Solar panel mystery continues

Hi there,

Finally got to measure readings for my 108 cell, mystery panel yesterday and guess what?...

I got the same volts open circuit (20.5 volts) - but the current measured 8.15 amps open circuit. I'm not sure this how it should be considering all I've been told. Do you think this still makes sense considering the stats on the back of the panel? 200 watts, 36 volts and 5.5amps.

Friend of mine (electrician) suggested that it may have been reconfigured to give 12 volts.

Haven't got to peeling away the potting on the back yet - it's a slow and delicate process. Should I continue, and would it clear up anything?
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Reply By: Maîneÿ . . .- Wednesday, Mar 17, 2010 at 21:59

Wednesday, Mar 17, 2010 at 21:59
Obviously you can't question the person you 'acquired' the panel from ?

Maîneÿ . . .
AnswerID: 409281

Follow Up By: SmokeyD - Wednesday, Mar 17, 2010 at 22:05

Wednesday, Mar 17, 2010 at 22:05
Ah...no - my son aquired it a few years ago and doubtless information from ebay sellers would not be reliable.
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FollowupID: 679250

Reply By: Glenndini - Wednesday, Mar 17, 2010 at 22:08

Wednesday, Mar 17, 2010 at 22:08
8.15 amps open cct? I don't think so. That's a contradiction in terms.
What are you actually measuring and what exactly are you trying to clear up?
AnswerID: 409283

Follow Up By: SmokeyD - Wednesday, Mar 17, 2010 at 22:14

Wednesday, Mar 17, 2010 at 22:14
Sorry, I meant short circuit current - I have been liasing with Ed C, Peter, Nomadic Navara and others re my mysterious panel, so I don't expect you'd know exactly where I'm coming from.
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FollowupID: 679252

Follow Up By: Battery Value Pty Ltd - Wednesday, Mar 17, 2010 at 22:21

Wednesday, Mar 17, 2010 at 22:21
ok SmokeyD,

you can only go one way and that is forward - just remove all this foam stuff from the junction box (there shouldn't be any because it'll make the diodes go belly up if they can't dissipate the heat if they have to).
Then remove one leg of every single diode and measure panel voltage/current again - in full sun.
Because there is no sun right now, you can grab an ohmmeter and measure the resistance of every single dioded (or pn forward voltage) in both directions.
Make a note of how the wires are connected, what colours, how many wire pairs, diodes and so on and get back to us.
Best would be to take a close up shot of what's in the box and post it here.

Best regards, Peter
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FollowupID: 679253

Follow Up By: SmokeyD - Wednesday, Mar 17, 2010 at 22:48

Wednesday, Mar 17, 2010 at 22:48
Thanks Peter,

I'll do all that after I wake up tomorrow.

At present, I am watching over the aircraft at Adelaide airport and will be until 5am tomorrow morning!!! It's a night shift of 12 hours and this site helps to keep me sane during the long hours. :)

I'll let you know what I find and post a photo.

Thanks again.
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FollowupID: 679256

Follow Up By: Battery Value Pty Ltd - Wednesday, Mar 17, 2010 at 22:54

Wednesday, Mar 17, 2010 at 22:54
no worries SmokeyD,

an air traffic controller with a secret addiction to EO aren't you ;)

Best regards, Peter
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FollowupID: 679257

Follow Up By: SmokeyD - Thursday, Mar 18, 2010 at 18:51

Thursday, Mar 18, 2010 at 18:51
Hi there,

Took a couple of pictures of the back of my panel as suggested - but haven't the resolve to de-solder the diodes yet. Not sure that it won't destroy anything. Still getting 20.5ish volts and near to 8 amps - do you think I should still go ahead and separate diodes?

Regards SmokeyD

Ahh... can't submit photos unless you're a member - that's a shame. Well, I'll try and describe it. The negative lead is attached to a connection (on left) which appears to disappear via a metal strap under the connection box. A diode is connected to it and its other leg to another terminal (no strap or wire that I can see on that terminal - unless it's connected via the retaining screw). Then that terminal has another diode attached to it, same deal re strap or wire (none obvious) then it concludes with the same setup as the negative - positive terminal has diode attached and the positive wire. Three diodes connected to 4 terminals - in series I believe.

