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Battery Management Systems for Large Lithium-Ion Battery Packs

Section 6.1.1.1.1 Addenda

Cells in parallel vs. strings in parallel

Here is one more reason why cells in parallel are better than strings in parallel.

In order to estimate cell resistance, the cell current must be known. With strings in parallel, the cell current is the string current, not the pack current. The typical BMS can only measure one value for current, and that would be the pack current. There is no way to infer string current from the pack current, because in real world applications the pack current will not divide equally among the series strings. Therefore, in systems that use strings in parallel, and a BMS that reads a single value of current, the only way to estimate resistance is to have one BMS per string.

Therefore, the only way to do IR compensation in a system that uses strings in parallel is to have one BMS for each string.

Cells in parallel

Additional section.

Having shown that cells in parallel perform better than strings in parallel, one must also consider the effect of paralleling cell of differing resistances on the life of the cells.

Consider 2 cells of equal capacity but significantly different resistance.

First, let's only look at a simple model for the cells, without any dynamic components.

When not in use, should the two cells have differing SOC levels, current flows naturally from the cell with the highest SOC (because it has the slightly higher voltage) to the other cell, until the two cells are balanced.

At low currents, the voltage drop across the cells' resistances is low, so both cells are used equally: the current is divided roughly equally between them.

At high currents, though, initially the top cell (with the lowest resistance) does most of the work. But that discharges the top cell more than the bottom cell. Consequently, the top cell's OCV drops more that the bottom cell's. Therefore, the portion of the output current sourced by the top cell decreases; at the same time, the bottom cell takes up more and more of the slack. After some minutes the current is divided nearly equally.

The time that it takes for the currents to settle until the 2 cells share the load equally depends on the ratio of the resistance of the 2 cells. Assuming that a 1 Ah cell has a resistance of 100 mOhm (and therefore a 10 Ah cell a resistance of 10 mOhm, etc.), then the following graph shows how log that will take.

As you can see, if the bottoms cell's resistance is twice, then it will take 30 minutes for the currents to equalize. If 5 times, it will take 100 minutes. This is regardless of the cell capacity, or of the discharge current; it only depends on the assumption that a 1 Ah cell has a resistance of 100 mOhm. Though, of course, should the current be sufficiently high, the top cell will be completely discharged before the currents have a chance to equalize.

Let's look as some examples, using the circuit above: the top cell has a resistance of 10 mOhm, and the total discharge is 20 A.

First, assume that the bottom cell is also 10 mOhm.

The equalization time is 0. After 1 hour, both cells are completely discharged.

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Now, let's assume that the bottom cell is 20 % higher: 12 mOhm.

The equalization time is 15 minutes. After 59 minutes, the top cell is completely discharged, and the bottom cell still has a 2 % SOC left.

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Now, let's assume that the bottom cell is 50 % higher: 15 mOhm.

The equalization time is 25 minutes. After 58 minutes, the top cell is completely discharged, and the bottom cell still has a 4 % SOC left.

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Now, let's assume that the bottom cell is twice: 20 mOhm.

The equalization time is 35 minutes. After 57 minutes, the top cell is completely discharged, and the bottom cell still has a 8 % SOC left.

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Now, let's assume that the bottom cell is 5 times: 50 mOhm.

The equalization time would be 100 minutes, but, before then, after 47 minutes, the top cell is completely discharged, and the bottom cell still has a 37 % SOC left.

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Finally, let's assume that the bottom cell is 10 times: 100 mOhm.

Long before the currents would have a chance to equalize, after 40 minutes the top cell is completely discharged, and the bottom cell still has a 63 % SOC left.

So, you can see that, with 2 cells, each at a 1 C discharge, if the ratio of their resistances is more than a factor of 2, then the performance is severely limited.

This graph shows the unused, remaining SOC in the bottom cell versus its resistance.

You can see that, if one cell's resistance in more than 7 times larger than the other one, less than half of the energy in that cell gets used.

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In the previous examples we kept the total current constant to 20 A and varied the resistance. Now let's keep the resistance constant at a 2:1 ratio, and vary the current.

You will note that in all these examples, the time for the cell currents to equalize is about 30 minutes, regardless of the output current. Though, at a current above 1.5 C (per cell), the top cell gets discharged before the cell currents have a chance to equalize.

Let's start with a 1/3 C current per cell (6.67 A total).

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Now let's increase the current to 2/3 C current per cell (13.3 A total).

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Now let's increase the current to 1 C current per cell (20 A total). (We already saw this curve above.)

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Now let's increase the current to 1.5 C current per cell (30 A total).

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Finally, let's increase the current to 2 C current per cell (40 A total).

You will have noticed how the SOC remaining in the bottom cell goes up as the current increases.

This graph shows that relationship. Note that the unused SOC remaining in the bottom cell never goes above 38 %, no matter what the current.

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In the above examples we assumed LiPo cells. If we considered LiFEPO4 cells instead, whose OCV vs SOC curve is flatter (meaning that the OCV changes less as the cell is discharged), then the cell currents will take much longer to balance, exacerbating the problem.

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So far we used a simple model for the cells, without any dynamic components.

If we add cell relaxation to the model (an additional parallel RC in series with the terminal), to model the cells more accurately, we find that the initial wave shape of the cell currents is slightly different. However, after the RC circuit settles, the cell currents end up the same as what we saw above.

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So what does this mean?

First, it means that when you have cells in parallel of differing resistance, the cell with the lowest resistance does most of the work, which is bad; though later the load tends to be shared more evenly, which is good, even though not all the charge in the high resistance cells is available, which, again is bad. So it's better if cells in parallel are matched in resistance.

A secondary effect is on the cells' life. Because the low resistance cell works harder than normal, it will age more rapidly. That means that its resistance will increase more rapidly as time goes by, which implies that the unbalance in resistance tends to be self correcting, which is good. But, the aging of the low resistance cell is more rapid than if the resistances of the cells were equal, which is bad: a high resistance cell tends to drag down the performance of the cell in parallel with it.

So, not only is a high resistance cell bad for the performance of the battery, but it also tends to degrade a cell in parallel with it, because it is not doing its own share of the work, leaving the other cell to do more work.

In many cases there are more than just 2 cells in parallel, so the effect of 1 high resistance cell is not as noticeable.

book/6.1.1.1.1.add (last edited 2011-02-02 18:39:26 by DavideAndrea)