU type being used by gig, then I can tell you who is the actual winner; otherwise, solution of post # 4 is dubious.
I assume:
Arduino GIGA R1 WiFi with 3.3V Operating voltage.
9V Ni-Cd Battery, 8000 mAh capacity.
Final volt of Battery: 8.25 V //Float/Final Volt = 1.2V/1.1V per cell
Sytem has trip circuit to cut power to GIGA when Battery arrives at Final volt.
Estimated Current (from 6 mA active and 4.3 uA sleep)
Average current consumed by GIGA = (6 mA + 4.3 uA)/2 = 3002.65 3.00265 mA.
By definition: 8000 mAh capacity Battery will take 1 hr time to arrive at final voltage if 8000 mA current is drawn.
So, by simple calculation, it can be shown that the GIGA will run only for about 23 minutes.
Run Time = 1 hr/(8000)*3.00265 ~= 1.35 seconds
If someone wants the Battery to be fully exhaused (Battery will not regain its life) and be allowed to arrive at terminal volt 3.3V, then the ---
Run Time = (1.35/0.1)*(9.0 - 3.3) ~= 77 seconds
My answer/calculation is based on Rule-of-Thumb and subject to experimental verification.
Now that it is drawing less current than in your original calculation, it discharges the battery much faster.
Doesn't that ring alarm bells in your head?
I'd calculate it this way:
Time(hours) = 8000mAh/3.00265mA = 2664.3h
2664h/24 = 111days
