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I tried to measure the voltage between positive/negative terminals of a PV system and ground while the inverter of the system is switched off, I found that the voltage decreases with time (the PV cells were discharging).

I know that this is because the PV cells are acting like capacitors. So we need to disconnect the connection between any terminal and ground to recharge the cells.

Will the electrons run out if I keep connecting and disconnecting the terminals with the ground? This maybe happen because we have a limited number of electrons(phosphorous) in N-part and a limited number of holes(Boron) in P-part!

And what about the charging process for N and P parts, do we have any difference in speed of that process?

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    \$\begingroup\$ Do the PV cells even have a connection to ground or are they floating? Please provide a schematic. \$\endgroup\$ – winny Jul 22 '16 at 8:54
  • \$\begingroup\$ The PV cells don't have a permanent connection with the ground, I make that connection to measure the voltage only. \$\endgroup\$ – To Above Jul 22 '16 at 8:56
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    \$\begingroup\$ Can we have a circuit diagram of this system. \$\endgroup\$ – JIm Dearden Jul 22 '16 at 9:42
  • \$\begingroup\$ You are "violating" KVL. You must have a return/ground path from your PV panel or you can't measure anything. \$\endgroup\$ – winny Jul 22 '16 at 11:00
  • \$\begingroup\$ @winny It's been my experience that you can measure a voltage without having your system connected to ground. Even if it was connected to ground on one side, a high impedance multimeter would still cause the voltage from a PV panel to discharge slowly. \$\endgroup\$ – Voltage Spike Aug 16 '16 at 15:28
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With no ground connection, the intrinsic capacitance of the solar cell causes it to charge up. It reaches it's open circuit voltage across the + and - terminals. You got that part right.

What you possibly didn't think of is that it's actually a giant network of capacitors, rather than just a single one.

Also remember that in materials "open circuit" doesn't quite exist, and even insulators leak a tiny amount of current, so each one of those capacitors is being simultaneously being charged by the sun and leaking current to discharge the capacitors too. If the solar panel is uniformly constructed, you would expect half it's normal open circuit voltage on the +ve terminal, half on the -ve terminal (in the other direction), and your virtual ground to be somewhere in the middle of the panel.

Your multi meter is working by draining a little current - and equating the voltage to the current drained. Your reading is effectively moving that "virtual ground" you had before in the middle of your panel towards your measurement point because you are discharging some of the capacitors (and thus allowing others in the middle to charge).

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  • \$\begingroup\$ But what about the electrons in cell, will they run out after connecting the cell to the ground for long time? because we have a limited amount of electrons! \$\endgroup\$ – To Above Jul 22 '16 at 19:35
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    \$\begingroup\$ You don't run out of electrons, because when you have a circuit they are by definition going back in on the other side. \$\endgroup\$ – pjc50 Aug 15 '16 at 21:20
  • \$\begingroup\$ I understood that when we measure the voltage between positive and ground for example, we measure the voltage of the capacitance between positive and ground. And you are saying that the electrons will go back to the other side, but I can't imagine what is the complete path of the electrons! \$\endgroup\$ – To Above Aug 16 '16 at 6:13
  • \$\begingroup\$ Stop thinking of your insulation as perfect, and start thinking of it as just an extremely high resistance. It should make sense then. \$\endgroup\$ – DonkeyOaty Aug 22 '16 at 8:01

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