Enphase 3-Phase Load Calc

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Sparks.4.All

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Location
Oregon
I have 75 Enphase 215 Micro Inverters that will be installed on a 120/208 WYE system. Each one is .9 amps. The will be installed in 5 strings of 15. They are calculated as a continous load at 125%. They will be balanced in a dedicated PV combiner panel. This calculation is for the OCP devise and feeder sizing. Each phase will have 25 Micro inverters on it.

25 x .9A x 1.25 = 28.125A per phase

28A/1.73= 16.26A

Am I doing this corect? It just seems very low to me!
 

Sparks.4.All

Member
Location
Oregon
Each M215 is connected to 2 phases with 208. Drew it out in Excel and the numbers were wrong. I have 56.25 Amps on each of the 3 phases.

So it would be a 60 Amp breaker using #6 THHN for the feeder.
 
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Smart $

Esteemed Member
Location
Ohio
Looked up M215 literature...

They are rated 1.0A @208 (3?)

75 ? 1.0A ? 208V ? 125% = 19,500VA

19,500VA ? 208 ? √3 = 54.1A
 

david luchini

Moderator
Staff member
Location
Connecticut
Occupation
Engineer
Each M215 is connected to 2 phases with 208. Drew it out in Excel and the numbers were wrong. I have 56.25 Amps on each of the 3 phases.

Don't add the phase currents together (ie, 25 * 0.9 * 1.25 = 28.125A. 28.125A (A-B) + 28.125A (A-C) = 56.25A on phase A.) Instead, in your balanced load you should multiply the current by 1.732 to get the phase currents. 28.125A * 1.732A = 48.71A on each phase.

Or using 1.0A, as Smart$ has suggested: 25 * 1.0 * 1.25 * 1.732 = 54.1A on each phase.
 

Sparks.4.All

Member
Location
Oregon
That makes sense but when I change from .9 to 1A in the Excel spreadsheet I get 12.5A per string per phase.

1.0A x 1.25 x 10 (number of inverters per phase per string) = 12.5
12.5A (per string) x 5 (# of strings) = 62.5A
This should be the true load

Am I missing something?
 

david luchini

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Staff member
Location
Connecticut
Occupation
Engineer
That makes sense but when I change from .9 to 1A in the Excel spreadsheet I get 12.5A per string per phase.

1.0A x 1.25 x 10 (number of inverters per phase per string) = 12.5
12.5A (per string) x 5 (# of strings) = 62.5A
This should be the true load

Am I missing something?

You said you had 15 inverters per string.

1.0A x 1.25 x 15 (number of inverters per phase per string) = 18.75
18.75A (per string) x 5 (# of strings) = 93.75A total.

93.75/1.732 (total load is balanced on a 3phase system) = 54.1Amps.
 
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