PV Circuit Wiring

Status
Not open for further replies.

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
Am I correct in reading the circuit sizing for parallel modules output when you take the 125% requirement in 690.8(A) and calculate in the overcurrent device requirements of 690.8(B) will be 156% (125% of 125%) of the module short circuit current unless you are dealing with device rated at 100% continuous ?
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
Yes. The first 125% is to account for the possibility that in extreme conditions the cells could output more than their nameplate short circuit current at STC. (see: edge of cloud effect) The second 125% is for it being continuous.

(It's kind of overkill because I doubt that the first factor would apply continuously. But it is what it is.)
 

Carultch

Senior Member
Location
Massachusetts
Am I correct in reading the circuit sizing for parallel modules output when you take the 125% requirement in 690.8(A) and calculate in the overcurrent device requirements of 690.8(B) will be 156% (125% of 125%) of the module short circuit current unless you are dealing with device rated at 100% continuous ?

The idea is that Standard Test Conditions is a pair of "nice round numbers" that the industry standardizes upon using for factory testing the solar modules. This is the cell temperature and light intensity that are used for flash testing the modules to determine the voltage and current specifications you see on the datasheet. It is 1000 Watts/meter^2, and 25 Celsius. So we are talking a day and time when you have a clear sky, sunlight that is directly perpendicular to the panels, and an air temperature before the summer heat kicks in. And surroundings that don't reflect enough sunlight to add to the irradiance.

The first 125% is an enhancement factor, that considers the possibility that the sunlight intensity exceeds our nice round number. Possibly due to background snow, a nearby glass building, or reflection off clouds elsewhere in the sky.

The second 125%, as discussed, is your continuous load factor. And you use it just as you would in calculations you do for other continuous loads. What is a bit counterintuitive, is that in 690.8, the second 125% factor only applies prior to your conditions of use factors. And then you apply it to usually the 75C column of the NEC, with 110.14(C) termination considerations being the weak link. The second 125% factor does not apply when you apply the conditions of use derate factors.

The way that I think about the sizing algorithm is as follows:
1.56*Total Isc -> Select termination ampacity to meet or exceed
1.25*Total Isc/(Total derate) -> Select conductor ampacity to meet or exceed
Previous OCPD must be less than termination ampacity, where 240.4(B) would apply. Otherwise actual OCPD in this criteria.
Previous OCPD/(Total derate) must be less than the table value for conductor ampacity, where 240.4(B) would apply. Otherwise actual OCPD in this criteria.
 
Last edited:

SolarPro

Senior Member
Location
Austin, TX
It's kind of overkill because I doubt that the first factor would apply continuously. But it is what it is.)

Totes. For systems over 100 KW, NEC 2017 will allow engineers to simulate system performance using computer modeling to determine the maximum 3-hour current value for conductor sizing purposes.
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
The idea is that Standard Test Conditions is a pair of "nice round numbers" that the industry standardizes upon using for factory testing the solar modules. This is the cell temperature and light intensity that are used for flash testing the modules to determine the voltage and current specifications you see on the datasheet. It is 1000 Watts/meter^2, and 25 Celsius. So we are talking a day and time when you have a clear sky, sunlight that is directly perpendicular to the panels, and an air temperature before the summer heat kicks in. And surroundings that don't reflect enough sunlight to add to the irradiance.

The first 125% is an enhancement factor, that considers the possibility that the sunlight intensity exceeds our nice round number. Possibly due to background snow, a nearby glass building, or reflection off clouds elsewhere in the sky.

The second 125%, as discussed, is your continuous load factor. And you use it just as you would in calculations you do for other continuous loads. What is a bit counterintuitive, is that in 690.8, the second 125% factor only applies prior to your conditions of use factors. And then you apply it to usually the 75C column of the NEC, with 110.14(C) termination considerations being the weak link. The second 125% factor does not apply when you apply the conditions of use derate factors.

The way that I think about the sizing algorithm is as follows:
1.56*Total Isc -> Select termination ampacity to meet or exceed
1.25*Total Isc/(Total derate) -> Select conductor ampacity to meet or exceed
Previous OCPD must be less than termination ampacity, where 240.4(B) would apply. Otherwise actual OCPD in this criteria.
Previous OCPD/(Total derate) must be less than the table value for conductor ampacity, where 240.4(B) would apply. Otherwise actual OCPD in this criteria.

awesome, but a bit over this dinesour inspector's head... as is most Solar.
For conversations sake...and to teach me a bit... let's assume the combined Isc from my combiner box to my inverter is 60 amps. NEC states my conductor ampacity must be 1.56% or 94 amps. Assuming THWN-2 conductors I can use a #4. Derating my #4 for standard reasons (fill, ambient) and assuming 75° terminations I should be fine supplying a 60 amp overcurrent device ? (or would that OCP need to be 125 of my combined module Isc ?)
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
awesome, but a bit over this dinesour inspector's head... as is most Solar.
For conversations sake...and to teach me a bit... let's assume the combined Isc from my combiner box to my inverter is 60 amps. NEC states my conductor ampacity must be 1.56% or 94 amps. Assuming THWN-2 conductors I can use a #4. Derating my #4 for standard reasons (fill, ambient) and assuming 75° terminations I should be fine supplying a 60 amp overcurrent device ? (or would that OCP need to be 125 of my combined module Isc ?)

