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Solar charging with some shade

2oldman
Explorer II
Explorer II
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.
"If I'm wearing long pants, I'm too far north" - 2oldman
86 REPLIES 86

BFL13
Explorer II
Explorer II
Ok, so that leaves whether you would gain string watts by disconnecting the shaded panel. Will the amps increase in the remaining 5 panels make up for the loss in total voltage to make more watts?

6 x 135 = 810w, 6 x 17v = 102v

5 x 135 = 675w, 5 x 17v = 85v

OP display shows 77.8v x 7.3a = 568w with the reduced string amps.

We don't know what the watts would be for the 5 un-shaded panels in the same solar conditions as the OP situation (you would have to try it out), but there is room there between 568 as was with one shaded of 6, and possible 675 (minus some for heat loss) at STC with five un-shaded.
1. 1991 Oakland 28DB Class C
on Ford E350-460-7.5 Gas EFI
Photo in Profile
2. 1991 Bighorn 9.5ft Truck Camper on 2003 Chev 2500HD 6.0 Gas
See Profile for Electronic set-ups for 1. and 2.

2oldman
Explorer II
Explorer II
BFL13 wrote:
My question was sort of how much you lose in front compared with how much you gain in back and what gets the most string watts with some of each
I won't know here.. as I said, doesn't make much difference in the shadow size. I have no idea where the diodes are.

If a panel is tilted too far, but still getting sun, the diodes won't operate and its wattage will drop a LOT.
"If I'm wearing long pants, I'm too far north" - 2oldman

BFL13
Explorer II
Explorer II
I think the important thing is where the shade lies on the back panel wrt the diode sections. If you know where the sections are exactly (which columns of cells) you can maybe only adjust the front panel down a little to clear a whole section for a big difference.

My question was sort of how much you lose in front compared with how much you gain in back and what gets the most string watts with some of each
1. 1991 Oakland 28DB Class C
on Ford E350-460-7.5 Gas EFI
Photo in Profile
2. 1991 Bighorn 9.5ft Truck Camper on 2003 Chev 2500HD 6.0 Gas
See Profile for Electronic set-ups for 1. and 2.

2oldman
Explorer II
Explorer II
I just tried lowering the front panel. With the sun's lower angle I can't make any appreciable difference in the shadow until that panel is near flat. So I'm leaving it.

When I get time I may try adding a couple mounting brackets to be able to scoot the shadowed panel's bottom back a foot. That will increase its tilt angle, but not anything significant.
pianotuna wrote:
If you parallel connect panels and the voltage on one is 2 volts while the voltage on the other is 12 volts, the 2 volt panel would act as a load--unless there are blocking diodes.
Yes, in series it's important to match amps. In parallel, match voltage.
"If I'm wearing long pants, I'm too far north" - 2oldman

pianotuna
Nomad III
Nomad III
Further along that line of thought.

If you series connect panels the amps of output will be limited by the smallest panel amperage (small hole).

If you parallel connect panels and the voltage is identical then amps will 'add'.

If you parallel connect panels and the voltage on one is 2 volts while the voltage on the other is 12 volts, the 2 volt panel would act as a load--unless there are blocking diodes.
Regards, Don
My ride is a 28 foot Class C, 256 watts solar, 556 amp-hours of Telcom jars, 3000 watt Magnum hybrid inverter, Sola Basic Autoformer, Microair Easy Start.

pianotuna
Nomad III
Nomad III
Hi 2oldman,

If you series connect two six volt batteries where one has a capacity of 2 amp-hours and the other has 200 amp-hours, then the capacity at twelve volts is only 2 amp-hours. (that darned small hole gets in the way big time).

If you parallel connect two twelve volt batteries where one has a capacity of 1 amp-hour and the other has 100 amp-hours, then the capacity at twelve volts will be 101 amp-hours.

2oldman wrote:


When you connect electrical devices together, you can increase volts or amps, but not both. Like batteries.. you get 12v from 2 -6v, but the amp-hours (capacity) remain the same.

Not sure I get the 2nd part..
Regards, Don
My ride is a 28 foot Class C, 256 watts solar, 556 amp-hours of Telcom jars, 3000 watt Magnum hybrid inverter, Sola Basic Autoformer, Microair Easy Start.

pianotuna
Nomad III
Nomad III
For amps, think of a small hole in a 55 gallon drum of water. That limits the flow of the entire barrel.

For volts that same hole will allow more flow with greater "pressure".
Regards, Don
My ride is a 28 foot Class C, 256 watts solar, 556 amp-hours of Telcom jars, 3000 watt Magnum hybrid inverter, Sola Basic Autoformer, Microair Easy Start.

SCVJeff
Explorer
Explorer
I have the same problem but nowhere near as severe on the shadow, and of course it's only at the lowest sun angle of the year; durning Quartzsite ๐Ÿ˜ž

I have to dip the front panel so far back for it to clear the shadow off the rear panel, I just leave it flat and I still make 30A+

Won't be climbing up there this year tho
Jeff - WA6EQU
'06 Itasca Meridian 34H, CAT C7/350

BFL13
Explorer II
Explorer II
Thanks, I do get that in series the voltages add and the amps will be the same as the lowest one's amps (I think)

The idea is to get the highest string-watts to the controller's input (where the controller sets the Vmp and the Imp is an accidental sort of number that doesn't matter here--nothing to do with panel amps)

Then you want the controller output watts to be high so when divided by battery voltage you get the most charging amps.

