Battery cell pouch

[dnmun]

you don't know there is anything wrong with the BMS. it is functioning normally now. it is turned off because the one cell is so low. it has prevented all the other cells from overcharging so it is doing what it is supposed to do.

the top cell, #15 which you call #1 is still at 2V? you were not able to charge it up with the cell phone charger?

you can measure the voltage right on the terminals of the battery cells and that is where you have to use the alligator clips to attach the cell phone charger.

charge up #15 then let it settle, if it drains down to 2V again then you will have to open the BMS to see if the BMS is draining that one cell down, but first you gotta get the cell phone charger and clip it onto the terminals of that cell and charge it up. i think it will turn on the BMS as soon as you can get charge into it but you wanna charge it all the way up like all the others at 3.9V. but not more. you do have to watch it like a hawk after it climbs above 3.55V.

 

[Milou]

you don't know there is anything wrong with the BMS. it is functioning normally now. it is turned off because the one cell is so low. it has prevented all the other cells from overcharging so it is doing what it is supposed to do.

the top cell, #15 which you call #1 is still at 2V? you were not able to charge it up with the cell phone charger?

you can measure the voltage right on the terminals of the battery cells and that is where you have to use the alligator clips to attach the cell phone charger.

charge up #15 then let it settle, if it drains down to 2V again then you will have to open the BMS to see if the BMS is draining that one cell down, but first you gotta get the cell phone charger and clip it onto the terminals of that cell and charge it up. i think it will turn on the BMS as soon as you can get charge into it but you wanna charge it all the way up like all the others at 3.9V. but not more. you do have to watch it like a hawk after it climbs above 3.55V.

Hi dnmum: I charged my number 1 cell as instructed. I used a 5V 0.73Amp charger, including alligator clips and all. I charged it for about 20 hrs (20hrs x 0.73Amps = 14.6 Amps). When first put on charge the voltage climbed rather quickly to 3V, it then stabilized at around 3.3V throughout till the end. I shut off the charger after 3.490V. The cell then went down to 3.35 where it seem to settle and remain (after 9 hrs). I am now in the process of charging my cell number 15. Tomorrow I will know more.

Question:

You had indicated I numbered the sensing wires/cells backward. What determines the right way and why? I like to make the appropriate corrections.

I am now confused which 10 gauge wire color is the positive and the negative? From my meter when I put red probe unto red 10 gauge wire and black probe unto 10 gauge black it show no negative sign on the volt setting. So, I assumed all one side (bottom of drawing) of the flat cells where positive and negative at the other end (top of the drawing).

I will report tomorrow on the end results of my battery. Thanks again to all for the fantastic, very useful help. Regards JJ

 

[dnmun]

if you wanna, we can lead you through procedure to bench test the capacity of the pack. but your BMS should function normally now and spin the wheel. you should see full pack voltage on the BMS P- to B+ leads.

 

[999zip999]

Be careful not to over charge as it gets full it will climb fast and over charge.

 

[Milou]

Hi: I finally got the last weak cell charged. After all-and-said here are the present results of my battery pack:

Pins Volts
1 - 2 3.34
2 - 3 3.54
3 - 4 3.40
4 - 5 3.57
5 - 6 3.52
6 - 7 3.43
7 - 8 3.51
8 - 9 3.24
9 - 10 3.34
10 - 11 3.49
11 - 12 3.50
12 - 13 3.35
13 - 14 3.33
14 - 15 3.32


Total 47.9V

Total before BMS at battery pack 51.6V

Total after BMS at the plug 50.7V

I tested the cells from the sensor pins at the battery side black connector. All cells appear to be holding their charge. At first when I started the charge on my cell number 15 the voltage was only 2.9mV. After over 20 hrs charge (Charger was 5V 0.73Amp) it made it up to 3.400V after stopping the charger it settled to 3.304V. I thought for sure this cell was a goner but, it was the "Little Engine that could"! I have not done anything else to the pack. I have not run the bike motor with it yet (tomorrow). I am not sure what I should do next?

