Installing serial solar panels

[Paul Elliott]

Originally Posted by ramblinrod

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.

In some cases you are right, but with my 100W, 2-diode panels, there's no way that an MPPT controller will ever load the unshaded panel so heavily that the voltage drops to where the parallel shaded panel can deliver current. With my panels, the unshaded panel would have to be loaded to where its voltage drops to below 11 Volts before the shaded panel can participate. The maximum power point is well above that (it's about 19V).

Perhaps with a higher-voltage panel having more bypass diodes, or with some series/parallel panel combinations you would have a panel with a shaded string contributing current. It's certainly possible. It's *not* possible with my commonly-used 2x18-cell panels wired in parallel.

I will go back and look for that assertion regarding killing the unshaded panel output -- I don't remember reading it that way. But you are right, this will not happen.

[Paul Elliott]

Originally Posted by ramblinrod

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.

Is this what you are referring to?:

Originally Posted by s/v Jedi

Really?! What is not clear about strings with their own bypass diode? Where a shaded parallel panel loses 100% power output with even minor shading, in series the panel will still produce 50% or even 66% of it's power. You don't seem to grasp that.
hint: power is not just current, voltage was allowed in the game as well.

If so, that's not at all what he said.

I suspect if we all gathered around a chalkboard and explained ourselves we would all be in agreement.

[ramblinrod]

Originally Posted by Paul Elliott

In some cases you are right, but with my 100W, 2-diode panels, there's no way that an MPPT controller will ever load the unshaded panel so heavily that the voltage drops to where the parallel shaded panel can deliver current. With my panels, the unshaded panel would have to be loaded to where its voltage drops to below 11 Volts before the shaded panel can participate. The maximum power point is well above that (it's about 19V).

Perhaps with a higher-voltage panel having more bypass diodes, or with some series/parallel panel combinations you would have a panel with a shaded string contributing current. It's certainly possible. It's *not* possible with my commonly-used 2x18-cell panels wired in parallel.

I will go back and look for that assertion regarding killing the unshaded panel output -- I don't remember reading it that way. But you are right, this will not happen.

Thank You.

I don't understand why the voltage of your shaded panel is dropping so low.

It should only drop in voltage equal to one string contribution, plus the bypass diode drop, which will be considerably lower than the unshaded panel, but should not be as low as 11 Vdc.

But it could certainly be below the MPPT voltage (a PWM controller wouldn't care other than the supplied voltage has to be above the input cut-off voltage).

I concur that if the shaded panel voltage is below the MPPT value that the controller is holding the node voltage at, or is below the minimum input of a PWM controller, the shaded panel will not contribute current.

[DotDun]

Originally Posted by ramblinrod

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.

1) There are no blocking diodes in the panels Paul used -
https://www.renogy.com/template/file...ifications.pdf

2) Since there are no blocking diodes, the situation is now (2) panels, each with a different voltage at your 'connection node'. This is something you have been adamant about that can't happen. Will you now admit it can happen?

3) You seem to understand that the shaded panel won't deliver any current until the voltage at the 'connection node' is lowered to match the voltage of the shaded panel. This is what I stated back in post #86 -

... in parallel power sources, when there is a difference in voltage, the lower voltage source loses the race as there isn't enough potential to push it's current into the circuit (a simple explanation).

This quote applies to both DC and AC power sources, hence my analogy to grid tie inverters.

[Paul Elliott]

Originally Posted by ramblinrod

Thank You.

I don't understand why the voltage of your shaded panel is dropping so low.

It should only drop in voltage equal to one string contribution, plus the bypass diode drop, which will be considerably lower than the unshaded panel, but should not be as low as 11 Vdc.

But it could certainly be below the MPPT voltage (a PWM controller wouldn't care other than the supplied voltage has to be above the input cut-off voltage).

I concur that if the shaded panel voltage is below the MPPT value that the controller is holding the node voltage at, or is below the minimum input of a PWM controller, the shaded panel will not contribute current.

The unloaded unshaded panel output voltage is 22.5V, so each of the two strings are 11.25V unloaded. Subtract 0.25V for the diode and you get 11V panel output when one string is shaded and bypassed. The panel will deliver no current until the voltage on the output drops below that.

This more or less matches what my tests were showing, so I'm pretty confident in my analysis.

[Paul Elliott]

Originally Posted by DotDun

2) Since there are no blocking diodes, the situation is now (2) panels, each with a different voltage at your 'connection node'. This is something you have been adamant about that can't happen. Will you now admit it can happen?

This "connection node" is where the two panels are connected together in parallel, right?

If so, then this node is at a single voltage (how could it be otherwise???). The shaded panel will have its high-impedance shaded cells forced to a higher voltage than they would otherwise be generating with zero illumination, but there is no appreciable current flow through them.

