# Pinout of / how to charge that certain battery (Lithium-ion, 3.7 V, 2 Cells, parallel, 4400 mAh)

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

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.

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.

• post a picture of the label on the one that you have – jsotola Apr 23 '18 at 0:31