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I would like to charge the following rechargeable Lithium-ion battery outside the device where it is normally placed in (and charged):

https://buy.garmin.com/en-US/US/p/pn/010-12456-06

I don't know the pin-out and don't know much about charging Lithium-ion batteries, especially if they are put in parallel.

What I have researched so far:

The battery type is 1ICR19/65-2, meaning that the battery consists of 2 Lithium-ion cells which are put in parallel. Each cell has a nominal voltage of 3.7 V and a capacity of 2200 mAh.

From the link above, we see that the battery has four connectors. Let's denote the right connector as C1 and the left connector as C4. I have charged the battery (in the device) until the device said charging was finished, and then have measured the voltage between any pair of connectors (6 possible combinations). The results:

C1-C2: 3.7 V
C1-C3: 3.7 V
C1-C4: 4.1 V
C2-C3: 0.0 V
C2-C4: 0.0 V
C3-C4: 0.0 V

In case it matters: I have done the measurements using a reliable measurement device (Fluke).

In a discussion with some friends who are in electrically powered model aircrafts (which use cells which are put in series, not in parallel), the following assumptions were made:

  • C1 and C4 are the "real" power source for the device (nearly sure).
  • C2 and C3 might be connected to a NTC to monitor the battery's temperature for safety reasons.
  • C2 and C3 might be connected to some sort of "balancer" which might compensate for (slightly) different voltages of the two cells.
  • C2 and C3 might just be used for measuring the voltage of each cell during charging.

My specific questions now are:

  • Does anybody know the pin-out of that specific battery (or can anybody at least make a sound guess)?

  • Could anybody in short explain what might be the reason that there is 3.7 V between C1 and C2/3 and 4.1 V between C1 and C4, but there is 0 V between C2/3 and C4 (according to my understanding, there should be 0.4 V)?

  • Is it correct that C1 and C4 actually provide the power the device consumes?

  • If the answer to the previous question is yes, I believe that I can "ignore" C2 and C3 when charging that battery outside the device. I then could use one of those universal Lithium-ion chargers with moveable contacts to charge that battery. Is this correct?

EDIT 1

In the meantime, I have done a further test. I have fully charged the battery and then have let run the device until it turned off due to low battery. Then I have repeated the voltage measurements as described above. The results:

C1-C2: 2.8 V
C1-C3: 2.8 V
C1-C4: 3.2 V
C2-C3: 0.0 V
C2-C4: 0.0 V
C3-C4: 0.0 V

As we can see, the voltage from C1 to the middle connectors (C2 and C3) has dropped by the same amount as the voltage from C1 to C4. This eventually means that C2 and C3 are connected to C4 by diodes which produce a voltage drop about 0.4 V, and it eventually means that the voltage at C2 and C3 is just used for monitoring when charging. What do you think?

EDIT 2

Due to @jsotola's request, I have added an image of the label of the battery. I apologize for the bad quality. The text should be readable, though.

This is the label of the battery in question.

EDIT 3

Just in case anybody is wondering: I am asking this because the manufacturer does not offer a charger for that battery. Obviously, they expect us to always charge the battery in the device which is not acceptable to me because I need to use some more of that batteries which I'd like to charge without inserting (and removing) each battery in (from) the device just for charging.

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  • \$\begingroup\$ post a picture of the label on the one that you have \$\endgroup\$ – jsotola Apr 23 '18 at 0:31
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Try usinga TP4056 based Lithium battery charger module with built-in battery protection. They are very cheaply available in bulk on popular shopping websites, and the datasheet has a table of resistor values for programming the charging current (if you can't find a module pre-made for your required charging current, see below).

The chip handles the charging of the battery completely on its own, with a single 5V power supply (most if not all those modules have a USB socket on them). If the cells are parallelled in side the battery, like in most powerbanks, they're just hardwired together without any balancing means.

Charging a lithium cell is easy, as long as you consider the maximum input current of the battery. If there is no such value available, use the half-C-rule (charging current does not exceed 0.5 times the mAh rating of the battery).

The battery is first charged with a constant current, until it reaches 4.2 volts. The charger then switches to constant voltage mode, until the charging current has dropped below a threshold value. The TP4056 has all the means to do this on its own, including a "soft start" for charging deeply discharged batteries, starting with a very low current. Connecting a temperature sensor is not required, but possible with some modification of the modules.

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  • \$\begingroup\$ I am accepting your answer although it actually is not an answer to my question (which was (among others) about the pin-out of that battery). Nevertheless, you have put me on the right track (searching for a charger IC and doing it myself or buying a readily made board with that IC). I won't use the TP4096, though. At the moment, I am after MAX1737 or MAX8731A. \$\endgroup\$ – Binarus Apr 26 '18 at 5:50
  • \$\begingroup\$ An additional question: I have difficulties in understanding the half-C-rule if two batteries are in parallel. From a naive point of view, we could say that we should add the capacities of the two cells and apply the half-C-rule to the sum. But I think that there are cases where one cell is dead and doesn't take any current any more, and therefore the second cell would get charged too fast. What do you think about that problem? \$\endgroup\$ – Binarus Apr 26 '18 at 5:54
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    \$\begingroup\$ @Binarus, if a battery is made out of two cells in parallel, manufacturer makes all efforts to combine cells with identical characteristics, specifically to address the possibility you are concerned with. So the scenario you envision would not happen in nearest future. So the charge/discharge current will be nearly equally divided between two cells, and the 0.5C rule still applies. \$\endgroup\$ – Ale..chenski Apr 26 '18 at 6:53

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