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Author Topic: Lowest Arduino Power consumption  (Read 2285 times)
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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.
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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.
« Last Edit: October 21, 2012, 11:48:45 am by Wrend » Logged

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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, smiley

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

smiley

Should those numbers include time?
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They already do include time.

http://en.wikipedia.org/wiki/Ampere
« Last Edit: October 21, 2012, 04:08:26 pm by Wrend » Logged

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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
« Last Edit: October 21, 2012, 05:23:35 pm by Coding Badly » Logged

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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.
« Last Edit: October 21, 2012, 04:54:32 pm by Wrend » Logged

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Should those numbers include time?

The time component cancels out. For example:

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

Units are:

Code:
(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:
0.097 * (24 * 30) * 5 = 349.2

Units are:

Code:
mA * hours/month * months = mA * hours

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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
« Last Edit: October 21, 2012, 05:22:54 pm by Coding Badly » Logged

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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, smiley

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?
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Hours in one year: 525,946

Ouch!
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525,946 hours in a year / 24 = 21,914 days in a year. (About 60 years.) smiley-cool
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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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