How much time the load would take to drop 1 volt of source battery?

Question

I am trying to figure out how much time a load takes to drop 1 V of the power source. I have a load (vehicle tracker) of 12 V 0.1 A (as per the datasheet) and a car battery of 12 V whose Ah I don't know but let's assume it is 40 Ah.

My take is that it will take 200 hours to drop 1 V (i.e. 200 hours to bring 12 V to 11 V). Here is how:

40 A load  -> battery lasts for 1 hour
1 A load   -> battery lasts for 40 hours
0.1 A load -> battery lasts for 400 hours (after 400 hours, battery is fully discharged)
Assuming at discharge, the battery's voltage is 10 V. So the time is, 400/(12-10)=200 hours for 1 V drop.

Now, am I on the right track? If yes, then here comes the tricky part. The load has a power save mode in which it consumes 70 mA and it enters in this mood after 25 minutes. What would be the time now? The math gets dirty here and this is where I don't know how to proceed. It even has a third mode (sleep mode) but let's forget that one. Does C-rating has to do anything with the calculation?

Answer 1

You are making some assumptions that are reasonable enough to get you into the right ballpark, but not good enough to make precise calculations.

Assumption 1) 40 Ah means 1 hr at 40 A or 400 hours at 100 mA

No. Capacity is not constant with discharge rate. Lead acid batteries in particular, and most batteries in general, provide a higher capacity if you discharge them slowly. That battery is likely to have been rated at the 20 hour rate, so 2 A for 20 hours. At 100 mA, it should take a bit more than 400 hours to fully discharged, with a new battery that is, but I don't have a factor for how much more, probably less than 2x. If you stick with the 400 hours, that will be conservative.

Assumption 2) 400 hours to drop down to 10 V means 200 hours to drop to 11 V

No. You are making the assumption that voltage versus charge state is linear, which it isn't. It tends to hold up for longer than you would expect in the middle voltages, and then drop quickly at the end of discharge. You would need to look at the manufacturer's discharge curve, or do some experiments, to get it exactly, but if it's 400 hours to 10 V, you would expect the time to 11 V to be nearer to 400 hours than to 200 hours. Again, your error is conservative.

Assumption 3) dividing by 70 mA is trickier than dividing by 100 mA.

Not really, I use a calculator. The difference between 70 mA and 100 mA on how long the battery will last is smaller than your other errors on discharge curve and capacity constancy. The most practical thing to do would be to compute an upper and a lower bound, one at 70 mA and one at 100 mA, and say the correct answer is somewhere between those.

If it was worth making an exact calculation, then you would need to know how often it entered the power save mode, to know how long it spent in each mode on average. That would give you an effective average current between 70 mA and 100 mA, and a new lifetime. As I said, the difference between 70 mA and 100 mA is almost not worth bothering about. Do the sums for 100 mA, and treat power save as a bonus.

Now if sleep mode was much lower consumption than 70 mA, say 1 mA or 100 uA, that would be worth treating properly. You handle it the same way, how much time does it spend in sleep, power save and full power, on average? Compute an average current, and calculate with that. That might take you out to thousands of hours, where you start worrying about the self-discharge rate of the battery.