How Big are Todays Solar Panels?

[Cruisin Cat]

Thanks everyone for your comments...some really great info here!

I have one additional question about the exact mechanism of output loss when part of a panel is in shade. Is this due to....

a) the shaded cell(s) allow the current from other non-shaded cells connected in parallel to run backwards through the shaded cells?

or

b) the shaded cell(s) act as an electrical block to the current from other non-shaded cells that are connected in series with the shaded cell?

Or some other mechanism that I haven't thought of?

[s/v Jedi]

Originally Posted by roetter

I am not sure what you mean by "the best we got was 130a/h" Do you mean you got 130Ah in one day? That would make more sense.

I also don't understand how they measure this.. I can't. What voltage would it be? If 13.5V then it would be 130*13.5=1.75kWh which would be super. My record is up to 2.7kWh with a 660W array, so I guess his voltage was a bit lower.

[s/v Jedi]

Originally Posted by Cruisin Cat

Thanks everyone for your comments...some really great info here!

I have one additional question about the exact mechanism of output loss when part of a panel is in shade. Is this due to....

a) the shaded cell(s) allow the current from other non-shaded cells connected in parallel to run backwards through the shaded cells?

or

b) the shaded cell(s) act as an electrical block to the current from other non-shaded cells that are connected in series with the shaded cell?

Or some other mechanism that I haven't thought of?

It's your option B: a cell that is shaded puts itself in a state of high impedance, blocking the current. The bypass diode then takes the whole string out. I don't know how many cells and strings this panel has, but orient it so that a re-occuring shade aboard like from the boom or mast, does not go across all strings which would kill them all. The strings should be oriented in parallel with the shades, minimizing the problem.

This is exactly what many don't do or don't understand. Planning for this can improve daily output by 25%.

[Paul Elliott]

Originally Posted by s/v Jedi

It's your option B: a cell that is shaded puts itself in a state of high impedance, blocking the current. The bypass diode then takes the whole string out. I don't know how many cells and strings this panel has, but orient it so that a re-occuring shade aboard like from the boom or mast, does not go across all strings which would kill them all. The strings should be oriented in parallel with the shades, minimizing the problem.

This is exactly what many don't do or don't understand. Planning for this can improve daily output by 25%.

This is 100% correct.

As for shadow orientation, the advantage to optimal panel orientation can be much better than 25%. My three panels are on top of the dodger, under the boom. This is one of the worst locations I could have chosen, but I did so with my eyes open. The three are oriented port, mid, and starboard. On average one of the three is badly shaded. I generally get 2/3 of the potential panel output.

Had I installed the panels fore-to-aft, the boom shadow would typically be running across all three panels, reducing the output horribly. In most conditions I would still get perhaps 10 or 20% of the potential output because the shadows are rarely "black" -- cloud cover and water vapor diffuse the light enough to wash out the shadows a bit.

[gspeak]

Originally Posted by roetter

I am not sure what you mean by "the best we got was 130a/h" Do you mean you got 130Ah in one day? That would make more sense.

Sorry, yes the best I got was 130ah in one day from the one Sunpower 327W panel.

Obviously this is dependant on how depleted the batteries were also. As I said this was just over 4 days over Easter break. This is reported by the data logging in the LiveView web interface from the Morningstar MPPT controller which I have connected by ethernet to my wifi router & viewed on my laptop. It also reports the kwh but I don't remember what that was.

Dave