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Topic: Lowest Arduino Power consumption (Read 4043 times) previous topic - next topic

tim7

I took a quick look at some datasheets for a few 3V lithium coin cells.  The ones I looked at quoted self-discharge rates around 0.1uA.

Duracell claim a self-discharge rate equivalent to 10uA in their Alkaline AA cells.

Wrend

#16
Oct 21, 2012, 06:45 pm Last Edit: Oct 21, 2012, 06:48 pm by Wrend Reason: 1
For low self discharge NiMHs, check out Eneloops. They're the only AA and AAA cells I use anymore.

For a cell that you want to last at least several years, the lithium primary ones are probably the way to go.

dhenry

Quote
I could get a 16MHz 5V Arduino down to approx 1uA.


If you could get the avr to run on a 16Mhz crystal with less than 1ua, you should contact Atmel, :)

Getting an avr to run below 50ua is easy, even on a crystal. Getting it to run below 10ua on a crystal is difficult. Getting it to run below 1ua on a crystal is out of this world.

dhenry

Quote
Shouldn't those numbers include time?  0.1 uA per second / day / year?


:)

Should those numbers include time?

Wrend

#19
Oct 21, 2012, 11:02 pm Last Edit: Oct 21, 2012, 11:08 pm by Wrend Reason: 1
They already do include time.

http://en.wikipedia.org/wiki/Ampere

Coding Badly

#20
Oct 21, 2012, 11:24 pm Last Edit: Oct 22, 2012, 12:23 am by Coding Badly Reason: 1

Assuming I did the math correctly this time, Rayovac AA Alkaline batteries have a self-discharge of 8.676 uA.

http://www.rayovac.com/Technical-OEM/~/media/Rayovac/Files/Product%20Guides/pg_battery.ashx

Wrend

#21
Oct 21, 2012, 11:52 pm Last Edit: Oct 21, 2012, 11:54 pm by Wrend Reason: 1
That seems too low.

I think the new 1800 cycle Eneloops, for example, average about 13.7µA, loosing 30% charged capacity (600mAh) after 5 years since they were charged.

Nick Gammon


Should those numbers include time?


The time component cancels out. For example:

Code: [Select]

Type        Capacity mAH  Discharge %/month   Self discharge (uA)

NiCD AAA         350             20                 98


If it discharges 20% in a month then it discharges (20 / 100) / (24 * 30) in an hour (assuming a 30-day month).

So, in milliamps:

Code: [Select]
350 * (20 / 100)  / (24 * 30) = 0.097

Units are:

Code: [Select]
(mA * hours) * (1/months) / (hours/month) = mA

Effectively, it represents an instantaneous current usage of 0.097 mA (97 uA).




To re-arrange the figures, assuming we draw 97 uA then this is how long it will last:

Code: [Select]
0.097 * (24 * 30) * 5 = 349.2

Units are:

Code: [Select]
mA * hours/month * months = mA * hours

Please post technical questions on the forum, not by personal message. Thanks!

More info:
http://www.gammon.com.au/electronics

Coding Badly

#23
Oct 22, 2012, 12:07 am Last Edit: Oct 22, 2012, 12:22 am by Coding Badly Reason: 1

Starting capacity: 2535 mAh
Capacity remaining after 1 year: 97%
Ending capacity: 2458.95 mAh
Difference: 76.05 mAh

Days in one year: 365.24
Hours in one year: 525,946  8,766

76.05 mAh / 525,946 h = 0.000144597 mA = 0.145 uA

76.05 mAh / 8,766 h = 0.008675802 mA = 8.676 uA

tim7


Quote
I could get a 16MHz 5V Arduino down to approx 1uA.


If you could get the avr to run on a 16Mhz crystal with less than 1ua, you should contact Atmel, :)

Getting an avr to run below 50ua is easy, even on a crystal. Getting it to run below 10ua on a crystal is difficult. Getting it to run below 1ua on a crystal is out of this world.



We're talking about the power saving modes, in which the CPU is stopped.  What are you talking about?


Wrend

525,946 hours in a year / 24 = 21,914 days in a year. (About 60 years.) 8)

Coding Badly


No doubt, ouch!  This is a bit more reasonable...

Starting capacity: 2535 mAh
Capacity remaining after 1 year: 97%
Ending capacity: 2458.95 mAh
Difference: 76.05 mAh

Days in one year: 365.24
Hours in one year:  8,766

76.05 mAh /  8,766  h = 0.008675802 mA = 8.676 uA

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