Inverter AC Disconnect 690.14

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tortuga

Code Historian
Location
Oregon
Occupation
Electrical Design
You'll want to check 705.20 in the 2020 NEC but If I am reading (7) correctly it gets sized to 'maximum circuit current' among other things.
in the 2017 NEC its 705.22 (4) 'Have ratings sufficient for the maximum circuit current, available short-circuit current, and voltage that is available at the terminals'
 

Carultch

Senior Member
Location
Massachusetts
Hi all,

I am installing an AC disconnect for an inverter output to comply with the NEC 690.14. The inverter's maximum AC output is 36.2 A. Can I just to a 40A disconnect (https://www.bentek.com/wp-content/uploads/ds/991-004409-000-B_Bentek-Disc-AC-DS.pdf) or should I do the 1.25 derate (36.2 x 1.25 =45.25 A) and use a 50A disconnect?

Is there an NEC code that explains if the 1.25 derate is required.

Thanks!

The disconnect requires a continuous duty rating, for you to not have to apply the 1.25 factor. This is common for unfused units. Ask the manufacturer and get it in writing for your records.
 

tortuga

Code Historian
Location
Oregon
Occupation
Electrical Design
The disconnect requires a continuous duty rating, for you to not have to apply the 1.25 factor. This is common for unfused units. Ask the manufacturer and get it in writing for your records.
Good point, that is said another way any disconnect listed to UL 98 is rated as described in the UL whitebook:
SWITCHES (WFXV)
RATINGS
Ratings of enclosed or nonenclosed pullout switches are limited to 600 V
or less, 400 A or less.
Switches without fuseholders (unfused) have been tested to determine
their acceptability for continuous operation at their marked rated load.
 

TheElectrician

Senior Member
You'll want to check 705.20 in the 2020 NEC but If I am reading (7) correctly it gets sized to 'maximum circuit current' among other things.
in the 2017 NEC its 705.22 (4) 'Have ratings sufficient for the maximum circuit current, available short-circuit current, and voltage that is available at the terminals'

Thanks for the input, I am in a state that still follows NEC 2008 with 2011 amendments that states, "Having ratings not less than the load to be carried and the fault current to be interrupted." under 705.22 (4). how do I interpret this? the maximum load is basically 36.2A, so can I use a 40A (the one attached to post #1)?
 

TheElectrician

Senior Member
The disconnect requires a continuous duty rating, for you to not have to apply the 1.25 factor. This is common for unfused units. Ask the manufacturer and get it in writing for your records.

Do you mean that if the manufacturer says that the disconnect is "Rated for continuous duty" then I have to multiply by 1.25 factor? Can you link me to a source/code/article that supports this? Thank you
 

TheElectrician

Senior Member
Good point, that is said another way any disconnect listed to UL 98 is rated as described in the UL whitebook:
SWITCHES (WFXV)

The datasheet I attached says the continuous current rating to be 40A, does this mean I have to use the 1.25 derate for the inverter's output (36.2 x 1.25 = 45.25A) and use a 50A disconnect instead?
 

jaggedben

Senior Member
Location
Northern California
Occupation
Solar and Energy Storage Installer
My opinion is that you can use that switch, with the caveat that i can't speak to your states ammendments without seeing more complete language.

In general, believe it or not, the NEC doesn't specify ratings for switches. In my experience the switch rating is usually determined by its label, which will require it to be protected by a certain size OCPD, which in turn is proscribed by the NEC for the circuit. However from what I see of the switch documentation it is rated for the use you have in mind, according to its own literature.
 
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