OK back to the string and how to get the most watts from it. We want all the voltages to be as high as possible to add to the most, and the lowest amps to be as high as possible.

When you tilt back the panel from "aimed" it will lose voltage and amps. You lose more amps in proportion to the loss of voltage when you tilt it up and down. The difference between flat and tilted depends on how high the sun ever gets, so it is greatest in the winter like now.

Meanwhile the shaded panel has lost amps but I am unclear on how much voltage it loses. Also the shading has to be all in the diode sections to leave any section clear, so the shape of the shade matters.

So if the panel in front is shading up into the next diode section it is killing that section too, but if lowered a bit, it might leave the whole section clear?

The contest I was asking about is whether the lowered panel in front would lose amps or volts more than what you gain from un-shading the back panel. (I suppose the loss of amps doesn't matter because they are already lost from being equal to the shaded one's amps?)

Then there is the whole math thing of whether the string watts will be higher by just disconnecting the lower amp panel so the amps of the remaining panels now fully count towards the string wattage. I think that would turn on how many remaining panels there are, since most of the watts come from the voltage product in the VA.
1. 1991 Oakland 28DB Class C
on Ford E350-460-7.5 Gas EFI
Photo in Profile
2. 1991 Bighorn 9.5ft Truck Camper on 2003 Chev 2500HD 6.0 Gas
See Profile for Electronic set-ups for 1. and 2.

CA_Traveler
Explorer III
Explorer III
Yes in series. For my setup I have the effect of 9 panel sections and if any 2 or more get sun I have charging. ie 20V out of 90V. Lets say 2 sections on one panel and 1 section on another panel have sun. That gives me 3/9 of the voltage or 30V from the panels.

There is no real advantage to disconnecting a shaded panel because:
1. In partial sun the sections can produce power.
2. The sun moves and the shadows are soon gone and who wants to climb back on the roof?

The best approach is to tilt to the optimal angle for the latitude and orient the rig for maximum solar power. Or add another panel. Or raise the flat panels as I did and not have any rig created sun shadows. Lots of options.
2009 Holiday Rambler 42' Scepter with ISL 400 Cummins
750 Watts Solar Morningstar MPPT 60 Controller
2014 Grand Cherokee Overland

Bob

2oldman
Explorer II
Explorer II
BFL13 wrote:
I know this has been explained several times about how it works with a string, but I am still confused. ...

How is tilting back an un-shaded panel in the string different from shading one, and whether it would be better to disconnect the back one and leave the tilted one up the same tilt as the others in the string or leave as is.
In a series string, amps are the same everywhere in the circuit - they are not additive like parallel panels. Reduce one panel in the string to near zero amps, and the entire string is near zero. I think of it as a series of water pumps, each contributing to the pressure, but not the volume.

When you connect electrical devices together, you can increase volts or amps, but not both. Like batteries.. you get 12v from 2 -6v, but the amp-hours (capacity) remain the same.

Not sure I get the 2nd part..
"If I'm wearing long pants, I'm too far north" - 2oldman

BFL13
Explorer II
Explorer II
2oldman wrote:
BFL13 wrote:
if it hurts the string to lower the front panel, why doesn't it hurt the string the same way to have the back panel shaded so much?
Wouldn't it be better to just disconnect that back panel when you are facing West with the fronts all tilted up?
Bypass diodes. It would appear the diodes are effectively doing that.


I know this has been explained several times about how it works with a string, but I am still confused. (A "string" means they are in series AFAIK) Perhaps somebody would be kind enough to try and explain this again.

How is tilting back an un-shaded panel in the string different from shading one, and whether it would be better to disconnect the back one and leave the tilted one up the same tilt as the others in the string or leave as is.
1. 1991 Oakland 28DB Class C
on Ford E350-460-7.5 Gas EFI
Photo in Profile
2. 1991 Bighorn 9.5ft Truck Camper on 2003 Chev 2500HD 6.0 Gas
See Profile for Electronic set-ups for 1. and 2.

2oldman
Explorer II
Explorer II
CA Traveler wrote:
Do I need to post that you need MPPT of course?
No. ๐Ÿ™‚ Outback 60 does a nice job.
"If I'm wearing long pants, I'm too far north" - 2oldman

CA_Traveler
Explorer III
Explorer III
Do I need to post that you need MPPT of course?
2009 Holiday Rambler 42' Scepter with ISL 400 Cummins
750 Watts Solar Morningstar MPPT 60 Controller
2014 Grand Cherokee Overland

Bob

CA_Traveler
Explorer III
Explorer III
Although I don't tilt I've posted several threads on the positive effect of bypass diodes and several kinds of shade.

Basically with my 3 panels, 9 diodes and 90V array I can easily see 10V array voltage steps and the resulting power. Charging starts with 20V to the controller.
2009 Holiday Rambler 42' Scepter with ISL 400 Cummins
750 Watts Solar Morningstar MPPT 60 Controller
2014 Grand Cherokee Overland

Bob