LiFePO4 charging, 4S24P times 26 in Series

Thread Starter

BJNKM

Joined Aug 3, 2018
14
I'm "test" charging LiFePO4 battery modules, (4S24P), with a 12 volt lead acid, (car battery), charger. After a complete charge, the parallel groups are 3.4 to 3.9 volts. How can I limit each parallel group to 3.6 volts without a BMS? I'm fine with simply dissipating any excess voltage over 3.6 volts. I tried 3.6 volt zener diodes in series with a 10 ohm resistor, + to -, on a parallel group, but they are not dropping the higher parallel packs down.
This method worked on my 12 volt lead acid batteries to prevent over-charging.
I have built 26 of these 12.8 volt, 60 AH packs to replace the 26 lead acid batteries in my 1997 Chevy S10Electric truck. This is a factory built truck using the EV-1 drive train, (FWD), and charges with a paddle charger. There is a Battery Pack Control Module that monitors charging and I can't remove or bypass it. A BMS would conflict with this and I've never heard of a 104S24P BMS.
I'm sure there is a lot more I may need to explain.
Thanks
 

Hymie

Joined Mar 30, 2018
819
Using a 3.6V zener & 10Ω resistor to achieve a uniform voltage across the parallel packs, would seem a waste of potential battery capacity.

Could you not measure each pack’s fully charged voltage, and then parallel them by matched fully charge voltage?

Admittedly the parallel packs voltage could range from 3.6V – 3.9V, but within the paralleled batteries there would not be a mix of batteries with differing charged voltages.
Or are you saying that each of the paralleled packs must be 3.6V to be suitable for the charge control you have?
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
Using a 3.6V zener & 10Ω resistor to achieve a uniform voltage across the parallel packs, would seem a waste of potential battery capacity.

Could you not measure each pack’s fully charged voltage, and then parallel them by matched fully charge voltage?

Admittedly the parallel packs voltage could range from 3.6V – 3.9V, but within the paralleled batteries there would not be a mix of batteries with differing charged voltages.
Or are you saying that each of the paralleled packs must be 3.6V to be suitable for the charge control you have?
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
The 24 cells in a parallel pack will stay the same voltage. But 4 of these packs in series do not finish at an equal voltage. If it was .1 or .2 volts I wouldn't worry about it, but .5 volt is a bit too much of a difference.
I installed Lead Acid UPS batteries several years ago. They worked great with normal charging at around 360 volts, about 13.8 volts per battery. But my truck will occasionally do a "Balance charge" and send 15 volts to each battery. About 390 volts to the 26 batteries in series. Too much for these batteries. So I installed "Zener Diode Regulators". (See attached). It's difficult to see but a 6.8 volt, 5 watt zener diode is installed in series in both the positive and negative terminals. I had 1 of these attached to each battery. When voltage exceeded 13.6 volts, the zeners would dump to a small 2 volt bulb and 10 ohm resistor. During charging, you could read higher voltage on several batteries. But once charge was complete, the regulators would bleed of excess voltage and they all settled to 13.6 volts. "Top Balanced".
Although some people told me it won't work, I got 12K miles and 5 years out of the pack.
So I figured I could do the same with each parallel pack of the lithium batteries using 3.6 volt 1 watt zener diodes. Obviously I'm missing something?
 

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Hymie

Joined Mar 30, 2018
819
With a 3.6V zener diode in series with a 10Ω resistor across each pack should limit the voltage to around 3.6V.

I would recommend you try reducing the resistance value to say 2.2Ω - even if the charging voltage across the cell was to reach 4.2V, the current through the zener diode (and resistor) would be 270mA with zener diode power dissipation of less than 1 Watt.

With a higher wattage zener diode, you could reduce the resistance further if required.
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
With a 3.6V zener diode in series with a 10Ω resistor across each pack should limit the voltage to around 3.6V.

I would recommend you try reducing the resistance value to say 2.2Ω - even if the charging voltage across the cell was to reach 4.2V, the current through the zener diode (and resistor) would be 270mA with zener diode power dissipation of less than 1 Watt.