Regards again
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FollowupID: 679334

Reply By: Member - Ed C (QLD) - Thursday, Mar 18, 2010 at 01:01

Thursday, Mar 18, 2010 at 01:01
Well, I'm going for the simple answer ;-)

Going by the dimensions you have given in a previous post, and the open / short circuit readings posted here, I'd say that what you have there is a 125W (or possibly 130W) 12V panel....

The 20.5 Voc should give a 'loaded' voltage or Vmp (Voltage @ maximum power) of around 17.5...
The 8.15A short circuit current will probably give around 7 - 7.5A under load...

17.5 (V) x 7.15 (A) = 125.125, or near enough to 125W ;-)

Connect it to a battery with (at least) a 7.5A load on the battery, and see what sort of readings you get....

The size of the panel means that it simply cannot be 200W, therefore the label on the back is totally worthless, so you may as well rip that off & burn it...

There is at least 1 ebay seller (that we know about) who is known to have been doing exactly what appears to have been done here, putting grossly over-specced labels on panels, and selling at prices according to what the label says, not what the panel actually is....
Not much you can do about that now, except to make the most of what you have...

That's my take on it anyway....

Hope this is of some help:)

:)

Confucius say.....
"He who lie underneath automobile with tool in hand,
....Not necessarily mechanic!!"

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AnswerID: 409297

Follow Up By: SmokeyD - Thursday, Mar 18, 2010 at 01:17

Thursday, Mar 18, 2010 at 01:17
Hi Ed C,

Yup, kinda thinking that way too - my son would have been none the wiser when he bought it. It's a pity that happens... still, I guess that's the nature of institutions like that - you do take a risk.

It makes sense to do a test run as you say, I am still however waiting for my regulator from the States (the Morning Star MPPT) to arrive, then I'll check.

Do you think that the panel and 2 x 75Ah batteries will be able to run a 60 watt fridge and a couple of LED lights for any length of time? Just be good to know if I am any where near the mark.

Thanks for your input.
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Follow Up By: Member - Ed C (QLD) - Thursday, Mar 18, 2010 at 03:06

Thursday, Mar 18, 2010 at 03:06
Well, the upside is that the MPPT controller will extract the maximum power available from the panel, and if my theory is correct, and the panel is indeed 125 (or so) W, then under ideal conditions you can expect to see up to 10A (@ 12.5V) available to the battery / load...
Yes, it is indeed possible to see more Amps going out of the (MPPT) controller than what is going into it (depending on load, of course;-))...

Assuming your 'fridge is a compressor type,and not running continuously, then I'd say the panel, combined with your 150A/hr battery bank, would be adequate for the job at hand..
Not overkill, but 'adequate' ;-)

We run 2 Engels (one as a freezer usually) on 160W, and have found this to be quite satisfactory, though we are seldom in one place for more than a few days at a time.. In any case, I turn the fridges off overnight.. after all, the nights are cool (sometimes freezing!), and the fridges aren't being opened anyway, so may as well give the batteries a rest.. that way, they'll charge up fairly quickly in the morning ;-)

I'm sure you'll work it out as you go along;-)

:)

Confucius say.....
"He who lie underneath automobile with tool in hand,
....Not necessarily mechanic!!"

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My Profile  Send Message

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Follow Up By: SmokeyD - Thursday, Mar 18, 2010 at 18:53

Thursday, Mar 18, 2010 at 18:53
Hi there,

Took a couple of pictures of the back of my panel as suggested - but haven't the resolve to de-solder the diodes yet. Not sure that it won't destroy anything. Still getting 20.5ish volts and near to 8 amps - do you think I should still go ahead and separate diodes?

Regards SmokeyD

Ahh... can't submit photos unless you're a member - that's a shame. Well, I'll try and describe it. The negative lead is attached to a connection (on left) which appears to disappear via a metal strap under the connection box. A diode is connected to it and its other leg to another terminal (no strap or wire that I can see on that terminal - unless it's connected via the retaining screw). Then that terminal has another diode attached to it, same deal re strap or wire (none obvious) then it concludes with the same setup as the negative - positive terminal has diode attached and the positive wire. Three diodes connected to 4 terminals - in series I believe.

