Solar DC wiring

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bcm

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Atlanta, Georgia
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Engineer
Having little DC power experience has me really scratching my head on the project I was just handed (due tomorrow, of course!).

I have a group of solar panels that generate 45KW DC. I am tasked with determining the wire size (and ground!?) between the panels and the inverter. I'm struggling to turn the 45KW into volts & amps since I have a bunch of different info about the panels:

*Performance Under Standard Test Conditions*
Maximum Power = 230 Wp (what's a Wp?)
Open Circuit Voltage = 36.9 V
Maximum Power Point voltage = 29.6 V
Short Circuit Current = 8.42A
Maximum Power Point Current = 7.76A

*Performance at 800 W/m^2, NOCT, AM 1.5*
Maximum Power = 164 Wp (what's a Wp?)
Open Circuit Voltage = 33.4 V
Maximum Power Point voltage = 26.6 V
Short Circuit Current = 6.96A
Maximum Power Point Current = 6.18A

Once I get DC watts, do I just use the tables in NEC Article 310 (310-16 to 21) to determine the wire size? Do I need a ground or is that just for AC?

Thanks so much guys! You are life-savers!
 

iwire

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Staff member
Location
Massachusetts
Having little DC power experience has me really scratching my head on the project I was just handed (due tomorrow, of course!).

I have a group of solar panels that generate 45KW DC. I am tasked with determining the wire size (and ground!?) between the panels and the inverter. I'm struggling to turn the 45KW into volts & amps since I have a bunch of different info about the panels:

*Performance Under Standard Test Conditions*
Maximum Power = 230 Wp (what's a Wp?)
Open Circuit Voltage = 36.9 V
Maximum Power Point voltage = 29.6 V
Short Circuit Current = 8.42A
Maximum Power Point Current = 7.76A

*Performance at 800 W/m^2, NOCT, AM 1.5*
Maximum Power = 164 Wp (what's a Wp?)
Open Circuit Voltage = 33.4 V
Maximum Power Point voltage = 26.6 V
Short Circuit Current = 6.96A
Maximum Power Point Current = 6.18A

Once I get DC watts, do I just use the tables in NEC Article 310 (310-16 to 21) to determine the wire size? Do I need a ground or is that just for AC?

Thanks so much guys! You are life-savers!

You must read 680, those DC conductors need 125% added to them twice and must take roof top temperatures into consideration as well.


You also need to know your string configuration, how many panels in series and how many stings in parallel
 

bcm

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Location
Atlanta, Georgia
Occupation
Engineer
Solar DC Wiring

Solar DC Wiring

I think you meant article 690, iwire. I have been looking through that and I do understand the 125% requirement. I can't apply it though until I can figure out amps.

From what I'm being told, we have:
Series strings will be 13 modules
15 strings per inverter
195 modules per inverter (we're supposed to be using 3 inverters)
44.84 KW per inverter

Does this help?

Thanks!
 

iwire

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Staff member
Location
Massachusetts
I think you meant article 690, iwire. I have been looking through that and I do understand the 125% requirement. I can't apply it though until I can figure out amps.

From what I'm being told, we have:
Series strings will be 13 modules
15 strings per inverter
195 modules per inverter (we're supposed to be using 3 inverters)
44.84 KW per inverter

Does this help?

Thanks!

Yeah 680, 690 one or the other. :grin:

Yeah if you look at 690 it will tell you how to add up the strings.
 

OTT2

Senior Member
Location
Orygun
Having little DC power experience has me really scratching my head on the project I was just handed (due tomorrow, of course!).

I have a group of solar panels that generate 45KW DC. I am tasked with determining the wire size (and ground!?) between the panels and the inverter. I'm struggling to turn the 45KW into volts & amps since I have a bunch of different info about the panels:

*Performance Under Standard Test Conditions*
Maximum Power = 230 Wp (what's a Wp?)
Open Circuit Voltage = 36.9 V
Maximum Power Point voltage = 29.6 V
Short Circuit Current = 8.42A
Maximum Power Point Current = 7.76A QUOTE]

I think you meant article 690, iwire. I have been looking through that and I do understand the 125% requirement. I can't apply it though until I can figure out amps.

