Ok I set things up temporarily outside to get a bit of light.
- Panels connected in parallel
- Panels connected to controller
- Batteries connected in parallel
- Batteries connected to controller
Really, it's not too hard.
And we have charging going on. It's not very sunny, and I am standing in front of the panels, so I'm sure that we can do better than .3A P(W) = I(A) × V(V) so P = 12V*.3A = 3.6W . We have 60W of panels, so I expect we will see more than that eventually.
Our potential should be I(A) = P(W) / V(V) = 60W/12V = 5A. But reality should show less as there is loss for various solarific reasons.
Discussions
Become a Hackaday.io Member
Create an account to leave a comment. Already have an account? Log In.