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I wonder calculation of the battery life, depends of its voltage level. Basic calculation of battery life is given below:

Battery Life = Battery Capacity in mAh / Load Current in mAh

However in this formula we are not using any voltage levels. I know that formula is very basic formula, there has to be a constant (0.7 or 0.85), although we can't use any voltage level in this formulas. How we can read the datasheets to making exact calculation? For an example if I use 3.7 V battery, what is the lasting voltage? What if my limit is 2.7 V in my system, I mean that system will passive under 2.7 V, how will be the calculation? What if I use the limit as 2.4 V? In the basic calculation formula we couldn't use voltage level.

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  • \$\begingroup\$ This formula is complete on its own. There is no need of using voltage in it. You can use the voltage to guess the battery percentage as a related but separate process. \$\endgroup\$
    – paki eng
    Dec 22 '20 at 12:10
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    \$\begingroup\$ The mAh rating is already defined by discharging the battery from 100% full to 0% empty, whatever the battery chemistry and voltages for full and empty battery for given battery chemistr are. Different batteries have wildly diffrent voltage curves, so battery state of charge may not be predicted accurately from the voltage. \$\endgroup\$
    – Justme
    Dec 22 '20 at 12:18
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    \$\begingroup\$ Where did you get that incorrect equation? Load current shouldn't be in mAh. It should be in mA. Then the division gives a result in h ... which is the life. \$\endgroup\$ Dec 22 '20 at 12:32
  • \$\begingroup\$ The equation works fine if you have a constant current load. But for a DC-DC converter, the load may not be constant. Usually what people do is use the average voltage for the battery over its life, and use the average current for the load (at the average voltage). This is usually close enough. Otherwise you measure the battery life directly by running the load from full battery to end of life. \$\endgroup\$
    – mkeith
    Dec 23 '20 at 20:29
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Determining the remaining battery charge (State of Charge) from its voltage is tricky; the discharge curves aren't straight lines, and for some chemistries, the first part of the discharge curve is very flat and this makes the battery voltage almost independent of the battery's charge level.

Also, the discharge curves vary depending on temperature, age, and load current.

You can get a very rough estimate of remaining charge looking at a battery's voltage at a known, constant discharge current by checking the discharge curves in the battery's documentation and implementing a calculation based on the linear part of the chosen curve (if any).

If you need a prediction of State of Charge that is more-or-less accurate, you can use a Fuel Gauge IC, which keeps track of "what goes in and what comes out" and derives a State of Charge from that.

This Application Note provides further reading, should you want any.

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Actually the battery voltage levels are defined by the chemistry, not its charge-carrying capacity. For example, a lithium-ion battery (most likely what you are referring to as a 3.7 V battery) would be considered "100%" at 4.2 V open circuit voltage and "0 %" at about 3.5 V, though there may be disagreements on the exact values of these voltage levels. For a lithium-ion battery, you don't want to discharge it below 3.5 V because it will hurt its internal chemicals permanently.

With these numbers in mind, you can guess the state-of-charge of a battery by linearly relating 100 - 0 % to 4.2 - 3.5 V.

For different battery chemistries, the voltage levels are different. For example, some types of lead-acid batteries are used from 13.2 - 10.5 V. A nickel-cadmium cell is used from 1.25 - 1.1 V and so on. You can use these voltage ranges to approximately get the battery percentage from 100 - 0 %.

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  • \$\begingroup\$ 3.5 Volts is not the minimum voltage for Li-Ion or Li-Po batteries. I can use the battery in 2.5 Volts. There could be a system , for example CC430 uC uses 1.7 Volts as minimum and my battery can powers the system when it become 1.7 Volts. \$\endgroup\$ Dec 22 '20 at 12:02
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    \$\begingroup\$ If you use the battery at a voltage much lower than 3.5 V, it will get damaged soon. You will see that it wouldn't be able to hold much charge. \$\endgroup\$
    – paki eng
    Dec 22 '20 at 12:05
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    \$\begingroup\$ Discharging a Li-ion battery that far can damage it, and will at least reduce its cycle life. See electronics.stackexchange.com/questions/219222/… \$\endgroup\$
    – ocrdu
    Dec 22 '20 at 12:05
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    \$\begingroup\$ No doubt a lot of microcontrollers and FPGAs can be powered from a voltage as low as 1.2 V or even lower. But that doesn't mean that you have to get a Li-ion battery down to that voltage. The battery has its own personality and it doesn't like voltages below 3.5 V very much. There are easy ways in which you can use a lithium battery within it normal voltage range (4.2 - 3.5 V) to power up a device that uses much lower voltage (2.5 V or whatever you like). \$\endgroup\$
    – paki eng
    Dec 22 '20 at 12:09
  • \$\begingroup\$ In this manner could we say that manufacturers takes the lowest limit voltage as referance and mAh(capacity) is calculating with using this limit? As an example assume 3.5 Volts is a low limit and we have 3000 mAh capacity , do you mean in 3.5 Volts level all of the energy will be used? Or is there any recommendation to read datasheets get exact values? \$\endgroup\$ Dec 22 '20 at 12:14
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For an example if I use 3.7 V battery, what is the lasting voltage? What if my limit is 2.7 V in my system, I mean that system will passive under 2.7 V, how will be the calculation? What if I use the limit as 2.4 V? In the basic calculation formula we couldn't use voltage level.