I did say the shaded panel would be "reverse-biased" in a previous post, but I don't think that's quite right. Let me ponder that. In any case, it makes no difference in how the panels behave.

[Paul Elliott]

Originally Posted by ramblinrod

I don't understand why the voltage of your shaded panel is dropping so low.

One more thought:

In my testing and analysis, I am considering the case where one or more cells in a string are *completely* shaded.

In the real world, shading is seldom 100%. Even with the mast shadow on the panel, there will still be ambient light hitting the cells, and so the open-circuit voltage will not be seriously affected. Instead, the available current will drop proportional to the shading. In this scenario, a "gently shaded" panel will still contribute current when in parallel with an unshaded panel.

[ramblinrod]

[QUOTE=DotDun;2404030]1) There are no blocking diodes in the panels Paul used -
https://www.renogy.com/template/file...ifications.pdf

2) Since there are no blocking diodes, the situation is now (2) panels, each with a different voltage at your 'connection node'. This is something you have been adamant about that can't happen. Will you now admit it can happen?

3) You seem to understand that the shaded panel won't deliver any current until the voltage at the 'connection node' is lowered to match the voltage of the shaded panel. This is what I stated back in post #86 -

Dot, referring to an erroneous quote or previous
Post does not Make an incorrect statement correct.

When 2 voltage sources are connected to a load, the source connection node voltage is one voltage. Has to be.

The open circuit circuit voltage of the individual sources, is either sunk or raised to a common voltage. Has to be.

The only way this may not be, is if the conductors between the sources was very high impedance in relation to the load. Which in the case of our
solar panel discussion, SHOULD NOT BE.

Like it or not, this is the way it is.

[TeddyDiver]

Originally Posted by ramblinrod

When 2 voltage sources are connected to a load, the source connection node voltage is one voltage. Has to be.

The open circuit circuit voltage of the individual sources, is either sunk or raised to a common voltage. Has to be.

The only way this may not be, is if the conductors between the sources was very high impedance in relation to the load. Which in the case of our
solar panel discussion, SHOULD NOT BE.

Like it or not, this is the way it is.

Your connection nodes should have a blocking diodes. If not the shaded panel will draw current instead of being a voltage source and the outcome is even worse.. The only way your scenario would work if they were connected to a low impedance draw without a solar charge controller. Thou having a boiler full of hot water is not a bad thing even if it takes more time compared to properly connected solar panels with MPPT controller.

BR Teddy

[ramblinrod]

Originally Posted by TeddyDiver

Your connection nodes should have a blocking diodes. If not the shaded panel will draw current instead of being a voltage source and the outcome is even worse

There's a revelation, for things to work properly, they must be connected properly.

Most panels are inherently capable of blocking reverse current. Those that aren't need blocking diodes to be connected in series with the panel on installation.

So regardless, the connection mode between the two panels, with or without blocking diodes in series, is necessarily one voltage.

The controller must have a relatively low impedance input, or it would impede current flow from the panels, which is not what we want.

If there is a significant voltage drop between the connected parallel panels, it can mean one of two things:

A) the controller is presenting high impedance (like disconnected or turned off).
B) The cable between the panels is high impedance, (likely
A bad connection)

Either of these scenarios is irrelevant to the discussion as we should only consider what happens in a system installed properly and turned on.

Few things turned off or installed wrong work well.

In conclusion, the connection node between the panels, must be one voltage.

[DotDun]

Originally Posted by ramblinrod

....snip

What is the rule about voltage sources connected in parallel?

Oh yeah, THEY MUST BE EQUAL.

...snip....

If you have two voltage sources in parallel, they must be the same voltage. It's the law.

...snip...

Originally Posted by DotDun

1) There are no blocking diodes in the panels Paul used -
https://www.renogy.com/template/file...ifications.pdf

2) Since there are no blocking diodes, the situation is now (2) panels, each with a different voltage at your 'connection node'. This is something you have been adamant about that can't happen. Will you now admit it can happen?

....snip

Do explain your earlier claim......

[ramblinrod]

Originally Posted by Paul Elliott

This "connection node" is where the two panels are connected together in parallel, right?.

Yes.

If so, then this node is at a single voltage (how could it be otherwise???).

Precisely, thank you.

The shaded panel will have its high-impedance shaded cells forced to a higher voltage than they would otherwise be generating with zero illumination, but there is no appreciable current flow through them.

Exactly, thank you.

I did say the shaded panel would be "reverse-biased" in a previous post, but I don't think that's quite right. Let me ponder that. In any case, it makes no difference in how the panels behave.

I'm OK with "reverse biased", each solar cell is a p/n junction with a twist.

[ramblinrod]

Originally Posted by DotDun

1) There are no blocking diodes in the panels Paul used -
https://www.renogy.com/template/file...ifications.pdf

2) Since there are no blocking diodes, the situation is now (2) panels, each with a different voltage at your 'connection node'. This is something you have been adamant about that can't happen. Will you now admit it can happen?
.....

