How to calculate mAh draw from a battery?

Question

I'm trying to calculate the "drained" capacity from the battery each second so I can get the nearest possible capacity used.

This is how I'm getting this.

• dischargingCurrentSum - Is the sum of the "current" which I get every second, milliamps-second

For example

• 1s = 300mA
• 2s = 380mA
• 3s = 540mA

When I sum this it's 1220mAs

• dischargingUpdateCount - Is the count of entries, to be precise each second is one entry, in this case, it will be 3, it helps me to determine the arithmetic mean.

• currentTimeUnix - This is the current time, it the exact same time of calculation in milliseconds (each second)

• dischargingStartTime - This is the time when discharging started also in milliseconds

• /1000 - is to convert milliseconds to seconds

• /3600 - is to get the mAh

Formula:

drainedMah = dischargingCurrentSum / dischargingUpdateCount * (currentTimeUnix - dischargingStartTime) / 1000 / 3600

I'm not sure what I have missed here, at the first time it counts normally, but after a while, it became way more than the actual battery capacity is.

Answer 1

And when I sum this it's 1220mA

More precisely, since you are taking the measurement every second you have 1220 mA seconds. Capacity is measured in units of current times time -- for instance, milliamp hours. A typical AA battery has a capacity of around 2000 mA hours.

1220 mA seconds equals 0.34 mA hours.

drainedMah = dischargingCurrentSum / dischargingUpdateCount * (currentTimeUnix - dischargingStartTime) / 1000 / 3600

I'm not sure what's going on in the formula.

Suppose you take a current reading in milliamps every second and let \$S\$ be that sum of all of the current readings. Then

• \$S\$ is the total charge extracted from the battery in milliamp-seconds
• \$S/3600\$ is that value in millamp-hours
• \$S/3600/1000\$ is that value in amp-hours

In response to your comment...

If you take one reading per second, the average discharge current is either:

• sumOfAllReadings / timeElapsedInSeconds, or
• sumOfAllReadings / numberOfReadings

The formula, as it is written, is multiplying by elapsed time, not dividing by it.

Note this will produce a current which has units of amps, not a capacity.

Answer 2

When you take a current sample each second, and plot the sample on a graph against time, then the area under that sample is the current measured times one second. This is a rectangle and all you are doing is adding up the area in each sample rectangle which approximates the discharge of the battery in terms of ampere seconds. If you know the sample time is one second then you don't need a timestamp to find the discharge time - the length of records in a file is the elapsed time in seconds.

In comment the time change is not constant so the area approximation depends on the difference between respective timestamp values.

Scroll down and review this reference which describes the use of finite difference methods to approximate the area under a curve:

https://openstax.org/books/calculus-volume-1/pages/5-1-approximating-areas

In your application the most accurate approximation would be to calculate a rectangular area for each current sample and timestamp difference and then add up those areas and then convert to ampere hour. You could take an average over any span of time, however, it will be no more accurate than the sum of actual sample data unless you think there is noise in the samples that must be compensated by averaging. Taking an average over any time window you choose may be useful for statistical analysis but would not improve precision of the area approximation. Accuracy and precision are determined by the sample equipment and sample rate and presence or absence of errors in measurement.