>for two kaneka wired in parallel,
we have a ISC of 1.22 and a series fuse rating of 7.
first take ISC of 1.22 x 1.25 (over irradiance) x 1.25 (3hr continuous rule) = 1.90a
1.90 x 2 (for two modules in parallel) = 3.81
4a ocpd (overcurrent protection device) would allow 4a of fault current to reach the faulted module/string.
this is the tricky part. then you must add the possible current of what one module in parallel could contribute to the other in our case. the above calculations were assuming a fault from downstream (controller, inverter, etc). we must add what the second of our pair COULD be putting THROUGH its next door neighbor if it were light when the fault is still occuring.
so we must add to our 3.81 another 1.25x ISC (in our case 1.22)
1.22 x 1.25 = 1.53
3.81 + 1.53 = 5.34 so for two kaneka in parallel, a 6a or 7a OCPD is acceptable.
if we try to do 3, it wont work because
3 x 1.56 x 1.22 = 5.7
but two other modules in parralel connected to the first in the event of a fault would give us 2(modules) x 1.25 (irradiance) x 1.22 (ISC) = 3.05
5.7 + 3.05 = 8.75
which is over the series fuse size of 7a.nothing like code in the morning
all of that can be found on page 132 as well
- james - Alt-E staff