Explanation:

If there were no blocking diodes and the panel did not have inherent ability to block reverse current, (ie, presented a high impedance), it would sink current and short the unshaded panel reducing its output to zero (or very close).

It doesn't.

So because the shaded panel presents a high impedance, the unshaded panel pulls up the voltage across the shaded panel to the unshaded panel voltage.

You can not have two voltages sources, connected in parallel, at different voltages.

If this were possible, you could not boost your car battery when it was dead. You would connect the charged battery at 12.8 Vdc, to the dead battery (lets say 8.0 Vdc) just to have the dead battery stay at 8.0 Vdc.

It doesn't.

The charged battery pulls up the voltage of the dead battery.

Because the dead battery does sink some current, the charged battery voltage sags, but not to 8.0 Vdc.

If you measure the voltage at each battery terminal, they are one and the same, likely around 12.0 Vdc, dropping to 11.0 Vdc while cranking the engine.

But no matter what, under any condition, as long as the batteries are connected in parallel with low impedance conductors and connections, the voltage across the sources is one and the same.

Has to be.

Kirchoff's law.

Two solar panels in parallel are exactly the same scenario, except the shaded panel does not sink the current of the unshaded panel.

Effectively, there is a short circuit between the two positives, and two negatives, so the voltage between the positive and negative of each has to be the same, regardless which panel it is measured at, because they are directly connected to each other.

What's that rule about the voltage in each branch of a parallel circuit?

Oh yeah, they are equal.

If you disconnect one of the parallel connections, the two go back to their respective voltages. Connect them, back to a common voltage.

You can do this 50,000 times and will get the same results.

2 voltage sources may be different values when disconnected, but the instant they are connected in parallel, they must come to a common voltage.

[DotDun]

Originally Posted by ramblinrod

Explanation:

If there were no blocking diodes and the panel did not have inherent ability to block reverse current, (ie, presented a high impedance), it would sink current and short the unshaded panel reducing its output to zero (or very close).

It doesn't.

So because the shaded panel presents a high impedance, the unshaded panel pulls up the voltage across the shaded panel to the unshaded panel voltage.

You can not have two voltages sources, connected in parallel, at different voltages.

If this were possible, you could not boost your car battery when it was dead. You would connect the charged battery at 12.8 Vdc, to the dead battery (lets say 8.0 Vdc) just to have the dead battery stay at 8.0 Vdc.

It doesn't.

The charged battery pulls up the voltage of the dead battery.

Because the dead battery does sink some current, the charged battery voltage sags, but not to 8.0 Vdc.

If you measure the voltage at each battery terminal, they are one and the same, likely around 12.0 Vdc, dropping to 11.0 Vdc while cranking the engine.

But no matter what, under any condition, as long as the batteries are connected in parallel with low impedance conductors and connections, the voltage across the sources is one and the same.

Has to be.

Kirchoff's law.

Two solar panels in parallel are exactly the same scenario, except the shaded panel does not sink the current of the unshaded panel.

Effectively, there is a short circuit between the two positives, and two negatives, so the voltage between the positive and negative of each has to be the same, regardless which panel it is measured at, because they are directly connected to each other.

What's that rule about the voltage in each branch of a parallel circuit?

Oh yeah, they are equal.

If you disconnect one of the parallel connections, the two go back to their respective voltages. Connect them, back to a common voltage.

You can do this 50,000 times and will get the same results.

2 voltage sources may be different values when disconnected, but the instant they are connected in parallel, they must come to a common voltage.

Originally Posted by ramblinrod

..snip...

What is the rule about voltage sources connected in parallel?

Oh yeah, THEY MUST BE EQUAL.

..snip...

If you have two voltage sources in parallel, they must be the same voltage. It's the law.

...snip...

In your rhetoric, you stated voltage sources, which are devices, which means, in this context, solar panels. My point the whole time has been they don't have to be the same voltage, the lower voltage source may not participate in the application, but no police will show up due to breaking a law! I went as far as to show you an example of an application where engineers play with voltage at a parallel source to provide varying results. When you connect (2) sources together in parallel that have different voltages and measure with your DVM, you are reading the higher voltage source, not the lower voltage source, it is still the lower voltage source which reacts differently in the circuit from the higher voltage source until something changes. The lower voltage source does not magically become an equal to the higher voltage source just because they are connected together.

So, my point the whole time has been you can connect sources with different voltages together if the application dictates doing so. No law broken!!!

[TeddyDiver]

If one has two 100W panels connected in series, you can't have 2.5 A going through the shaded panel and 5 A going through the non-shaded. They are in series. If one is 2.5 A, both are 2.5 A.

Sometimes you make perfect sense and sometimes not, like above. But can you recognize what's wrong?