Li-Ion automated charge equaliser

I have built a simple clockwork logic charge equaliser for a series connected battery of 8 Li-Ion cells. The design is based upon the familiar flying capacitor charge equaliser but this uses flying cells rather than flying capacitors.

Testing the unit required an automated battery management device to periodically measure the terminal voltage of each cell whilst alternately applying a load and then charging the battery to see how well the equaliser works.

I have chosen to refer to the device as an equaliser to distinguish it from the more common technique of ‘balancing’ which usually involves the application of dump loads across individual cells should the terminal voltage exceed some pre-defined limit.

This approach has several advantages:

1. No wasted power

2. No software

3. Low cost

4. Increases the battery capacity by (2 * cell count)-1

5. Maximises energy transfer by reducing outliers

The reason for this post is to see what other engineers think about the basic idea. As far as I can tell this is not something that is commonly done, in fact I have not seen anything published regarding this approach.

No doubt most people will realise that 2018 will see the start of an explosion in the use of battery technology and currently Li-Ion is one of the leading chemistry types so this is a very topical subject with huge application potential.

As this is my first post I shall not overload it with schematics and circuit descriptions but simply provide some empirical data obtained by operating the device.

In the following graphs the cells are reclaimed 18650 cells taken from old laptop batteries. The charge current is 405mA and the discharge rate is 1.3A. The horizontal axis is time with a tick interval of 10 Seconds. The vertical axis is in volts. The top two traces indicate the application of charge current or load.

The BMS is set to apply a 1.3A load until the lowest cell voltage reaches 3.5V. It then turns off the load and switches on the charger. The charger stays on until one of the cells reaches 4.0V when the charge is switched off and the load is re-applied.

The first graph is without the equaliser. You can see that cell 7 has more charge than the others and cell 1 is a bit low. As expected, the performance remains the same for multiple cycles because there is no cell balancing.

The second graph simply carries on where the first graph stops but this time the equaliser is switched on. The initial discharge looks similar to the first but as charging takes place the charge on cell 7 is being equalised with cell 6 which flattens the rate of rise on cell 7.

You will also notice that without the equaliser charging terminated after 65 minutes but with the equaliser charging took 188 minutes on the second cycle.

Static cells 1 and 8 have only one flying cell to share power with therefore I would expect them to take longer to stabilise than the other static cells.

The power required to charge to 3.8V without the equaliser fitted is 13.5Wh. The power drawn is 13.4Wh to reach 3.5V

The power required to charge to 3.8V with the equaliser fitted is 29.5Wh. The power drawn is 29.3Wh to reach 3.5V

So, the equaliser provides an additional 15.9Wh or + 118.7% for + 87.5% extra cells.

My conclusion so far is that the equaliser is highly beneficial. Not only are the cells now working in harmony the energy available from the pack is increased by 118.7% with no wasted power.

I intend to carry out more tests and will be interested to hear what others think. Also, it would be great if anyone has data they can share using alternative methods.

It's clear that the approach works, but I don't think that it would be as good as balancing resistors that are switched on when needed.

The test at the start showed that the cells weren't getting better balanced, but nor were they getting worse. There would be quite a lot of energy lost getting the cells balanced using resistors, but once they were balanced, the power loss would be minimal.

I don't think that you can justify the claim of no wasted power. The wasted power might be a lot less than would be the case with resistors, but using a cell to transfer energy from a higher voltage cell to a lower voltage one will inevitably waste some power, because the energy taken from the higher voltage cell is the higher voltage times the charge, and the energy passed to the lower voltage cell will be the lower voltage times the charge. The charge will be just about the same.

The percentage by which the capacity is increased depends entirely on how badly the cells were balanced to start with, so there could be a vast percentage increase if the some cells were completely flat and others completely charged, while a balanced pack would see little improvement.

Having what is in effect a second battery pack just for balancing seems like overkill and a waste of a lot of cells. You could possibly have just one balancing cell, which could be smaller that the others, and more elaborate switches to connect it to each cell of the main pack in turn. That would slow the equalising a lot, but there is no rush.

Thank you Mr Diver, that is the most thorough analysis of my circuit that I have so far received and I am very grateful.

Diver300 said:

t's clear that the approach works, but I don't think that it would be as good as balancing resistors that are switched on when needed.

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I agree if you only consider the balancing aspect. If you also consider that the flying cells increase the pack capacity then this approach has additional merit.

Diver300 said:

The test at the start showed that the cells weren't getting better balanced, but nor were they getting worse. There would be quite a lot of energy lost getting the cells balanced using resistors, but once they were balanced, the power loss would be minimal.

Click to expand...

I agree with that.

Diver300 said:

I don't think that you can justify the claim of no wasted power. The wasted power might be a lot less than would be the case with resistors, but using a cell to transfer energy from a higher voltage cell to a lower voltage one will inevitably waste some power, because the energy taken from the higher voltage cell is the higher voltage times the charge, and the energy passed to the lower voltage cell will be the lower voltage times the charge. The charge will be just about the same.

