Inverter Output Circuit conductor ampacity

[Kyle J]

As I'm preparing for the upcoming NABCEP exam, I'm trying to figure out the correct calculation to determine ampacity of the inverter output circuit.

See Mike Holt's book, "Understanding NEC requirements for Solar Photovoltaic Systems", pg. 348 Figure 690-43 and pg. 350 Figure 690-48.

Mike has a reference for Figure 690-43 ampacity: *690.8(A)(3) which states, "(3) Inverter Output Circuit Current. The maximumcurrent shall be the inverter continuous output currentrating. "
However, in Mike's example he is taking the inverter continuous output nameplate rating x 1.25.

690.8(B) states, "Circuit conductors shall be sized to carry not less than the larger of 690.8(B)(1) or (2)."
(B)(1) states to add 125% to max currents in 690.8(A)

As I read the NEC, I believe the calculation to add 125% to inverter output (in the above examples, 24A) is correct.
However, Mike appears to reference 690.8(A)(3) on both pages, but does not add 125% to the example on pg. 350. Is Mike's reference incorrect? Should the reference be 690.8(B)(1)?

Thanks for all who read & reply to this, either way in the book's examples the conductor size is OK for the calculated current but I'm wondering, "What if…?" in the NABCEP exam there is a question similar that has some tighter ampacity calculations. Do I use (A)(3) or the higher of (B)(1) or (2)?

 

[Carultch]

As I'm preparing for the upcoming NABCEP exam, I'm trying to figure out the correct calculation to determine ampacity of the inverter output circuit.

See Mike Holt's book, "Understanding NEC requirements for Solar Photovoltaic Systems", pg. 348 Figure 690-43 and pg. 350 Figure 690-48.

Mike has a reference for Figure 690-43 ampacity: *690.8(A)(3) which states, "(3) Inverter Output Circuit Current. The maximumcurrent shall be the inverter continuous output currentrating. "
However, in Mike's example he is taking the inverter continuous output nameplate rating x 1.25.

690.8(B) states, "Circuit conductors shall be sized to carry not less than the larger of 690.8(B)(1) or (2)."
(B)(1) states to add 125% to max currents in 690.8(A)

As I read the NEC, I believe the calculation to add 125% to inverter output (in the above examples, 24A) is correct.
However, Mike appears to reference 690.8(A)(3) on both pages, but does not add 125% to the example on pg. 350. Is Mike's reference incorrect? Should the reference be 690.8(B)(1)?

Thanks for all who read & reply to this, either way in the book's examples the conductor size is OK for the calculated current but I'm wondering, "What if…?" in the NABCEP exam there is a question similar that has some tighter ampacity calculations. Do I use (A)(3) or the higher of (B)(1) or (2)?

There has been same re-organization in NEC690.8, so it is difficult to keep track of exact references without knowing whether you have NEC2011 or NEC2014 applying. In any case, both should give you the same results.

The way that it is now written is that you start by calculating the maximum continuous current (Imax) of the circuit in question. For anything that is current controlled, such as an inverter or dc-dc converter, this value comes directly from a datasheet. For any circuit that comes directly from PV modules, this value is Isc*1.25, multiplied by quantity in parallel if applicable.

First step is Imax*1.25. Then select both conductor and termination ampacities that meet or exceed this. OCPD needs to meet or exceed this value as well.
Next step is Imax/total derate. Then select conductor ampacity that meets or exceeds this.

Now check the non-derated termination ampacity, and the derated conductor ampacity, and make sure that any required OCPD protects this wire per Article 240.

 

[jaggedben]

Kyle,

Take the inverter nameplate rating and multiply by 1.25.

I don't have Mike's book to review if he made a mistake or not, but the above sentence is correct.

Note that nowadays the same requirements are also contained in article 705.

(Also, regarding where the code says "Inverter Output Circuit Current. The maximumcurrent shall be the inverter continuous output currentrating. " The word 'continuous' in this context is to distinguish from nameplate ratings that might be max 'surge' current or similar. This does not mean the 1.25 multiplier is already applied to the nameplate rating.)

 

[Smart $]

FWIW, you do not add 125%... the result would be 225%.

