Author Topic: connecting panels of different wattage together (Read 11752 times)

super windy · « **on:** July 06, 2009, 08:47:09 AM »

Hi guys

Need a bit of advice here, I am in the middle of buying a 140 watt panel, and connect it to my 50 watt panel, what is the best way of doing this?? run the 50 watt to the 140 watt panel, or the other way, and how would you connect it, + to + then battery etc..

I am also due to lack of space and storage thinking of connecting all the batteries together into one massive battery bank of 7 105amp batteries giving me a total of 735 amps, what is the best way of doing this also.

As some of you guys know I have a 40 watt and a 50 watt solar panel, and if i connect all the batteries together, I would ideally connect the 40 watt to the 50 watt to the 140 watt one into the battery bank is this feasible?, I know in the ideal world all the panels would be the same wattage, same make etc.., but unfortunatelly I have to work with what I have got, how would you tackle this issue.

cheers guys

super windy

electrak · « **Reply #1 on:** July 06, 2009, 06:06:28 AM »

I'm guessing 12 volts, connect all + and all - on all panels, do the same with the batteries, keep a close eye on the batteries with so many in paralel, if one goes it will take the rest of them. I would charge and use batteries in smaller groups to start with. use what you need, then make it bigger as you need, try to find the right balance panels to battery to use.

others will get in more detail, and or search and read.

wooferhound · « **Reply #2 on:** July 06, 2009, 07:50:01 AM »

This is how I would do it . . .

super windy · « **Reply #3 on:** July 06, 2009, 11:11:58 AM »

cheeers guys

I knew it could be possible, I like your sketch wooferhound, just need to get a mental picture of how to do it in practise, I am abit worried about the comments made about having all the batteries going to waste if one of them goes, should I put fuses in each battery or is this a useless idea, the system is 12v, maybe looking at this video someone can tell me what's missing or what extra equipment I will need:

http://www.youtube.com/watch?v=kR5ouOKNejU

I have a 600 watts generator and an ametek 30v so do i connect all of this to the batteries I need a charge controller able to handle 60 amps+, for the wind turbines, and my juta 16 amp charge controllers should be o.k for the solar panels.

thank you once again

super windy

www.windchasers.eu

ghurd · « **Reply #4 on:** July 06, 2009, 11:36:31 AM »

The fuse between the batteries is pointless, as far as saving a bad battery from draining down the others. The amps will go through the fuses both ways.

Every so often, the batteries will need disconnected from each other and load tested with maybe a 5-10A load for a while to see if one is bad.

Do not forget the blocking diodes in Woof's sketch if the panels will be subjected to shading at different times. I am not entirely convinced it makes a lot of difference, but it won't hurt.

Getting to be too many amps for a 60A controller.

Probably want to leave the solar through the existing controller,

then connect a 60A diversion/dump controller for the wind amps.

G-

Madscientist267 · « **Reply #5 on:** July 06, 2009, 02:16:30 PM »

just need to get a mental picture of how to do it in practise

One thing I'd like to add comment/correction-wise to Woof's otherwise excellent diagram -

When using multiple batteries, particularly when there are higher currents involved, physical arrangement of the connections becomes a concern. With the diagram as is, the upper battery will recieve more charge than the bottom, and each one in between going from more to less, top to bottom. Likewise, the upper battery will give up more than the ones below it during discharge as well. This is due to the voltage drops in the interconnects.

The super-simple solution for this in an all parallel arrangement such as in the diagram is to connect one of the master leads (either positive or negative, doesn't matter) and connect them to the battery at the opposite end of the bank. This distributes the charge/load more evenly throughout the bank since they will all see an equal length of cable.

For example, in Woof's diagram, just take the negative where it comes into the negative terminal on the top battery and connect it to the negative on the lower battery instead.

Steve

DoctorCon · « **Reply #6 on:** July 07, 2009, 06:39:37 AM »

Hmm - nice diagram,

but one little question...

??PERPETUAL MOTION, Perpetual Motion, perpetual motion, perpetual,perp, per, pe,p.....

Ungrounded Lightning Rod · « **Reply #7 on:** July 08, 2009, 11:38:32 AM »

When they're in parallel they're all at the same voltage. But if you push current into a fully charged battery it cracks the water into gas and uses up the energy, coming back to its "fully charged" voltage when the charging is over and it sits a while. So if you have one cell in one battery go low or the cells of one battery running at a slightly lower voltage due to chemistry variations, that battery keeps the other batteries in the parallel group from reaching full charge.

Leaving a battery undercharged for a month or more causes it to "sulphate". The plates turn to lead sulphate as it discharges, which is normal - they convert back to lead and lead peroxide as you charge them. But if you leave them uncharged for a while the sulphate cr - in two steps. After the first step you CAN break the crystals back with an overvoltage (if you do it carefully and don't overheat the battery, and if the crystals aren't too big, which breaks up the plates.) But normal chargers won't do this very well, so a sulphated battery is usually considered "dead". After the second step the third crystal form can't be broken up electrically and your battery is dead for good. Thus a failing battery pulling the voltage down will cause the good ones to fail after a while.

