Solar charging with some shade

[2oldman]

I'm facing west today, so I'm getting a pretty good shadow on one of six series-connected panels. Panels are all around 135w @ 12v.

Shade is on the far-left panel - about half of it:




Not a bad charge for that big of shade- 51.9v x 10.6a



The shaded panel is the last in a series string. It's a Kyocera. Looks like the diodes are working well.

[2oldman]

Salvo wrote:
I didn't know we were discussing your string in particular.

Some of us are discussing mine, some of us are not.

[Salvo]

I didn't know we were discussing your string in particular. If your string isn't showing the massive current reduction then it's due to bypass diodes. There shouldn't be any confusion.

2oldman wrote:

Salvo wrote:
if one of the batteries malfunctions and can only supply 6V at 5A then the string outputs 36V at 5A, or 180W

Of course, but what he's questioning is the fact that with partial shading, my string isn't showing a massive reduction in amps, like from 8 to 3.

[2oldman]

Salvo wrote:
if one of the batteries malfunctions and can only supply 6V at 5A then the string outputs 36V at 5A, or 180W

Of course, but what he's questioning is the fact that with partial shading, my string isn't showing a massive reduction in amps, like from 8 to 3.

[2oldman]

BFL13 wrote:
Are the pigtails so long they reach the J box? I was thinking there would be an MC4 at the end of one of the pig tails you could undo.

Nope. No MC4 anywhere. All hard-wired to enclosed screw terminals, one long continuous #10 standard extension cord connected to two #10 wires on the roof down to basement.

[Salvo]

Forget solar and look at 6V batteries. These batteries are current limited to 10A each.

If you put 6 batteries in series you get 36V. The maximum current the series string can output is 10A. Max power is 360W.

If one of the batteries malfunctions and can only supply 6V at 5A then the string outputs 36V at 5A, or 180W.

If we have a switch that can sense a battery malfunction and bypasses it from the string, then the string outputs 30V at 10A, or 300W.

2oldman wrote:

BFL13 wrote:
If that shaded panel is half by-passed, then it is putting out half its current? (I really am confused about the current amount)

Me too.. and good question!

[BFL13]

But with the blanket on, it is still in the string, so how can you compare with a 5 panel (unshaded)string?

Are the pigtails so long they reach the J box? I was thinking there would be an MC4 at the end of one of the pig tails you could undo.

Interesting readings. I wonder if as the shade gets less whether the diode(s) will go below trigger amount and stop doing any good and readings will crash before getting better after shade is gone.

[2oldman]

BFL13 wrote:
Would covering it completely with a blanket be like more shading (bad) or should you disconnect it from the string (maybe good) ?

The best you're gonna get is covering it! Wire connections are in jct box and sealed up with caulk.

85.6 x 7.0. Only the bottom row is now shaded.

[BFL13]

2oldman wrote:
Tomorrow. Interesting discussion for sure.

The constant amps argument makes sense, since all the cells in a panel are in series, and by now we should all know that the current is the same everywhere in series configuration.

83.8 x 7.3 right now with about 40% shading.

Would covering it completely with a blanket be like more shading (bad) or should you disconnect it from the string (maybe good) ?

[2oldman]

Tomorrow. Interesting discussion for sure.

The constant amps argument makes sense, since all the cells in a panel are in series, and by now we should all know that the current is the same everywhere in series configuration.

83.8 x 7.3 right now with about 40% shading.

[BFL13]

2oldman wrote:
You keep this up and I'm going to go up there and put a blanket on that panel!!

I would sure want to try that and see what happens. Value also depends on how long it is shaded during the day of course, as somebody else already mentioned.

If you have time and a second person handy, you could play with covering that back panel by different amounts up there and see what the controller's input Imp reading does.

[2oldman]

You keep this up and I'm going to go up there and put a blanket on that panel!!

[BFL13]

CA Traveler wrote:

BFL13 wrote:
If that shaded panel is half by-passed, then it is putting out half its current? (I really am confused about the current amount)

Because you keep returning to the same incorrect assumptions.

Try this: Assume you can magically rewire a shaded bypassed section of a panel so that the section is no longer in the circuit. Would you not agree that the panel is now producing the same current but with less voltage - say 1/3 less voltage with a typical 3 diode panel?

Also we're discussing array voltage and array amps. Of course the controller output amps are now less as it now has less input power due to the reduced input voltage and it has to maintain the output voltage to the battery.

Ok I did ask that before, so it is confirmed that a single cell in a panel has the same current as the whole panel's current and it is just the individual cells' voltages adding up in the "internal string" that makes its watts using that single cell's amps as the A in the VA.

So shading that one panel does not reduce its amps and so not string amps either; there is no amps bottleneck.

So why does the shaded panel knock down string watts so much? It would be like having an 18 cell panel in series with five 36 cell panels so the voltage of those 18 cells would still add to the string's voltage and it should do better than having just the 5 36s un-shaded.

But that is not the case. In the OP, his input watts is 77.8 x 7.3 = 568W If he disconnected that shaded panel, he would have (using near full Vmp 17.6 x 5= 88 Vmp x that 7.3 he was getting at that insolation = 642w

Why did the shading make his Vmp 77.8 with all 6, when it would have been 88 with 5 un-shaded?

EDIT--that single cell might have the amps but it can't do them until it is connected to some tiny thing with less voltage than it has.

[NinerBikes]

Moral of the story, for full output capabilities of your solar panels, design so it's impossible to shade your cells. Park accordingly, adjust accordingly, or have your batteries fully juiced up to full charge by the time shade hits one of your solar panels. Everything else is moot.

[2oldman]

CA Traveler wrote:
The current output from the panels is dependent upon the amount of sun and can vary from 0 to the panels Imp. Shadows that activate bypass diodes reduce the voltage and don't effect the current from the panels.

This explanation makes sense to me..with a bit of editing.. 'The voltage output..' (not current - meaning amps)

I haven't done much research yet, but I'm thinking that you may not lose BOTH amps and volts. When you connect batteries up, you gain volts or amps, not both. So it would stand to reason the same is true when shade is killing part of a panel.

It's easy to see how it works when 100% of the panel is shaded, but what isn't clear is when 2/3 of it is. If the amps are cut by that amount, the current drop would be substantial, and drag down the entire string. So, one may cautiously conclude that it's the voltage that's cut, not the amperage.

BFL13 wrote:
If that shaded panel is half by-passed, then it is putting out half its current? (I really am confused about the current amount)

Me too.. and good question!

[CA_Traveler]

The above is basically what a bypass diode does when it bypasses a panel section.