It is either factor or multiply by 125%... or add 25%.

 

[Kyle J]

Thanks!

Thanks!

Thanks for the good replies, and yes a good catch from Smart $. I mistyped "adding 125%" and understand it's multiplying by 125%.

I think I found my misunderstanding:
690.8 = Circuit Sizing & Current
690.8(A)(3) = Inverter continuous output current is the maximum current.

690.8(B) = Conductor Ampacity, Circuit conductors shall be sized to carry not less than the larger of 690.8(B)(1) or (2) before application of adjustment and correction factors.
B1 says to take current calculation in 690.8(A) and multiply by 125%
B2 takes into account adjustment and correction factors.
(No multiplication noted in B2. - kj)

 

[SolarAlternativesDesign]

There is a third calculation also.
2014 NEC 690.9(A) also requires ... inverter output circuit... conductors shall be protected in accordance with the requirements of Article 240 [overcurrent protection]. This requires that the conductor ampacity with conditions of use applied be as great as the OCPD protecting it. However 240.4(B) permits using the next smaller standard OCPD rating in most cases.
I don't know whether the NABCEP exam would throw in the OCPD rating in its questions on inverter output circuit conductor sizing.

 

[Smart $]

There is a third calculation also.
2014 NEC 690.9(A) also requires ... inverter output circuit... conductors shall be protected in accordance with the requirements of Article 240 [overcurrent protection]. This requires that the conductor ampacity with conditions of use applied be as great as the OCPD protecting it. However 240.4(B) permits using the next smaller standard OCPD rating in most cases.
I don't know whether the NABCEP exam would throw in the OCPD rating in its questions on inverter output circuit conductor sizing.

No, no, no. 240.4 (B) permits using the next greater standard OCPD rating above the conductor ampacity when not more than 800A.

Stated another way, the values of both 690.8(B)(1) and (B)(2) must exceed the ampacity of the next smaller standard OCPD rating.

 

[SolarAlternativesDesign]

I agree. That's what I tried to state.

 

[ggunn]

No, no, no. 240.4 (B) permits using the next greater standard OCPD rating above the conductor ampacity when not more than 800A.

Stated another way, the values of both 690.8(B)(1) and (B)(2) must exceed the ampacity of the next smaller standard OCPD rating.

I disagree. For 90 degree rated conductors only the 90 degree ampacity derated for conditions of use needs to be greater than the rating of the next size down OCPD. The 75 degree ampacity derated for continuous use can be less.

 

[Carultch]

I disagree. Only the 90 degree ampacity derated for conditions of use needs to be greater than the rating of the next size down OCPD. The 75 degree ampacity derated for continuous use can be less.

I understand it that both the derated 90C ampacity for conditions of use, and the non-derated 75C (or other if applicable) terminal ampacity that must exceed the previous size down OCPD. And per NEC rounding conventions, I believe it should exceed it by a half an amp greater than the previous size (I've wondered just how close it can be for a long time, and only recently learned this rounding convention).

The intent is that the OCPD is the "weak link" in the chain, so that it trips before the wire or terminations overheat.

 

[SolarAlternativesDesign]

I disagree. Only the 90 degree ampacity derated for conditions of use needs to be greater than the rating of the next size down OCPD. The 75 degree ampacity derated for continuous use can be less.

Right. The OCPD requirement only applies to the derated ampacity with the conductor temp rating, not the connector temp rating or the 1.25X1.25 factor.
Question for you - is the OCPD sizing method intended only to protect from a fault current from the service to the inverter? Is there any other situation in which a current of that magnitude would flow in that conductor?

 

[ggunn]

Right. The OCPD requirement only applies to the derated ampacity with the conductor temp rating, not the connector temp rating or the 1.25X1.25 factor.
Question for you - is the OCPD sizing method intended only to protect from a fault current from the service to the inverter? Is there any other situation in which a current of that magnitude would flow in that conductor?

Well, no, not that I can think of. Solar grid tied inverters are current limited to their maximum rating. We design inverter output conductors such that the inverter is incapable of endangering them from overcurrent.

Offgrid inverters that run off of batteries are possibly another story but I don't have much experience with them.