The 'lasting voltage' (discharge cutoff voltage) is specified by the manufacturer. If you use a higher cutoff voltage the realizable capacity will be less. How much less? That depends on the particular battery and how it is used. If you use a lower cutoff voltage the capacity will be increased, but probably not by much and the battery could be damaged.

Battery manufacturers tend to rate their products with the most optimistic specifications that they can get away with, which often involves running the battery down to the lowest possible voltage that doesn't immediately destroy it. If they say 2.7 V, don't go below it!

As a practical example, here is a test of two samples of a Samsung INR18650-20Q 2000mAh Li-Ion cell:-

enter image description here

The manufacturer rates this cell for discharge down to 2.5 V. These tests only took it down to 2.8 V, but it is clear from the discharge curves that the battery is running out of active material even before then. The difference between 2.8 V and 2.5 V is so small that there is no point in continuing the tests down to a lower voltage.

At higher discharge current the voltage is generally lower (due to internal resistances and limited charge mobility) so the realizable capacity may be much less if a high cutoff voltage chosen in a device that draws high current.

The manufacturer should specify the discharge current at which the capacity is rated. Depending on the type of cell and intended application, this could be 1C (where 'C' is the current for a nominal 1 hour discharge, in this case 1 x 2000 mA = 2 amps) or a lower C rate for 'high capacity' cells that are designed for slower discharge.

In this test the 1C discharge (bright cyan line) shows a voltage 'knee' at ~3.3 V. Beyond this point there is virtually no capacity left, and no point continuing to suck out the tiny bit of energy left. At 0.1C (0.2 A) the 'knee' voltage rises to ~3.4 V, but the capacity at this lower discharge rate is not much higher.

Now compare that graph to this test of a 'high capacity' cell of the same size, also rated for discharge down to 2.5 V:-

enter image description here

The thinner plates in this cell have higher resistance, so it doesn't handle high discharge currents well. The voltage 'knee' is also much softer, with significant capacity remaining below 3.3 V. For this cell a cutoff voltage of 3.0 V would be realistic, while discharging all the way down to 2.5 V might just meet its rated capacity of 4 aH.

Another thing to note about these discharge curves is the 'bump' that occurs before the final nosedive, particularly noticeable at low current. This indicates changes in the chemical reactions occurring in the cell, which may result in increased wear and reduced cycle life beyond this point. In both of these examples the 'bump' occurs just below 3.5 V. This suggests 3.5 V might be a better cutoff voltage for low drain applications.

Device manufacturers often choose a higher "empty" voltage to increase cycle life. This reduces realizable capacity, but total capacity over the cell's lifespan is often dramatically improved because the cell gets damaged less on each discharge.

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If there is a rating given in mAh, the voltage is already considered. You can multiply the value with the battery voltage to get the watthours the battery contains. Watthours (or rather wattseconds) are the same unit as Joule, aka energy.

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You are correct that voltages do matter. The formula is accurate for cases where the load on the battery has constant current. Also, the capacity of the battery in mAh is given by the manufacturer unders specific conditions. It may vary a bit in your application if the current is different or if you use a different end voltage. But usually these variations are not too large for lithium ion batteries.

The other problem is that if you use a DC-DC converter, your current will not be constant. As the voltage decreases, the load current will increase. As a practical matter, what is usually done is to use the battery average voltage (3.7 in your case) and the current at average voltage for the life calculation. This will give an approximately correct result. But in my experience, battery life is actually measured for confirmation. You run the device continuously (sometimes with special software to simulate user interaction) and record the voltage and current with logging instruments until shutdown.

Consumers are very unhappy if the battery life claimed is misleading. So after you measure, you subtract a little and use that as the claimed battery life. Under-promise and over-deliver on battery life. But that is not engineering, I guess.

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