From what I'm being told, we have:
Series strings will be 13 modules
15 strings per inverter
195 modules per inverter (we're supposed to be using 3 inverters)
44.84 KW per inverter

Does this help?

Thanks!

To determine your voltage use 690.7. Use the value for open circuit voltage to get this. 13 x 36.9 = 479.7 volts d.c. You still have to compensate for the coldest recorded temperature in your area, which will increase the voltage even higher. For this you will use the information provided with the pv module, or table 690.7 if not provided with the pv modules. Typically you can expect to add an additional 10% - 20% to your voltage.

Keep in mind to check the specs on the inverter to make sure your not exceeding its rating.

Use 690.8 to determine your conductor ampacity and max OCP. Short circuit current x 15 strings x 1.56 will equal your minimum circuit ampacity from your combiner box to the inverter.

Use 690.45 to determine your equipment grounding conductor size.

Do the inverters provide Ground fault protecton? This may make a difference.
 

zip1

Member
Location
Massachusetts
I'm struggling to turn the 45KW into volts & amps since I have a bunch of different info about the panels:

You don't turn KW into amps for PV, you use the rated short circuit current of the panels (8.42Amps in your example.) For each string you add up the module currents and multiply by 125% to get the Max PV source current. You must then multiply that by 125% to get the required ampacity of the wiring. (So 13x8.42x1.25x1.25 would be the required ampacity for each string.)

Are you using a string combiner between the arrays and the inverter? If so, the conductors from the combiner to the inverter would be sized by adding up the Max PV source currents and multiplying by 125% (or (13x8.42x1.25)x15x1.25 is the required ampacity from the combiner to the inverter.

You then use NEC 310, as you suggest, but don't forget to derate for ambient temperature as Bob noted.
 

zip1

Member
Location
Massachusetts
You don't turn KW into amps for PV, you use the rated short circuit current of the panels (8.42Amps in your example.) For each string you add up the module currents and multiply by 125% to get the Max PV source current. You must then multiply that by 125% to get the required ampacity of the wiring. (So 13x8.42x1.25x1.25 would be the required ampacity for each string.)

It just dawned on my that I presumed a parallel-module connection, but the question didn't specify parallel. For series connect modules, the string ampacity would be 8.42x1.25x.1.25. (No adding up of the module currents.)
 

bcm

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Location
Atlanta, Georgia
Occupation
Engineer
soalr dc

soalr dc

Nobody mentioned anything specifically about a combiner, but then when I asked, I'm being told we are using combiners.

Combiner info:
Min PP V Max Open V
Max Temp 319.49
Min Temp 455.0

I was able to use 690.7 to determine the voltage and got 497.7 v as you state.
 

bcm

Member
Location
Atlanta, Georgia
Occupation
Engineer
solar dc

solar dc

zip1:
Presuming parallel-module connection is correct.
Can somebody define a "string" and a "module" for me? I'm not finding much that's helpful but a lot that adds to the confusion! The paperwork I have is terrible! It uses "module" to mean several different things!

I used 8.42A (sht ckt current) x 15 (strings) = 126.3A (sht ckt current for all strings)
126.3A x 1.25 (per NEC 690-8(a)(1)) = 157.88A (max PV source current)
157.88A x 1.25 = 197.34A (required ampacity) Is this for all 15 strings or what?

13 modules x 8.42A (sht ckt current) x 1.25 x 1.25 = 171.03A which is supposed to be the ampacity for each string as you say.

I didn't know before but when I asked, lo & behold, we ARE using combiners.

13 modules x 8.42A x1.25 x 15 strings x 1.25 =2565.46A is the required ampacity from the combiner to the inverter. That's HUGE. I'm supposed to come up with conductors for over 2500 amps using NEC 310 tables? That's going to be like 8 sets of 400kcmil (alum @90 deg C, )!!! Is this going to be the same between the panels and the combiners?

That seems like a lot of cable to get ~300A of AC power. Then I have to derate for temps between 4 degF and 95 degF (provided min/max temps) too, right?

OTT2:
I calculated 13 modules x 36.9V (open volts) = 479.7 v (dc open volts).

