Solar charging with some shade

CA Traveler wrote:

BFL13 wrote:
If that shaded panel is half by-passed, then it is putting out half its current? (I really am confused about the current amount)

Because you keep returning to the same incorrect assumptions.

Try this: Assume you can magically rewire a shaded bypassed section of a panel so that the section is no longer in the circuit. Would you not agree that the panel is now producing the same current but with less voltage - say 1/3 less voltage with a typical 3 diode panel?

Also we're discussing array voltage and array amps. Of course the controller output amps are now less as it now has less input power due to the reduced input voltage and it has to maintain the output voltage to the battery.

Ok I did ask that before, so it is confirmed that a single cell in a panel has the same current as the whole panel's current and it is just the individual cells' voltages adding up in the "internal string" that makes its watts using that single cell's amps as the A in the VA.

So shading that one panel does not reduce its amps and so not string amps either; there is no amps bottleneck.

So why does the shaded panel knock down string watts so much? It would be like having an 18 cell panel in series with five 36 cell panels so the voltage of those 18 cells would still add to the string's voltage and it should do better than having just the 5 36s un-shaded.

But that is not the case. In the OP, his input watts is 77.8 x 7.3 = 568W If he disconnected that shaded panel, he would have (using near full Vmp 17.6 x 5= 88 Vmp x that 7.3 he was getting at that insolation = 642w

Why did the shading make his Vmp 77.8 with all 6, when it would have been 88 with 5 un-shaded?

EDIT--that single cell might have the amps but it can't do them until it is connected to some tiny thing with less voltage than it has.