Regards again
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FollowupID: 679335

Follow Up By: Battery Value Pty Ltd - Thursday, Mar 18, 2010 at 19:11

Thursday, Mar 18, 2010 at 19:11
SmokeyD,

before ripping anything apart, measure the voltages across each individual diode and report back, if you don't mind.

Also check, if the cathode ring marks are visible on the diodes, or arrow or similar (point of the arrow indicates cathode)
If you measure voltage across diodes, the meter's red (positive) lead has to go on the diode's kathode for the meter to read a positive voltage. This can be used to determine the kathode/anode if no marks are visible.
If you only get around 0.7 to 1.0 volt across a diode, it's been mounted the wrong way around.

Best regards, Peter
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FollowupID: 679339

Follow Up By: SmokeyD - Thursday, Mar 18, 2010 at 20:20

Thursday, Mar 18, 2010 at 20:20
Hi Peter,

Yes will do... having the diodes still connected won't give incorrect reading?

SmokeyD
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FollowupID: 679349

Follow Up By: Battery Value Pty Ltd - Thursday, Mar 18, 2010 at 20:35

Thursday, Mar 18, 2010 at 20:35
Hello SmokeyD,

leaving the diodes in circuit while measuring the voltage across them gives you an idea about the state of the cell string they're wired in parallel to, an idea about the state of the diodes themselves, and if they were put in the right or wrong way (polarity).
So I guess it's another 20 hours for you to test this under full sun :)
BTW, you could touch them in full sun and see if they're getting hot - which they will if been mounted the wrong way around (in which case there will be a small voltage of 0.7 to 1.0V across them)

Best regads, Peter
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Follow Up By: SmokeyD - Thursday, Mar 18, 2010 at 21:15

Thursday, Mar 18, 2010 at 21:15
Hey Peter,

Thanks a lot Bro - sounds like you have a full time job just keeping up with space cadets like me :)

Yes I'll be out of action for another 20 odd hours... still on the bright side, when my shifts finish, I'll have 4 days off!! More time to play outside!!!!!!

Cheers, Paul
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FollowupID: 679362

Follow Up By: SmokeyD - Monday, Mar 22, 2010 at 05:47

Monday, Mar 22, 2010 at 05:47
Hi Peter,

Well I finally got to do what you suggested - I measured the voltage across the diodes. I found the silver ring at each end of the diode and placed the positive on that end and came up with the following...

From the negative lead end - the first diode was .48 volts; the second was .22 volts and the last one was .36 volts. Interestingly, the diodes did not get hot at all in the time I left it in the sun.

So according to what you said, there must be a problem somewhere with the system. Are these diodes necessary for the proper function of the panel? I read somewhere that they prevent the power going back the other way during the nigh (no sun).

Is there a way I could sent you a photo of the panel?

So now what is the next step yo reckon?

Regards SmokeyD.
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FollowupID: 679811

Follow Up By: Battery Value Pty Ltd - Monday, Mar 22, 2010 at 10:03

Monday, Mar 22, 2010 at 10:03
ok SmokeyD,

here is the summary:

what you've got are three bypass diodes wired into a series string.
This diode string is wired in parallel with 3 parallel 36 cell strings.

Thus, your 36V panel seems to be configured as a 12V panel, with the three bypass diodes wired in series, instead of parallel.
This effectively creates one diode with a three times higher forward voltage, and one third of the current carrying capability compared to a 3-diodes-in parallel configuration.
It would be better to have the three diodes each wired in parallel to the three parallel 36 cell strings.

But the existing diode configuration won't hurt anything, as long as you don't put this panel into a 24V or higher array.

One reason why you didn't get full output current of 11.5A (200W 12V panel) could be that you didn't have the panel face the sun fully when taking the current reading of 8.1A.
The current output is a function of the sine and angle of incidence, so I guess your panel was 44 degrees (inv sine (8.1/11.5)) out of whack during your measurement.

The easiest way to achieve max current output is by doing the following: put the panel flat on the ground, then rotate it flat on the ground until one edge faces the sun fully, and then tilt it up into the sun until the length of the shadow on the ground becomes a maximum.

Try this, and your current will most likely increase to higher than 8.1A which would make this a fully functioning 12V 200W panel.

And even if you won't achieve full current, you can still use this panel to harvest energy from the sun.

