Installing serial solar panels

[john61ct]

Should be pretty easy to demo who's right with a YouTube video. . .

[hellosailor]

Never forget: #1 Best Cheapmost Export Quality Panel must be built Cheapmost!

And it forces everyone on the mass market to compete more cheaper.

Every bypass diode adds cost. Installing it adds more cost.

Every bypass diode adds another potential failing component. Cheapmost supplier add more failings. Exponentially, not just additively. Every diode connection/solder point adds more labor, adds more failure point again.

Three diode, six connection, many more cost, many more failure point. Wide-eyed barbarian only care price, make him happy give good price no more failure things.

And every diode make 0.3VDC forward voltage drop best case. Less voltage less power, make hard claim for "best power" for panel.

So yes, diodes have their reasons, but there is plenty of reason for the mass market to omit them. Save money, make the rubes happy, have fewer failure points.

Back to the original question: To disconnect a 60V source, you can use one cheap SWITCH even a cheap KNIFE SWITCH aka Frankenstein or any old theatrical lighting panel. No big deal to buy something better. So you turn off the power, and THEN you can disconnect the panels with whatever terminal connections keep you happy.

No one but Peter Pan got made happy with a shadow.

[ramblinrod]

Originally Posted by TeddyDiver

Sorry, trying just use simple consepts to make you understand a shaded panel doesn't produce any current while another nonshaded installed parallel with it..

BR Teddy

Actually, TD I admire the way you are all serious until unequivocally proven wrong, and then rather than admit error, you just pretend it was a different argument altogether.

So the argument I objected to presented by Obi Wan, was that if 2 panels are connected in paralelll and one panel is shaded, it kills the unshaded panel output, which of course, is total nonsense.

He will never admit it, after calling others ignorant and casting aspersions on their intelligence, because to do so would be to admit fault, and that just isn't a possibility for him. How 'bout you? How 'bout Dot? Got it in ya men? Can you admit you were wrong and I'm right?

[downunder]

Originally Posted by ramblinrod

Incorrect.

Larger panels, ~ 300W may very well have bypass diodes.

Smaller panels, ~100W rarely have bypass diodes.

Larger panels are great in a residential installation because you get more W/$, but IMHO they are not so great on a cruising boat, where you may need to take them down and stow them away due to a pending storm, hurricane, or when laying up on the hard.

Just try handling a ~300W panel in 25 knots. Fun.

Now try stuffing it down the companionway.

I generally recommend an array made up of 100W panels to my customers, so that they are more manageable, and if one is damaged, it's not such a big deal, until it can be replaced.

In my experience, shading on smaller panels connected in series is devastating to performance, parallel, no biggy, and locating/mounting large panels is too limiting, and one is forced to mount in location that will be prone to shading.

Fair enough using tiny 100W panels on tiny boats.

However seems to me Best Practise for many is to use approx 300w panels where vessel real estate allows. There are such panel setups in the linked thread.

http://www.cruisersforum.com/forums/...ts-179638.html

[Hkalan]

Hello,

I made my own combiner boxes with the breakers that isolate the negative and positive.

It can be done DIY and that is one part of owning a boat that I do enjoy so very much... especially when it works :-)

https://youtu.be/pPjW3MIjSF4

Alan

[ramblinrod]

Originally Posted by downunder

Fair enough using tiny 100W panels on tiny boats.

However seems to me Best Practise for many is to use approx 300w panels where vessel real estate allows. There are such panel setups in the linked thread.

http://www.cruisersforum.com/forums/...ts-179638.html

Interesting that as I scanned the link thread, by chance I'm sure l, the 4 I picked were all small panels. It really doesn't have to do with the size of boat. It is all to about energy consumption and replenishment.
If a skipper on a 100 ft yacht is very energy conservative, and consumes only 100 A-hrs per day, 400 W of solar should be plenty. If he is not so energy conscious and consumes 200 A-hrs per day, and brings the bank up to 80% by other means every morning, and wants solar just to top up the bank on nice days to keep the batteries functioning nicely, he still only needs 400 W.

4x100W panels, in an unshaded location is a nice system for just about any size boat. Putting on several more hundred Watts in locations prone to shading, doesn't really give anyone any more.

I've seen lots of boats with large panels on the lifelines or
side rails. Doesn't make it smart. At least one is in the shade most of the day. I laughed watching a Youtube where the guy was touting his rail mounted panels he could angle toward the sun with an elevation adjustment only. Of course this only worked for the unshaded panel and when the sun was a abeam the boat. Several episodes later he had only one panel; tacked the boat and forgot to raise the panel. In a later episode yet, the panels were moved up
onto an arch aft of the boom. Much better for solar production, but
now the boat heeled more due too all the weight being higher.

Yup, there are pros and cons to every desgn decision. Action before considering them properly, usually means sub optimal performance and dissatisfaction, when one learns how good it could have been.

[Paul Elliott]

Originally Posted by ramblinrod

Everything was perfect until that last sentence.

All panels block reverse bias current flow (with either internal or external blocking diodes). (Note to kids following this, don't put blocking diodes in series with panels that don't require them, most don't, or you add an unnecessary voltage drop).