With a higher wattage zener diode, you could reduce the resistance further if required.
I have located a 5 watt, 3.6 volt zener diode.
To try to explain my situation better.... I have 104 parallel packs of 24 -2500 mah cells, (60 AH), each. These parallel packs are all connected in series. Charging is up to 390 volts/16 amps.
 

Hymie

Joined Mar 30, 2018
819
With a 5W, 3.6V zener diode, if you consider that the maximum cell charging voltage should be 4.2V, then with a 0.6Ω in series with the zener diode will result in a maximum zener/resistor current of 1A. The zener diode will then be dissipating 3.6W (allowing some margin) and the resistor dissipating 0.6W (recommend 1W rating to allow some margin).

But you will need 104 zener diodes and resistors.

If you have the data sheet for the zener diode, you might find that you do not need any additional series resistor – with the resistance within the zener diode being sufficient for your circuit.
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
With a 5W, 3.6V zener diode, if you consider that the maximum cell charging voltage should be 4.2V, then with a 0.6Ω in series with the zener diode will result in a maximum zener/resistor current of 1A. The zener diode will then be dissipating 3.6W (allowing some margin) and the resistor dissipating 0.6W (recommend 1W rating to allow some margin).

But you will need 104 zener diodes and resistors.

If you have the data sheet for the zener diode, you might find that you do not need any additional series resistor – with the resistance within the zener diode being sufficient for your circuit.

My cells are LiFePO4. 3.65 volt max charge voltage. 24 in parallel. I used a 5 watt 3.6 volt zener diode, (anode to plus), in series to a 5 watt 2.2 ohm resistor to minus. During a charge my cells do go over 3.6 to about 3.8. But after a charge, my cells drop to 3.3 volt. Why are my 3.6 volt zener diodes still dissipating voltage below 3.6 volts?
 

Hymie

Joined Mar 30, 2018
819
Last edited:

Thread Starter

BJNKM

Joined Aug 3, 2018
14
It looks like your zener diodes are at the low end of their voltage tolerance – I can’t think of any easy way of increasing them by 0.3V.

With the values you have chosen (3.6V zener and 2.2Ω resistor), with a 4.2V battery voltage, both the resistor & zener will each be dissipating less than 1W.

https://www.ebay.co.uk/itm/Zener-Diode-DIP-1W-3V-9-1V-12-Values-Choice-Free-P-P-New/401271357116?hash=item5d6da2febc:m:mMZJD1IfgbBjBafy2w6FZhg&var=670884093079
I guess I'm just going to live with the 3.3 volt charge. The good thing here is every one of the 104 parallel packs are within .03 volts.
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
So I haven't found a solution to my passive balancing system. 2 electronic engineers have said it's easy but haven't offered any ideas. The zener diodes are not working. I'm trying my best to be patient and I don't want to hurt these LiFePO4's.
When looking for an answer on-line, I found a very basic passive balancing diagram. It's simply a switch, in line with a resistor between the + and - of the parallel packs. At 3.65 volts, the switch in on and any excess current/voltage is burned off through the resistor.

So maybe a new question. What activates this switch?
To re-cap... I have parallel packs of 24 x 26650, 2500 mah LiFePO4 cells. 104 of these in series. (Actually I have 4 of the parallel packs in series to replace each of the 26 lead acid batteries. And all 26 of these in series).
Thanks, Kevin
 

Hymie

Joined Mar 30, 2018
819
It would be relatively easy to configure a voltage monitoring circuit that switched in a load once the cell voltage reached a pre-set level (3.6V). You would need 4 such circuits for each of your 14V battery sets.

If you believe such a circuit might work, I’m willing to post a circuit – I would then recommend you build 4 and see how it performs across one 14V battery (balancing the 3.6V cells) before building a sufficient number for the complete vehicle battery pack.
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
It would be relatively easy to configure a voltage monitoring circuit that switched in a load once the cell voltage reached a pre-set level (3.6V). You would need 4 such circuits for each of your 14V battery sets.

