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Is it possible to work out the current or power a device is drawing/using, based on the following information:

  • Maximum capacity of a battery (48 Ah)
  • A table of voltage readings over time (starting at ~13v ending at ~11v over a period of 40 days, sampled every day at 2pm, ignoring the end part of the test where the battery voltage drops off non-linearly)

I am doing this to try and estimate how long the device would work on batteries of different capacities to the one tested, and I have a feeling it's not as simple as just dividing the battery's rated ampre-hours by the number of days it lasted and extrapolating...!

Thanks in advance for your help.

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    \$\begingroup\$ Why not measure the actual current? What battery technology? How acurate do you want to be? \$\endgroup\$ – Grant Nov 5 '14 at 12:25
  • \$\begingroup\$ I could measure the actual current, but I'm trying to find out if it's possible from just this information for now. Sealed lead-acid battery. I want to be able to work out how long the device will last on different capacity batteries of the same tech to the nearest day. \$\endgroup\$ – Dunstan Langrish Nov 5 '14 at 13:32
  • \$\begingroup\$ However you end up measuring the capacity, also consider things like environmental conditions such as temperature. In general, temperature tends to accelerate chemical reactions (such as that in a battery), so if you know the highest temperature you would expect this system to exist in, you could find (Theoretically) a maximum battery life at high temperature, which would be much fewer Ampere-hours than that at ambient temperature (i.e. 25 degrees C). \$\endgroup\$ – cowboydan Mar 20 '15 at 22:07
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Measure the current directly. Put a low ohms value resistor in series with the load and then measure the voltage drop across this resistor. Trying to do this by the schemes you are trying are at best an experiment.

You are in selection mode not high volume production mode. A few extra components and a little extra circuitry to do it the direct and correct way is the best. The cost is inconsequential.

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  • \$\begingroup\$ Thanks for the reply! Current is 28 mA. How low is low ohmage? a few milli-ohms? ~1 ohm? \$\endgroup\$ – Dunstan Langrish Nov 5 '14 at 14:16
  • \$\begingroup\$ Low value of the current measurement resistor is determined by several factors. The main one is that you select it low enough so that the voltage drop across it is not so high as to significantly affect your load. Too big a voltage change at your load can change its current draw and if too much could cause some loads to malfunction. On the other hand if the current shunt resistor is too small in ohms you may have difficulty measuring the voltage across it. In some cases it becomes necessary to add a opamp circuit to boost the voltage (continued) \$\endgroup\$ – Michael Karas Nov 5 '14 at 16:30
  • \$\begingroup\$ (continued from above) drop across the sense resistor to the point that it can be measured with reasonable resolution. So you can see it becomes a trade off in the selection. The third factor to consider is what resistors that you may already have available. I've been known to take 20 quarter watt axial lead resistors and twist them into a parallel bundle to create a quick low ohm sensing resistor for measurements like this. \$\endgroup\$ – Michael Karas Nov 5 '14 at 16:34
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In principle you can, but you need to know the discharge rate of the battery depending on the load. Also, its series resistance will cause a drop that depends on the load current, so you will need to compensate for some factors (ISR, discharge rate) in order to get a reasonably accurate estimation.

Note that the battery capacity will change depending on the load (not only the duration, the actual capacity). You can see several rate-of-discharge graphs, and that should give you an idea.

If you have a fairly constant load or an accurate enough model, you can indeed predict the performance of your battery depending on voltage measurements. That's what I suppose is done in mobile devices to estimate battery lifetime.

If you can measure the voltage and current together, your estimation will be a bit more accurate. Also, a higher sampling frequency will help. Do you have any reason to limit it to once a day?