To dnmun: I am embarrassed to admit I did not realize my cells where connected in series. The reason for the negative/positive confusion on my part. I am clear about it now. FYI, attached are images of my BMS.

Regards JJ

 

[dnmun]

just leave it charging, no problem, everybody learns it by doing it. guess how many voltmeters i have blown up laying on my table.

you can adjust the charger voltage up also. you want the final voltage to be 15 X 3.65 at least. 54.75V and if you can set it to 55V or 56V then it will eventually balance all the cells to 3.65V+ for each cell. that is when it will be fully charged. you can do capacity test then.

these cells that were over discharged may not have been damaged if there was not a lot of current flowing through them.

however, they may not ever hold a resting voltage up around the 3.65V level and will wanna stay around the 3.35V level. but you wanna get the charger voltage up enuff to find out. you can actually use the cell phone charger too.

 

[Milou]

just leave it charging, no problem, everybody learns it by doing it. guess how many voltmeters i have blown up laying on my table.

you can adjust the charger voltage up also. you want the final voltage to be 15 X 3.65 at least. 54.75V and if you can set it to 55V or 56V then it will eventually balance all the cells to 3.65V+ for each cell. that is when it will be fully charged. you can do capacity test then.

these cells that were over discharged may not have been damaged if there was not a lot of current flowing through them.

however, they may not ever hold a resting voltage up around the 3.65V level and will wanna stay around the 3.35V level. but you wanna get the charger voltage up enuff to find out. you can actually use the cell phone charger too.

Hi: Thanks for the information. Sorry to hear you have a multimeter graveyard on your desk. I though one could replace the fuse when they go bad????

How do I increase the volts on my charger? It does not have an option to increase it, (as I can see)??? It is only rated to output 48V (see image).

I am assuming one leaves each cell separately on charge until the voltage increases to the 3.65V with the phone charger, It will reach that level because the phone charger is rated at 5V (above the required)????

 

[dnmun]

nope, that is not how the BMS works. the BMS functions to balance the cells by allowing the current to bypass the full cells and continue to fill up the unfilled cells.

you have to raise the charger voltage high enuff so that when several of the cells are up around 3.9V like yours, then the bulk charger needs enuff 'headroom' to continue charging up the low cells. you can open your charger and take a picture inside and we can show you where to adjust the voltage up to 55V.

 

[Milou]

nope, that is not how the BMS works. the BMS functions to balance the cells by allowing the current to bypass the full cells and continue to fill up the unfilled cells.

you have to raise the charger voltage high enuff so that when several of the cells are up around 3.9V like yours, then the bulk charger needs enuff 'headroom' to continue charging up the low cells. you can open your charger and take a picture inside and we can show you where to adjust the voltage up to 55V.

Here are the images (not much to show) But, I'll take what I can. Regards JJ

 

[dnmun]

do you see the circle with R29 printed next to it? do a close up of that and the back side, that is where the cermet goes to adjust the voltage. they just have a resistor there now. you might be able to figure out which resistor, look for the resistor that spans some of the spots on the circle where the cermet goes.

 

[Milou]

do you see the circle with R29 printed next to it? do a close up of that and the back side, that is where the cermet goes to adjust the voltage. they just have a resistor there now. you might be able to figure out which resistor, look for the resistor that spans some of the spots on the circle where the cermet goes.

Here as requested. One resistor spans the R29 circle (see yellow highlighted area). The resistor's marking:

Blue
red
black
green

Resistor just above to the left of above (next to R19). The marking:

green
yellow
black
red

Regards JJ

 

[dnmun]

can you read the resistor value of the one in the box and the one just left of it with the blue body and the one just above it with the brown body and red black.

to read up u can go to wikipedia and read under 'resistor color codes'


is it R20 that is to the left? what does it measure? and R24?

 

[Milou]

can you read the resistor value of the one in the box and the one just left of it with the blue body and the one just above it with the brown body and red black.

to read up u can go to wikipedia and read under 'resistor color codes'


is it R20 that is to the left? what does it measure? and R24?