Don't derate twice. If you are using 1.56 for Isc you don't derate the conductors by 0.8. To keep things straight, I use 1.25 * Isc to get max current and then treat it like any other continuous current.
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
awesome, but a bit over this dinesour inspector's head... as is most Solar.
For conversations sake...and to teach me a bit... let's assume the combined Isc from my combiner box to my inverter is 60 amps. NEC states my conductor ampacity must be 1.56% or 94 amps. Assuming THWN-2 conductors I can use a #4. Derating my #4 for standard reasons (fill, ambient) and assuming 75° terminations I should be fine supplying a 60 amp overcurrent device ? (or would that OCP need to be 125 of my combined module Isc ?)

The overcurrent device would need to be 100A (unless listed for 100% continuous, then 75A) if required, and the conductor needs to be protected by the OCPD. But keep in mind that if you are not paralleling that combined output with another output, you don't need an OCPD on it (unless part of the inverter listing). So usually what ggunn said about conductor sizing is straightforwardly correct.

A system with around 60A combined Isc would typically only have fuses on each string output (e.g. 15A fuses on 6 strings, each 9A Isc).
 

augie47

Moderator
Staff member
Location
Tennessee
Occupation
State Electrical Inspector (Retired)
Don't derate twice. If you are using 1.56 for Isc you don't derate the conductors by 0.8. To keep things straight, I use 1.25 * Isc to get max current and then treat it like any other continuous current.

Still trying to grasp.... sorry, sometimes I'm a bit slow....
The 1st 1.25 is to satisfy 690.8(A)(1), the second is to satisfy the "continuous load" factor so the ampacity would need to be 1.25 X 1.25 or 156% ( Assuming no corrections factors for fill or ambient ) Correct ?

Once I select my conductor, I can apply the adjustment factors if needed.
(adjustments to the 90° rating if applicable, the result not to exceed the 75° rating).

Is that correct ?

(I didn't grasp ignoring the second 1.25)
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Still trying to grasp.... sorry, sometimes I'm a bit slow....
The 1st 1.25 is to satisfy 690.8(A)(1), the second is to satisfy the "continuous load" factor so the ampacity would need to be 1.25 X 1.25 or 156% ( Assuming no corrections factors for fill or ambient ) Correct ?

Once I select my conductor, I can apply the adjustment factors if needed.
(adjustments to the 90° rating if applicable, the result not to exceed the 75° rating).

Is that correct ?

(I didn't grasp ignoring the second 1.25)

1) Multiply Isc by 1.25. This is your max current (Imax).

2) Pick a conductor.

3) Assuming it is 90 degree wire derate the 90 degree ampacity for conditions of use (temperature, heat adder, current carrying conductors in conduit).

4) For the same sized conductor, derate the 75 degree ampacity for continuous use (multiply by 0.8). There is your second 1.25.

5) Both derated ampacities must be greater than Imax.

Notice that the second 1.25 only applies in step 4.

For OCPD, multiply Isc by 1.25 to get Imax, and then by 1.25 to get minimum OCPD like you would with any other current.

You can use 1.25 * Imax to pick a minimum conductor size (step 2) from the 75 degree column (it's the reciprocal of step 4 but it does not give you actual ampacity) , but it's not part of the actual calculation.
 
Last edited:

Carultch

Senior Member
Location
Massachusetts
awesome, but a bit over this dinesour inspector's head... as is most Solar.
For conversations sake...and to teach me a bit... let's assume the combined Isc from my combiner box to my inverter is 60 amps. NEC states my conductor ampacity must be 1.56% or 94 amps. Assuming THWN-2 conductors I can use a #4. Derating my #4 for standard reasons (fill, ambient) and assuming 75° terminations I should be fine supplying a 60 amp overcurrent device ? (or would that OCP need to be 125 of my combined module Isc ?)

Taking your example:
Given that standard test conditions total Isc is 60 A.

Max continuous current = 1.25*Isc = 75A
Sizing calculation for OCPD's = 1.56*Isc = 94A

This means we'd need 100A worth of overcurrent device if required. And we'd need 94A of termination ampacity.

Most terminations are rated 75C. For 100A and less, you need to check. Assuming such, this would mean a #3 Cu conductor is required for terminations, given that 1.25*1.25*Isc = 94A.

Now the conditions of use. A temperature correction factor of 0.82 is very common. We're talking 33C ambient and a durablock's height above the roof. Another common one when not in direct sunlight is 0.96, which I call the "basic derate" for simple ambient temperatures, and if this is all that there is, it usually will not affect conductor selection.

Standard 30C ampacity of #3 Cu THWN-2 in the 90C column = 110A
Adjust for 0.82 temperature correction: 0.82*110A = 90.2A

Now you might think that this calculation has to at least be 94A, and so did I at one point. But it doesn't. You do not apply both 1.25 factors when doing derates, only the first enhancement factor. Once you apply the enhancement factor, you treat it as any other continuous load.