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  • \$\begingroup\$ Cell phones use much more advanced algorithms like coulomb counting. \$\endgroup\$ – ACD Nov 5 '14 at 13:17
  • \$\begingroup\$ How do I find out discharge rate? I am fairly sure the load is reasonably constant. I probably didn't make it clear, but the test has already been done (for a different purpose), I'm just trying to see if I can work out something else from the same data. How will better resolution help? Over the period I quoted the line is nearly perfectly straight, so if I just take it's average slope it's -0.05 volts/day. Any idea how I actually apply the knowledge that this device discharges a 48 Ah battery @ -0.05 volts/day to work out how many days it will last on a battery of a different capacity? \$\endgroup\$ – Dunstan Langrish Nov 5 '14 at 13:39
  • \$\begingroup\$ There are quite a few other parameters that will affect the usable energy in the battery - self discharge rates (since we are measuring in days or weeks here) and temperature (if it's not in a temperature controlled environment) \$\endgroup\$ – Grant Nov 5 '14 at 14:19
  • \$\begingroup\$ @DunstanLangrish, the best estimate of the load, given the information you have is just 48 Ah / (40 days * 24 hrs/day) = 50 mA. It would sure be nice to measure it, if that is possible. \$\endgroup\$ – mkeith Nov 6 '14 at 19:10
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First I have some comments / caveats.

Batteries have self-discharge, and discharge due to external load. Also, the capacity depends on the load. Larger loads will give you less Ah than smaller loads. But as the load gets smaller, at some point the capacity becomes nearly constant. And your load qualifies as very small, since the battery was able to support it for 40 days. Lithium and NiMH battery capacities are slightly less dependent on load than lead acid batteries, but all of them are in the "small load" category when discharged over 40 days.

However, in a 40 day period, depending on battery type, there may have been some significant self-discharge. I assume it is a lead acid battery, but what is the subtype? flooded, or gel cell, or absorbed glass mat (AGM)? The latter two have relatively lower self-discharge.

I told you that to tell you this: Very often, if self-discharge is negligible (not sure in this case), and the load is small, you can absolutely just estimate the life of the second cell based on the capacity of the first cell (assuming same type or subtype). In other words, if you get 40 days from 40 Ah, you will very likely get 60 days from 60 Ah, and 20 days from 20 Ah as long as the battery type is the same.

One last thing: self-discharge has a strong dependency on temperature. A battery which will hold 80 percent of its charge for a year at 25C, may discharge completely in 3 months at 60C. Those may not be the real numbers. I am kind of making that up for illustration purposes. But it is very dramatic, and applies to many battery types.

-McKenzie

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The Amp-Hour rating of any battery is provided by the manufacturer and reported as a result following a very specific test documented within the datasheet, typically this tests consists of a constant load over time. If your load is not constant (a periodic, pulsatile load), the Amp-Hour rating provided by the manufacturer should be viewed as guidance to a ball park estimate of its performance.

Since you said the test has already been done, and you are trying to extract meaningful data from old logs, you will need to identify the load which contributed to the readings overtime. If you still have control of taking new data I would suggest to apply a small series resistor and measure the voltage drop across of it. This will provide you with Current and Power information.

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Maybe this isn't what you're asking, but if you have a battery with a given ampere-hour capacity and - with a constant load resistance - you know how long it takes it to go from fully charged to fully discharged, to a first approximation a battery using the same chemistry but with twice the capacity serving the same load will last twice as long, one with half the capacity will last half as long, and so on...

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The right way to do this, is to use a Coulomb (charge) counter. This circuit is available as IC and typically is very tiny, leading to a very small foot print.

The rationale is that most likely there will be spikes in current consumption (especially if the load is some uC/uP which alternates between states of high activity and low power idleness).

So you will most likely miss key events and underestimate the battery current, if relying solely on sampled values.

The Coulomb counter performs integration over time of the charge transferred, hence allowing for low frequency sampling. All you need is a stable time base.

Then the reads can also happen a-periodically, as long as you keep track of the time passed inbetween successive reads.

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TRY a hall effect current sensor to drive the coloumb counter this will give good accuracy AND good dynamic range AND low volt drops If your currents are lower than the halls rating by a factor of 2 or more then run 2 turns etc

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A lot of answers here are saying that you should include some current measurement or coulomb counting, which doesn't seem to answer the question, in my opinion. If you only have periodic voltage measurements and the load current is small, you can approximate the state of charge of the battery with a SOC-OCV (state of charge - open circuit voltage) graph. You can probably find this graph for whatever chemistry battery you have and find the SOC that corresponds to a given voltage.

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