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How does the resistance theoretically behave for most commercially available photovoltaic modules, when an external DC voltage is applied to them, with and without illumination?

It's common to wire solar panels of the same voltage in parallel, in order to provide greater current or greater resilience to partial shade. Presumably, it can be inferred from this that solar panels consistently have considerable resistance (relative to their rated voltage) when not illuminated— otherwise, having different light intensities on the parallel modules would cause significant current and waste heat to go through the panels at a lower voltage.

Is this correct? Do solar panels always/generally have enough resistance to prevent an external voltage around their nominal voltage from inducing a current in them when they're not illuminated?

If so, what is the behavior of commercially available photovoltaics as that resistance is challenged and overcome?

Does their output power decrease when external voltage is applied to them, and as that voltage is increased?

What happens to their resistance (and at what rate) as light starts shining on them and they start producing their own voltage?

How does one determine the maximum external voltage to which a solar panel can safely be subjected? Presumably, at some point, you'll overcome its resistance and either send a lightning scorch mark across it or melt it (if there aren't any quantum or metallurgical shenanigans before then). Is that point a product of its material, open-circuit voltage, or manufacturer decisions? And what exactly happens when you reach it— What is the failure case from shorting (an external voltage) across a solar panel, and how do you get there?

And lastly, how non-linear or asymmetric are these behaviors? Without illumination, do solar panels have higher resistance in one direction than the other? Are they more vulnerable to being burned by external voltage in one direction than the other?

The specific motivation behind this question is to be able to use solar arrays in parallel with other DC power sources without modifying the load (and ideally while gaining the benefits of both and damaging neither), but information about the mechanistic basis of such a setup would be much more valuable.

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    \$\begingroup\$ Solar panels are essentially very large photodiodes. They have a reverse breakdown voltage like all diodes, but it is not very large, and for this reason a separate "blocking diode" is recommended if there is a risk the panel may be subjected to reverse voltage. \$\endgroup\$
    – pjc50
    Commented Sep 2, 2020 at 8:03
  • \$\begingroup\$ Individual per-panel diodes are usually added either in single or parallel use. || A deeply unilluminated panel will draw very little reverse current when eg used to charge a battery - voltage wise the same as the parallel panel source. I|| If you have N panels in parallel without diodes and one is shaded it's Voc will be not much lower than fully illuminated panels. As the other panels are loaded their Vloaded will be about 80-85% of their Voc. If you apply a voltage to the shaded panel it will be driven up its load curve to a point where it makes less current at the applied voltage. | \$\endgroup\$
    – Russell McMahon ♦
    Commented Sep 2, 2020 at 12:14
  • \$\begingroup\$ Covering just one cell in a large panel will increase its resistance to the point where it produces 10% of its current or less. If you are operating partly shaded solar panels, look for ones with bypass diodes across each string within the panel to prevent this. In larger installations with several panels in series, you can fit external bypass diodes across each whole panel. \$\endgroup\$
    – user16324
    Commented Sep 2, 2020 at 12:48
  • \$\begingroup\$ @WillChen you might want to reduce the number of questions in one post. (I count 13) Thanks \$\endgroup\$
    – Voltage Spike ♦
    Commented Sep 3, 2020 at 20:40
  • \$\begingroup\$ @Voltage Spike Those "13" questions are all subsets of the single, specific question of how resistance behaves in solar panels. If you were to answer that single question with a complete equation, then that equation would also entail answers to those "13" sub-questions. I listed them out separately to indicate that yes, in fact, I would appreciate that level of comprehensiveness and precision (as opposed to a general rule of thumb or application-focused suggestion). \$\endgroup\$
    – Will Chen
    Commented Sep 4, 2020 at 7:32

2 Answers 2

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Individual per-panel diodes are usually added either in single or parallel use so this is not usually an issue.

A panel with almost no illumination (= dark) will draw very little reverse current when eg used to charge a battery - voltage wise the same as the parallel panel situation.

What you actually asked about :-) : If you have N panels in parallel without diodes and one is shaded it's open-circuit-voltage (Voc) will be not much lower than fully illuminated panels. Voc does not start to drop appreciably until illumination is around 5%-10% of maximum.
If the other panels are optimally loaded their Vloaded will be about 80-85% of their Voc. Even non-optimally loaded panels will probably have Vloaded of not more than 90% of Voc.
If you apply a voltage to the shaded panel it will be driven up its load curve to a point where it makes less current at the applied voltage so no reverse current will occur - it will simply provide ABOUT the proportion of full current that the most shaded cell can make. Current output is very close to linear with insolation.

It would be difficult under normal conditions to shade a panel to the extent that it was receiving say less than 5% of the insolation of other panels in an array. If you covered a panel with a heavy tarpaulin or similar this may occur. In the situation between 5% and 0% illumination as Voc falls below Vloaded you MAY get some back current - but I'm not certain how much. At most I'd expect it to be no more than the %illumination - if so a say 350W 36V loaded panel may pass somewhere under 0.5A. I'd not expect that to be damaging. I may be wrong. Reverse bias may be a greater risk.

NB in the situation above I'm suggesting possible reverse current into a diode. Sounds wrong. I'm note sure it is. I'll have to think about it.

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  • \$\begingroup\$ It seems unlikely that someone asking this question knows what "Voc" is. You may want to make it clear. \$\endgroup\$
    – Matt
    Commented Sep 2, 2020 at 18:05
  • \$\begingroup\$ @Matt 2+ years on - I just saw and implemented your suggestion :-) \$\endgroup\$
    – Russell McMahon ♦
    Commented Nov 14, 2022 at 4:07
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Is this correct? Do solar panels always/generally have enough resistance to prevent an external voltage around their nominal voltage from inducing a current in them when they're not illuminated?

It's a bit simpler than the ideas described in the OP

It's not resistance a solar panel has a bypass diode between cells to shunt current away from the cells (or cell groups) that are not producing sufficient voltage. If you didn't have the bypass diode, the shaded cell could sink current which would heat it up and degrade or destroy it.

enter image description here
Source: https://www.pveducation.org/pvcdrom/modules-and-arrays/bypass-diodes

As far as nonlinear effects:

The conducting diode around the shaded portion then consumes a fraction of the cells power output and changes the IV curve (dropping the max voltage by a fraction and creating another 'knee' in the IV curve). If multiple cells are shaded, then it can drop the voltage even further, which could have implications for cells connected to a converter with an MPPT controller.

enter image description here Source: https://www.pveducation.org/pvcdrom/modules-and-arrays/bypass-diodes

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