Calculating Amp/Hours over 24 hour Period

[Veramacor]

Hello again,

Mental Block time.  I have a PIC circuit that will draw 35 milliamps for 300 seconds out of every 24 hour day.  I put the PIC to sleep for the the rest of the time so It draws about .1 microamps so for the sake of making the calculation easier I am ignoring that for now.

I am trying to figure out battery life in days.  Is it correct to estimate the effective milliamp hour draw by doing:

60 seconds * 60 minutes * 24 hours = 86400 seconds in a day

86400 seconds - 300 seconds = 86100 seconds in a day the PIC is asleep (ignoring for now)

Spreading 35 milliamps over 24 hours:

300/86400 = .00347

.00347 * 35 milliamps = .121 milliamps or 121 microamps.

So is this a valid way to spead the millamps over a 24 hour period?  In other words,  instead of having a 35 millamp burst for 300 seconds,  can I say I have a constant current over a 24 hour period of 121 microamps?

If this is the case, If I had a 2000 millamp/hour AA battery, and I had a constant current draw of 121 microamps:

2000 ma/h / .121 ma = 16528.9 hours or 688 days.   FYI I'm using 2000 ma/h from the following website for a AA battery:

http://www.techlib.com/reference/batteries.html

[w2aew]

35ma for 300 seconds out of each day...

35ma * 300sec * 1min/60sec * 1hr/60min = 2.9mAhr consumed each day.

Assuming 2000mAhr is the right capacity, then the battery would last:
2000mAhr / 2.9mAhr per day = 685 days

Yeah, so you got it basically right.  However, you'll find that the amphour rating for batteries will be load dependent.  It will likely be greater than 200mAhrs.  But, given this long operational life, there will probably be self-discharge too.

[alm]

You appear to be confused about units. Battery capacity is measured in A * h, for example 2000 mAh. Your circuit draws 35 mA * 300 s = 10500 mA * s = 2.9 mAh per 24 h period, or an average current of 2.9 mAh / 24 h = 122 uA. The battery can deliver 2000 mAh, so it should supply 122 uA for about 2000 mAh / 122 uA = 16457 h.

As a first approximation this is correct. Other factors that you may have to take into account:
- self discharge, forget 1 year of battery life with regular (non-LSD) NiMH batteries, self discharge is higher than this.
- battery voltage will drop as the battery discharges, depending on your power supply you may be able to use more or less of this capacity.
- loss due to output impedance. Dave discussed this in his very early video about the HP 20b calculator (EEVblog #4). I wouldn't expect this to be an issue since 35 mA is a very small current for an AA cell.