Best regards, Peter
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FollowupID: 679826

Follow Up By: SmokeyD - Monday, Mar 22, 2010 at 10:24

Monday, Mar 22, 2010 at 10:24
Hi Peter,

Thanks for all your help mate - it's been a good learning process.

One last question - if I wanted more out put/power, can I simply add another panel (assume the panel would have to be a 12 volt one) into the system. I would need to make sure the wiring could handle it re amps I know - but would there be any issue in adding further panels?
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FollowupID: 679830

Follow Up By: Battery Value Pty Ltd - Monday, Mar 22, 2010 at 12:04

Monday, Mar 22, 2010 at 12:04
Yes, you'd need another 12V panel(s) wired in parallel to this one if you wanted to increase your output current/power.

No issues with this whatsoever.

Best regards, Peter
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FollowupID: 679839

Follow Up By: Mike DiD - Monday, Mar 22, 2010 at 14:32

Monday, Mar 22, 2010 at 14:32
> "From the negative lead end - the first diode was .48 volts; the second was .22 volts and the last one was .36 volts. "

- It seems that the panel has been set up with the three separate 36 cell sections are connected in parallel using diodes. The low voltage drop shows that Schottky diodes have been used. The difference in voltage shows that two of the sections are not delivering full current - they may have gone faulty.


> "Interestingly, the diodes did not get hot at all in the time I left it in the sun."

- If the section working best is putting out 4 amps then that diode is only dissipating 2 watts and that heat will be conducted away by the wires.


> "Are these diodes necessary for the proper function of the panel? I read somewhere that they prevent the power going back the other way during the nigh (no sun). "

- that's right, if one section is partially shaded or goes faulty, it will have no effect on the output from the other sections. If you removed the diodes you would notice no drop in current from the reduced voltage drop, except at very low insolation levels. This is because solar panels are constant-current sources - than can source up to 18 volts in sunlight - plenty of reserve for charging 12 volt batteries.


> "So now what is the next step yo reckon? "

- You can't repair solar panels. All you can do is check the connections coming out of the panel and make sure all three sections have good connections so each one is putting out the same current.
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FollowupID: 679854

Follow Up By: Battery Value Pty Ltd - Monday, Mar 22, 2010 at 15:08

Monday, Mar 22, 2010 at 15:08
Mike,

what I got Paul to measure was the reverse voltage across each diode.
All three diodes are in reverse bias as can be seen from his positive voltage results, when connecting the positive test lead with the cathodes.

Therefore there is no current through them, which is confirmed by Paul's observation that they stay cool (2 Watts would be very hot to touch on an axially wired diode).

No current through them means they're wired to act as bypass diodes.

There is no need to 'repair' this solar panel as it performs quite normal in this configuration of 3x36 cells parallel with a bypass diode across it.
It just happens that this bypass diode has been wired up from three individual diodes in series, creating one diode with a forward voltage three times that of a single diode.
But this doesn't matter because the diodes are never going to see any forward current through them as long as he only uses the panel in a 12V system.
Under these conditions, he could just as well cut these diodes out alltogether with no effect on the panel's performance or any ill effects.
Or, he could wire them in parallel, and then in series with the panel, so that they act as a blocking diode if that's required at all in his setup.

Best regards, Peter
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FollowupID: 679857

Follow Up By: Lex M (Brisbane) - Monday, Mar 22, 2010 at 15:47

Monday, Mar 22, 2010 at 15:47
SmokeyD posted
" I found the silver ring at each end of the diode and placed the positive on that end"

I'm with Mike on this one. The three diodes were not in reverse bias according to the data above.

Battery value posted
"what you've got are three bypass diodes wired into a series string.
This diode string is wired in parallel with 3 parallel 36 cell strings."
This statement makes no sense to me.

I agree with Mike. Effectively (three 12V arrays with series diode) wired in parallel.

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FollowupID: 679858

Follow Up By: SmokeyD - Monday, Mar 22, 2010 at 17:10

Monday, Mar 22, 2010 at 17:10
Hi all,

I must say this is more than confusing. Is it possible that I do have a faulty panel, or is it the effect of the diodes? If so, should I go ahead and unsolder the diodes? Is there a way to find out if my panel is indeed faulty? I have only two obvious connections and cannot access anymore "wiring" or see any "strings" to see how they are configured.

If I use the panel as is, will it be a useful power supply?