When shaded, and connected to an unshaded panel, the connection node voltage is pulled up to the unshaded panel voltage.

Each panel supplies the current produced to the connection node at the connection node voltage.

Has to be.

A few hours ago I did some testing using a pair of Renogy 100W solar panels. These have two strings of 18 cells, with per-string bypass diodes (two diodes total).

First, I connected the two panels in parallel. I don't have a precision load handy, so I used two 12V/20W incandescent bulbs in series. I was using flimsy clipleads to connect everything, so there were voltage drops there of a few tenths of a volt. I used a clamp-on meter to measure DC Amps at various points in the circuit, and a DVM to simultaneously measure voltage.
Obviously the panels were lightly loaded, but the results were still obvious. The sky was lightly overcast.

With no shadows, each panel was delivering about 0.7A at 20.3V.
If I shaded one cell in string on one of the panels, the unshaded panel delivered 1.32A at 17.6V, while the shaded panel delivered 0.04A (virtually nothing). If I shaded the entire string the current dropped to zero.

This shows that a partially-shaded panel, even with bypass diodes, will deliver no current when in parallel with an unshaded panel.

I then connected both panels in series. I used three 12V/20W bulbs in series as a load.
No shading: 1.56A @ 33.5V
One string shaded: 1.30A @ 24.0V
One panel shaded: 1.05A @ 15.9V
1-1/2 panels shaded: 0.68A @ 7.0V

With this nonlinear load it's not obvious how much power could have been drawn in either configuration, but the results show that partially-shaded panels deliver no power in a parallel connection, but can still deliver some power when connected in series (with bypass diodes).

Of course, the MPPT regulator input characteristics are important too -- if the voltage drops below the input range you won't get power in any case.

So can we agree that a partially-shaded panel does not pump out current when connected in parallel with an unshaded panel of similar design?

[downunder]

Originally Posted by ramblinrod

Interesting that as I scanned the link thread, by chance I'm sure l, the 4 I picked were all small panels. It really doesn't have to do with the size of boat. It is all to about energy consumption and replenishment.
If a skipper on a 100 ft yacht is very energy conservative, and consumes only 100 A-hrs per day, 400 W of solar should be plenty. If he is not so energy conscious and consumes 200 A-hrs per day, and brings the bank up to 80% by other means every morning, and wants solar just to top up the bank on nice days to keep the batteries functioning nicely, he still only needs 400 W.

4x100W panels, in an unshaded location is a nice system for just about any size boat. Putting on several more hundred Watts in locations prone to shading, doesn't really give anyone any more.

I've seen lots of boats with large panels on the lifelines or
side rails. Doesn't make it smart. At least one is in the shade most of the day. I laughed watching a Youtube where the guy was touting his rail mounted panels he could angle toward the sun with an elevation adjustment only. Of course this only worked for the unshaded panel and when the sun was a abeam the boat. Several episodes later he had only one panel; tacked the boat and forgot to raise the panel. In a later episode yet, the panels were moved up
onto an arch aft of the boom. Much better for solar production, but
now the boat heeled more due too all the weight being higher.

Yup, there are pros and cons to every desgn decision. Action before considering them properly, usually means sub optimal performance and dissatisfaction, when one learns how good it could have been.

Tiny boats or 100ft vessels are not what many are talking about.

The thread suggests many are aiming for 1kw or more of panels and in my part of the world many have 1000w or more.

http://www.cruisersforum.com/forums/...uide-to-solar-
installations-on-boats-179638-2.html

Post 28 38ft Lagoon 4x320w LG's
Post 16 38ft Island Packet 1kw
Post 47 Lagoon 450 1962 W
Post 54 Jenateau 42 1054 W
Post 92 Catalina 470 2x327 W
post 98 beneteau 45 600 W
Post 59 65ft Schooner 4 x 327w
Post 111 Jenateau 409 700w
Most catamarans are adding over 1000 w solar.

[aXis]

Originally Posted by Paul Elliott

So can we agree that a partially-shaded panel does not pump out current when connected in parallel with an unshaded panel of similar design?

Yep that makes complete sense.

The voltages on parallel panels need to be close in order to allow load sharing. If some cells on a panel get shaded, the internal voltages will drop significantly and the blocking diodes will become reverse biased. No current will flow. This is what ramblinrod has been saying.

[TeddyDiver]

Originally Posted by aXis

Yep that makes complete sense.

The voltages on parallel panels need to be close in order to allow load sharing. If some cells on a panel get shaded, the internal voltages will drop significantly and the blocking diodes will become reverse biased. No current will flow. This is what ramblinrod has been saying.

Quote:
Originally Posted by s/v Jedi
Let's say we have two panels and each has 3 strings with bypass diodes. We first connect them in parallel and shade 1 cell of 1 panel. This will knock out 1/2 (1 panel) of total capacity.

Now we do the same for series. This will knock out 1/6 (1 string of 1 panel) of capacity. Series connection is 3x as efficient as parallel.



And RamblinRods answer
This is just completely incorrect.

Show me how you arrived at this conclusion.

