Help in explain schematic

[Litch]

Looking at a application note from Microchip HERE. It details how one could calculate the maximum power from the PV cell by utilising a buck converter with current and voltage feedbacks to adjust the the PV's load.

I'm trying to adopt this to make a 12V Lead Acid Battery charger, but I'm having difficulty understanding the how/what/why of a couple areas in the schematic.

Schematic rotated and attached as PNG (Easier to read, and has my references)

Two questions:
1. In the red box, we see the regulated output (Vout) pin (which I assume goes to the battery), and then the J2-OUT header... What is J2-OUT? If it's closed circuit then it essentially feeds the whole Vout to ground via shunt resistors R5 and R6 (Which, granted, would allow the OpAmp to measure maximum output current) - but what purpose is that when we're trying to charge a battery? Is this jumper designed to be a momentary thing?

2. Yellow box; J1-VPV makes no sense to me. If we close circuit it, it would essentially short circuit the PV cell... Again we could read the current value - but for what purpose?

Shouldn't these shunts be in-line with the load? IE. Measure between the +ve of the PV cell and the buck, and between the Vout and the battery?

Can someone please help explain?

Thanks in advance.
-KB

 

[KJ6EAD]

In both cases, the current sense resistor networks appear to be in series with the ground leg of the source or load. That seems reasonable.

 

[Litch]

Can you elaborate? That doesn't answer my questions as far as I can tell.

 

[Arouse1973]

J2 looks like it is your output connector from the regulator to your battery. It is not a jumper. And oh J1 probably the input. It needs the current sense on both to maximise the PV curve.
Adam

 

[(*steve*)]

The current shunt can be in either the ground or +V. It doesn't make any difference usually. Current shunts are often in the ground rail because it can make circuit design easier.

Is J1VPV the input voltage? If so it also has a ground side current sensing and it doesn't appear to short anything out.

 

[Litch]

J2 looks like it is your output connector from the regulator to your battery. It is not a jumper. And oh J1 probably the input. It needs the current sense on both to maximise the PV curve.
Adam

Ok, so J2 is basically the header to plug the battery in. Sweet. It seems the Vout tag north of that threw me off (as I though that was the output "pin" to the battery and J2 served another function).

The current shunt can be in either the ground or +V. It doesn't make any difference usually. Current shunts are often in the ground rail because it can make circuit design easier.

Ah - thanks for clarifying.

Is J1VPV the input voltage? If so it also has a ground side current sensing and it doesn't appear to short anything out.

VPV is used throughout the circuit as a reference to the +ve rail of the solar cell (which is the circuit input voltage) so I'd say yes.

But, with that in mind, if pin 2 of J1-VPV goes to +ve rail of the PV cell and pin 1 goes to GND.... How does that achieve measuring the current when it essentially doesn't even utilise the +ve rail, and neither does it utilise a return current path for pin 1 - like J2 does?

 

[Litch]

AH HA!

I'm looking at J1 the wrong way. I'm thinking, how does it measure the current from the -ve rail of the PV cell when it just goes to earth?

But so does everything else in the circuit - the current path with respect to J1 is FROM earth, through the shunts, and out pin 1 of J1.

Man, that had me so confused.

Thanks to all for your help.

 

[Arouse1973]

Yeah you got it.