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Comments on Mismatch for Cells Connected in series


Hi could you please tell a bit more on how you calculate the Isc of the combination? why is Cell 1 reflected across the current axis? (and not cell 2 for instance). Thanks

Hi Leocalau,

I've spent a LONG LONG time trying to understand mismatch, despite being heavily educated I have been struggling with mismatch and bypass diodes more than any other concept..

I think I've finally figured it out. Let's look at parallel first then series.

For parallel it is useful to realise that:
If you want to figure out the Voc combined of the parallel configuration then notice that it occurs when the combined current is zero (by definition open circuit implied zero current) - so to do this you have to find a point where the I1 and I2 sum to give zero. This can only occur when one current is positive and the other is negative.

For the good cell - I1 is the upper right quadrant of the graph, for the bad cell the negative I2 is the lower right quadrant of the graph, to visualise this on a graph we can easily you can flip one of the cells - it doesn't really matter which one you flip. We know in our head that we've flipped, we're just doing this to see where the values of I are identical for I1 + I2 = 0.

And so the result is that voltage for combined current equal to zero is actually slightly higher than Voc2 (bad cell).

NOW for series:
You do the same thing, by identifying the current where the voltages sum to give zero (short circuit definition). For the good cell we use V2 in the top right quadrant and a voltage from the bad cell in the top left quadrant - you can then visualise this by flipping one of the cells across the current axis. Again we know in our heads that one is positive and one is negative - we do this just to see where the lines cross for a combined voltage of zero (ie. the short circuit condition)

You will notice that the combined short circuit current is actually slightly higher than the Isc of the bad cell.


Okay.... now let's think - we thought the bad cell limited the current of the system and we could not get a current higher than the current of the bad cell - YES this is true for typical operation however this is no longer true in the short circuit position (when combined voltage = zero).

So what happens next... This bit I'm not too sure on but I will try to explain and perhaps someone can correct me if I am wrong.

We need to think of 3 things - the bad cell, the good cell and the combined cells.

Let's consider the combined cell (system) being operated at Short Circuit conditions.. The good cells are not actually operating at their own personal short circuit conditions, they can can actually operate at the combined new
Isc whilst remaining in positive voltage without any problem.

The bad cell, however, can be forced to experience this combined current Isc but to do so it is driven at a negative voltage. Remembering that it must operate on the IV curve. As I is positive and V is negative then the bad cell is now consuming power.

K.Gibson (under construction)

Thanks for the better explanation !!

It doesn’t matter which cell is flipped. There will only be one intersection point that the is the operating point of the combination of the cells.

This is great, but the animations do not work..

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