So the trick is to try to keep paralleled batteries "balanced" so they all come to full charge at the same voltage (even a small fraction of a volt difference means one is partly charged when another is full) and to keep track and remove any bad ones from the parallel group before the others are permanently harmed.

It's easiest to balance paralleled batteries by buying a bunch that are by the same manufacturer and from the same manufacturing lot, then wiring them so the wiring resistance causes them to all "work" the same amount and mounting them so they are all at the same temperature and get the same amount of cooling (because temperature changes voltage). If you connect batteries that are different ages or manufacturers they're likely to drift apart more quickly so you have to watch them more carefully. DON'T connect batteries of different types (flooded, adsorbed-glass-matt, and/or jell-cell) because they have different operating voltages and different maintenance requirements.

super windy · « **Reply #8 on:** July 09, 2009, 03:05:36 PM »

Guys

JUst a quck reply to thank everyone for their input, I am gathering materials and bits and will do it soon, I have learned alot just from your replies.

cheers

super windy

www.windchasers.eu

Ungrounded Lightning Rod · « **Reply #9 on:** July 09, 2009, 11:42:12 PM »

The super-simple solution ... is to connect one of the master leads (either positive or negative, doesn't matter) and connect them to the battery at the opposite end of the bank. This distributes the charge/load more evenly throughout the bank since they will all see an equal length of cable.

This exactly balances two paralleled batteries. And for more than two it is a HECK of a lot closer to balancing their charge and discharge currents. (A similar hack is used on bridges to keep paralleled "street lamp" fixtures at about the same brightness across the span, rather than having them get dimmer as you get farther from the end that's fed.)

But for three or more batteries it's not exact. The batteries closer to the middle "loaf" a little bit, and it gets worse as the number of paralleled cells increases.

To be exact with the kiddy-korner arrangement the jumpers would have to be of different thicknesses. With 4 batteries and equal-length jumpers the "exact balance" arrangement would be:

(+)..1..(+)--2--(+)::3:+)--feed--'

and:

--feed--(-)::3:-)--2--(-)..1..(-)

But making jumpers with one, two, three, etc. "in hand" bundles or varying wire sizes is a big pain. A simpler way to get the equivalent arrangement is to have a + and a - terminal block or bolt and run a separate line from each battery post to the appropriate block, and run the feeds from the blocks. Make sure the SUM of the lengths of the + and - runs for each battery are the same as the sum of the lengths for each of the other batteries (though some batteries will have a short - lead and a long + or vice-versa).

Madscientist267 · « **Reply #10 on:** July 10, 2009, 09:56:48 AM »

"To be exact with the kiddy-korner arrangement the jumpers would have to be of different thicknesses."

Not true; if the jumpers are all the same length, the resistance between each battery and the charge source/load will be identical.

Lets say you have 5 batteries, all in parallel, connected by 1 foot jumpers. There will be 4 feet of cable seen by each terminal of each battery regardless of it's position in the bank.

Battery 1 - Neg: 0 ft Pos: 4 ft Total: 4 ft

Battery 2 - Neg: 1 ft Pos: 3 ft Total: 4 ft

Battery 3 - Neg: 2 ft Pos: 2 ft Total: 4 ft

Battery 4 - Neg: 3 ft Pos: 1 ft Total: 4 ft

Battery 5 - Neg: 4 ft Pos: 0 ft Total: 4 ft

I do agree however that the more accepted approach is a common bus, with equal length jumpers going to each terminal of each battery, located as close to each other as possible on the bus bars.

The reason for this however, is because the overall resistance is lower since the jumpers are effectively in parallel with each other, netting a 'larger' connection to the bank as a whole, reducing losses in the cabling.

Steve

Ungrounded Lightning Rod · « **Reply #11 on:** July 11, 2009, 10:05:36 AM »

Yeah, that's what I thought a while back. And it took two iterations with somebody else on the board before I got it, too. (Though I do have the excuse that I'd seen the bridge street-lamp hack in the literature so I had a preconception to get over.)

What matters isn't the LENGTH or the RESISTANCE "seen" by each battery. It's the VOLTAGE DROP. When two batteries share a wire and put the same current down it they see twice the voltage drop as when only one is driving current through it (or being driven). Three batteries produce three times the voltage drop, and so on. The closer to the middle of the string the battery is, the more sharing it "sees".

In your five battery case an end battery sees 1 + 2 + 3 + 4 = 10 units of voltage drop and the middle battery sees 4 + 3 + 3 + 4 = 14. (Actually the voltage drop is a tad lower because the voltage drop results in the more center-ward battery driving lower current - but that's what we were trying to prevent, isn't it?)

Of course this is better by a factor of five than feeding one end battery (where the far end battery sees (4 + 3 + 2 + 1) * 2 = 20 and the near end sees zero). And it's better by a factor of 1/3 compared to feeding the middle of a string of five (where the far ends see (2 + 1) * 2 = 6 and the center sees zero). But it's not perfect.