Click to expand...

That is what I expected to happen but the losses due to energy conversion efficiency is extremely small for these cells so a tiny increase in a small loss becomes insignificant.

Diver300 said:

The percentage by which the capacity is increased depends entirely on how badly the cells were balanced to start with, so there could be a vast percentage increase if the some cells were completely flat and others completely charged, while a balanced pack would see little improvement.

Click to expand...

I agree with that.

Diver300 said:

Having what is in effect a second battery pack just for balancing seems like overkill and a waste of a lot of cells. You could possibly have just one balancing cell, which could be smaller that the others, and more elaborate switches to connect it to each cell of the main pack in turn. That would slow the equalising a lot, but there is no rush.

Click to expand...

The flying cells effectively become part of the pack so are not wasted at all. It is common practice for applications such as power tools batteries to be constructed with multiple cells in parallel. This is not so very different.

Brian Drury said:

The flying cells effectively become part of the pack so are not wasted at all. It is common practice for applications such as power tools batteries to be constructed with multiple cells in parallel. This is not so very different.

Click to expand...

I may not have understood your circuit. I was under the impression that there were 8 main cells and 7 flying cells. If that were the case I couldn't see how you could get more than the AmpHour rating of one cell from the set, while still supplying the voltage from 8 cells.

A good start will be a simplified diagram plus circuit description:

S1, S2 & S3 are static cells wired in series.

F1 & F2 are flying cells.

Q1 – Q8 are MOSFET’s.

The MOSFET’s are switched in pairs. Q1 & Q3, Q2 & Q4. (Q1 & Q5 are P type)

The top 14047 controls Q1, Q2, Q3 & Q4.

So, if Q1 & Q3 are ON then F1 is in parallel with S1. If S1 has more charge than F1 then current will flow from S1 into F1. Alternatively if F1 has more charge than S1 then current will flow from F1 into S1.

When Q2 & Q4 are ON then F1 is in parallel with S2. If S2 has more charge than F1 then current will flow from S2 into F1. Alternatively if F1 has more charge than S2 then current will flow from F1 into S2.

The bottom 14047 controls Q5, Q6, Q7 & Q8.

The basic action is as above so charge is distributed S2 S3 & F2. Also, because the two 14047 are not synchronised there will be occasions when F2 is in parallel with S2 and F1.

If we start with S1 charged and S2, S3, F1 & F2 discharged then eventually the charge from S1 will be evenly distributed between all 5 cells.

The flying cells are switched at about 1Hz and the switch transition time is nanoseconds therefore the flying cells can be considered to be in time division parallel with the static cells. The ‘OFF’ time is a tiny fraction of the ‘ON’ time.

The 8 static, 7 flying cell version is simply a scaled up version of the example shown.

Conceptually, we have 8 static cells in parallel with 7 flying cells. Taking this one step further an individual electron may consider the static cells to be in series and in parallel at the same time. This takes a bit of thinking about.

Hi Brian
How are your efforts with this circuit going? Any new insights? Any unforeseen shortcomings?

I've come across an IC designed to operate on lead acid cells in a similar manner: LTC3305 (@US$15/pc!).
The circuit uses a 1:2 to 1:4 topology, and can be stacked as required.

A few observations about your design:
- You've indicated that the overall capacity of this topology is (2N-1). This relationship doesn't apply to the power capability nor charge rate of the overall bank.
If power and/or charge rate are critical to the application, you may reduce the capacity of the flying cells wrt the static ones.
Hence there is flexibility in balancing capacity vs power vs cell cost.
- Your circuit has no provision for limiting current through the transistors. Extreme imbalances may violate the transistors' maximum drain current spec.
- I've (mentally) added intrinsic diodes to the transistors of your circuit.
It looks to me that during the phase when S2 & F1 are paralleled, S1 would self-discharge through Q1's parasitic diode and open Q2.
Without obsessing too deeply over your circuit (nor simulating) it, this would seem to be a show-stopper.
What am I missing?

Gaudeamos said:
You've indicated that the overall capacity of this topology is (2N-1). This relationship doesn't apply to the power capability nor charge rate of the overall bank.

I have included a circuit diagram of what I eventually built. It is not easy to appreciate but because the circuit is constantly asynchronously switching, the cells are effectively in parallel in a time division multiplex fashion. You need to really think about it to understand.

I applied fusing to each section using narrow width tracking that bridges 1206 footprints.

Thanks for the updated schematic, Brian.
This one's a bit more difficult to figure than the Jan '18 version! I'll need a few more moments to wrap my brain around it .
I notice that the flying cells are now isolated from each other. Why did you find that necessary?