This is really an excessive for your eyes!!! I used several magnifiers for this exercise. I tried to get the correct values regarding the color bands. The values may be off due to inability to determine the correct order for the bands and also the precise color identification. View images. Regards JJ

 

[dnmun]

yes, the green band means it is a high precision .5% tolerance resistor. you did good to learn it fast. i think the one in parallel with the 200k is not that big. list the colors, remember the green ring is last, and is the tolerance.

the 62 ohm? resistor is tied to ground. you can see the trace as it comes around the side and the resistor is soldered into it.

i was wrong, it is not R24 i was thinking, it is the resistor right above the R in R20. that resistor is soldered to the positive side of the charger output. you can follow it over to where that heavy trace comes out of the middle of the schottky diode. that is a dual cathode so there is a trace that comes from the transformer and splits to go to the two outer legs, the cathodes, of the schottky diode and the current flows through and out to the charger plus output.

that resistor above R20 connects to charger plus output and you can follow that resistor to where it contacts the solder pad that connects to the resistors in parallel that connect to R29. that pad has a leg that goes up to the inside of the pwm IC, second pin on the left, so that is how the pwm turns on to push current and push the voltage up. we want to change the voltage setpoint at this pin. on that pad.

you can just read the colors, the red black yellow is 200k, is it 63 (blue orange black?) ohms for the one highlighted, if you can get good light and a closeup we could read the others. but basically we are gonna try to reduce the resistance of the lower section of that resistor divider bridge between plus and ground. to raise it 10% from 50 to 55V will mean reducing the fraction of low side to total resistance by 10%.

actually it is the resistor right above R23 that connects plus through the bridge to ground. looks like a precision resistor. black red black black yellow? i don't believe R20 is 1.1 meg. can you measure it with a voltmeter? the others too.

 

[Milou]

yes, the green band means it is a high precision .5% tolerance resistor. you did good to learn it fast. i think the one in parallel with the 200k is not that big. list the colors, remember the green ring is last, and is the tolerance.

the 62 ohm? resistor is tied to ground. you can see the trace as it comes around the side and the resistor is soldered into it.

i was wrong, it is not R24 i was thinking, it is the resistor right above the R in R20. that resistor is soldered to the positive side of the charger output. you can follow it over to where that heavy trace comes out of the middle of the schottky diode. that is a dual cathode so there is a trace that comes from the transformer and splits to go to the two outer legs, the cathodes, of the schottky diode and the current flows through and out to the charger plus output.

that resistor above R20 connects to charger plus output and you can follow that resistor to where it contacts the solder pad that connects to the resistors in parallel that connect to R29. that pad has a leg that goes up to the inside of the pwm IC, second pin on the left, so that is how the pwm turns on to push current and push the voltage up. we want to change the voltage setpoint at this pin. on that pad.

you can just read the colors, the red black yellow is 200k, is it 63 (blue orange black?) ohms for the one highlighted, if you can get good light and a closeup we could read the others. but basically we are gonna try to reduce the resistance of the lower section of that resistor divider bridge between plus and ground. to raise it 10% from 50 to 55V will mean reducing the fraction of low side to total resistance by 10%.

actually it is the resistor right above R23 that connects plus through the bridge to ground. looks like a precision resistor. black red black black yellow? i don't believe R20 is 1.1 meg. can you measure it with a voltmeter? the others too.

Hi dnmun: Thanks to give me some time to look over your data. I had to review what was what. I am not familiar with lots of these terms and their location on the board. I was not sure if you were talking about the top or bottom of board. In any case, view images to see hopefully the correct results??? I have included another close up of the top of the board (perhaps it will help?). Regards JJ

 

[dnmun]

your right arrow on the bottom of that last picture that says 1.81k is pointing right to the leg of another resistor at right angles to R20, that goes over to where it say R23. the trace that resistor goes to is the positive output of the charger. if you follow the trace back over to the side, you can see where there are two legs that come off and go over to the schottky diodes. that is how i know it is the positive.

so between positive and the ground there is a resistor chain. you measured 1.81k on one part, so measure that last resistor at R23.

it looks like you can add about 20kR in parallel for the 10% so look in you junk pile for something close to that.

maybe you can unsolder the two resistors in parallel (red black and brown green brown) and measure them by themselves too. do you have a soldering iron too? you only have to unsolder one leg of each to measure it by itself. you can leave one leg in the solder and put the probe on that end and the other on the open leg.