Given the coldest ambient temp is 4 deg F., so per NEC table 690.7 I believe that
should be 479.7V x 1.17 = 561.25 v My inverter's ratings are 300v min, 600v max.

8.42A (sht ckt current) x 15 (strings) x 1.56 = 197.34A as calculated above. Ok, so I need a 200A OC device... at each inverter?

I'm struggling to use 690.45. "The gnd conductor shall not be smaller than the required size of the circuit conductors when sht ckt current < 2x OC device rating." Ok, sht ckt current = 197.34A which is < 2 x 200A OC device. So the ground conductor must then be rated for 400A or greater (i.e. #3 from NEC table 250-122). Is that correct?

The inverters do not seem to provide ground fault. (Sunny Tower ST36).
 

zip1

Member
Location
Massachusetts
zip1:
Presuming parallel-module connection is correct.
Can somebody define a "string" and a "module" for me? I'm not finding much that's helpful but a lot that adds to the confusion! The paperwork I have is terrible! It uses "module" to mean several different things!

I used 8.42A (sht ckt current) x 15 (strings) = 126.3A (sht ckt current for all strings)
126.3A x 1.25 (per NEC 690-8(a)(1)) = 157.88A (max PV source current)
157.88A x 1.25 = 197.34A (required ampacity) Is this for all 15 strings or what?

13 modules x 8.42A (sht ckt current) x 1.25 x 1.25 = 171.03A which is supposed to be the ampacity for each string as you say.

I didn't know before but when I asked, lo & behold, we ARE using combiners.

A module is each individual PV panel. A string is the combination of modules together. So you have 13 module strings in your example.

Your max PV source current (from string to combiner) would be 8.42 x 13 x 1.25. (not 15.) This is basically the "string" current. The ampacity require for the string as you note it above is correct.

The Max PV output circuit current (from combiner to inverter) would be 15 x Max PV source current. Multiply that by 1.25 to get the required ampacity of that feeder (I get 2565A.)

I calculated 13 modules x 36.9V (open volts) = 479.7 v (dc open volts).

Given the coldest ambient temp is 4 deg F., so per NEC table 690.7 I believe that
should be 479.7V x 1.17 = 561.25 v My inverter's ratings are 300v min, 600v max.

8.42A (sht ckt current) x 15 (strings) x 1.56 = 197.34A as calculated above. Ok, so I need a 200A OC device... at each inverter?

I'm struggling to use 690.45. "The gnd conductor shall not be smaller than the required size of the circuit conductors when sht ckt current < 2x OC device rating." Ok, sht ckt current = 197.34A which is < 2 x 200A OC device. So the ground conductor must then be rated for 400A or greater (i.e. #3 from NEC table 250-122). Is that correct?

The inverters do not seem to provide ground fault. (Sunny Tower ST36).

Are you sure you're parallel connected? If so, you're Max PV system voltage will only be 39.6, before you apply the voltage correction factor. I'd think for a system of this size, you'd be series connected.
 
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bcm

Member
Location
Atlanta, Georgia
Occupation
Engineer
solar dc

solar dc

According to the Sunny Tower inverter installation manual:

"The maximum allowable fuse size for the DC string fuses is 15 amps.... 15 amp fuses are provided with the Sunny Tower...The proper size PV string fuse is determined by 1.56 x Isc (of the PV module) and then rounding up to the next standard fuse size. Example: If the Isc of the PV module is 6.9 Adc then the fuse size is determined by 1.56 x 6.9 = 10.76. The next standard fuse size would be 12A, 600Vdc fuse."

The above directions & example don't seem to be NEC 690 compliant though. The above directions & example don't take into account that I have 15 strings of 13 modules. It seems like the calculation should be 1.56 x 6.9 x 13 = 139.93Adc. That's WAY larger than the 15A fuse size allowable!!