Thanks to all once again.
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FollowupID: 679869

Follow Up By: Battery Value Pty Ltd - Monday, Mar 22, 2010 at 17:31

Monday, Mar 22, 2010 at 17:31
Hello Paul,

no need to get confused, and yes you can use your panel 'as is'.
You measured the short circuit current of 8.1A, and the open circuit voltage being a touch over 20V.
This represents a normal 12V panel, albeit the short circuit current is a bit lower than a 200W rating would suggest - but I've already given you an idea what you can try to achieve a higher output current.

One last test you could do to clear things up 100%:

short circuit the panel, put it under some light (sun or strong artificial will do).
Then grab your voltmeter and measure the voltages across each diode.

Place the positive test lead on the anode of diode one, and the negative test lead on the cathode (the 'ringed' end). Jot down the voltage.
Do the same test with all three diodes.

See how this goes and report back.

Best regards, Peter
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FollowupID: 679875

Follow Up By: Mike DiD - Monday, Mar 22, 2010 at 17:50

Monday, Mar 22, 2010 at 17:50
Bypass Diodes are used in Solar Panels to allow the other 35 CELLS in the string of 36 to keep feeding current if one CELL is shaded.

Bypass Diodes are only useful if they're wired directly across each cell - they are inside the panel - not at the interconnect panel. You cannot measure the voltage across Bypass Diodes unless you open up the panel.

When a cell has sunlight on it, it will have 0.5 volts across it. If it's shaded the other cells will try to push current through it, but it shoes a high-resistance to this current. A diode across every cell limits the voltage drop to MINUS 0.8 volts (or 0.4 volts for a Schottky diode.)

Diodes to block nightime discharge current (or shaded panels in multi-panel systems) could be at the panel or at the regulator.
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FollowupID: 679882

Follow Up By: Battery Value Pty Ltd - Monday, Mar 22, 2010 at 22:03

Monday, Mar 22, 2010 at 22:03
Hello Paul,

thought to upload this for you because a picture says more than a thousand words.....

Image Could Not Be Found

The 'blocking diode' possibilities A and B can be ticked off because they don't conform with your findings of the diodes wired in series.

The only configuration fitting all the criteria derived from the Q&A is C.
The circles in the drawing represent a 36 cell string.
The three diodes in C have no purpose if you use the panel in a 12V system, you can either leave them in there, or cut them out.

Best regards, Peter

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FollowupID: 679923

Follow Up By: SmokeyD - Tuesday, Mar 23, 2010 at 10:30

Tuesday, Mar 23, 2010 at 10:30
Hello Peter,

Yes, that picture sheds a bit of light on the picture. I understand the set up, but not why they bothered to put in the diodes. Perhaps cutting them out would be an intersting exercise anyway. I haven't had the time to retest as you suggested as yet, but will when I can. I guess the underlying question is... is this panel going to perform at a reasonable level to be useful considering my requirements?
Panels are not cheap, and I would rather utilise and augment what I already have. The panel sounds enigmatic and I certainly would not buy another one - so the proof will be in the pudding when I get to try out the system.

What ever the outcome, I would like to thank you for your time Peter - this forum is a gem for sure.

Just had a thought - would having the auxillary batteries being charged by the car system affect anything else? By that I mean, the regulator is connected to the battery constantly, would it (the regulator) be affected when the batteries are being charged by the car whilst travelling?
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FollowupID: 679961

Follow Up By: Battery Value Pty Ltd - Tuesday, Mar 23, 2010 at 12:48

Tuesday, Mar 23, 2010 at 12:48
no worries Paul,

the diodes are sitting there in series, because it's easier for the workers to connect the interconnecting wires/studs to the individual 36 cell strings if the panel needs to be configured for 36V use.

Yes no worries, the panel will perform at least like a 140W 12V panel, and probably better. You can find out how good, if you fine tune your current test measurements as discussed earlier.

Now to your batteries' charging by alternator:

Presuming you're asking about the solar regulator: no, it won't get negatively affected by the alternator charging the van batteries at the same time.
The regulator will just see an increase in battery voltage when it takes the additional charge from the alternator. This slowly pushes the battery voltage up to the level where the regulator cuts the solar current flowing through the battery.

Glad to be of help.

Best regards, Peter
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FollowupID: 679977

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