Help in explain schematic

Thread Starter

Litch

Joined Jan 25, 2013
85
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.

I'm trying to adopt this to make a Lead Acid Battery charger.

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, and then J2-OUT. 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 would allow the OpAmp to measure 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 allow the measurement of the maximum current output of the PV cell, but drop the voltage to near zero.

Can someone please help explain this schematic in practical terms?

Thanks in advanced.
 

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Thread Starter

Litch

Joined Jan 25, 2013
85
J is also for Jumper... Hence the confusion.

So, where would one expect these jacks to go? I don't see any purpose for them off the board...
 

ErnieM

Joined Apr 24, 2011
8,377
J is also for Jumper... Hence the confusion.

So, where would one expect these jacks to go? I don't see any purpose for them off the board...
If they intended to designate a jumper they would have drawn one.

I could only hazzard a guess here as you never linked to the app note itself, but since this device is intended to connect between a solar cell and a battery I would say it is safe to connect J1 to the solar cell and J2 to the battery. Do observe proper polarity.

That way J1 is connected as VPN to the cell to mearure it's voltage, while R1/R2 will sense the current. Same thing with VOUT on J2.
 

kubeek

Joined Sep 20, 2005
5,795
J1 is the input connector for the panel. J2 is the output connector for your load. Shunt resistors measure the current on input and output and the pic acts accordingly to find the MPP.
 

Thread Starter

Litch

Joined Jan 25, 2013
85
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.
 

kubeek

Joined Sep 20, 2005
5,795
The input current amp is inverting, so current goes in J1-2 - the positive input terminal - circles through the load and comes back through the shunts to J1-1.
 
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