Anybody have ideas what I should do?
 

david luchini

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Location
Connecticut
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Engineer
According to the Sunny Tower inverter installation manual:

"The maximum allowable fuse size for the DC string fuses is 15 amps.... 15 amp fuses are provided with the Sunny Tower...The proper size PV string fuse is determined by 1.56 x Isc (of the PV module) and then rounding up to the next standard fuse size. Example: If the Isc of the PV module is 6.9 Adc then the fuse size is determined by 1.56 x 6.9 = 10.76. The next standard fuse size would be 12A, 600Vdc fuse."

The above directions & example don't seem to be NEC 690 compliant though. The above directions & example don't take into account that I have 15 strings of 13 modules. It seems like the calculation should be 1.56 x 6.9 x 13 = 139.93Adc. That's WAY larger than the 15A fuse size allowable!!

Anybody have ideas what I should do?

It sounds like your inverter has a fused string combiner internal to it. The fuses you mention are protecting each individual string, not the total inverter input current. From your original example, 8.42x1.25x1.25=13.2A. The 15A fuses would be fine.
 

bcm

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Location
Atlanta, Georgia
Occupation
Engineer
solar dc power

solar dc power

It sounds like your inverter has a fused string combiner internal to it. The fuses you mention are protecting each individual string, not the total inverter input current. From your original example, 8.42x1.25x1.25=13.2A. The 15A fuses would be fine.

I can't find anything in the inverter data I have to confirm that there is a fused string combiner in the inverters.

That aside, don't we still have to take into account that each string has 13 modules (at 8.42Adc each)? Thus 8.42 x 1.25 x 1.25 x 13 modules = 171.03Adc per string. How can we leave out the # of modules involved?

Thanks!
 

david luchini

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I can't find anything in the inverter data I have to confirm that there is a fused string combiner in the inverters.

That aside, don't we still have to take into account that each string has 13 modules (at 8.42Adc each)? Thus 8.42 x 1.25 x 1.25 x 13 modules = 171.03Adc per string. How can we leave out the # of modules involved?

Thanks!

If the modules in each string are connected in series, then the current on each string will be 8.42x1.25=10.5A. The voltage will be 36.9Vx13=479.7V, which then must be corrected for ambient. 479.7x1.18=566V (at 4degF ambient.)

So each string would be rated for 10.5A and 566V. When the strings are combined, the total load is 157.5A and 566V.

If you have multiple fuses to connect multiple strings within the inverter, then that is a string combiner. It combines all of the strings into one output.
 

bcm

Member
Location
Atlanta, Georgia
Occupation
Engineer
If the modules in each string are connected in series, then the current on each string will be 8.42x1.25=10.5A. The voltage will be 36.9Vx13=479.7V, which then must be corrected for ambient. 479.7x1.18=566V (at 4degF ambient.)

So each string would be rated for 10.5A and 566V. When the strings are combined, the total load is 157.5A and 566V.

If you have multiple fuses to connect multiple strings within the inverter, then that is a string combiner. It combines all of the strings into one output.

Ahhh!!! Now I see what the series question affects!

So what I would have are multiple sets of conductors with 10.5Adc, 566V on each of them going to the combiner(s) that are inside the inverter, right? So I have 15 sets of 2#12 (CU) + 1#12ground in a 1/2" conduit, all going to the inverter's combiner(s) because I have 15 strings. Am I catching a clue now?
 

david luchini

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Ahhh!!! Now I see what the series question affects!

So what I would have are multiple sets of conductors with 10.5Adc, 566V on each of them going to the combiner(s) that are inside the inverter, right? So I have 15 sets of 2#12 (CU) + 1#12ground in a 1/2" conduit, all going to the inverter's combiner(s) because I have 15 strings. Am I catching a clue now?

Yes, I think you're getting it.
 

BillK-AZ

Senior Member
Location
Mesa Arizona
Solar DC wiring

You can use the software tools for the SunnyBoy inverters at
http://www.sma-america.com/en_US.html
to make sure that the configuration of the PV modules is correct for the PV module you are using and the inverter.
See 'Design Tools' under 'Service'

An ST36 has six SB6000 inverters in the tower. Your configuration does not appear to be correct as three of the inverters will need to have 3 strings and the three strings will exceed the the Maximun Recommended PV array for a SB6000.

You may need a ST42 Sunny Tower. Your PV dealer should be helping you on this.
 
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