If you believe such a circuit might work, I’m willing to post a circuit – I would then recommend you build 4 and see how it performs across one 14V battery (balancing the 3.6V cells) before building a sufficient number for the complete vehicle battery pack.
That would be greatly appreciated.
Since the 26 batteries are in series, the system charges at a max of 390 volts and 20 amps.
Thanks, Kevin
 

Hymie

Joined Mar 30, 2018
819
Attached is my proposed circuit to balance the 3.6V across your sets of secondary lithium cells in your vehicle. I would recommend you study the diagram to understand how it works. If you have a bench power supply, I’d recommend you build one example of the circuit to check its performance. As stated, you will need 4 sets of these for each 14V battery.

The circuit works by monitoring the voltage across the paralleled cells (B1, B2,.. Bn) and if during charging the voltage exceeds a pre-set level then transistor Q1 is energised, sinking the current rather than it flowing through the batteries.

The component values have been chosen to allow a pre-set voltage in the range 3.3V – 4.0V.

Circuit operation
The battery voltage is divided in half by R1 & R2 and fed into the non-inverting input of op-amp IC1 via a 10k resistor. The inverting input of the op-amp is adjusted by VR1 from 1.65V to 2.0V (achieved by the 2V zenner diode together with the two 1N4001 diodes).
The gain of the op-amp has been set at 100 by the values of R4 and R5.
There is an error in the diagram in that both inputs to the op-amp should be shown with a capacitor to 0V to reduce electrical noise impacting performance.
Should ½ the battery voltage exceed the voltage set on the op-amp inverting input (set by VR1), then the output of the op-amp will go high switching on transistor Q1.

The op-amp is a 7611 (or 7612), this op-amp can operate with a supply rail as low as 1V (this is needed since your supply to the op-amp could be less than 3V).

The op-amp output is fed to the base of the darlington transistor Q1 with the resistor R3 (0.1ohm) in the collector path. If transistor Q1 is a TIP121 then the transistor can pass up to 8A and the resistor R3 should be rated at 10W. If a higher current is required then Q1 can be formed using a 2N3055 transistor combined with a suitably rated driver transistor to form a darlington – the current would then be increased to a maximum of 15A with R3 rated at 25W. Don’t forget that Q1 will need to be mounted on a suitable heatsink.

No fusing has been shown in the diagram, consideration should be given to adding fusing at appropriate locations given the current that could flow in the event of a fault.

There is a further error in my circuit, in that the op-amp feedback resistors R4 & R5 loading the input (more specifically R5) – therefore you should omit resistors R4 & R5 which will give a high open loop gain for the op-amp; but it will still work without R4 & R5 fitted.
 

Attachments

Last edited:

Thread Starter

BJNKM

Joined Aug 3, 2018
14
Attached is my proposed circuit to balance the 3.6V across your sets of secondary lithium cells in your vehicle. I would recommend you study the diagram to understand how it works. If you have a bench power supply, I’d recommend you build one example of the circuit to check its performance. As stated, you will need 4 sets of these for each 14V battery.

The circuit works by monitoring the voltage across the paralleled cells (B1, B2,.. Bn) and if during charging the voltage exceeds a pre-set level then transistor Q1 is energised, sinking the current rather than it flowing through the batteries.

The component values have been chosen to allow a pre-set voltage in the range 3.3V – 4.0V.

Circuit operation
The battery voltage is divided in half by R1 & R2 and fed into the non-inverting input of op-amp IC1 via a 10k resistor. The inverting input of the op-amp is adjusted by VR1 from 1.65V to 2.0V (achieved by the 2V zenner diode together with the two 1N4001 diodes).
The gain of the op-amp has been set at 100 by the values of R4 and R5.
There is an error in the diagram in that both inputs to the op-amp should be shown with a capacitor to 0V to reduce electrical noise impacting performance.
Should ½ the battery voltage exceed the voltage set on the op-amp inverting input (set by VR1), then the output of the op-amp will go high switching on transistor Q1.

