Solar install turning out really bad - looking for some help/advice please.

[Volusiano]

Some additional information on the maximum DC current, saw this label on the side of the inverter.

So knowing this information, what is the proper size wire for a 50' run from the combiner box to the inverter?

This Wiki AWG link shows a resistance of 0.3951 milliOhm per ft for 6 awg copper. You have 50 feet so that's 0.3951*50=19.755 milliOhm. Your max input current is 34.5A DC. So the max dc voltage drop over your 50' 6 awg run would be 34.5*0.019755=0.68V dc. This is not even 1Vdc, a drop in a bucket. If the inverter is operating at minimum voltage of 270V dc, that's 0.68/270=0.2%. If the inverter is operating at max 500Vdc then it's 0.68/500=0.14%.

You can go through the same math for 4 awg and 8 awg and 10 awg if you want. But given how tiny the voltage drop seems to be even at max input dc current as seen above for your 6 awg 50' long wire run, is it even a concern anymore? Especially if your setup typically operates at 350Vdc like you said (if I remember what you said correctly), the less than 1V drop across your 6 awg wire is not going to take your inverter out of range (270Vdc min input) any time soon.

And the 6 awg wire should be able to handle the 34.5A dc current just fine. 34.5*1.25=43.125A and the 6 awg ampacity at 60C is 55A according to that AWG Wiki link. 8 awg might not have been sufficient if you go by 60C (40A<43.125A), but if you go by 75C then it might have been sufficient since 50A>43.125A. This is all based on that Wiki awg table from that link above.

And once again, disclaimer: I don't have any expert knowledge of code so I'm not citing code numbers here to support my finding of the wiring gauge discussion above. The opinion above is strictly mine based on the calculation I did above and AWG Wiki table I used as shown in the link. Anyone can check my work above in case I made any mistake. But it's pretty simple calculation I think. Listen to my opinion or not at your own risk and if you're not sure, you'd be best consult the codes to be sure.

[bcroe]

Some additional information on the maximum DC current, saw this label on the side of the inverter.

So knowing this information, what is the proper size wire for a 50' run from the combiner box to the inverter?

OK, that is more good info. The inverter has a max 34.5 ADC. Just maybe, your
max is 3 times a panel Impp? What did it read out?

I'd worry that a string of 16 panels would have a Vmpp on the order of 576 VDC.
Do the Optimizers somehow avoid that issue for the 500 VDC max inverter? What
limits do Optimizers placed on number of panels?

A guess is, the more difference in the number of panels between strings, the
more difficult for the Optimizers to operate the strings in parallel.

Still trying to figure the algorithm Optimizers use. Practicality dictates that they
are used in series strings, so all the Optimizers in a string are putting out the same
current. How about, every panel tries to deliver its maximum current starting up?
The panels with less capability (shaded, etc) can't keep up, so they are bypassed,
zero output voltage? The Optimizer starts multiplying the panel output current (by
stepping down its panel voltage) till the current matches the current from the strongest
panels? This "current multiplication" is adjusted to keep the Optimizer's panel on
its present Vmpp? So each Optimizer puts out the same current, but since some
panels are shaded, they may have lower voltages? So the string total voltage
could go ANYWHERE depending on sun? But this doesn't allow parallel strings?
And certainly doesn't set any particular overall string voltage?

A problem with something like this, is you have a bunch of closed loop systems
trying to work together. They could start hunting for an operating point, or
oscillation. So how does it REALLY work?

The output voltage limits printed on your inverter look like the standard ones for
my Fronius, except mine are programmable. Should be OK unless your line
voltage is too high. Mine was 254 VAC, and pushing back inverter power
raised it even higher. Bruce Roe

[HX_Guy]

Some additional information on the maximum DC current, saw this label on the side of the inverter.

So knowing this information, what is the proper size wire for a 50' run from the combiner box to the inverter?

[Volusiano]

Volusiano... your posts do not show a good understanding of the code involved. While I respect your enthusiasm and ability to logic your way through it, you should refrain from posting specific conductor or OCPD sizing unless you can justify it with a code citation.

I'm not offended by what you said, and I admit I don't have expert understanding of the code, so I'd like you to educate me on what I said so far that's not in accordance to code for my own education.