 

[Milou]

your right arrow on the bottom of that last picture that says 1.81k is pointing right to the leg of another resistor at right angles to R20, that goes over to where it say R23. the trace that resistor goes to is the positive output of the charger. if you follow the trace back over to the side, you can see where there are two legs that come off and go over to the schottky diodes. that is how i know it is the positive.

so between positive and the ground there is a resistor chain. you measured 1.81k on one part, so measure that last resistor at R23.

it looks like you can add about 20kR in parallel for the 10% so look in you junk pile for something close to that.

maybe you can unsolder the two resistors in parallel (red black and brown green brown) and measure them by themselves too. do you have a soldering iron too? you only have to unsolder one leg of each to measure it by itself. you can leave one leg in the solder and put the probe on that end and the other on the open leg.

Sorry for the late reply. Presently, it is a busy work time for me. I do have a soldering iron and have the means to use it. The firs image above show the meter resistance reading on each resistor. I measure from one leg to the other. If solder or unsolder, I did not think it made a difference? In any case, The R23 (90 degrees from R20) reads 17.50 K Ohms. I will have to get back to you for the reading unsolder after this weekend.

I do not have any spare resistors but I have Radio Shack not far. If I do add the "20kR, 10%" Where would I put it? Parallel to what?

Just curious would I not be better off getting a higher end multi-volt charger? Although, after looking over in the internet I am not sure what to look for to buy what I need???

 

[dnmun]

those two resistors have the same resistance because they are connected in parallel. you can see how their legs are tied to the same traces underneath. one has the red brown and what i thought was yellow, so it would started something like 21 (x10^4)? not likely since the equivalent parallel resistance you measured is 1770 ohms. so it would help to know what R20 really is too.

so maybe the red is the precision band instead. who knows unless they get measured, i can't read them. but they are attached to the input of the big 14 pin IC so that may affect the resistance measurement, maybe.

so 17.5k and 1.81k total 19.3k. that is the resistance between the output voltage to ground. we want the ratio of 1.81/19.3 to be reduced by 10% to 84.4x10^-3.

times 19.3k=1629 ohms for the total equivalent resistance of those three resistors below the input and the new resistor R2 added in parallel.

(R1XR2)/(R1+R2)=1629 ohms. solve for R2=16,290 ohms. so something around 16-17k is ok. from the ground where R29 is soldered to the end of R20 where it connects to that 17.5k ohm resistor. that should change the output voltage about 5V to 55V from 50.

i have some kingpan chargers in that range too if you wanna do that. they have a trimpot to adjust the output voltage. i even have some of the 48V headway chargers i might be able to hack to change the voltage too in the same way as this, but these are big resistors and easier to work on than the little surface mount resistors in the headway chargers.

but first do this to get the pack fully charged so it will balance properly.

i could mail you a resistor cheaper than driving there to radio shack i bet. can wait until you pick up those resistors to verify. maybe we could swap out one or both of them to get there.

also, double check the output voltage before unsoldering the resistors. i have assumed it is 50V and that we are trying to get to 55V.

 

[Milou]

those two resistors have the same resistance because they are connected in parallel. you can see how their legs are tied to the same traces underneath. one has the red brown and what i thought was yellow, so it would started something like 21 (x10^4)? not likely since the equivalent parallel resistance you measured is 1770 ohms. so it would help to know what R20 really is too.

so maybe the red is the precision band instead. who knows unless they get measured, i can't read them. but they are attached to the input of the big 14 pin IC so that may affect the resistance measurement, maybe.

so 17.5k and 1.81k total 19.3k. that is the resistance between the output voltage to ground. we want the ratio of 1.81/19.3 to be reduced by 10% to 84.4x10^-3.

times 19.3k=1629 ohms for the total equivalent resistance of those three resistors below the input and the new resistor R2 added in parallel.

