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I'm using two of the following rechargeable batteries http://www.renata.com/fileadmin/downloads/productsheets/lithium_polymer/ICP501233PA.pdf in a parallel configuration to power up a small development board that reads data from different sensors and saves it on an sd card.

The device works for around 10 hours and then turns off (battery drained). Recharged for a couple of hours and again measurement.

I've measured the voltage (on the board, so with both batteries connected in parallel) and usually it shows 4.1V (fully charged) and above 3V after discharge. Still, I've also noticed that some of the batteries show voltages between 1.7V and 2.1V after working ok for 10 or more hours (also measured in parallel).

I assumed that the protection circuit is defect and tried to replace the batteries.

My issue is that if I measure the voltage on the disconnected battery, the value is again 4.x V. (measured together in parallel OR each separately)

Can anyone explain me this ?

Are the batteries broken or the protection circuit is broken or it just needs a hard reset ?

Note 1: The batteries are soldered on the board and never disconnected.

Note 2: The batteries have already more than 300 charge/discharge cycles.

Thank you.

(Pretty specific question but I need to understand what's happening)

Note 3: The firmware installed on the boards on which I've measured the unusually low voltages doesn't start anymore. I've disconnected the battery and soldered back, the voltage is again 4.0x volts and everything seems fine. Having several boards in use, I am still concerned that it will fail again.

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    \$\begingroup\$ While they are in the board is something still trying to draw current from them? Try connecting a small load (something like a 10k to 1k resistor) and measure again. Recovering 3V if disconnected from load is a bit atypical but not unheard of. \$\endgroup\$ – Arsenal Oct 4 '16 at 15:45
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    \$\begingroup\$ How did you measure those 1.7 and 2.1V? With the batteries on the board? Was there any load connected to them at the time? \$\endgroup\$ – anrieff Oct 4 '16 at 16:45
  • \$\begingroup\$ @anrieff I've measured with the batteries on the board, but the board was turned off (on/off switch) \$\endgroup\$ – ossx Oct 5 '16 at 6:42
  • \$\begingroup\$ As stated in Note 3, after disconnecting and reconnecting the rechargeable battery the voltage is again by 4V. Should I just perform this "hard reset" on all the boards or should I just replace all the batteries ? \$\endgroup\$ – ossx Oct 5 '16 at 6:49
  • \$\begingroup\$ "some of the batteries show voltages between 1.7V and 2.1V after working ok for 10 or more hours" - What voltage is your LVC (low voltage cutoff) set to? What is the peak current draw of your device? \$\endgroup\$ – Bruce Abbott Oct 5 '16 at 8:20
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Your description is rather unclear, but what is happening is that the two batteries are not perfectly matched. After 300 cycles, you should expect at most maybe 90% capacity, and potentially 10% or even less depending on previous abuse. Best case, you might get another few hundred cycles from them.

If you see significant differences between on and off load voltages, this is another sign that the cells are old. A good cell will have an internal resistance of maybe 100 m\$\Omega\$, this ingreases with age.

Taking the batteries down to 1.7V will reduce their lifetime (or so I understand). Anything below 3V per cell isn't really useful. If a protected cell is dropping below its cut-off voltage, this is probably self-discharge (or a chemistry effect) occuring in the hours after the cell reaches the disonnect threshold. Unless, of course you're measuring on the outside of the protection circuit (in which case you will observe that the cell is mostly isolated).

A battery protection circuit is designed to prevent you charging a cell which has been over-discharged. This is because the cell may be damaged, and charging (particularly at the normal 0.5C rate) is likely to cause a fire.

  • Measure the individual cell capacities.
  • Discard the weak cells (maybe anything below 60% rated value, I don't know).
  • Consider how your charge/discharge application will be affected by individual cells becoming short-circuit.
  • If you don't understand the specific protection circuit you have, you have to assume it doesn't protect anything (cirrent, over voltage, under voltage)
  • Remember that lithium batteries frequently catch fire, especially when they're not used in carefully controlled scenarios.
  • Remember lithium batteries are not designed to be float charged (and cell-level protection doesn't monitor this)
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  • \$\begingroup\$ Is there an explanation to why after disconnecting the batteries from the board they show again 4V ? \$\endgroup\$ – ossx Oct 5 '16 at 8:06
  • \$\begingroup\$ Internal resistance. Your batteries are old. \$\endgroup\$ – Sean Houlihane Oct 5 '16 at 8:45

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