Please note the conductor ampacity must be the larger of 690.8(B)(1) or (B)(2)... which for a 38A output rating is: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?
1) 125%*38A=47.5A
2) 38A/0.82=46.3
47.5A > 46.3A
#8 THWN-2 is too small to be compliant in this scenario because its derated ampacity is less than required per 690.8(B).