I believe I said the inverter's max output current rating is 47.5A so the 60A breaker and 6 awg wiring should be good. This is not based on my direct knowledge of the code. This is based on my observation that my 60A breaker has 6 awg wire and it passed inspection so the assumption is that it passed codes. I think you agreed with this in one of your posts (you said "The SE11400A has a 47.5 A rating, and 125% of that is just under 60 A. Once the breaker is set at 60 A, the wire size follows.").

I believe the other thing I said without any code knowledge is that usually common practice for ground wire is to go 1 size lower (in this case 8 awg for ground if main wires are 6 awg). This is again not based on my knowledge of code but based on my observation of how my solar system wiring set up is, which passed code (8 awg for ground wire and 6 awg for main wires to 60A breaker in my setup).

My statement about guessing 10 awg for ground wire maybe OK (for 6 awg main wires) because the ground wire doesn't have to carry full continuous load before the breaker trips is neither based on my observation of my system nor based on code which I don't have expert knowledge of. It's just an educated guess on my part and that's why I only said "I think it's OK" and I didn't say "I'm sure it's OK".

I looked at the link you gave about "More discussion about AC wiring sizes for this system in this thread" but I didn't see any discussion about ac wiring sizes there. Did you give the wrong link? I'd be interested in seeing a link that discusses what the ground wiring size needs to be and what all the rules pertaining to it are.

If I made any other references to wiring sizes beside the 2 citations above, and did it so erroneously without an understanding of code, please point it out so I'm aware of it so I can retract it by editing my post(s) to take it out.

Again, I'm not offended by your comment. My response here is more in the interest of learning for clarification on what I said wrong (per code) for self education. Thanks.

[HX_Guy]

I am trying to figure out just what your DC voltage & current are. IF you had 4 strings of
10 panels (of 72 cells), they would nominally run about 360V and 8.61A peak per string,
or 34.4A total (but only if all panels could see best sun at the same time). A loop of 6
gauge 100' long would be about 0.04 ohms, giving worst case 1.38V drop. That is only
0.38% loss, and with varied panel orientations, even that number is too high.

HOWEVER, your optimizers change the numbers. I don't know how your strings are
arranged, and have yet to hear an exact explanation of the algorithm optimizers use
in operation. So I can't come up with accurate numbers. Perhaps checking the inverter
readout could show what the DC numbers really are? Bruce Roe

I believe the current fluctuates in order to keep the DC voltage around 350V or so (in reality it looks like it keeps it more around 370VDC looking at the inverter LCD display). Here is the spec sheet on the inverter, mine is the P400.

http://www.solaredge.com/files/pdfs/products/power-optimizers/p-series-us-application-note.pdf

Worst case it looks that it would output 15A, which is what the fuse in the combiner is too. I've got 3 strings wired in parallel (1 of 16 panels, 1 of 13 panels and 1 of 11 panels), so worst case scenario that would be 45A but I don't think that would ever happen...more realistically like you said, 8.61A x 3 strings = 25.83A.

The permit however was designed for 4 strings of 10 panels, or 60A total...which is maybe why it asked for 2AWG wire. After talking to SolarEdge however, I changed the strings down to 3 and they said 6AWG would be more than plenty, they actually said even 10AWG would be fine and I could go with 8AWG to be safe.

[sensij]

I am trying to figure out just what your DC voltage & current are. IF you had 4 strings of
10 panels (of 72 cells), they would nominally run about 360V and 8.61A peak per string,
or 34.4A total (but only if all panels could see best sun at the same time). A loop of 6
gauge 100' long would be about 0.04 ohms, giving worst case 1.38V drop. That is only
0.38% loss, and with varied panel orientations, even that number is too high.

HOWEVER, your optimizers change the numbers. I don't know how your strings are
arranged, and have yet to hear an exact explanation of the algorithm optimizers use
in operation. So I can't come up with accurate numbers. Perhaps checking the inverter
readout could show what the DC numbers really are? Bruce Roe

SolarEdge systems operate at about 350 Vdc. The optimizers are capable of converting the output of the panel as needed to maintain that voltage in the string. String size doesn't matter in the way it does for other string inverters or charge controllers. The voltage drop difference of 2 AWG or 6 AWG on the DC side is a couple tenths of a percent. Probably not enough to justify the higher cost of the larger wire.

More discussion of the AC wire size for this system is in this thread. The ground wire sizing is covered by different rules, and SolarEdge (with an ungrounded PV circuit) will have different requirements than the transformer based inverter.

