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.

[BFL13]

That doesn't seem right to me. (current staying the same) Shading does reduce the current and it also reduces the voltage a little. I have some numbers in my notes somewhere from when I was playing with a towel on parts of my panel.

Imp is just what you get when you divide input watts by the Vmp. I see it as a result not a cause. (Maybe that is backwards though) I was wondering where the input watts (VA) came from in the first place and how it is affected by shading one panel.

I don't really like the current bottleneck explanation, because Imp happens after the array, not amongst the array where a bottleneck might exist at one panel. I am thinking it is two different sets of current. After a while it will come clear as these things do, but there is no reason for you guys to get all tangled up.

[CA_Traveler]

BFL13 wrote:
What I meant by "fixed" was being the same before and after connecting the controller. (if there is any wattage before connecting) Voc and Isc do change with insolation so they are also "dynamic"

No wattage.

Voc*0A = 0W

Also for a short circuit test.

0V*Isc = 0W

Both of these tests are useful for basic panel testing. With good sun both the open circuit voltage and short circuit amps will be near the rated values.

[CA_Traveler]

westend wrote:

But who says that is the same amps amount as what the Imp is when you divide the array output watts by the controller's Vmp?

You're trying to tie together two measurements?
The current produced by the panel array will be, at best, slightly more than the controller output because of the controller's less than total efficiency...and... by the amount of draw being dictated by the battery. In the case of shading, the current output of the panel array will be less and therefore, the controller's output would be less.

Does that explain a discrepancy?

I have no idea what the two of you are discussing but perhaps I missed something.

Let's go back to basics: Serial matched panels with bypass diodes and MPPT. 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.

So with the same current and reduced voltage the controller seeks a new maximum power point - that is maximum watts for those conditions. Do not confuse this with the panels rated Imp.

It then converts this input power to the output power by reducing the voltage and increasing the amps. I've observed 49A from my Imp rated panels of 8.2A. Yes the wiring and controller losses are a small factor. The battery acceptance for charging is a big factor on the actual controller output.

Simple example: My array is producing 5A at 90V and 5*90/14=32A which the battery is accepting. A shadow actives 2 bypass diodes of the 9 and now the controller finds a new maximum power point at 5A and 90*7/9V and the output is now 5*90*7/9/14=25A to the 14V battery.

[BFL13]

What I meant by "fixed" was being the same before and after connecting the controller. (if there is any wattage before connecting) Voc and Isc do change with insolation so they are also "dynamic"

Now I forget what the question was 🙂 Never mind this, the OP needs to get up there and shuffle all his panels. 🙂

[westend]

Once you connect, the controller picks its Vmp and you get an actual flowing Imp, which depends on the wattage. If wattage is fixed, and Voc and Vmp are different, then Isc and Imp will be different amounts. But before you connect there is no wattage because no current?

In this statement lies the answer for understanding the relationships of power produced and how it's effected by shading. When the series string is shaded the wattage is not "fixed" but at a dynamic level due to the shading.

You can toss around all the solar nomenclature but the output of the shaded array will be diminished, especially so if the array is a series string.

I hope I'm helping with the chicken and the egg but I get a headache from theoretical things, occasionally. Come on over and talk to my kid about whether photons, which have no mass, are effected by gravity. :E
Instant headache on that.

[BFL13]

Thanks. Yes, before the controller is attached, the array has no current, just an Isc. However, you can measure Voc and Isc and those are affected by shading. But there is no current flowing so how can there be an Isc current bottleneck for a string at the shaded panel?

Once you connect, the controller picks its Vmp and you get an actual flowing Imp, which depends on the wattage. If wattage is fixed, and Voc and Vmp are different, then Isc and Imp will be different amounts. But before you connect there is no wattage because no current?

Anyway, never mind. I am having a chicken and egg which came first problem or something, which I will eventually figure out. No need for anybody else to get tangled up in my confusion. 🙂

[westend]

Where does it say that amps amount is the same as Imp

In the situation of measuring the total array, the IMP or amperage at maximum power is the measured current between the array and the controller. I think you're getting hung up on IMP being the rating of the panels and how that is effected for delivering the rated watts.

The bottom line is that in the case of shaded panels, IMP is a dynamic measurement of power, not static. Shaded cells affect the IMP of a panel. If the panels are wired in series, shading one will effect the IMP of the total array.

This website explains shading clearly.

[BFL13]

I am not clear on how the Imp can be the same as the " array panel amps" that are affected by having one panel shaded so the other panels are down to the same amps as the shaded one. Where does it say that amps amount is the same as Imp?

The voltages in series add up panel by panel including the voltage of the shaded panel to get "array voltage." There is an "array wattage" based on that total voltage and the lowest panel amps ( I imagine )

But then the controller jumps in and sets an "array voltage" by using a chosen Vmp.

The total array voltage before that happened is not???? the same as the chosen Vmp ???? so why should the "input wattage" be the same, and if not the same, this would mean the "wattage" divided by the chosen Vmp would give you a different amps amount as Imp than the lowest amps on any of the panels in the string--- or might even be the same by coincidence?

I am confused as you can tell 🙂

[westend]

But who says that is the same amps amount as what the Imp is when you divide the array output watts by the controller's Vmp?

You're trying to tie together two measurements?
The current produced by the panel array will be, at best, slightly more than the controller output because of the controller's less than total efficiency...and... by the amount of draw being dictated by the battery. In the case of shading, the current output of the panel array will be less and therefore, the controller's output would be less.

Does that explain a discrepancy?

[BFL13]

westend wrote:
Those freakin' IMP's...
AFAIK, the current at maximum power could be measured between the panels and the controller. The rated IMP of the panel doesn't change, of course, but the current changes with shading.

Yes you get the Imp along the wires from the array to the controller by actual ammeter measurement.

But that is just what "goes with" the array output watts and the controller's choice of Vmp.

What I don't understand is the array watts components in V and A before you divide by Vmp.

If the shaded panel's amps is the "array" amps as the lowest amps in the string, then the array watts should be using that low amps amount for its A in the VA.

But who says that is the same amps amount as what the Imp is when you divide the array output watts by the controller's Vmp?

[westend]

Those freakin' IMP's...
AFAIK, the current at maximum power could be measured between the panels and the controller. The rated IMP of the panel doesn't change, of course, but the current changes with shading.

[BFL13]

Can't leave it like that anyway. Plus you said you usually park facing East but they tilt for facing West? Weird! 🙂

So all you solar experts, we know in series the individual panel amps can be different in the string from shade or whatever, and the controller tells us input watts and what it has chosen. So we can derive Imp if we want or measure it on the array to controller wiring and some controllers even display Imp as in the OP.

However, AFAIK, Imp has nothing to do with the actual panel amps which is based on the lowest one's amps. So Imp is not the lowest amps of a panel in the string, it is just what you get when you divide the array watts by the controller's chosen Vmp.

Or not quite or even close?

[2oldman]

I can bolt the shaded panel to the top of the one in front!

[BFL13]

2oldman wrote:

BFL13 wrote:
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?

I don't know.. it gets complicated. 🙂

If I do anything it will be to move that shaded panel's bottom back.

Not enough work. You need to move them all back some with some vertical instead of sideways, so they can all be in the front row 🙂

[2oldman]

BFL13 wrote:
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?

I don't know.. it gets complicated. 🙂

If I do anything it will be to move that shaded panel's bottom back.