I'm currently working on a project that involves monitoring the remaining battery charge percentage of a device.

specifications of the Forte ER34615 battery:

1. Chemistry: Lithium-thionyl chloride (Li-SOCl2)
2. Nominal Voltage: 3.6V
3. Capacity: Typically around 19,000 mAh (milliampere-hours)

Remaining charge percentage = (Current capacity / Maximum capacity) * 100

Assume the load of your choice

the typical maximum capacity for the Forte ER34615 battery is around 19,000 mAh (milliampere-hours).

I am not sure I am using correct formula and i don't understand how can we get current capacity.

 

I'm currently working on a project that involves monitoring the remaining battery charge percentage of a device.

specifications of the Forte ER34615 battery:

1. Chemistry: Lithium-thionyl chloride (Li-SOCl2)
2. Nominal Voltage: 3.6V
3. Capacity: Typically around 19,000 mAh (milliampere-hours)

Remaining charge percentage = (Current capacity / Maximum capacity) * 100

Assume the load of your choice

the typical maximum capacity for the Forte ER34615 battery is around 19,000 mAh (milliampere-hours).

I am not sure I am using correct formula and i don't understand how can we get current capacity.

You need to use a technique known as a Coulomb counter. It is a current-controller oscillator which counts down from full capacity. That's what is in most of the "fuel-gauge" ICs

 

The datasheet for the expensive Chinese battery has no details except its capacity is rated at a very low current of only 2mA and when its 3.6V has dropped down to only 2V. Its maximum continuous current is rated at only 200mA.

 

You need to use a technique known as a Coulomb counter. It is a current-controller oscillator which counts down from full capacity. That's what is in most of the "fuel-gauge" ICs

First of all I want to know whether the formula which I am using is correct or not. ? @Audioguru again

If it's correct than i am interested to know how can we get current capacity.?

 

The math fractional result is correct but your simple formula gives a number that's only as accurate as the numbers given to it.

https://www.analog.com/media/en/technical-documentation/technical-articles/p356_en-soc.pdf
Know a Battery’s State-of-Charge By Counting Coulombs

 

There may be some confusion here over the term 'current capacity' It is a reference to the remaining charge available; not to the flow of charge (electrical current).

 

There may be some confusion here over the term 'current capacity' It is a reference to the remaining charge available; not to the flow of charge (electrical current).

Would "present capacity" be a better choice of phrase?

 

The rated capacity occurs only with a tiny 2mA load and when the battery voltage has dropped down to 2V.
Is 2mA and 2V good enough for powering your circuit?

 

To perhaps clarify a "coulomb counter" measures the current into or out of the battery and multiplies by the time, to estimate the mAh left in the battery.
It does not look at the voltage to make the estimate.

 

Hmm... what you seem to want to know when you say "current capacity" is the SoC (State of Charge).

Different cell chemistries allow different methods for determining SoC. Some allow a pretty good estimate based on one terminal voltage. Oddball gaseous hydrogen cells sometimes used in satellites are particularly easy because the internal pressure is directly proportional to SoC.

But you have a lithium cell. It is impossible to reliably know SoC for a lithium cell without accounting for all the charge that went in and how much has come out. That's what a coulomb counter is all about.

A coulomb (C), is the SI unit for quantity of charge. It is defined as the amount of charge that moves past a point in a conductor in 1 second when the current is 1A.

A coulomb counter is the name for the device at the heart of a battery fuel gauge like the one in a laptop. The coulomb counter watches the charging process and counts the coulombs of charge that go into the battery. When the battery begins to discharge, it subtracts the number of coulombs leaving the battery from the total charge it saw go in. This is a close approximation of SoC.

It is not prefect due to losses and unknown charge/discharge. This is why for a laptop fuel gauge to be as accurate as possible, you must fully discharge the battery, then fully charge it. At that point the coulomb counter has a good accounting of the charge in the battery and can add and deduct as charge-discharge cycles happen.

So, to get the "current capacity" (say SoC, instead) you will need to use a coulomb counter. Fortunately options are abundant and some are very easy to use. You can get breakouts of counters from places like Adafruit and Sparkfun for learning and evaluation, or clones of those breakouts from the usual sources.

 

The OP is lucky to have picked a very linear (energy delivered per coulomb circulated) primary battery to monitor SoC where a simple a coulomb counter is effective. The details for tracking Flooded Lead Acid batteries SoC (where some coulombs are more equal than others) , with any type of precision, with dynamical load and charging profiles, between full charge reset points is a good way to explore exactly how batteries are not electronic devices with lots of nth degree deviations from linearity.

https://github.com/nsaspook/mbmc_k42/blob/master/mbmc_k42.X/bsoc.h
https://github.com/nsaspook/mbmc_k42/blob/master/mbmc_k42.X/bsoc.c

 

I am providing some values and interested to know how the battery performs within the voltage range of 3.6V to 3.3V.

Battery provide maximum 3.6 V.
Load require minimum 3.3 V and maximum 3.6V.
Battery discharge when it drop voltage from 3.6 to 2V.
load draw current of 24mA.

how the battery performs within the voltage range of 3.6V to 3.3V.
as the battery voltage decreases 3.6V to 3.3V during , does current delivered by the battery may also decrease? Does battery provide constant of 24mA?

 

how the battery performs within the voltage range of 3.6V to 3.3V.
as the battery voltage decreases 3.6V to 3.3V during , does current delivered by the battery may also decrease? Does battery provide constant of 24mA?

That depends entirely on the load.
If the load is resistive, the current will decrease proportional to the voltage.
If the load takes a constant power (for instance a LED with a switched-mode current regulator) the current will increase as the battery voltage decreases.

 

I am providing some values and interested to know how the battery performs within the voltage range of 3.6V to 3.3V.

Battery provide maximum 3.6 V.
Load require minimum 3.3 V and maximum 3.6V.
Battery discharge when it drop voltage from 3.6 to 2V.
load draw current of 24mA.

how the battery performs within the voltage range of 3.6V to 3.3V.
as the battery voltage decreases 3.6V to 3.3V during , does current delivered by the battery may also decrease? Does battery provide constant of 24mA?

Do the experiment and let us know what the results are.

 

The Chinese battery you found has a datasheet without detailed specs. Maybe because it has poor performance.
I found a German battery of the same size and chemistry that has detailed specs. Its voltage does not drop during its discharging time but its capacity at your high load of 24mA kills it after about 600 hours.