The file path of the posted schematic implies that you're building a powerwall.
So I'm guessing that the entirety of your bank consists of multiple 8s strings paralleled.
Will there be a balancing unit (PCB+7 fliers) wired to each of these 8s strings?
I see that you've omitted the actual cells from your schematic in favour of connectors.
Is this because (a) you'll connect the balancer to individual 8s strings as part of a periodic maintenance procedure? or (b) you ran out of real estate on your schematic? or ...?

Brian Drury said:

You need to really think about it to understand.

Click to expand...

Ok I'm trying hard. Here's my reasoning sofar:
It's clear that a bank of (2N-1=15) cells configured in this fashion will yield the full summed capacity, over time: (1+7/8)C=1.875C Ah (C = Ah capacity of each cell), or 15*3.7V*C=55.5C Wh. ... ignoring all reality-based imperfections!
OTOH: Instantaneous power (P=VI) and instantaneous current of the entire bank is as it would be for a simple 8S bank.
That's because current capability and discharge/charge tolerance characteristics of a series is limited to the characteristics of the weakest cell in the string.
Since there is one less flying cell than there are stringed cells, there'll always be at least one cell in the series that isn't paired with a flier.
This unpaired cell represents the weakest link in the chain. Hence my comment:

Gaudeamos said:

This relationship doesn't apply to the power capability nor charge rate of the overall bank.

Click to expand...

Keep up the awesome engineering
Ewald

Brian Drury said:

I applied fusing to each section using narrow width tracking that bridges 1206 footprints.

Click to expand...

How will you proceed whenever one of these PCB fuses opens?

I made a short video a while back. Here is a link. I shall try to respond to your questions later today.
https://drive.google.com/file/d/1OJ7YnMEtfl8liz8pLs0RxxNHbE1Rwuh6/view?usp=sharing

Another video:
https://drive.google.com/file/d/1TrQzT21WmnxgkYu_fgmOeHQMZztvcL4o/view?usp=sharing

This drawing should answer your questions regarding the battery configuration. The balancer is not included in the diagram.

Gaudeamos said:

How will you proceed whenever one of these PCB fuses opens?

Click to expand...

The PCB fuses are not expected to open (famous last words) but should they do so the plan is to simply bridge the 1206 pads with a small piece of wire.

Agreed, flying cells increase the battery capacity but not the instantaneous current capability. Good job I only require 0.5A per cell.

You asked about the resulting current for the worst case voltage differential between flying and static cells.

Did you try some numbers?

Say max V = 4V2, min V = 2V5, difference is 1V7
Typical cell internal resistance 0R1
So, using I = V/R 1V7/(0R1 X 2) = 8.5A

No problem

I have built this thing and tested it in several different ways and it does indeed do all that I thought which is encouraging but now I have some more facts to work with I am able to reach a conclusion.

I received some good advice from other engineers who suggested that even if the equaliser can achieve all that I claim it is simply not worth the cost and effort because the dissipative method using an ‘intelligent’ BMS is much easier to do and is probably cheaper.

My tests have shown that even using reclaimed 18650 cells the degree of imbalance is small and can be easily corrected using resistive loads.

So, I conclude that the current industry standard dissipative methodology is probably the best way to balance a series connected pack of 18650 cells. Even Tesla use this approach in their cars!

Brian Drury said:

I made a short video a while back.

Click to expand...

I'm not able to view (to load?) these videos on my Chromium/Linux Mint platform.

Brian Drury said:

This drawing should answer your questions regarding the battery configuration.

Click to expand...

Brian Drury said:

Say max V = 4V2, min V = 2V5, difference is 1V7
Typical cell internal resistance 0R1
So, using I = V/R 1V7/(0R1 X 2) = 8.5A

No problem

Click to expand...

I see 16 x isolated/OR'ed 8s strings of 18650's. Sitting back in my chair, in front of the computer , I was picturing a somewhat larger scale.
This guy's HBPowerwall is based on "cell packs" consisting of 18650 x 80p cell modules, stacked to 7s/24V banks (eventually 80p 14s/48V)... a lot more energy, lower resistance there.
That basic 80p unit enable him to leverage a single balancing circuit over 80 x 7/14 = 560/1120 18650's.
I'm not sure if he ended up with active balancing; for the longest time he was satisfied with a periodic manual balancing procedure.

"Total 100 cells at roughly £3 ea" - what is your source at that cost?

Brian Drury said:

So, I conclude that [...]

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Good on ya for sharing what you've learned... and for sharing how you arrived at that wisdom.
My hat's off to you for your efforts in determining the practicality of it all.
May others benefit from your work!

cheers Ewald

Gaudeamos said:

I'm not able to view (to load?) these videos on my Chromium/Linux Mint platform.

Click to expand...

I have no idea what a Chromium/Linux Mint platform is. Maybe you could try a Windows computer or an iPad or iPhone or anything that runs a browser like Firefox.

Failing that download the files and use something that can view divx format.

Brian Drury said:

I have no idea what a Chromium/Linux Mint platform is.

Click to expand...

Chromium = open-source version of Google Chrome
Linux Mint = my preferred variant of the Linux O/S family