You only need 1.25*Isc worth of conductor ampacity to satisfy the second condition. In this case, we need 75A worth of conductor ampacity, and we have 90.2A. So far, the #3 wire is still good.


Finally, the OCPD, where required, shall protect the wire as sized. This means that both the termination ampacity and the derated conductor ampacity must round up to the actual OCPD (for 800A and less), or be greater than the actual OCPD for over 800A. An obscure rule, thanks to 240.4(B), but it certainly is strategic to make use of it. Where OCPD is not required, and not part of the design, you don't need to think about this. Where not required, but somehow there anyway, you also don't need to apply this rule. I suppose in concept you could have 100A worth of wire landing on an unnecessary 200A breaker where OCPD is not required, but it would raise questions when reviewed by anyone else.

So. We've determined a 100A OCPD, with #3 wire. Does the 100A OCPD protect it?

Its termination ampacity is 100A. Good
Its derated conductor ampacity is 90.2A. OOOOOO!!!! this is interesting. We've exceeded the previous OCPD by just a fraction of an ampere. But it still rounds up to 100A. Therefore, the 100A OCPD still protects the #3 wire, and thus the #3 wire is sufficient for local factors only.

Next comes voltage drop, which is a story for another day. And not as firm of an NEC requirement as the local factors.
 
Last edited:

Carultch

Senior Member
Location
Massachusetts
The overcurrent device would need to be 100A (unless listed for 100% continuous, then 75A) if required, and the conductor needs to be protected by the OCPD. But keep in mind that if you are not paralleling that combined output with another output, you don't need an OCPD on it (unless part of the inverter listing). So usually what ggunn said about conductor sizing is straightforwardly correct.

A system with around 60A combined Isc would typically only have fuses on each string output (e.g. 15A fuses on 6 strings, each 9A Isc).

In the special case of paralleling exclusively two source circuits, or exclusively two combiner output circuits, you do not need OCPD when doing so, provided that each feeder has enough ampacity for the larger of the two.

So you can "bare tap" two strings together, and you can also "bare tap" two identically-sourced combiner outputs together. Or if you have to disproportionately sourced combiners, you can "bare tap" their outputs together provided that both feeders have enough ampacity for the larger of the two. By "bare tap", I mean that you connect the two together, without protecting each with an overcurrent device.

The point of the rule for requiring OCPD when combining more than two DC sources, is to prevent multiple "healthy" circuits from feeding a faulted circuit in excess of its ampacity.
 

Carultch

Senior Member
Location
Massachusetts
Totes. For systems over 100 KW, NEC 2017 will allow engineers to simulate system performance using computer modeling to determine the maximum 3-hour current value for conductor sizing purposes.

So what you are telling me, is that you can take the peak day's 3-hour average current centered on solar noon, and use that as your maximum continuous current, instead of 1.25*Isc?

Rather than the present Murphy's law approach in assuming that it is perpetually solar noon and that it is short circuited.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Finally, the OCPD, where required, shall protect the wire as sized. This means that both the termination ampacity and the derated conductor ampacity must round up to the actual OCPD (for 800A and less), or be greater than the actual OCPD for over 800A.
I don't think that is true. I'm pretty sure that the derated conditions of use ampacity must be greater than or equal to the next size down OCPD from the one used, but not the continuous use ampacity. We had quite a detailed discussion about that a while back and that was the consensus.
 

Carultch

Senior Member
Location
Massachusetts
I don't think that is true. I'm pretty sure that the derated conditions of use ampacity must be greater than or equal to the next size down OCPD from the one used, but not the continuous use ampacity. We had quite a detailed discussion about that a while back and that was the consensus.

Exclude the "or equal" part from what you said, because 240.4(B) specifically says that ampacity cannot correspond to a standard OCPD size for you to use the next size up. I do wonder just how rigid this rule is, because in Augie's example, the ampacity was just a fraction of an ampere higher than 90A, which I concluded that the wire was sufficient for a 100A breaker.

Maybe what I should say instead of "round up", is that it should "ceiling up". "Ceiling" is the term used in Excel for the action of forcing a number to round up, even if it is closer to the number to which it would round down.

Can you give me an example of what you mean, that would show the error in what I said?
 
Last edited:

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Exclude the "or equal" part from what you said, because 240.4(B) specifically says that ampacity cannot correspond to a standard OCPD size for you to use the next size up. I do wonder just how rigid this rule is, because in Augie's example, the ampacity was just a fraction of an ampere higher than 90A, which I concluded that the wire was sufficient for a 100A breaker.

Maybe what I should say instead of "round up", is that it should "ceiling up". "Ceiling" is the term used in Excel for the action of forcing a number to round up, even if it is closer to the number to which it would round down.

Can you give me an example of what you mean, that would show the error in what I said?
Inverter Imax = 38A
(1.25)(38A) = 47.5A
Choose 50A OCPD

Choose #8 THWN-2
90 degree ampacity 55A, conditions of use derated to 50.1A
50.1A > 38A
75 degree ampacity 50A, continuous use derated to 40A
40A > 38A

50.1A > 45A, the wire is protected by the OCPD
 
Status
Not open for further replies.
Top