BMS questions

[greasypants]

I have not posted on here for a long time but I have been busy building up a new bike over the winter. The new bike will have 60 psi cells configured in a 30s2p pack. They will be charged as 2 15s2p packs. I have built two 15 cell Goodrum BMS boards and will be mounting them in a small pelican case bolted to the top tube.I am using 20 gauge silicon wire for the cell monitoring connections. For charging I am using two 60 volt 3 amp cc/cv variable bench supplies.

First questions is can I leave the two packs connected in series while charging? I am hoping I can just disconnect the main power to the controller and then open up the pelican case, hook up the two power supplies and charge both battery packs. 42 cells with be in one box inside the frame and the other 18 with be mounted in another box suspended below the downtube so there will be one connection between the two packs sealed inside the main box.I could run wires outside the box to easily disconnect it but I would prefer to avoid that if possible.

Second question is regarding the length of the monitoring wires. As some cells will only need a short piece of wire because they are close to the BMS and some will need almost 3 feet of wire because they are much farther away. Using a wire resistance calculator from the internet it says 3 feet of 20 gauge copper wire will have a voltage drop of .0156 @.5 amps. Would this mean the cells farther away would be charged to 3.67 volts and have a low voltage cutoff of 2.12 volts? Could this cause a problem over time and would it be better to make all the wires the same length so every cell is charged to exactly the same voltage?

Thanks
Greg

 

[GGoodrum]

If your supplies have isolated outputs, which most do I think, you should be okay to leave the packs connected in series. You just need to adjust the outputs of both so that they are a bit above the sum of the shunt voltages, which should be 3.68V per channel, or 55.2V. If you can set the supplies to 56V, or so, both BMS boards should be happy.

I wouldn't worry about the different length tap wires. The shunt voltages could all vary a bit anyway, just due to the 1% tolerances on the voltage divider resistors for each channel. Also, the LVC function shouldn't matter either. Basically, all you are trying to do is keep the cells from dumping all the way down to zero, which they can do at the end of capacity pretty quick. Whether it is 2.1V or 2.5V or anything in between isn't going to make a big difference. Maybe more so with lower C-rated cells, but even there, whether it is 2.10V or 2.20V won't matter. The big advantage of cell-level LVC protection is really for higher quality, more powerful cells. With 10C PSIs and 30C a123s, the problem is that you really don't get an indication that you are at the end of the capacity, because you get so little voltage drop change throughout the capacity. It can feel just as strong 20 seconds from the cells dumping as it does fresh off the charger. Because of this, without protection, you could have a cell kill itself and still have the pack voltage above a controller set LVC point.

-- Gary

 

[greasypants]

Thanks Gary, that's exactly the info I was looking for.

 

[fechter]

*edit* read post first good idea... Yes, you can charge as planned as long as the charger outputs are isolated from the AC ground. You can test with an ohmmeter from the ground pin on the AC plug to the output terminals. It will either be near zero (not isolated), or some value above 10k ohms (isolated).

If the cells are reasonably balanced, there is practically no current in the sensing wires. The worst case condition still does not cause a cell to get outside of the 'safe' voltage range.

The two main pack connections are the exception to this. These wires always have a lot of current, so you don't want the voltage drop in these to cause a problem. I think there's a way to use two wires on each end to keep the charging current off the sensing wire.

In practice, you notice the two end cells tend to behave a bit differently than the rest due to the voltage drop.