Breaker Sizing for Solar Project

Status
Not open for further replies.

AWinston

Member
Location
Murrieta, Ca
I am working on a solar project and I am making sure the client selected the correct circuit breaker. The system is as follows:

8 inverters (Imax = 36A/Ifault=42.72A) connected to a combiner (50A CB)

4 combiners connected to a final combiner (2kA rated/each combiner going to a 500A CB)

My thought is that the final combiner circuit breakers are undersized (500A). The worst case output of each combiner would be 42.72A x 8 = 341.76A. Using the 80% rule (341.76A x 1.2% = 410.1A), the breaker to be used would be 450A.

With using the 500A CB currently purchased, would the system be adequately protected?
 

Carultch

Senior Member
Location
Massachusetts
I am working on a solar project and I am making sure the client selected the correct circuit breaker. The system is as follows:

8 inverters (Imax = 36A/Ifault=42.72A) connected to a combiner (50A CB)

4 combiners connected to a final combiner (2kA rated/each combiner going to a 500A CB)

My thought is that the final combiner circuit breakers are undersized (500A). The worst case output of each combiner would be 42.72A x 8 = 341.76A. Using the 80% rule (341.76A x 1.2% = 410.1A), the breaker to be used would be 450A.

With using the 500A CB currently purchased, would the system be adequately protected?

The fault current of the inverters is insignificant compared to the fault current of the grid. The latter shall govern, and the easiest strategy is to match the KAIC rating of existing equipment on the premises.

Inverter manuals might specify the breaker to use, but if desired, you usually can connect to breakers in excess of the minimum, provided that the conductors are protected per Article 240. Once you connect the inverter to its dedicated breaker on the first combiner, it doesn't matter how large the breakers elsewhere in the circuit are, provided that they meet the minimum. For instance, you can plan for future expansion and put a much larger main breaker and main feeder on your panel than you need for present scope of work, and that will not affect individual inverter requirements.

While the NEC isn't clear on whether or not you are required to accumulate rounding errors when adding up the breakers in a dedicated PV system panelboard, I do not see any reason to do so. IMO, all you should have to do is add up the total inverter operating current, and size a master breaker per the 1.25 continuous duty factor, and then size a feeder as required for 1.25*total current which is properly protected by that OCPD, taking advantage of 240.4(B) where you can. Be aware that some AHJ's might disagree, and require you to accumulate rounding errors of each inverter branch breaker (ridiculous, in my opinion).

In this example, you'd need:
45A branch breakers for each inverter (50A is OK as well)
400A main busbars & breakers for each AC combiner that combines a group of 8
400A branch breakers with a 1600A main for the master AC combining switchboard
 
Last edited:

AWinston

Member
Location
Murrieta, Ca
The fault current of the inverters is insignificant compared to the fault current of the grid. The latter shall govern, and the easiest strategy is to match the KAIC rating of existing equipment on the premises.

Inverter manuals might specify the breaker to use, but if desired, you usually can connect to breakers in excess of the minimum, provided that the conductors are protected per Article 240. Once you connect the inverter to its dedicated breaker on the first combiner, it doesn't matter how large the breakers elsewhere in the circuit are, provided that they meet the minimum. For instance, you can plan for future expansion and put a much larger main breaker and main feeder on your panel than you need for present scope of work, and that will not affect individual inverter requirements.

While the NEC isn't clear on whether or not you are required to accumulate rounding errors when adding up the breakers in a dedicated PV system panelboard, I do not see any reason to do so. IMO, all you should have to do is add up the total inverter operating current, and size a master breaker per the 1.25 continuous duty factor, and then size a feeder as required for 1.25*total current which is properly protected by that OCPD, taking advantage of 240.4(B) where you can. Be aware that some AHJ's might disagree, and require you to accumulate rounding errors of each inverter branch breaker (ridiculous, in my opinion).

In this example, you'd need:
45A branch breakers for each inverter (50A is OK as well)
400A main busbars & breakers for each AC combiner that combines a group of 8
400A branch breakers with a 1600A main for the master AC combining switchboard

Thanks. That was very helpful. The 50A selection was driven by the manufacturer.
 
Be aware that some AHJ's might disagree, and require you to accumulate rounding errors of each inverter branch breaker (ridiculous, in my opinion).

Hmm, what about this situation? A 20kw inverter at 480V comes out to 24.08A.
I've seen it listed as 24A and 24.1A-- 24.1 *1.25 is a 30.125A breaker. (The maker says 24A, so is 24.1A even worth considering?)

So are you saying that with 3 of those inverters @ 90.375A, a 90A breaker would be ok, but if you had 5 of them @120.5A, the AHJ might require a 200A breaker, as 120.5A * 1.25 is 150.625A?
That does seen a bit ridiculous.

