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I am working on integrating single 4.2V Lithium battery(400mAh capacity), charger IC, and battery fuel gauge for an embedded project.

I have watched several TI training videos about battery management, and understood the following:

  1. The battery has internal resistance, and what we will measure on the battery terminals should always be below 4.2(Vmax). E.g. if max voltage deliverable by battery is 4.2, we will have 0.1V internal drop, and thus measure 4.1 on battery terminals.
  2. As the battery ages the internal resistance increases, hence, after few years, I should read even less voltage on battery terminals.
  3. Majority of charger ICs terminate the charging once the battery terminals reach fixed voltage threshold, usually 4.2V. It is the point where charger IC understands that battery is now fully charged.

I have 3 questions:

  1. Are my conclusions above correct?
  2. Let's say we are charging an aged battery. During fully charged state, since it is aged, it should have less than 4.2V on terminals due to high internal resistance . Let's take 4.0V as an example as a fully charged terminal voltage(when it is 4.0, the 0.2V is lost on internal resistance->4.2Vmax). Charger IC will keep charging the battery until the terminal voltage reaches 4.2V, but it had to charge it until 4.0. So it will actually overcharge it. Is my understanding correct?
  3. Many battery fuel gauges have an alarm interrupt for low battery voltage. For the most batteries, the 3.0V is considered as 100% discharged. Let's take 3.1V as an alarm condition. Battery fuel gauge is reading voltage from the battery terminals. Let's say it is reading 3.1V. In my understanding alarm should not be generated, because when terminal voltage is 3.1V, the actual overall voltage would be around 3.2-3.3V for example. I think the correct behavior would be for the fuel gauge to approximate internal voltage based on the externally sensed terminal voltage, and generate alarm based on approximation of internal voltage level. I know it is a general question, but I believe majority of fuel gauges should have common behavior for this. Do they generate alarms according to internal or external voltage?

I would appreciate your help.

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  • \$\begingroup\$ for resistance to affect voltage, there must be a flow of current. this means that for the series resistance you would read the same voltage pulling smaller current even if the resistance increased. Also the membranes separating sections of a battery deteriorate over their life too. \$\endgroup\$
    – Abel
    May 6, 2021 at 12:13
  • \$\begingroup\$ Thank you for the comment. Yes, I know this. Since batteries are typically under load, and charging chip will be charging while device is in use, there will be current flow. But in the charger IC that we use, there is a feature called power-path - during charging battery stops feeding the system, and power is fed directly from adapter. In that case since there won't be current draw from the battery, there shouldn't be drop in voltage and charger chip would work ok. \$\endgroup\$
    – UserRR
    May 6, 2021 at 12:36
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    \$\begingroup\$ A good setup will not drain the battery while charging and will power the load from the charger. \$\endgroup\$
    – Abel
    May 6, 2021 at 12:39
  • \$\begingroup\$ Yes, we actually were intending to use this feature to save battery power. But I realized the advantage relevant to proper voltage level just now. \$\endgroup\$
    – UserRR
    May 6, 2021 at 12:41
  • \$\begingroup\$ Furthermore, the charger IC has a timer setting for charge termination. If I understood it correctly, then it serves as a workaround for the problem in question 3, in case the battery is being drained while being charged. \$\endgroup\$
    – UserRR
    May 6, 2021 at 12:50

1 Answer 1

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The battery has internal resistance, and what we will measure on the battery terminals should always be below 4.2(Vmax). E.g. if max voltage deliverable by battery is 4.2, we will have 0.1V internal drop, and thus measure 4.1 on battery terminals.

What you will measure across the terminals of the battery will depend upon what is happening to the battery. It may be

  • charging
  • discharging
  • neither supplying nor storing up power (quiescent)

When then battery is charging, the voltage across the terminals will be higher than when the battery is quiescent.

When the battery is discharging, the voltage across the terminals will be lower than when the battery is quiescent.

This is because the "voltage drop of the internal resistance" depends upon power flow. When there is no current and therefore no power flow, there is no "voltage drop of the internal resistance". When power is flowing into the battery, the "voltage drop of the internal resistance" will give the terminals a higher voltage. When the power is flowing out of the battery, the "voltage drop of the internal resistance" will give the terminals a lower voltage.

As the battery ages the internal resistance increases, hence, after few years, I should read even less voltage on battery terminals.

That will be true when the battery is being used, i.e. discharging, and power is being drawn from the battery.

Charger IC will keep charging the battery until the terminal voltage reaches 4.2V, but it had to charge it until 4.0. So it will actually overcharge it.

No, the charger will stop at 4.2V. The quiescent voltage of the battery is only 4.0V, so the battery will not be overcharged.

Do they [i.e. low battery detectors] generate alarms according to internal or external voltage?

They generate an alarm when the terminal voltage is at a certain level. This can either be when the battery is quiescent, or in use. While quiescent, the battery voltage may be higher than when in use, but if the voltage falls below the alarm threshold while in use, it should trigger an alarm.

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  • \$\begingroup\$ Thanks a lot for explanation. It is clear now for me. \$\endgroup\$
    – UserRR
    May 6, 2021 at 13:34

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