Answering some solar panel with MPPT controller ?

Remember, a dime an hour income difference ain't squat.

A dime for 10,000 hours is......squat...

I've got a headache! :( I did manage to follow it all though! :)

Thanks for all the work. It answered some of the puzzling aspects of my system. Or rather "collection of parts strung together and hoping for the best". :S

I really have to re-think and modify my setup, I have way too much voltage drop. . . on the output. I guess that is the "exercise for the student". :h

I already made my panels tiltable 4 ways and shortened the input cables. The output though has serious issues. I'm about to post those on a new thread.

Salvo wrote:
You may have a measurement error. Where are you measuring Vmp? It should be directly at the panel, not at the controller.

In any event, the controller is starved for volts. This system should show better results if battery voltage is dropped to 12.5V.

ktmrfs wrote:
Post #5


Parallel connection with 90ft of cable
Vmpptin=14.60V (TX3DMM)
Impptin=8.3A (DM330)
MPPT power in=14.60*8.3=121W
Vbat= 13.50V (TX3DMM)
Ibat=9.0A
Power to battery= 121.5W

No signficant measurement error. Math and calculations are very close.
As noted in the writeup I'm NOT measuring voltage at the panel, I'm measuring the voltage and current AT the MPPT controller input (Vmpptin, Impptin) to determine available power to the controller. Yes, it is getting starved for voltage, the 90ft of approx #8 cable has a 1.3V drop across it, so parallel connected panel output voltage is 14.6+1.3=15.9V. That 90 ft of cable in parallel connection is about a 1A disadvantage in charging current compared to a series connected set of panels with 90ft of cable. Driven by system losses between the panel and controller that are noticeably higher for the parallel case than the series case with the same cable length.


Below spec'd peak Vmp & Imp, yes, but it's fall, low sun angle, high latitude.

And battery voltage during charging is determined by the battery state of charge, and charge current. In this case with batteries at near 80% SOC 9.0A battery input current, battery terminal voltage rises to 13.5V. Using a 12.5V voltage source, battery acceptance current would likely go down, not up.

Now if the batteries were near 50% SOC, I suspect I would see higher current, because I would also see a lower battery voltage, but the power (watts) to the battery wouldn't appreciably change, it would still come out near 120W.

As to lowering the voltage to say 12.5V, sorry, when charging batteries you can choose the charging current or the charging voltage you CANNOT set both of them. The battery has some internal impedance. Solve the loop equation, since you can't change battery impedance, if you SET the voltage, the current will be determined by the loop equation solution. If you SET the current, the voltage will be set by the loop equation. You CANNOT set both arbirarily. Setting the voltage to 12.5 V would LOWER the charging current, not raise it.

think about it. If I "lowered" the converter voltage to 12V I wouldn't get 10A going into the battery, The Battery would likely be supplying current TO the controller and probably blow the fuse inline with the converter!!

The controller is behaving during bulk charging mode as a constant current device.

You may have a measurement error. Where are you measuring Vmp? It should be directly at the panel, not at the controller.

In any event, the controller is starved for volts. This system should show better results if battery voltage is dropped to 12.5V.

ktmrfs wrote:
Post #5


Parallel connection with 90ft of cable
Vmpptin=14.60V (TX3DMM)
Impptin=8.3A (DM330)
MPPT power in=14.60*8.3=121W
Vbat= 13.50V (TX3DMM)
Ibat=9.0A
Power to battery= 121.5W

Salvo wrote:
WTF?

12thgenusa wrote:

:R

It means, give it a rest. You made your point in the other thread. Let this thread be what it is.

WTF?

12thgenusa wrote:

:R

Thanks for the further clarification.

I'm sure everyone that reads your report will appreciate the effort you put into it, the questions you answered and the homework exercises for the rest of us!

CA Traveler wrote:
In Post #3 and Post #9 ktmrfs observed that serial panels produce the current set by the highest output current while the second shaded panel has reduced voltage.

I expected the current to limited by the shaded panel. Ie At 1000W a 8A panel would produce 4A at 500W based on the typical IV curves.

So how did ktmrfs see 4.1A in direct sun and 4.1A with one panel in the shade and reduced voltage for the shaded panel?

CA Travel. Great questions.

In retrospect, I probably should have made some MPPT voltage measurements, it would have helped explain what is likely going on with series panels that are shaded. sorry for the oversight.

And at the same time had I done a parallel shaded output, it also in retrospect would have given better insight in how panels operate in series vs. parallel.

Here is what I believe is happening.

If it was a SINGLE panel, or two panels in parallel with one completely shaded, then yes, I would have expected panel output current to drop roughly in half, just as you outlined.
Effect would be the same, roughly 1/2 the current to the battery. That's what I've qualitatively observed in use with parallel panels.

I didn't directly measure the second shaded panel voltage, but the voltage set by the MPPT controller for max power dropped. I suspect it dropped from about 30V down to 15V or so into the MPPT controller. Since the panels are in series the current remained 4.1A, but the panel combined voltaged dropped probably to what a single panel would output so power dropped roughly in half. Output impedance of the shaded panel is still low enough to not impact the current output capability.

So series or parallel, shading 50% of the total panel area cuts charging current in half. In the parallel case, or a single panel case, current to the contoller will drop by 50%, but panel output voltage remains almost constant. in the series case panel ouput voltage drops by 50% but current remains almost constant and ends with a roughly 50% drop in charging current. In either case panel output power drops by about 50%.

Note, this is when a whole series panel is shaded, or how one of two parallel panels is completely shaded. how partial shading of a panel or panels effects output I didn't test.

Interesting note. If instead of shading a panel, you completely cover it with say a thick blanket, that is another story entirely, at least with the series case. in that case power and charging current drops to almost zero. Even in shade the panel will get some solar energy, enough to keep the panel output impedance "low enough" but not supply much current. When NO solar energy reaches the panel it's output impedance must go quite high. My first attempt at shading was t throw a big thick towel over the panel, and current went to less than 1A. removed the towel, rotated the panel 180 degrees so it was shaded, then current dropped by 50%. Picked the panel up and placed it in the shade between trailers, current stayed at the 50% level.

However, in a parallel panel case I suspect putting a towel over one panel will just drop the current in half. The covered panel has high output impedance but is in parallel rather than series and so it's effect is minimal.

But this kind of shading isn't real life either, so in a practical sense, it's a moot point.

In Post #3 and Post #9 ktmrfs observed that serial panels produce the current set by the highest output current while the second shaded panel has reduced voltage.

I expected the current to limited by the shaded panel. Ie At 1000W a 8A panel would produce 4A at 500W based on the typical IV curves.

So how did ktmrfs see 4.1A in direct sun and 4.1A with one panel in the shade and reduced voltage for the shaded panel?

Salvo wrote:

ktmrfs wrote:
Post #5

For series, cable power loss is I^2R, or V*I or V^2/R. To satisfy Salvo I’ll use V*I to calculate the power loss.

You haven't understood the previous discussion. I don't care how you calculate cable power loss. That's never been an issue.

Note that the parallel case with double the current has 4x the power loss, not double the power loss. (11.6W vs. 2.9W)

Abain, that's never been the issue. If you double current, cable power loss will be 4 times has high; no matter what configuration you use.

The intelligent way to figure out what size cable is needed is to use a 1% or 2% voltage drop rule of thumb. Whatever percent voltage drop is used, battery charging current will drop by same amount, not 4 times that amount.

Sal

:R