Also, what did you mean by "taking advantage" of 240.4(B) "where you can"?
Thanks.
 

ggunn

PE (Electrical), NABCEP certified
Location
Austin, TX, USA
Occupation
Electrical Engineer - Photovoltaic Systems
Hmm, what about this situation? A 20kw inverter at 480V comes out to 24.08A.
I've seen it listed as 24A and 24.1A-- 24.1 *1.25 is a 30.125A breaker. (The maker says 24A, so is 24.1A even worth considering?)

So are you saying that with 3 of those inverters @ 90.375A, a 90A breaker would be ok, but if you had 5 of them @120.5A, the AHJ might require a 200A breaker, as 120.5A * 1.25 is 150.625A?
That does seen a bit ridiculous.

Also, what did you mean by "taking advantage" of 240.4(B) "where you can"?
Thanks.

In the first place, if the inverter manufacturer says the rated output is 24A, you use 24.0A per 690.8(A)(3). In the second place, the next breaker size up from 150A is 175A, not 200A.
 
In the first place, if the inverter manufacturer says the rated output is 24A, you use 24.0A per 690.8(A)(3).

So if you divide 20,000w by 480V and then by 1.73 and get 24.08 amps.... the actual rule is round down to 24A?
I'm just asking because of the AHJ discretion thing- and because as far as I can tell the "rounding rules" are somewhat open to interpretation.

And yes, 175A is next up, but thanks, you would have to round up there.
 

Carultch

Senior Member
Location
Massachusetts
So if you divide 20,000w by 480V and then by 1.73 and get 24.08 amps.... the actual rule is round down to 24A?
I'm just asking because of the AHJ discretion thing- and because as far as I can tell the "rounding rules" are somewhat open to interpretation.

And yes, 175A is next up, but thanks, you would have to round up there.

There is no round down rule. Go straight off the inverter's datasheet. If it says 27.3, don't use 27. Use 27.3. It is possible that Watts/Volts/sqrt(3) is slightly above what it says on the inverter datasheet or slightly below. Or there may be a much more significant discrepancy. One reason in particular why there might be, is that the inverter is built to operate at nameplate power at lower voltages than nominal. In any case, the inverter datasheet shall govern.

The rounding errors I was talking about, is when your breaker is in excess of 1.25*current, do you need to accumulate the inconsistency in sizing your master circuit? Example: 13 inverters at 27.3 A. Each breaker = 35A. Adding up breakers gets you 35*13 = 455. But adding up amperes first, then applying 1.25, gets you 27.3*13*1.25 = 443.625A. So is that a 450A main or a 500A?
 
Last edited:

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
From [2011] on you add up the numbers of the breakers closest to each inverter, with no additional accumulation of size rounding as you go upstream.
With [2014] you add up the nominal output amps instead, all the way up the chain.
 

GoldDigger

Moderator
Staff member
Location
Placerville, CA, USA
Occupation
Retired PV System Designer
The 20,000W number is marketing, so rather than calculate amps from that you take the amps from the spec sheet.
In addition to the current potentially varying with voltage if you try to calculate back from the Watts number.
 
There is no round down rule. Go straight off the inverter's datasheet.

ggunn No. Read 690.8(A)(3).

GoldDigger The 20,000W number is marketing, so rather than calculate amps from that you take the amps from the spec sheet.


Well...what do ya do then when then spec sheet says 24A and the installation manual says 24.1A?
Both linked to from the company's page, for the same model. Kinda weird there.
 

iwire

Moderator
Staff member
Location
Massachusetts
ggunn No. Read 690.8(A)(3).

GoldDigger The 20,000W number is marketing, so rather than calculate amps from that you take the amps from the spec sheet.


Well...what do ya do then when then spec sheet says 24A and the installation manual says 24.1A?
Both linked to from the company's page, for the same model. Kinda weird there.


You have been here a while, how about learning how to use the quote button?

Anyway, I would never use the cut sheets or the installation manual as the source for calculations. Go by the label on the unit. This stuff is changed and updated so often I don't count on cut sheets to keep up.
 
The rounding errors I was talking about, is when your breaker is in excess of 1.25*current, do you need to accumulate the inconsistency in sizing your master circuit? Example: 13 inverters at 27.3 A. Each breaker = 35A. Adding up breakers gets you 35*13 = 455. But adding up amperes first, then applying 1.25, gets you 27.3*13*1.25 = 443.625A. So is that a 450A main or a 500A?

So by 2011 it's 500A, but by 2014 it's 450A.
That's what I'm understanding- but I'm still asking to be sure.
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
I don't agree that the 2011 code requires one to add breaker sizes to get the combined output breaker size. 450A would be okay in the last example given.

There is a difference in how you calculate panel busbar ratings, among several other significant differences in busbar requirements between the code cycles.
 
Status
Not open for further replies.
Top