Volusiano... your posts do not show a good understanding of the code involved. While I respect your enthusiasm and ability to logic your way through it, you should refrain from posting specific conductor or OCPD sizing unless you can justify it with a code citation.

[bcroe]

What's even more confusing though is that for the DC run from the combiner box to the inverter (50' run), they show 2AWG wire! Now that's a little nuts because the inverter won't even take that big of a wire, the largest it will allow is 6AWG which is what we used.

I am trying to figure out just what your DC voltage & current are. IF you had 4 strings of
10 panels (of 72 cells), they would nominally run about 360V and 8.61A peak per string,
or 34.4A total (but only if all panels could see best sun at the same time). A loop of 6
gauge 100' long would be about 0.04 ohms, giving worst case 1.38V drop. That is only
0.38% loss, and with varied panel orientations, even that number is too high.

HOWEVER, your optimizers change the numbers. I don't know how your strings are
arranged, and have yet to hear an exact explanation of the algorithm optimizers use
in operation. So I can't come up with accurate numbers. Perhaps checking the inverter
readout could show what the DC numbers really are? Bruce Roe

[Living Large]

Per several online wiring size calculators, 4AWG is what is called for, though the Southwire spec sheet says 6AWG is good for up to 65A @ 75º which is what the system is spec'd for. So there might have been some confusion there from the designer that designed the wiring diagrams.

What's even more confusing though is that for the DC run from the combiner box to the inverter (50' run), they show 2AWG wire! Now that's a little nuts because the inverter won't even take that big of a wire, the largest it will allow is 6AWG which is what we used.

Yes, but there are adapters which will allow you to make that connection. The problem is the voltage drop across a length of 50' at the maximum current. That is a long run if you are pushing a fair amount of current (say 60A), and you could lose 2% or more of your voltage with 6AWG. I believe you should be more like 0.5 to 1%. So you may need a larger wire not to carry more current, but to lose less voltage across the 50'. In other words, the inverter manufacture doesn't plan for everyone to require oversized wire, so they provide a termination capability most can use without an adapter. In my case, to achieve what Amy at AltE calls the Dance of Joy, which loosely translates to meeting the NEC requirements, I also had to go to 2AWG, and plan for a terminal adapter. AltE has an excellent YouTube that guides the viewer through the determination of the combiner to load center wire size.

[HX_Guy]

Oh...and what the heck is with this overcast weather today! Booooo!

[HX_Guy]

The only thing weird is why the permit calls for 4 awg wiring when only 6 awg is needed due to the 47.5A rating. Maybe somebody got confused between the DC STC power rating and the AC power rating so they over calculated the wiring size thinking incorrectly that it's 12.4kw AC.

Per several online wiring size calculators, 4AWG is what is called for, though the Southwire spec sheet says 6AWG is good for up to 65A @ 75º which is what the system is spec'd for. So there might have been some confusion there from the designer that designed the wiring diagrams.

What's even more confusing though is that for the DC run from the combiner box to the inverter (50' run), they show 2AWG wire! Now that's a little nuts because the inverter won't even take that big of a wire, the largest it will allow is 6AWG which is what we used.

[Volusiano]

Thanks Volusiano, the electrical stuff is so over my head, I love when it's explained well like that so thanks for that.

The 12,400 is DC STC (40 x 310w panels). For AC the permit says 11.4kW AC though that seems like the maximum amount, not for my specific system.
I say that because looking at the SolarEdge spec sheet, it shows a max DC STC of 15,350 with a nominal AC output of 11,400.

http://www.solaredge.com/files/pdfs/products/inverters/se-single-phase-us-inverter-datasheet.pdf

Going by what you said, 11.4kw (AC) / 240V = 47.5A...with the 80% rule, that's 59.4A...but again that's the max the inverter will put out if I maxed out the DC input I assume, right?

So how do I figure out the AC system rating for my actual system if the 11.4kW seems to just be something they plugged in from the spec sheet?

The spec sheet shows max continuous current of 47.5A (I assume you have the SE11400A-US), so I think your installer is correct to go by that spec for the wiring rating and sizing.