 

[Carultch]

Right. The OCPD requirement only applies to the derated ampacity with the conductor temp rating, not the connector temp rating or the 1.25X1.25 factor.
Question for you - is the OCPD sizing method intended only to protect from a fault current from the service to the inverter? Is there any other situation in which a current of that magnitude would flow in that conductor?

The method is set up to avoid nuisance tripping due to the continuous load at nominal current.

 

[Smart $]

I disagree. For 90 degree rated conductors only the 90 degree ampacity derated for conditions of use needs to be greater than the rating of the next size down OCPD. The 75 degree ampacity derated for continuous use can be less.

Well, it's somewhat moot regarding inverter output circuits as both the conductor ampacity and OCPD rating are based 125% of the "inverter continuous output current rating"... and not each other.

OCPD is not less than 125% 690.8(A) per 690.9(B)... period.

Conductor Ampacity is not less than the larger of 690.8(B)(1) 125% 690.8(A) before adjustments and corrections vs. 690.8(B)(2) 100% 690.8(A) after adjustments and corrections.

Given the latter, you cannot say the (B)(2) value will always be larger than (B)(1).

 

[Smart $]

I should add that we would resort to general requirements of Article 240 if, after the above conductor ampacity determination, there is a conflict between OCPD rating and conductor ampacity, i.e. the conductor is not considered protected per Article 240.

We must also take into consideration that the requirement of 110.14(C) applies as it is not modified in Article 690. If you parse Chapters 1 & 2, you will see that circuit ampacity cannot exceed the termination limit ampacity, regardless of greater derated conductor ampacity. So when we have to resort to Article 240, we compare the OCPD rating to the smaller of the termination ampacity and the derated ampacity.

 

[ggunn]

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, so the wire is protected by the OCPD even though the 75 degree ampacity is < 45A

 

[Carultch]

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, so the wire is protected by the OCPD even though the 75 degree ampacity is < 45A

Here's my thought process on the same problem:
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, required to equal 125% of 38A due to a 38A continuos load. 50A > 47.5A.

50.1A conductor ampacity > 45A
50A termination ampacity > 45A
Therefore it is protected by the 50A OCPD, since 45A is the previous standard OCPD rating. Both the termination and derated conductor ampacities exceed 45A, and this is an instance where 240.4(B) can apply (even though it doesn't need to).

 

[Smart $]

Inverter Imax = 38A
OCPD minimum (1.25)(38A) = 47.5A
Choose 50A OCPD


Conductor Ampacity
690.8(B)(1): (38A)(125%) = 47.5
690.8(B)(2): (38A) (55/50.1*) = 41.7A, *You didn't state what the adjustment and correction factors, only the result.
690.8(B): 47.5A is the larger of the two

Choose #8 THWN-2

Verify 240.4(B) compliance:
90°C ampacity 55A, conditions of use derated to 50.1A, no continuous factoring

• 50.1A > 47.5A​

75°C ampacity 50A, 125% continuous, no derating

• 50A > 47.5A​

50A > 45A, so the wire is protected by the OCPD (NOTE: conductor ampacity cannot exceed termination temperature limitation ampacity; OCPD is not rated for over 75°C operation)

 

[Smart $]

...(NOTE: conductor ampacity cannot exceed termination temperature limitation ampacity; OCPD is not rated for over 75°C operation)

conductor should have been "circuit"

 

[Carultch]

Here's a more interesting example, that shows why these subtle rules matter:

Inverter Imax = 38A
Conditions of use derate factor: 0.82
(1.25)(38A) = 47.5A
Choose 50A OCPD

Start with #8 THWN-2.

90 degree ampacity 55A, conditions of use derated at 0.82 (direct sunlight) to 45.1A
45.1A > 38A

75 degree ampacity 50A, required to equal 125% of 38A due to a 38A continuos load. 50A > 47.5A.

45.1A conductor ampacity > 45A
50A termination ampacity > 45A
Therefore it is protected by the 50A OCPD, since 45A is the previous standard OCPD rating.

The good question to ask is, is 45.1A greater than 45A "enough", in order to be protected by a 50A OCPD? And how would you know?