Although in some posts he has indirectly hinted so he's original claim is above. Thou we can split hairs how much the difference is (3x or a smaller percentage) he argues it's totally wrong. Talking about nodes and Kirchoffs Law doesn't change the fact..

BR Teddy

[ramblinrod]

Originally Posted by DotDun

And yet, Paul just explained what happens when something breaks your [B]coveted LAW!

Dot, a grid tie inverter can be set so that open circuit voltage is above the voltage to be connected to. The instant it is connected, the connection node voltage is the connection node voltage. They cannot be different voltages.

Certainly the inverter open circuit voltage could not be less, or it would sink current. Why? Because there is nothing to block current flow from the circuit connected to. (Well there is, a big @$$ fuse or breaker, so that should this condition happen the inverter doesn't go into meltdown.)

This is where AC and DC differ.

A diode can only block DC current (or current in one direction).

So your analogy was just completely wrong and has nothing to do with solar panels or the way they work.

[DotDun]

Originally Posted by ramblinrod

Dot, a grid tie inverter can be set so that open circuit voltage is above the voltage to be connected to. The instant it is connected, the connection node voltage is the connection node voltage. They cannot be different voltages.

Certainly the inverter open circuit voltage could not be less, or it would sink current. Why? Because there is nothing to block current flow from the circuit connected to. (Well there is, a big @$$ fuse or breaker, so that should this condition happen the inverter doesn't go into meltdown.)

This is where AC and DC differ.

A diode can only block DC current (or current in one direction).

So your analogy was just completely wrong and has nothing to do with solar panels or the way they work.

You have no clue about what you are saying. Please stop digging yourself a deeper hole.

[ramblinrod]

Originally Posted by Paul Elliott

So can we agree that a partially-shaded panel does not pump out current when connected in parallel with an unshaded panel of similar design?

No!

Well, not normally, well, at least not when it matters.

Paul, thanks for taking the effort to perform and prepare this test and share the results.

I absolutely concur that what you described can happen and is repeatable under the same circumstances.

Unfortunately, your test methodology is flawed.

Fortunately, it did prove what I have been saying about the unshaded panel output.

The trouble with your test as it relates to the current contribution of the shaded panel, is that the resistive load is too low.

The load is preventing current flow in excess of 1.4 A.

A solar controller would never do this (unless the battery was fully charged and couldn't accept any more, in which case, it doesn't matter).

If you repeat this test with a higher load, (like a solar controller connected to a hungry battery) so that the load demands more current than the unshaded panel can deliver, the node voltage will drop to the voltage on the other side of the partially shaded panel blocking diode, and the shaded panel will start delivering current to the controller.

So all of this is fine and dandy, but doesn't change the fact that he who posted (Obi Wan) that a shaded panel kills the output of the unshaded panel connected in parallel, and all those who supported that notion, are totally and completely wrong, and this is in fact, defined by Kirchoff's first law AND WHAT HAPPENS IN REALITY, AS YOUR TEST JUST CONFIRMED.

[ramblinrod]

Originally Posted by DotDun

And yet, Paul just explained what happens when something breaks your coveted LAW! Is Paul a higher entity than Obi Wan??

No, and not to belittle Paul and his effort to explain his theory, but his theory and test was flawed, as I have explained in another post.

You are so obsessed with your partial knowledge about the topic, you refuse to accept there is more to the story than what you know.

On this too, you are mistaken. I openly admit there are lots of things I don't know. But I do know of what I am posting about in this thread.

I hinted to your lack of knowledge when I belittled you about implementations...

I feel sorry for you, and others in this thread, that feel a need to belittle others due to your perception of their lack of knowledge. It is especially ironic, when their knowledge proves to be greater than your own.

You very quickly jumped to accept an erroneous explanation that supported your mistaken belief.

I have since proven this wrong. I have not even waited for Paul's response, as I believe he is a stand up guy and will confirm my explanation of his results is correct.

Since you have declared it is acceptable to purposefully belittle others in a public forum, (it is absolutely not by the way), I have a question for you.

Now that I have proven your position wrong, will you admit error, and apologize for trying to belittle me?

This is not an attempt to belittle you, it is simply a direct test of your character, on a public forum.

[ramblinrod]

Originally Posted by DotDun

You have no clue about what you are saying. Please stop digging yourself a deeper hole.

OK Dot, you believe I have no clue. You are entitled to your belief (even if it is faulty).

However the post you are referring to, as copied below is 100% correct.

If you don't believe so, and you are so clued in, please explain, in your own words, based on your own knowledge, what is wrong with it, (other than I said it, it is correct, proved you wrong, and that is upsetting you)...

RR: Dot, a grid tie inverter can be set so that open circuit voltage is above the voltage to be connected to. The instant it is connected, the connection node voltage is the connection node voltage. They cannot be different voltages.

Certainly the inverter open circuit voltage could not be less, or it would sink current. Why? Because there is nothing to block current flow from the circuit connected to. (Well there is, a big @$$ fuse or breaker, so that should this condition happen the inverter doesn't go into meltdown.)

This is where AC and DC differ.

A diode can only block DC current (or current in one direction).

So your analogy was just completely wrong and has nothing to do with solar panels or the way they work.