Thickening the cables in proportion to the number of batteries sharing them gets you to the case where the voltage drops are proportional to the length and THEN it all works out right. But if you're going to do that, you might as well make it simpler and save on connectors and complexity by doing the terminal block / separate cables for each battery thing in the first place.

Other alternatives are to deliberately make some of the cables overly long to compensate for having fewer batteries sharing them (which costs you power) or branching designs (which only work for some numbers of batteries). But I'd go with the simple one of direct cables for each battery.

(Actually the voltage drop isn't quite as bad as this simplified computation because the voltage drop results in a lower current from the batteries that see more sharing. But that IS what you're trying to prevent.)

ghurd · « **Reply #12 on:** July 11, 2009, 11:49:39 AM »

"Ding!" First time anyone explained it so I agreed.

Still seems fairly insignificant to me.

If the battery bank and cables are suitably sized to the in and out currents, the voltage difference would be very tiny even with large currents.

And 99% of the time the current would be well below peak, so the voltage difference would be in the 1/100th of a volt 99% of the time?

Could reduce the difference with a jumper from Battery #1 Pos to Battery #6 Pos,

and Battery #10 Neg to Battery #4 Neg.

You wouldn't happen to have an explanation of why the center two 6V batteries on a 24V system would exhibit lower voltages? Single series string of 4 matching batteries.

I don't get it. 10A in/out = 10A in/out.

G-

Madscientist267 · « **Reply #13 on:** July 11, 2009, 02:25:33 PM »

Interestingly enough, I had thought of a potential exception after I had posted, but that wasn't it... LOL I got to really thinking about it, and of course theoretical (as all above) and real world don't usually line up. The differences I found would have come from junction resistances, something none of us hit on yet. A few milliohms in each connection would really start to add up at high current levels.

But for the issue of sharing the load, I see what you're saying, and it makes sense. The issue arises due to the internal resistance of all the batteries (once again theoretically) being equal. The end batteries take a hit for really the same reason that a set that is just daisy chained (such as in Woof's diagram), but it's just distributed differently. The additional resistance of the jumpers compounds the internal resistance of the inner batteries.

Add those two factors together, and eventually, yes I suppose the ends will see more 'use' than the center.

LOL now my disclaimer: Either way, the effect is less pronounced with the opposing corner layout than with a straight daisy chain, and besides, heheh I think the words I used were "the super simple solution"...

I ultimately agree with all of this - If I were laying out a large, high capacity bank, I would use the bus system instead of the opposing corner. It's more expensive, but more solid - in a number of ways.

G - Sounds like just simple imbalance, and they happen to be in the center. I can't think of any other reason for it either... shrug... Switch flop 'em around just for giggles and see what happens...?

One other thing comes to mind - if they have been cycled together several (ok, more like many) times, and they are side-by-together to the point that heat can build up a bit more in the center than on the ends. one possibility is thermal aging that isn't affecting the outer batteries as much. Switching the ends with the center in this case would eventually correct, then of course, reverse the issue. Maybe?

Steve

Ungrounded Lightning Rod · « **Reply #14 on:** July 11, 2009, 10:46:03 PM »

You wouldn't happen to have an explanation of why the center two 6V batteries on a 24V system would exhibit lower voltages? Single series string of 4 matching batteries.

I don't get it. 10A in/out = 10A in/out.

I'd assume the string was in need of equalization and it just happens that the low cells are in the middle batteries.

Maybe it's chance they were the ones that went down. Maybe (as madscientist267 points out) they got warmer from having less air circulation and this resulted in higher leakage so it wasn't chance that these were the ones that went low. Regardless: Try running an equalizing charge and see if they come back up.

isaac338 · « **Reply #15 on:** August 09, 2009, 03:37:57 PM »

Sorry for the rudimentary question - I was under the impression that hooking two panels of different wattages together would force them to both operate at the lower wattage? (IE a 140 and 50 watt panel connected together would only result in at most 100 watts of capacity)?

Am I wrong?

Thanks

ghurd · « **Reply #16 on:** August 09, 2009, 04:43:27 PM »

It is a 12V system.

G-

isaac338 · « **Reply #17 on:** August 09, 2009, 06:21:16 PM »

So you can hook different wattage panels to the same controller in a 12V system and the outputs just add up?

ghurd · « **Reply #18 on:** August 09, 2009, 08:28:16 PM »

Yes, with a normal controller, and 12V panels.

It is like filling a bucket with 2 hoses at the same time?

You are thinking about connecting the panels in series, either for 24V or a MPPT controller.

That situation will limit the output current to the lowest panels output current,

and a 12V 140W PV in series with a 12V 50W PV would be about 100W at 24V.

G-

isaac338 · « **Reply #19 on:** August 09, 2009, 09:35:28 PM »

Wow, in hindsight I should have figured that out myself - so simple! Thanks for your help.

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Author Topic: connecting panels of different wattage together (Read 11752 times)