The op-amp is a 7611 (or 7612), this op-amp can operate with a supply rail as low as 1V (this is needed since your supply to the op-amp could be less than 3V).

The op-amp output is fed to the base of the darlington transistor Q1 with the resistor R3 (0.1ohm) in the collector path. If transistor Q1 is a TIP121 then the transistor can pass up to 8A and the resistor R3 should be rated at 10W. If a higher current is required then Q1 can be formed using a 2N3055 transistor combined with a suitably rated driver transistor to form a darlington – the current would then be increased to a maximum of 15A with R3 rated at 25W. Don’t forget that Q1 will need to be mounted on a suitable heatsink.

No fusing has been shown in the diagram, consideration should be given to adding fusing at appropriate locations given the current that could flow in the event of a fault.

There is a further error in my circuit, in that the op-amp feedback resistors R4 & R5 loading the input (more specifically R5) – therefore you should omit resistors R4 & R5 which will give a high open loop gain for the op-amp; but it will still work without R4 & R5 fitted.
Thank you for the work you put into this. However, It's a bit over whelming to a beginner.
 

Hymie

Joined Mar 30, 2018
819
Given the amount of time & money you have invested in putting an electric motor in the vehicle powered by the re-chargeable lithium batteries – if you feel that building such a circuit is beyond your capabilities, perhaps you could find someone who could build the circuit for you.

Whilst to a beginner the circuit might seem quite complex, in reality there are only a dozen or so components – anyone with the basic ability to solder components on strip-board should be able to complete the project from start to finish within an hour.

Given the cost of the op-amp in the circuit – it would certainly be worth using an IC socket rather than directly soldering the IC.
 

Thread Starter

BJNKM

Joined Aug 3, 2018
14
Given the amount of time & money you have invested in putting an electric motor in the vehicle powered by the re-chargeable lithium batteries – if you feel that building such a circuit is beyond your capabilities, perhaps you could find someone who could build the circuit for you.
Given the amount of time & money you have invested in putting an electric motor in the vehicle powered by the re-chargeable lithium batteries – if you feel that building such a circuit is beyond your capabilities, perhaps you could find someone who could build the circuit for you.

Whilst to a beginner the circuit might seem quite complex, in reality there are only a dozen or so components – anyone with the basic ability to solder components on strip-board should be able to complete the project from start to finish within an hour.

Given the cost of the op-amp in the circuit – it would certainly be worth using an IC socket rather than directly soldering the IC.
Given the amount of time & money you have invested in putting an electric motor in the vehicle powered by the re-chargeable lithium batteries – if you feel that building such a circuit is beyond your capabilities, perhaps you could find someone who could build the circuit for you.

Whilst to a beginner the circuit might seem quite complex, in reality there are only a dozen or so components – anyone with the basic ability to solder components on strip-board should be able to complete the project from start to finish within an hour.

Given the cost of the op-amp in the circuit – it would certainly be worth using an IC socket rather than directly soldering the IC.
Whilst to a beginner the circuit might seem quite complex, in reality there are only a dozen or so components – anyone with the basic ability to solder components on strip-board should be able to complete the project from start to finish within an hour.

Given the cost of the op-amp in the circuit – it would certainly be worth using an IC socket rather than directly soldering the IC.
Thanks! I did find a friend that can assemble the first one for me.
BTW. My 1997 Chevy S10E is factory built. They share a similar front wheel drive train as the EV-1. Just under 500 trucks were produced. Most were leased but about 150 were sold to Military and utility companies. (The only ones that could afford them) The MSRP in 1997 was around $43K. The leased trucks, along with all of the EV-1's were destroyed. The last info I've heard was there's less than 35 of these trucks still on the road. I've had mine for over 10 years and it's a constant learning experience.
 

Hymie

Joined Mar 30, 2018
819
Thanks for posting that you have managed to get the circuit working as advised – it’s good to know that my circuit design worked.

Don’t forget to keep a copy of the circuit (and the description of how it works) with the vehicle; then should a fault or problems occur at a later date, you (or someone with more electronics knowledge) might be able to diagnose/correct the issue.
 
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