Hello All,
I hope this note finds everyone is well & enjoying today!
I have a battery powered lawnmower. The battery pack is labeled as 60 volts 5 Ah. But there is a mystery of how the pack rating is delivered by the cells inside.
I had to open the battery pack after smoke came out while charging. (Problem solved- Was a loose battery tab floating around inside. It must have momentarily shorted two opposite polarity terminals & thankfully acted like a fuse, burning in half before a cell was ruined.)
When I opened the pack's case, I saw that the cells are 2500 mAh. They are Samsung INR18650-25R M. The data sheet shows them to be 2500 mAh with a nominal 3.6 volt rating.
Therefore I expected 32 cells with parallel groups of two cells each for the advertised 5 Ah. I expected that each group of two would be wired in series with the next for 57.6 volts. This isn't what is in the pack.
In reality, there are 26 cells. I could not get in deep enough to see if they are 100% split into pairs for 5 AH without major surgery. The cells I can see are in paralleled groups of two. And groups I can see are in series.
To My Mind Something Isn't Correct:
Twenty-six cells divided by two (for parallel to get 5 Ah) would yield 46.8 volts. (13 x 3.6 = 46.8)
When charged, the pack voltage measures at 62.0 volts. Sixty-two divided by 3.6 equals 17 pair of two cells. That would require 34 cells- But we only have 26 cells.
If I put my brain that the 5 Ah is a marketing scam and the pack is really 2.5 Ah at 62 volts, our 3.6 volt cells in series would give us 93.6 volts.
Also note that there is a circuit board in the pack, but I don't know its purpose. Perhaps there is "voltage magic" happening from the board, but I would not know what the magic could be. One side of the board isn't visible, so I can't seek clues.
Do any of you know how this pack is wired to give 62 (peak after charging) volts at 5 Amp Hours? I apologize, but the cell count and amp hour rating versus cell actual mAh rating does not add up in any formula of which I can find.
Thanks Very Much For Educating!
Paul
I hope this note finds everyone is well & enjoying today!
I have a battery powered lawnmower. The battery pack is labeled as 60 volts 5 Ah. But there is a mystery of how the pack rating is delivered by the cells inside.
I had to open the battery pack after smoke came out while charging. (Problem solved- Was a loose battery tab floating around inside. It must have momentarily shorted two opposite polarity terminals & thankfully acted like a fuse, burning in half before a cell was ruined.)
When I opened the pack's case, I saw that the cells are 2500 mAh. They are Samsung INR18650-25R M. The data sheet shows them to be 2500 mAh with a nominal 3.6 volt rating.
Therefore I expected 32 cells with parallel groups of two cells each for the advertised 5 Ah. I expected that each group of two would be wired in series with the next for 57.6 volts. This isn't what is in the pack.
In reality, there are 26 cells. I could not get in deep enough to see if they are 100% split into pairs for 5 AH without major surgery. The cells I can see are in paralleled groups of two. And groups I can see are in series.
To My Mind Something Isn't Correct:
Twenty-six cells divided by two (for parallel to get 5 Ah) would yield 46.8 volts. (13 x 3.6 = 46.8)
When charged, the pack voltage measures at 62.0 volts. Sixty-two divided by 3.6 equals 17 pair of two cells. That would require 34 cells- But we only have 26 cells.
If I put my brain that the 5 Ah is a marketing scam and the pack is really 2.5 Ah at 62 volts, our 3.6 volt cells in series would give us 93.6 volts.
Also note that there is a circuit board in the pack, but I don't know its purpose. Perhaps there is "voltage magic" happening from the board, but I would not know what the magic could be. One side of the board isn't visible, so I can't seek clues.
Do any of you know how this pack is wired to give 62 (peak after charging) volts at 5 Amp Hours? I apologize, but the cell count and amp hour rating versus cell actual mAh rating does not add up in any formula of which I can find.
Thanks Very Much For Educating!
Paul