It looks like your installer did not just conveniently spec that rating into the permit to fit within the 6 awg 60A breaker system. They properly followed the inverter's spec. It's actually more like the manufacturer who intentionally limit their design to within that range as to not require more than 6 awg/60A breaker combo, or else it may be harder to sell the inverter if it exceeds that size combo. Even the name of the model SE11400A-US implies that it's designed for 11.4KW AC.

Although the spec sheet shows a max power output of 12kw AC and max voltage output of 260V AC, those 2 numbers are irrelevant and only the max current of 47.5A is relevant when it comes to sizing the wire and breaker. Also, when connected to the main panel, you're limited to the 240V on the grid anyway, so your inverter can never get to 12KW AC or 260V AC under that situation anyway. The only way it can hit that power or voltage level is if it's not tied to the grid but is used to drive a non-grid load instead.

In reality, those are all max rating numbers anyway and you probably won't hit those limits in real life because that inverter's max input DC (STC) power rating is at 15350, and you're only at 12,400 DC STC for your panels.

So I think it's all good. The only thing weird is why the permit calls for 4 awg wiring when only 6 awg is needed due to the 47.5A rating. Maybe somebody got confused between the DC STC power rating and the AC power rating so they over calculated the wiring size thinking incorrectly that it's 12.4kw AC.

[HX_Guy]

So it's a 11400W inverter.
And you just calculated:
"279.3 X 40 panels = 11,172W AC"

11400W - 11172W = 228W

So I think you can add ~228W worth (~250W DC nameplate) of panels.
And possibly you'd be starting to clip your peak production.
(but probably not because you have multiple angles and shading)

Yeah I don't think I'll ever see close to that with the multiple angles, much less with the shading. For example today it peaked at just under 7,100W AC. Looking at other SolarEdge accounts from the Phoenix area, it seems May is the best month for production and I won't have any shading problems so Ill have to see what it peaks at then, and subtract that from the 11,400 and see what I could add. As an example, there's a 13.42kW DC system that peaked at 11.54kW AC on May 10th and a 16.0kW DC that peaked at 13.476kW AC on May 13th. Seems to be about 85% average DC to AC production between those two systems...applying that to my system would mean 10.54kW AC or an additional 860W AC of panels (3 more Canadian Solar 310w DC would probably just about take care of it).

Good thinking though, I hadn't thought of looking at it that way...was only going by what the max DC input specified on the spec sheet.

[foo1bar]

it means I can max out the inverter (add another 2,950W worth of panels)

So it's a 11400W inverter.
And you just calculated:
"279.3 X 40 panels = 11,172W AC"

11400W - 11172W = 228W

So I think you can add ~228W worth (~250W DC nameplate) of panels.
And possibly you'd be starting to clip your peak production.
(but probably not because you have multiple angles and shading)


In any case - IMO sensij's right - "real world" is what the inverter can possibly put out - which is at most 47.5A - which means a 60A breaker, and appropriate wire size for that (although there's no reason you can't go with larger wire - it would even give you some very very tiny increase in production realized because of lower resistance - probably not enough to be economically worthwhile.)

If you really wanted to ignore NEC and really go by "real world" - you could probably go smaller since there's some pessimism in the code by necessity - heck even 12AWG would probably last quite a while before it melted, shorted and caught fire. (lets see - 20' of 12AWG would be ~0.03ohms; 47A would be ~1.5W) So only 1.5W more than what it gets from sunlight on the conduit... Might last a real long while. Of course if it does fail you're screwed since the insurance company will quite possibly not pay.

[sensij]

Correct...I think we were just trying to figure out based on the system we actually have, what would the real world needed breaker size be (not what NEC would require).

If you say so. I would think that "real world" calculation would also take into account the various angles your panels are at and the potential for shading, since they will all not be at peak power at the same time. NEC is real world enough for me.

[HX_Guy]

You guys are going down the wrong path. With respect to the breaker, 2011 NEC 690.8(A)(3) says:



and continues on in 690.8(B)



In other words, the number of panels has no effect on the breaker, only the continuous AC rating of the inverter. The SE11400A has a 47.5 A rating, and 125% of that is just under 60 A. Once the breaker is set at 60 A, the wire size follows.

Correct...I think we were just trying to figure out based on the system we actually have, what would the real world needed breaker size be (not what NEC would require).

Either way I like that the NEC goes by the max the inverter will put out, it means I can max out the inverter (add another 2,950W worth of panels) and I don't have to make any changes to the wiring or breaker.