(R1XR2)/(R1+R2)=1629 ohms. solve for R2=16,290 ohms. so something around 16-17k is ok. from the ground where R29 is soldered to the end of R20 where it connects to that 17.5k ohm resistor. that should change the output voltage about 5V to 55V from 50.

i have some kingpan chargers in that range too if you wanna do that. they have a trimpot to adjust the output voltage. i even have some of the 48V headway chargers i might be able to hack to change the voltage too in the same way as this, but these are big resistors and easier to work on than the little surface mount resistors in the headway chargers.

but first do this to get the pack fully charged so it will balance properly.

i could mail you a resistor cheaper than driving there to radio shack i bet. can wait until you pick up those resistors to verify. maybe we could swap out one or both of them to get there.

also, double check the output voltage before unsoldering the resistors. i have assumed it is 50V and that we are trying to get to 55V.

Hi dnmum:

Before un-soldering anything I tested the charger to see how much volt it puts out. I get 54.7V so, I charged the battery overnight and the battery went from 50.6V to 54.7V. I am not sure what to do next??? Is this good or should I go for something else as you described above?

 

[dnmun]

no, 54.7 is pretty good. i thought it was only putting out 50V.

15X3.65=54.75 so it is right where it should be if the pack will balance. 56V is better but go with this for now, imo.

 

[Milou]

Hi dnmun: I had some time to put the bike back up together, zip shut the battery, and gave the bike a try. It works wonderfully. The battery seem to keep its charge.

FYI, I have a very steep climbing driveway of approximately 30 ft to get out, and a steep declining road, going away from home. Obviously, going back home is the opposite. The bike does handle the climbing at a struggling, desperately, seeking help way. The battery gauge says zero but the bike does not stall and keeps going with no help from me. After up on the hill the battery gauge goes back to normal (full) quickly. I have gone up and down three times (at different times with one full charge) without any problems. On the flat the bike does reach 20MPH without problems or excessive battery drain.

I am assuming the battery are very good. My battery gauge is a very simple device that says "full, medium, or empty".

I really want to thank you and all the others for the help you guys provide. I admire your knowledge. Although, at times somewhat difficult, I truly appreciate the opportunity to pick your brains. I feel learning Chinese language is easier than learning electronic language (diodes, resistors, etc...). Regards JJ

 

[dnmun]

now you know how to keep track of the battery health so it should work ok. just keep charging. long term it will be best to get the charging voltage up a little more to 56V or so to help it balance.

you can short out the 63 ohm resistor at R29 with a wire, and the charger voltage should climb to 55.81V.

 

[Milou]

now you know how to keep track of the battery health so it should work ok. just keep charging. long term it will be best to get the charging voltage up a little more to 56V or so to help it balance.

you can short out the 63 ohm resistor at R29 with a wire, and the charger voltage should climb to 55.81V.

Shorting out a resistor with a wire??? Now, you are scaring the willies out of me. How do I know I will get 55.81V? Should it not be replaced with another Ohms resistor? I would be willing to buy one from you if you would like. Do you have a PayPal account? I would be willing to take on the modification. Regards JJ

 

[dnmun]

you can either short it with a wire or just replace it with a wire. same difference since it will still be zero ohms.

[(63ohms/1881ohms) + 1] X 54V = 55.81V

but that is just an estimate, it really should be the ratio of the resistance of the (1881-63) out of the total when you remove the resistor. but i don't have that total, so i just guessed.

it would be interesting to see what happens when you jumper over that resistor, to see what the voltage ends up at.

 

[Milou]

you can either short it with a wire or just replace it with a wire. same difference since it will still be zero ohms.

[(63ohms/1881ohms) + 1] X 54V = 55.81V

but that is just an estimate, it really should be the ratio of the resistance of the (1881-63) out of the total when you remove the resistor. but i don't have that total, so i just guessed.

it would be interesting to see what happens when you jumper over that resistor, to see what the voltage ends up at.

Do I have to remove the resistor first and pass a wire? Or can I just pass a wire around it without removing it? Thanks. Regards JJ