# Clarification on Effect of PV Module Bypass Diodes

#### wwhitney

##### Senior Member
Just wanted to double check my understanding of bypass diode behavior:

Without any bypass diodes, if a single cell in a string were shaded, it would become reverse bypassed at its breakdown voltage (20-30 V ?) and would pass full string current (could be 10A). So it would be a little heater of on the order of 200W-300W.

If you have a bypass diode for every 20 cells (e.g. 3 diodes on a 60 cell module), then only those 20 cells are affected by the one shaded cell. Each cell has a Voc of around 0.6V, and the 19 * 0.6 V = 11V the lit cells can generate is less than the breakdown voltage of the shaded cell. That means the shaded cell will only pass at most its Isc.

If the shaded cell is fully shaded, its Isc = 0, and the full string current will go through the bypass diode, with a voltage drop of around 0.7V. With 10A string current, the diode is dissipating 7W. No power is dissipated in the shaded cell.

But if the shaded cell is say, 50% shaded, its Isc will still be positive, maybe 5A. The bypass diode fixes the voltage drop along the 20 cell substring at 0.7V, but the string current could still split as 5A through the bypass diode and 5A through the partially shaded cell. The 19 lit cells could each generate about 0.5V for a total of 9.5V, so the shaded cell could be reverse biased at 10.2V. Meaning the shaded cell would still be dissipating 50W. That's less than the 200W-300W figure with no bypass diodes, but more than I was expecting.

Is the above basically correct (within +/-20% on the numbers), or is a significant error in my analysis?

Thanks,
Wayne

#### synchro

##### Senior Member
Your analysis looks correct to me. There is a lot of power dissipation in shaded cells that can affect long term power output and reliability.

#### pv_n00b

##### Senior Member
Just wanted to double check my understanding of bypass diode behavior:
Looks pretty good. Let's see if I can add bonus material.

Without any bypass diodes, if a single cell in a string were shaded, it would become reverse bypassed at its breakdown voltage (20-30 V ?) and would pass full string current (could be 10A). So it would be a little heater of on the order of 200W-300W.
The reverse bias breakdown voltage can vary based on the material and doping. It's something that would have to be specified by the manufacturer. The string Voc would be (N-1)*Voc(cell). String Vmp would be (N-1)*Vmp(cell) -Vbreakdown. The shaded cell would probably be permanently damaged if this goes on for too long.

If you have a bypass diode for every 20 cells (e.g. 3 diodes on a 60 cell module), then only those 20 cells are affected by the one shaded cell. Each cell has a Voc of around 0.6V, and the 19 * 0.6 V = 11V the lit cells can generate is less than the breakdown voltage of the shaded cell. That means the shaded cell will only pass at most its Isc.
If the shaded cell is fully shaded, its Isc = 0, and the full string current will go through the bypass diode, with a voltage drop of around 0.7V. With 10A string current, the diode is dissipating 7W. No power is dissipated in the shaded cell.
The voltage across the string of cells with the shaded cell will be the forward bias voltage of the diode, about -0.7V. The module Voc string would be 59*Voc(cell), the string Vmp would be 40*Vmp(cell)-0.7, and the 20 cells in the section with the shaded cell are bypassed by the diode. In the section with the shaded cell the Vmp total is 19*Vmp(cell)-Vshaded=-0.7 where Vshaded is the reverse voltage on the shaded cell which is less than the reverse breakdown voltage. The current through the shaded cell would be the very small reverse bias current. Kirchhoff still rules.

But if the shaded cell is say, 50% shaded, its Isc will still be positive, maybe 5A. The bypass diode fixes the voltage drop along the 20 cell substring at 0.7V, but the string current could still split as 5A through the bypass diode and 5A through the partially shaded cell. The 19 lit cells could each generate about 0.5V for a total of 9.5V, so the shaded cell could be reverse biased at 10.2V. Meaning the shaded cell would still be dissipating 50W. That's less than the 200W-300W figure with no bypass diodes, but more than I was expecting.
As long as the shaded cell is reverse biased but not past the reverse break down voltage it passes very little current, even if it is only partially shaded. Once the bypass diode is forward biased those cells no longer contribute to the module output. The current either goes through the forward biased diode or the cells, it won't get split.

Keep in mind that this changes when modules are connected in parallel. If the voltage applied to a module from the outside stays above the voltage that allows the diode on the shaded cell to bypass then the shaded cell will be forced into reverse breakdown. For instance if we have a 60 cell module in a string of modules in parallel with other strings and an inverter controlling the string voltage with MPPT this is what might happen; a cell is shaded and to activate the bypass diode the module voltage has to fall below 40*Vmp(cell)-0.7. If the inverter holds the voltage above this value using the MPPT circuit and the other parallel string voltages then the diode will remain open, the shaded cell will go into reverse breakdown, and the string current will be significantly reduced. The string voltage will drop slightly because of the reverse biased cell but the MPPT will still be on a local maximum, fat dumb and happy. Now if the inverter is smart it will lower the string voltage far enough to forward bias the diode and restore the full string current. This will be another MPPT local maximum at a higher power than the one it was on at the higher voltage. This is why we have shade smart inverters now, they search for other MPPT maximums and find the maximum of the local maximums.

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#### pv_n00b

##### Senior Member
Once you move to the right of U[SUB]L [/SUB]there is no current flow through the cell. In the graph quantities on I below the U axis is normal current flow through the cell and the module. U[SUB]L [/SUB]is Voc to most of us. Past that there is no current flow in the forward direction and the light is generating carriers that recombine and just create heat. This is a complex subject that is easy to misunderstand, I recommend looking to some internet resources that will do a much better job of exlaining it with graphs and equations than we can here.
Here's a start. https://en.wikipedia.org/wiki/Theory_of_solar_cells

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#### wwhitney

##### Senior Member
Once you move to the right of U[SUB]L [/SUB]there is no current flow through the cell.
That is not what the graph shows. Once you move to the right of U[SUB]L[/SUB], the direction of current flow through the cell changes to the opposite direction of the usual current flow, I*U changes sign, and the cell switches from being a power generator to a power dissipator.

But that's not the region of the graph that is important for the case of one cell that is partially shaded in series with other cells that are fully illuminated. Instead it is the region to the left of the I-axis. If you consider composing the I-U graphs of 20 cells, where 19 of them follow E[sub]5[/sub] above, and 1 of them follows, say, E[sub]3[/sub], then you will see that the behavior matches what I suggested in my OP. If it's not clear, let me know, and I can write it all out more explicitly.

Cheers, Wayne

#### pv_n00b

##### Senior Member
We can keep beating this to death but the fact is that the model you are using is only applicable under very specific conditions. If anyone really wants to nerd out about this here an accessible paper from NREL. It has a nice discussion of shading on Pages 30 which gives a detailed description of the shading effect in series and parallel connected cells. It's a student paper but has nice graphs and is fairly easy to follow even for non-physics folks. https://www.nrel.gov/docs/fy12osti/54601.pdf

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#### wwhitney

##### Senior Member
That paper is an interesting reference, I read/skimmed it. The section on shading discusses panel level performance when one cell is shaded on a module with or without bypass diodes.

However, the part of the OP under contention is specifically about what is happening in each cell of a substring in parallel with a bypass diode, when one cell is partially shaded and the others are fully illuminated, and the bypass diode activates. That is not addressed in the paper, and the information the paper provides is consistent with the details I worked out in my OP. When V on a cell goes negative, it will still pass I[sub]sc[/sub] or a bit more. So when the bypass diode activates, the substring is still a parallel current path, and the string current will divide between the two paths.

Cheers, Wayne