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I am using a charging and discharging circuit to do operations with a Li-ion battery.

When I charge the battery at 4.2V rated voltage and at 1 ampere current, I want to continuously measure the battery voltage to cut the circuit when the battery is fully charged at 4.2V. I have connected a voltage divider circuit at the battery terminal to measure the voltage, since 4.2V can't be directly fed to the microcontroller ADC pins. The problem is, when the charging circuit is open and I am measuring the voltage of battery using the voltage divider circuit then the battery voltage is correct as per voltmeter measurement, but when the charging circuit is on, then the battery voltage reading suddenly rises by 0.4V. The voltage reading is wrong.

Can anyone tell me that what should I modify in my circuit? Should I turn off the charging circuit every time I want to check the battery voltage?

The same problem happens with the discharging circuit. When I am discharging the battery at higher current, then the battery voltage reading drops by 0.3-0.5V. What could be the reason?

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  • \$\begingroup\$ The cell has internal resistance. When charging the terminal voltage = Vcell+ Ichg x R internal. When discharging the terminal voltage = Vcell - Ichg x R internal. || The cell is NOT fully charged at 4.2V. Standard LiIon practivce is to charge at constamt current (CC) to 4.2V/cell the hold the cell at constant voltage (CV) until Ichg drops to say 25% of Icc. \$\endgroup\$
    – Russell McMahon
    Jul 6, 2021 at 11:36
  • \$\begingroup\$ @VivekKarna: Does your voltmeter show the same changes in battery voltage? \$\endgroup\$
    – JRE
    Jul 6, 2021 at 11:39
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    \$\begingroup\$ Sounds like you are not charging lithium cells like they are supposed to be charged. How exactly you are charging them? You need to charge them properly or it can shorten the cell life and it can be dangerous even. Usually you use a charer chip that integrates safe charging features for you. \$\endgroup\$
    – Justme
    Jul 6, 2021 at 11:45
  • \$\begingroup\$ What makes you think it's wrong just because it jumps up by 400 mV when you supply it with 1 A charge current? \$\endgroup\$
    – winny
    Jul 6, 2021 at 11:52
  • \$\begingroup\$ @JRE: yes, the multimeter is also showing the same voltage value. \$\endgroup\$ Jul 6, 2021 at 11:57

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The cell has internal resistance which cause Vterminal to rise during charging due to IR internal voltage drop.

  • When charging the terminal voltage = Vcell+ Ichg x R internal.

  • When discharging the terminal voltage = Vcell - Ichg x R internal.

The cell is NOT fully charged at 4.2V.
Standard LiIon practice is to charge at constant current (CC) to 4.2V/cell and then to hold the cell at constant voltage (CV) until Ichg drops to say 25% of Icc.

V_terminal should NEVER exceed 4.2V.

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  • \$\begingroup\$ But the thing is how should I know the Rinternal so that I can accurately measure the voltage. \$\endgroup\$ Jul 6, 2021 at 12:02

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